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https://math.stackexchange.com/questions/tagged/analytic-geometry
Newest 'analytic-geometry' Questions - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Questions tagged [analytic-geometry] Ask Question Questions on the use of algebraic techniques for proving geometric facts. Analytic Geometry is a branch of algebra that is used to model geometric objects - points, (straight) lines, and circles being the most basic of these. It is concerned with defining and representing geometrical shapes in a numerical way. Learn more… Top users Synonyms (1) 7,004 questions NewestActiveBountiedUnansweredMore Bountied 0 Unanswered Frequent Score Trending Week Month Unanswered (my tags) Filter Filter by [x] No answers [x] No upvoted or accepted answers [x] Has bounty Days old Sorted by Newest Recent activity Highest score Most frequent Bounty ending soon Trending Most activity Tagged with My watched tags The following tags: analytic-geometry Apply filter Cancel 1 vote 1 answer 105 views What is the correct definition of "double tangent"? In the literature , there are two definitions of "double tangent". One of these is a single straight line which is tangent to a curve at two different points. I am not worried about this -- ... terminology definition analytic-geometry curves Prime Mover 5,254 asked 8 hours ago 2 votes 1 answer 68 views Locus of the foci of all the conics of given eccentricity In W. H. Besant's classic book Conic Sections: Treated Geometrically there is a question at the end of the first chapter: Find the locus of the foci of all the conics of given eccentricity which pass ... geometry analytic-geometry conic-sections locus designer0588 23 asked yesterday -2 votes 0 answers 30 views Euclidean norm for n>3 [duplicate] I'm studying linear algebra and have problems understanding the Euclidean norm for dimensions higher than 3. My professor talks about the size of the vector, but I don't understand how size exists for ... linear-algebra analytic-geometry Sharpe 9 asked Sep 23 at 9:53 -2 votes 0 answers 71 views Why does collapsing a 3D line into one equation create a plane? Where did we lose information? [closed] Imagine a 3D line given by the equations x=y=z. If we split this up we get x=y, y=z, and x=z x=y (given) x-y=0 (subtract y from both sides (1)) x=z (given) z=x (equality is ... geometry solution-verification analytic-geometry 3d Isaac Sechslingloff 581 asked Sep 20 at 19:37 0 votes 0 answers 33 views Confusion about the "dual" map between Archimedean and Catalan solids From what I've read, the Archimedean and Catalan solids are "dual" shapes. From an algebraic/graph-theoretic standpoint this is easy to understand: The number of faces in one Archimedean ... geometry graph-theory analytic-geometry Patch 4,647 asked Sep 16 at 13:15 2 votes 0 answers 42 views What makes one p-adic isometry be rational preserving, and another not? What makes one p-adic isometry rational-preserving, and another not? Consider the function f(x)=\dfrac{ax+b}{cT(x)+d} where a,b,c,d are 2-adic units. Definition: A rational-preserving 2-adic ... analytic-geometry p-adic-number-theory rational-functions open-problem Robert Frost 9,688 asked Sep 11 at 7:44 2 votes 2 answers 67 views Let A\equiv (3,5,4), B\equiv (4,3,5) and P\equiv (a,b,0). If P be such that \angle APB\in[0^{\circ},180^{\circ}] is maximum, find a and b Let A\equiv (3,5,4), B\equiv (4,3,5) and P\equiv (a,b,0). If point P be such that \angle APB\in[0^{\circ},180^{\circ}] is maximum, then find the value of a and b. My Attempt: If P lies ... geometry vectors analytic-geometry maxima-minima 3d Maverick 10.9k asked Sep 8 at 22:46 1 vote 0 answers 53 views Existence of four concyclic points on the graph of a real function under scaling of abscissa Suppose f:\mathbb R \to \mathbb R is a smooth function. Fix four distinct reals u_1<u_2<u_3<u_4. For each x_0\in\mathbb R,\lambda\in\mathbb R-{0}, define four points $$ \bigl(x_0+... geometry analytic-geometry circles plane-geometry curvature user1673563 184 asked Sep 8 at 19:48 0 votes 0 answers 68 views Existence of circle passing through four distinct points on y=x^a,x>0 I’m wondering if four distinct points can lie concyclic on the graph of y=x^a,x>0. For 1/2\le a\le 2 or a<0, no such circle exists, as the determinant’s sign is stable. For 0<a<1/2... determinant analytic-geometry circles user1673563 184 asked Sep 7 at 22:49 0 votes 1 answer 42 views The set of centers from which some circle has 4 intersections with y=x^3-ax expands to cover the whole plane as a → ∞ For all a\in\Bbb R let S_a be the set of centers from which some circle has 4 intersections with the graph of y=x^3-ax. For example, in the image, (1,0) is the center of a circle which has $... analytic-geometry circles osculating-circle user1673563 184 asked Sep 7 at 19:58 2 votes 1 answer 90 views Parabola touching the circle x^2+y^2=4 and the line 3x+4y=10 at the same point S'(x,y)=S(x,y)+t L^2(x,y)=0, t \in \Re represents a family of conic sections which touch the conic S(x,y)=0 and the line L(x,y)=0 at the same point. For example, the parabola touching a circle $... analytic-geometry conic-sections Z Ahmed 46.8k asked Sep 7 at 14:27 3 votes 2 answers 120 views The parabola passing through intersection of lines 2x+3y+1=0, 3x+2y-1=0 with the circle x^2+y^2=4 S'=S+t~ L_1 L_2=0, t\in \Re represents a family of conic sections passing through intersection points of the conic S(x,y)=0 with the lines L_1(x,y)=0 and L_2(x,y)=0. Here, in this example we ... analytic-geometry conic-sections Z Ahmed 46.8k asked Sep 5 at 6:43 5 votes 3 answers 177 views Circle passing through (3,4) and touching x+y=3 at (1,2) See Family of Circles Touching a Circle and a Line S'(x,y)=S(x,y)+t L(x,y)=0, t\in \mathbb{R} is a family of conics passes through the conic S=0 and cutting /touching the line L=0. In this ... analytic-geometry circles tangent-line Z Ahmed 46.8k asked Sep 2 at 4:42 2 votes 2 answers 70 views Equation of the quadratic curve if two tangents and corresponding chord of contact are given Given a quadratic curve/conic S(x,y)=0 and an outside point (x',y') we can get equation of chord of contact T(x,y)=0 and the combined equaion of the corresponding tangents as $$T^2(x,y)=S(x,y) S(... analytic-geometry circles conic-sections Dharmendra Singh 1,261 asked Sep 1 at 5:41 11 votes 1 answer 224 views When is most of an inner ball outside the hypercube which bounds the outer balls which bound the inner ball? There is a simple question where a unit square contains four circles of radius \frac14 with a fifth inner circle, and the aim is to find that the radius of the inner circle is $r_2=\frac{\sqrt{2}-1}{... geometry analytic-geometry spheres hyperspace Henry 171k asked Aug 28 at 13:51 153050per page 1 2345 … 467Next 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 Hot Network Questions Discussing strategy reduces winning chances of everyone! Why do universities push for high impact journal publications? Bypassing C64's PETSCII to screen code mapping Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results Do we need the author's permission for reference How to use \zcref to get black text Equation? 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Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" Clinical-tone story about Earth making people violent What can be said? Passengers on a flight vote on the destination, "It's democracy!" Newest analytic-geometry questions feed Subscribe to RSS Newest analytic-geometry questions feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547
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[Solved] A firm has production function What is the slope of the isoquant - Microeconomics (PUBLPOL 301A) - Studocu Skip to main content Teachers University High School Discovery Sign in Welcome to Studocu Sign in to access study resources Sign in Register Guest user Add your university or school 0 followers 0 Uploads 0 upvotes Home My Library AI Notes Ask AI AI Quiz New Recent A firm has production function 𝑄 = √(𝐾 ⋅ 𝐿). What is the slope of the isoquant curve if 𝐾 = 5 and 𝐿 = 5?Microeconomics (PUBLPOL 301A) My Library Courses You don't have any courses yet. Add Courses Studylists You don't have any Studylists yet. Create a Studylist Stanford University Microeconomics Question A firm has production function What is the slope of the isoquant Stanford University Microeconomics Question Anonymous Student 1 year ago A firm has production function 𝑄 = √(𝐾 ⋅ 𝐿). What is the slope of the isoquant curve if 𝐾 = 5 and 𝐿 = 5? Like 0 Related documents ECO 201 Project Microeconomics Assignments None Let us first consider the labour market where we deal with production function in which capital stock is fixed and labour is the variable input The Agroecology Of Farming Systems In Latin America Coursework None Winter 2019 PS3Solutions Introductory Economics A Assignments None Misalloc draft 100313 - Lecture notes sdaf Ger 2A Substitution (2Nd)Lecture notes None PMG Sem-2 UNIT-4: Closed Economy Short-Run Economic Models Analysis Introduction To Geology: California Desert Field Geology Tutorial work None 3070 PSet-ch3 Solutions Elementary Economics Lecture notes None Economics- Chapter 5 Flashcards (Mark)Economic Analysis I Lecture notes None Answer Created with AI 1 year ago Answer The slope of the isoquant curve, also known as the Marginal Rate of Technical Substitution (MRTS), is the rate at which a firm can substitute capital (K) for labor (L) while ... Like 0 This is a preview Go Premium and unlock full access to AI trained answers and study materials Optimised for study questions Access all study materials Get Unlimited Downloads Upload Share your documents to unlock Free Trial Get 7 days of free Premium Already Premium?Log in AI answers may contain errors. Please double check important information and use responsibly. Ask a new question Discover more from: Microeconomics PUBLPOL 301A Stanford University 5 Documents Go to course 5 Standard Utility Functions (Feir)Microeconomics 100%(1) 56 PSAT Reading Test QA October 2020 - Key Insights & Analysis Microeconomics 67%(3) 34 Uncertainty Notes for Econ 101: Choice Behavior Models Microeconomics None 4 Microbiology Concepts and Applications (3.3.5)Microeconomics None Discover more from: Microeconomics PUBLPOL 301AStanford University5 Documents Go to course 5 Standard Utility Functions (Feir)Microeconomics 100%(1) 56 PSAT Reading Test QA October 2020 - Key Insights & Analysis Microeconomics 67%(3) 34 Uncertainty Notes for Econ 101: Choice Behavior Models Microeconomics None 4 Microbiology Concepts and Applications (3.3.5)Microeconomics None 6 ECO 201 Project Microeconomics None Related Answered Questions 9 months ago Consider a monopolistically competitive industry in long-run equilibrium. Will this industry be productively efficient? Question 10 options: No. Firms are selling their output at a level where price exceeds marginal cost and thus, by definition, cannot be productively efficient. Yes. In long-run equilibrium, each firm is producing at an output level where price is equal to marginal cost. Yes. Since the firms are in long-run equilibrium, they will all be operating on their LRAC curves. Yes. Since the firms are in long-run equilibrium, they will all be producing at the minimum of their LRAC curves. No. Since firms are selling differentiated products and there is no industry-wide price, we cannot conclude that marginal cost will be equated across all firms. Microeconomics (PUBLPOL 301A) 9 months ago Why is monopoly allocatively inefficient? Question 7 options: because the price exceeds the marginal cost of the last unit produced because the marginal cost exceeds the average cost for the last unit produced because the opportunity cost exceeds the marginal cost of the last unit produced because lower costs could be achieved because the firm has no incentive to maximize profits Microeconomics (PUBLPOL 301A) 1 year ago Given the production function 𝑄 = (4√𝐾 ⋅ 𝐿)Where 𝑄 denotes the production volume, 𝐿 is the quantity labor and 𝐾 is the amount of capital. The company has a fixed capital stock of 𝐾 = 100. At the moment is production 𝑄 = 600. By approximately how many units does production increase if 𝐿 increases by 100 devices? a) 121 b) 10 c) 40 d) None of the above. Microeconomics (PUBLPOL 301A) 1 year ago Would it be possible to organize the whole society according to market economic principles? Based on your opinion, discuss the three biggest problems that would arise. For each case, try to explain why one market solution leads to an undesirable situation and how a state can improve the situation. Microeconomics (PUBLPOL 301A) 1 year ago According to the first welfare theorem, a free market leads to a Pareto-optimal distribution. For the result to apply, a number of conditions must be met and if not, market failures occur. Give an example of a market failure and explain why a free market in the example leads to a distribution that is not Pareto-optimal. Choose an unique example an as possible. Microeconomics (PUBLPOL 301A) 1 year ago Oris comes across a seller who sells diamonds and rubies. The diamonds cost 𝑝𝑥 = $20 while the rubies cost 𝑝𝑦 = $10. Oris has the utility function 𝑈(𝑥, 𝑦) = (𝑥^2 + 𝑦) where 𝑥 is his consumption of diamonds and 𝑦 his consumption of rubies. Oris has $40. The budgetequation: 20x + 10y = 40. Solving this: x = 1 or 1 diamond and y = 2 or 2 rubies. What's the marginal utility derived from the utility function? What utility does Oris get if he chooses to spend all of his income on diamonds and rubies so that: 𝑀𝑈𝑥 / 𝑀𝑈𝑦 = 𝑝𝑥 / 𝑝𝑦. Further information: px / py = 40 / 20 = 2? Microeconomics (PUBLPOL 301A) Ask AI Home My Library Discovery Discovery Universities High Schools High School Levels Teaching resources Lesson plan generator Test generator Live quiz generator Ask AI English (US) United States Company About us Studocu Premium Academic Integrity Jobs Blog Dutch Website Study Tools All Tools Ask AI AI Notes AI Quiz Generator Notes to Quiz Videos Notes to Audio Infographic Generator Contact & Help F.A.Q. Contact Newsroom Legal Terms Privacy policy Cookie Settings Cookie Statement Copyright & DSA English (US) United States Studocu is not affiliated to or endorsed by any school, college or university. 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https://thirdspacelearning.com/gcse-maths/geometry-and-measure/area-of-a-segment/
What is the area of a segment? How to calculate the area of a segment Area of a segment worksheet Area of a segment examples Example 1: using area of a triangle formula Example 2: using area of a triangle formula Example 3: using trigonometry Example 4: theta not given Example 5: given base angle of isosceles triangle Example 6: missing radius Common misconceptions Practice area of a segment questions Area of a segment GCSE questions Learning checklist Next lessons Still stuck? GCSE Tutoring Programme Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring. Teacher-trusted tutoring In order to access this I need to be confident with: Types of triangles Angles around a point Isosceles triangles Area of a triangle Parts of a circle Area of a circle Circumference of a circle How to work out perimeter Area of a sector Area of triangle trig (½ ab sin C) Substitution Rearranging formula This topic is relevant for: Introduction What is the area of a segment? How to calculate the area of a segment Area of a segment worksheet Area of a segment examples ↓ Example 1: using area of a triangle formula Example 2: using area of a triangle formula Example 3: using trigonometry Example 4: theta not given Example 5: given base angle of isosceles triangle Example 6: missing radius Common misconceptions Practice area of a segment questions Area of a segment GCSE questions Learning checklist Next lessons Still stuck? Area Of A Segment Here we will learn about the area of a segment including how to identify a segment of a circle and how to find the area of it. There are also worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck. What is the area of a segment? The area of a segment is the space inside the segment that is formed by an arc and a chord. A segment of a circle is the area enclosed by a chord and an arc of a circle. There are two types of segments, ‐ A minor segment is a segment where the arc length is less than half of the circumference of the circle. ‐ A major segment is a segment where the arc length is greater than half of the circumference of the circle. To work out the area of a segment we can use the formula for the area of a triangle and the formula for the area of a sector. We subtract the area of the triangle from the area of the sector. The formula for the area of a triangle is \text { Area of a triangle }=\cfrac{\text { base } \times \text { height }}{2}. Step-by-step guide: Area of a triangle Higher – There is also a trigonometric formula we can use to find the area of a triangle, two sides and the included angle. Useful for triangles where we don’t know the height and we are unable to calculate it. \text { Area of a triangle }=\cfrac{1}{2} \, a b \sin C Step-by-step guide: Area of a triangle trig = (\cfrac{1}{2} \, a b \sin C) The formula for the area of a sector is \text { Area of a sector }=\cfrac{\theta}{360} \, \times \pi r^2 where \theta is the angle of the sector and r is the radius of the circle. Step-by-step guide: Area of a sector What is the area of a segment? How to calculate the area of a segment In order to calculate the area of a segment: Find the length of the radius. Find the size of the angle creating the sector. Find the area of the sector. Find the area of the triangle created by the radii and the chord of a circle. Subtract the area of the triangle from the area of the sector. Clearly state your answer. Explain how to calculate the area of a segment Circles, sectors and arcs worksheet (includes area of a segment) Get your free area of a segment worksheet of 20+ circles, sectors and arcs questions and answers. Includes reasoning and applied questions. DOWNLOAD FREE x Circles, sectors and arcs worksheet (includes area of a segment) Get your free area of a segment worksheet of 20+ circles, sectors and arcs questions and answers. Includes reasoning and applied questions. DOWNLOAD FREE Related lessons on circles, sectors and arcs Area of a segment is part of our series of lessons to support revision on circles, sectors and arcs. You may find it helpful to start with the main circles, sectors and arcs lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include: Circles, sectors and arcs Equation of a circle Segment of a circle Sector of a circle Arc of a circle Parts of a circle Area of a segment examples Example 1: using area of a triangle formula Calculate the shaded area. Give your answer correct to 2 decimal places. Find the length of the radius. The length of the radius is labelled on the diagram. Radius = 3 \, cm 2Find the size of the angle creating the sector. The angle creating the sector is marked on the diagram. \theta=90^{\circ} 3Find the area of the sector. The whole shape is the sector. The area of the sector is \begin{aligned} & =\cfrac{\theta}{360} \, \times \pi r^2 \\ & =\cfrac{90}{360} \, \times \pi \times 3^2 \\ & =\cfrac{9}{4} \, \pi \\ & =7.0685… \mathrm{~cm}^2 \end{aligned} Note: We could have found the area of the sector, in this instance, by using the area of a circle formula to find the area of the circle as a whole, and then divided by four, because this sector has a 90^{\circ} angle, which means it is one quarter of the whole circle. 4Find the area of the triangle created by the radii and the chord of a circle. The triangle created by the radii and the chord of the circle is the white sections of the sector. It is a right-angled triangle. The area of the triangle is \begin{aligned}& =\cfrac{\text{ base } \times \text{ height }}{2} \\ & =\cfrac{3 \, \times \, 3}{2} \\ & =4.5 \mathrm{~cm}^2\end{aligned} 5Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =7.0685…-4.5=2.5685… 6Clearly state your answer. The question asks for the answer to 2 decimal places, so we write the answer as =2.57 \mathrm{~cm}^2 . Example 2: using area of a triangle formula Calculate the shaded area. Give your answer to 2 decimal places. Find the length of the radius. The length of the radius is labelled on the diagram. Radius = 5 \, cm Find the size of the angle creating the sector. The angle creating the sector is marked on the diagram. \theta=90^{\circ} Find the area of the sector. The whole shape is the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{90}{360} \times \pi \times 5^2 \\ & =\cfrac{25}{4} \pi \\ & =19.63495408… \mathrm{~cm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Find the area of the triangle created by the radii and the chord of a circle. The triangle created by the radii and the chord of the circle is the white sections of the sector. The area of the triangle is \begin{aligned}& =\cfrac{\text { base } \times \text { height }}{2} \\ & =\cfrac{5 \times 5}{2} \\ & =12.5 \mathrm{~cm}^2\end{aligned} Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =19.63495408…-12.5 =7.134954085… Clearly state your answer. The question asks for the answer rounded to 2 decimal places, so we can round it and give the answer as 7.13 \mathrm{~cm}^2. Example 3: using trigonometry Calculate the shaded area. Give your answer to 3 significant figures. Find the length of the radius. The length of the radius is labelled on the diagram. Radius = 4 \, cm Find the size of the angle creating the sector. The angle creating the sector is marked on the diagram. \theta=108^{\circ} Find the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{108}{360} \times \pi \times 4^2 \\ & =\cfrac{24}{5} \pi \\ & =15.07964474… \mathrm{~cm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Find the area of the triangle created by the radii and the chord of a circle. We do not know the height of the triangle created by the radii and the chord of the circle in this example. So, we use trigonometry for the area of a triangle, \text { Area of a triangle }=\cfrac{1}{2} \, a b \sin C, where a and b are radii, and the angle C is the angle creating the sector. The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 4 \times 4 \times \sin (108) \\ & =7.60845213… \mathrm{~cm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =15.07964474… - 7.60845213… = 7.47119261… Clearly state your answer. The question asks for the answer rounded to 3 significant figures, so we can round it and give the answer as 7.47 \mathrm{~cm}^2. Example 4: theta not given Below is a circle with radius 6 \, mm. Calculate the area of the shaded segment. Give your answer to 3 significant figures. Find the length of the radius. The length of the radius is given in the question. Radius = 6 \, mm. Find the size of the angle creating the sector. The angle creating the sector is not marked on the diagram. However, we can use the angle fact ‘angles around a point add to 360^{\circ} ’ to calculate the angle creating the sector. \theta=360-246=114^{\circ} Find the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{114}{360} \times \pi \times 6^2 \\ & =\cfrac{57}{5} \pi \\ & =35.8145625… \mathrm{~mm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Find the area of the triangle created by the radii and the chord of a circle. We do not know the height of the triangle created by the radii and the chord of the circle in this example. So, we use the trigonometric formula for the area of a triangle, \text { Area of a triangle }=\cfrac{1}{2} \, a b \sin C, where a and b are radii, and the angle C is the angle creating the sector. The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 6 \times 6 \times \sin (114) \\ & =16.44381824 \mathrm{~mm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =35.8145625… - 16.44381824… =19.37074426… Clearly state your answer. The question asks for the answer rounded to 3 significant figures, so we can round it and give the answer as 19.4 \mathrm{~mm}^2. Example 5: given base angle of isosceles triangle Below is a circle with radius 4 \, mm. Calculate the area of the shaded segment. Give your answer to 3 significant figures. Find the length of the radius. The length of the radius is given in the question. Radius = 4 \, mm Find the size of the angle creating the sector. The angle creating the sector is not marked on the diagram. We need to calculate it. The first thing to recognise is which sides of the triangle formed in the sector are radii and which is the chord. The two sides that meet in the centre of the circle are both radii, therefore, equal in length. The third side of the triangle cuts the circle from circumference to circumference, not passing through the centre, so is the chord. The fact that we have two equal sides in this triangle, means it is isosceles. The angle facts ‘base angles in an isosceles triangle are equal’ and ‘angles in a triangle add to 180^{\circ} ’ will help us to calculate the angle creating the sector, \theta . \theta=180-22-22=136^{\circ} Find the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{136}{360} \times \pi \times 4^2 \\ & =\cfrac{272}{45} \pi \\ & =18.98918226… \mathrm{~mm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Find the area of the triangle created by the radii and the chord of a circle. We do not know the height of the triangle created by the radii and the chord of the circle in this example. So, we use the area formula \text{ Area of a triangle }=\cfrac{1}{2} \, a b \sin C, where a and b are radii, and the angle C is the angle creating the sector. The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, ab \sin C \\ & =\cfrac{1}{2} \times 4 \times 4 \times \sin (136) \\ & =5.557266964… \mathrm{~mm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =18.98918226… - 5.557266964… =13.4319153… Clearly state your answer. The question asks for the answer rounded to 3 significant figures, so we can round it and give the answer as 13.4 \mathrm{~mm}^2. Example 6: missing radius Below is a sector. The area of the sector is 67.02 {~cm}^2. The radius of the sector, r, is an integer. Calculate the area of the segment labelled B. Give your answer to 3 significant figures. Find the length of the radius. The length of the radius is given in the question as r. So we need to calculate it. We are given the area of the sector, so we can use this to calculate the radius. Substituting what we know into the formula for the area of a sector we get 67.02=\cfrac{120}{360} \times \pi \times r^2. Rearranging this to make r^2 the subject gives r^2=\cfrac{67.02}{\cfrac{120}{360} \times \pi} \, . So, r^2=63.99938572… We want the radius, r, so we need to square root r^2, to give r=7.999961607… The question tells us that the radius, r, is an integer, so for this example we can say, r=8 \mathrm{~cm}. Find the size of the angle creating the sector. The angle creating the sector is marked on the diagram. \theta=120^{\circ} Find the area of the sector. The area of the sector is given in the question as 67.02 \mathrm{~cm}^2. Find the area of the triangle created by the radii and the chord of a circle. We do not know the height of the triangle created by the radii and the chord of the circle in this example. So, we use the area formula \text{ Area of a triangle }=\cfrac{1}{2} \, a b \sin C, where a and b are radii, and the angle C is the angle creating the sector. The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 8 \times 8 \times \sin (120) \\ & =27.71281292… \mathrm{~cm}^2\end{aligned} Note: Do not round your answer at this stage, for accuracy. Subtract the area of the triangle from the area of the sector. The area of the shaded area, which is the segment, is =67.02 - 27.71281292… =39.30718708… Clearly state your answer. The question asks for the answer rounded to 3 significant figures, so we can round it and give the answer as 39.3 \mathrm{~cm}^2. Common misconceptions Sector and segment Sometimes the two can be confused. Remember, A sector is formed when a circle is divided using two radii. A segment is the area enclosed by an arc of a circle and a chord. Recognising the angle creating the sector The angle creating the sector is the angle inside the sector you’re working with. For example, in the diagram below, it is the angle labelled 112^{\circ} that creates the sector, not the angle labelled 248^{\circ}. Rounding too early It is important to not round your answers in the workings until you get to the end where you can round your final answer as the question asks. Otherwise, you could make your final answer slightly inaccurate. Practice area of a segment questions Calculate the shaded area. Give your answer to 2 decimal places. 14 \mathrm{~cm}^2 13.98 \mathrm{~cm}^2 62.98 \mathrm{~cm}^2 63 \mathrm{~cm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{90}{360} \times \pi \times 7^2 \\ & =\cfrac{49}{4} \pi \\ & =38.48451001… \mathrm{~cm}^2\end{aligned} The area of the triangle is \begin{aligned}& =\cfrac{\text { base } \times \text { height }}{2} \\ & =\cfrac{7 \times 7}{2} \\ & =24.5 \mathrm{~cm}^2\end{aligned} Hence, the area of the segment is =38.48451001…-24.5=13.98451001…=13.98 \; (2 d p). Calculate the shaded area. Give your answer to 2 decimal places. 5.14 \mathrm{~mm}^2 5 \mathrm{~mm}^2 1 \mathrm{~mm}^2 1.14 \mathrm{~mm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{90}{360} \times \pi \times 2^2 \\ & =\pi \\ & =3.141592654… \mathrm{~mm}^2\end{aligned} The area of the triangle is \begin{aligned} & =\cfrac{\text { base } \times \text { height }}{2} \\ & =\cfrac{2 \times 2}{2} \\ & =2 \mathrm{~mm}^2\end{aligned} Hence, the area of the segment is =3.141592654…-2=1.141592654…=1.14 \; (2 d p). The circle below has a radius of 9 \, mm. Calculate the shaded area. Give your answer to 2 decimal places. 77.98 \mathrm{~mm}^2 179.77 \mathrm{~mm}^2 178 \mathrm{~mm}^2 78 \mathrm{~mm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. First we need to calculate \theta, the angle that creates the sector. The two sides that meet in the centre of the circle are both radii, therefore, equal in length. The third side of the triangle cuts the circle from circumference to circumference, not passing through the centre, so is the chord. The fact that we have two equal sides in this triangle, means it is isosceles. The angle facts ‘base angles in an isosceles triangle are equal’ and ‘angles in a triangle add to 180^{\circ} ’ will help us to calculate the angle creating the sector, \theta. \theta=180-18-18=144^{\circ} The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{144}{360} \times \pi \times 9^2 \\ & =\cfrac{162}{5} \pi \\ & =101.787602 \mathrm{~mm}^2\end{aligned} The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 9 \times 9 \times \sin (144) \\ & =23.80530272… \mathrm{~mm}^2\end{aligned} Hence, the area of the segment is =101.787602…-23.80530272…=77.98229928…=77.98 \; (2 d p). A circle has a radius of 3 \, cm. Calculate the shaded area. Give your answer to 2 decimal places. 17 \mathrm{~cm}^2 7.02 \mathrm{~cm}^2 17.39 \mathrm{~cm}^2 7 \mathrm{~cm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. First we need to calculate \theta, the angle that creates the sector. We can use the angle fact ‘angles around a point add to 360^{\circ} ’ to calculate the angle creating the sector. \theta=360-228=132^{\circ} The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{132}{360} \times \pi \times 3^2 \\ & =\cfrac{33}{10} \pi \\ & =10.36725576 \mathrm{~cm}^2\end{aligned} The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 3 \times 3 \times \sin (132) \\ & =3.344151715… \mathrm{~cm}^2\end{aligned} Hence, the area of the segment is =10.36725576…-3.344151715…=7.023104045…=7.02 \; (2 d p). A circle has a radius of 5 \, cm. Calculate the shaded area. Give your answer to 2 decimal places. 43.74 \mathrm{~cm}^2 44 \mathrm{~cm}^2 17 \mathrm{~cm}^2 16.69 \mathrm{~cm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. The area of the sector is \begin{aligned}& =\cfrac{\theta}{360} \times \pi r^2 \\ & =\cfrac{124}{360} \times \pi \times 5^2 \\ & =\cfrac{155}{18} \pi \\ & =27.05260341 \mathrm{~cm}^2\end{aligned} The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ &=\cfrac{1}{2} \times 5 \times 5 \times \sin (124) \\ & =10.36296966… \mathrm{~cm}^2\end{aligned} Hence, the area of the segment is =27.05260341…-10.36296966…=16.68963375…=16.69 \; (2 d p). Below is a sector. The area of the sector is 33.93 {~cm}^2. The radius of the sector, r, is an integer. Calculate the area of the segment labelled B. Give your answer to 3 significant figures. 16.81 \mathrm{~cm}^2 51.05 \mathrm{~cm}^2 51 \mathrm{~cm}^2 17 \mathrm{~cm}^2 We need to calculate the area of the sector and the area of the triangle. Then subtract the area of the triangle from the area of the sector. The length of the radius is given in the question as r. So we need to calculate it. We are given the area of the sector, so we can use this to calculate the radius. Substituting what we know into the formula for the area of a sector we get 33.93=\cfrac{108}{360} \times \pi \times r^2 . Rearranging this to make r^2 the subject gives r^2=\cfrac{33.93}{\frac{108}{360} \times \pi} . So, r^2=36.00084813… We want the radius, r, so we need to square root r^2, to give r=6.000070677… The question tells us that the radius, r, is an integer, so we can say, r=6 \mathrm{~cm}. The area of the sector is given in the question. So we need to find the area of the triangle. The area of the triangle is \begin{aligned}& =\cfrac{1}{2} \, a b \sin C \\ & =\cfrac{1}{2} \times 6 \times 6 \times \sin (108) \\ & =17.11901729 \mathrm{~cm}^2\end{aligned} Hence, the area of the segment is =33.93-17.11901729…=16.81098271…=16.81 \; (2 d p). Area of a segment GCSE questions The diagram shows a sector of a circle with centre Ο. The radius of the circle is 8 \, cm. ABC is an arc of the circle. AC is a chord of the circle. Angle AOC = 50^{\circ} Calculate the area of the shaded segment. (5 marks) Show answer \cfrac{50}{360}\times \pi \times 8^2 \, or \, \cfrac{1}{2}\times 8\times 8\times \sin(50) (1) Area of sector = 27.92 (1) Area of triangle = 24.51… (1) ‘Their area of sector’ – ’their area of triangle’. (1) 3.41(…) {~cm}^2 (1) 2. The diagram shows sector Ο ABC of a circle with centre Ο. OC = 5.2 \, cm Calculate the area of the shaded segment ABC. (6 marks) Show answer 360-240 (1) 120^{\circ} (1) \cfrac{120}{360}\times \pi \times 5.2^2 \, or \, \cfrac{1}{2}\times 5.2\times 5.2\times \sin(120) (1) Area of sector = 28.31… and area of triangle = 11.7 (1) ‘Their area of sector’ – ’their area of triangle’. (1) 16.6(…) {~cm}^2 (1) ABC is an arc of a circle centre O with radius 60 \, m. AC is a chord of the circle. Angle AOC = 35^{\circ}. Calculate the area of the shaded region. Give your answer to 3 significant figures. (5 marks) Show answer \cfrac{35}{360}\times \pi \times 60^2 \, or \, \cfrac{1}{2}\times 60\times 60\times \sin(35) (1) Area of sector = 1099.5(…) (1) Area of triangle = 1032.43(…) (1) ‘Their area of sector’ – ’their area of triangle’. (1) 67.1 {~m}^2 (1) Learning checklist You have now learned how to: Calculate the area of a segment given the angle creating the sector, \theta , and the radius, r Calculate the radius given the area of the sector Recognise isosceles triangles in sectors Use the properties of isosceles triangles to find the angle creating the sector, \theta The next lessons are How to work out perimeter Angle rules Angles in polygons Still stuck? Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors. Find out more about our GCSE maths tuition programme. Introduction What is the area of a segment? How to calculate the area of a segment Area of a segment worksheet Area of a segment examples ↓ Example 1: using area of a triangle formula Example 2: using area of a triangle formula Example 3: using trigonometry Example 4: theta not given Example 5: given base angle of isosceles triangle Example 6: missing radius Common misconceptions Practice area of a segment questions Area of a segment GCSE questions Learning checklist Next lessons Still stuck?
11403
https://secure-media.collegeboard.org/digitalServices/pdf/ap/apcentral/ap14_chemistry_q7.pdf
Question 7 (4 points) The half-life (t1/2) of the catalyzed isomerization of cis-2-butene gas to produce trans-2-butene gas, represented above, was measured under various conditions, as shown in the table below. Trial Number Initial Pcis-2-butene (torr) V (L) T (K) t1/2 (s) 1 300. 2.00 350. 100. 2 600. 2.00 350. 100. 3 300. 4.00 350. 100. 4 300. 2.00 365 50. (a) The reaction is first order. Explain how the data in the table are consistent with a first-order reaction. For a first-order reaction, the half-life is independent of reactant concentration (or pressure) at constant T, as shown in trials 1, 2, and 3. 1 point is earned for a correct explanation. (b) Calculate the rate constant, k, for the reaction at 350. K. Include appropriate units with your answer. 1/2 1 0.693 0.693 = = = 0.00693 s 100. s k t -1 point is earned for correct numerical answer with units. (c) Is the initial rate of the reaction in trial 1 greater than, less than, or equal to the initial rate in trial 2 ? Justify your answer. The initial rate in trial 1 is less than that in trial 2 because rate = k [cis-2-butene] or rate = k -2-butene cis P (with reference to values from both trials). OR because the initial concentration of cis-2-butene in trial 1 is less than that in trial 2 and k is constant. 1 point is earned for the correct answer with justification. (d) The half-life of the reaction in trial 4 is less than the half-life in trial 1. Explain why, in terms of activation energy. The temperature is higher in trial 4, meaning that the KEavg of the molecules is greater. Consequently, in this trial a greater fraction of collisions have sufficient energy to overcome the activation energy barrier, thus the rate is greater. 1 point is earned for a correct answer with justification. ©2014 The College Board. Visit the College Board on the Web: www.collegeboard.org. ©2014 The College Board. Visit the College Board on the Web: www.collegeboard.org. ©2014 The College Board. Visit the College Board on the Web: www.collegeboard.org. ©2014 The College Board. Visit the College Board on the Web: www.collegeboard.org. ©2014 The College Board. Visit the College Board on the Web: www.collegeboard.org. AP® CHEMISTRY 2014 SCORING COMMENTARY Question 7 Overview The question required students to interpret kinetic data in the isomerization reaction of cis-2-butene to trans-2-butene under various conditions in the gaseous state. Four trials were presented in a table that listed the initial pressure of cis-2-butene, the volume of the reaction chamber, temperature, and half-life of the reaction. Part (a) stated that the reaction is first order and asked for an explanation of how the data was consistent with this fact. In part (b) students calculated the rate constant k for the reaction at 350 K. Part (c) assessed students’ knowledge of the relationship between reaction rates, rate constant, and concentration by asking for a prediction, with justification, about whether the initial rate of reaction in trial 1 would be greater than, less than, or equal to that in trial 2. (Relative to trial 1, trial 2 had an identical volume, temperature and half-life, but twice the initial partial pressure of reactant.) In part (d) students needed to explain why, in terms of activation energy, the half-life of the reaction in trial 4 is less than the half-life of trial 1. This part examined the ability to logically associate a higher temperature with greater average kinetic energy of reactant molecules and thus a greater fraction of molecules able to collide with enough energy to overcome the activation energy barrier. Sample: 7A Score: 4 This response received all 4 possible points: 1 point in part (a), 1 point in part (b), 1 point in part (c), and 1 point in part (d). Sample: 7B Score: 2 Part (a) did not earn a point because it omits any mention of half-life and makes no clear reference to relevant data from the table. The answer to part (b) did not earn a point because, although it provides a valid setup with the correct numerical value of k, it does not include units. The points were earned in parts (c) and (d). Sample: 7C Score: 1 In part (a) the point was not earned because the response makes only vague references to the data and does not mention half-life specifically. Part (b) did not earn the point because, although the setup and numerical answer are correct, the answer has incorrect units. The point was not earned in part (c) because the student said the initial rates are equal. The point was earned in part (d). © 2014 The College Board. Visit the College Board on the Web: www.collegeboard.org.
11404
https://www.realestatenews.com/2024/12/07/new-nar-guide-explains-types-of-agency-to-consumers
Published Time: 2024-12-07T15:00:02.328Z New NAR guide explains types of agency to consumers Go Skip to Content Sign Up Sign InSubscribe Agents Brokerages MLS/Assn Tech Industry By the Numbers Consumer Shutterstock Agents New NAR guide explains types of agency to consumers A real estate professional can help homebuyers and sellers in a number of ways. Not all of them are legal everywhere, but all require disclosure and agreement. Stephanie Reid-Simons December 7, 2024 3 mins The National Association of Realtors' latest consumer guide tackles the question of what makes a real estate professional an agent, and — subtly — some of what makes an agent a Realtor. It starts with the basics: "A licensed real estate professional is your agent when you have hired them to represent you as you buy or sell a home." And while this may seem obvious, the guide takes pains to point out that sellers usually establish this relationship via a listing agreement, and many buyers now follow a different path than they would have before August 17. "Buyer's agents — those who are required as a part of the National Association of Realtors settlement of litigation related to broker commissions — will ask the buyer to sign a written buyer agreement establishing the terms of their relationship prior to touring a home," the guide says. Many flavors of agency: The guide starts with single agency, where an agent represents either a homebuyer or seller. It also explains dual agency, "which allows an agent to represent both the buyer and seller at the same time." It avoids any debate over the ethics of dual agency, but says that in states where it is permitted, some form of disclosure and informed consent is usually required from both the buyer and seller "because of the potential for conflicts of interest." The guide also explains subagency, where an agent "works with the seller's agent to help serve the seller's best interests and to bring potential buyers to the seller's property." A subagent can't help the buyer in any way that would be unfavorable to the seller. Like dual agency, subagency isn't permitted in every state. In some transactions, agents from the same brokerage may be assigned to represent the buyer and seller and "avoid potential conflicts of interest." States that allow this will likely require both sides to agree to this arrangement. Is agency required? The guide tells consumers that their relationship with a real estate professional can be "any type — agency or non-agency — that is permitted by state law." One example of a non-agency relationship is a "transaction broker," or facilitator, which is a neutral third party who helps a buyer and seller reach an agreement on the terms of a sale. As with other types of agency, this is not allowed in all states. Why NAR is doingthese guides: It's part of the ongoing series designed in part to help NAR get ahead of antitrust concerns after having the Department of Justice "implicitly indicting us for not educating the consumers," association CEO Nykia Wright told NAR's board of directors last month. Write toStephanie Reid-Simons. More Agent News Agents Nearly half of agents say local markets favor buyers: Real survey Real’s June survey also found that a majority of the growing brokerage’s agents are now putting AI tools to work every day to bolster their productivity. AJ LaTrace July 23, 2025 Agents Decoded: 5 things brokers can do to increase walk-in traffic From offering mortgage pre-qualifications to improving office signage, these simple strategies can help bring buyers and sellers back into your office. J. Philip Faranda July 21, 2025 Agents ‘Unfiltered’: Are agents ready for a buyers market? Watch the conversation with Dan Cooper as the tech entrepreneur reminds agents that they “have to put in the work” as the market shifts in buyers’ favor. Zephin Livingston July 18, 2025 Get the latest real estate news delivered to your inbox. Sign Up An editorially independent division of About us Our vision Editorial standards Our team Subscribe Why subscribe? Manage subscription Support FAQ Contact us Advertise Learn more © 2025 Real Estate News Privacy PolicyTerms of Use
11405
https://www.linguee.com/english-spanish/translation/rapprochement.html
Translate texts with the world's best machine translation technology, developed by the creators of Linguee. Look up words and phrases in comprehensive, reliable bilingual dictionaries and search through billions of online translations. Linguee Apps ▾ Dictionary English-Spanish rapprochement noun (plural: rapprochements)— acercamiento m (almost always used) (plural: acercamientos m) Examples: greater rapprochement n— political rapprochement n— process of rapprochement n— gradual rapprochement n— cultural rapprochement n— ▾ External sources (not reviewed) | | | --- | | However, in many situations the absence of a political will to see implementation through, and the intrusion of minority and religious divisions [...] into the social dialogue, make it particularly difficult to achieve [...] a broad climate ofrapprochementand communication. eur-lex.europa.eu eur-lex.europa.eu | No obstante, en muchos casos falta la voluntad política de aplicar realmente este marco, y la irrupción en las negociaciones sociales de diferencias [...] vinculadas a las minorías o a las religiones hace especialmente difícil [...] que se cree un clima de acercamiento y conciliación. eur-lex.europa.eu eur-lex.europa.eu | | The West'srapprochementwith Russia should be better used [...] to lever Russia to cooperate on the frozen conflicts in the common EU-Russian neighbourhood. fride.org fride.org | Hay que aprovechar mejor el acercamiento de Occidente a Rusia [...] para influir en ésta a fin de que coopere en los enfrentamientos [...] congelados en las fronteras comunes. fride.org fride.org | | Efforts at facilitating and enhancing the implementation of [...] the commitment made by both countries in pursuit of the agreed [...] goals of their recentrapprochementare also continuing. daccess-ods.un.org daccess-ods.un.org | Los esfuerzos por facilitar y reforzar la [...] aplicación del compromiso hecho por ambos países para lograr los objetivos [...] acordados de su mutuo acercamiento también prosiguen. daccess-ods.un.org daccess-ods.un.org | | With this in mind, he notes with satisfaction the [...] proposal in the explanatory note to dedicate the Games in 2010 to the [...] International Year for theRapprochementof Cultures. unesdoc.unesco.org unesdoc.unesco.org | Habida cuenta de ello, toma nota con satisfacción [...] de que en la Nota explicativa se propone dedicar los Juegos de 2010 al [...] Año Internacional de Acercamiento de las Culturas. unesdoc.unesco.org unesdoc.unesco.org | | Ongoing violence by the security forces [...] against activists hindersrapprochementbetween the government [...] and human rights defenders. crisisgroup.org crisisgroup.org | La violencia persistente de las fuerzas de seguridad [...] contra activistas impide el acercamiento entre el gobierno y los [...] defensores de derechos humanos. crisisgroup.org crisisgroup.org | | A moment ofrapprochementbetween intellectuals with [...] a humanist bent was over, but the questions it had raised remained. obreal.org obreal.org | El momento de acercamiento entre intelectuales con [...] una inclinación humanista finalizaba, pero las cuestiones planteadas seguían vigentes. obreal.org obreal.org | | Rapprochementis a deliberate mutual [...] political action by two equal countries. cidob.org cidob.org | El acercamiento es una deliberada acción [...] política mutua por parte de dos países iguales. cidob.org cidob.org | | And if so, are [...] there any sign ofrapprochementbetween both actors? fride.org fride.org | Y si lo están ¿Hay algún signo de acercamiento entre [...] ambos actores? fride.org fride.org | | But has there beenarapprochementof the Cubans towards [...] the Church in the last years? miradaglobal.com miradaglobal.com | ¿Pero ha habido un acercamiento de los cubanos a la Iglesia [...] en los últimos años? miradaglobal.com miradaglobal.com | | Nevertheless, these congresses [...] constituted a moment ofrapprochementbetween Europe and Latin [...] America before the United States [...] emerged as the great leader during World War II. obreal.org obreal.org | No obstante, estos congresos [...] constituyeron un momento de acercamiento entre Europa y América [...] Latina antes de que Estados Unidos [...] emergiera como el gran líder durante la Segunda Guerra Mundial. obreal.org obreal.org | | There was therefore norapprochementpossible between [...] North and South as was bitterly observed on the issue of cotton, a highly symbolic dossier. europarl.europa.eu europarl.europa.eu | No había pues apertura posible entre el Norte [...] y el Sur, como se comprobó amargamente con respecto al algodón, un asunto altamente simbólico. europarl.europa.eu europarl.europa.eu | | More than ever before, the international community was needed [...] to facilitatearapprochementbetween the parties. daccess-ods.un.org daccess-ods.un.org | Más que nunca, se necesita que la comunidad [...] internacional promueva una reconciliación entre las partes. daccess-ods.un.org daccess-ods.un.org | | His arrest is one of the main conditions set by the EU on Serbia for the [...] continuation of the stabilisation and association agreement that will enable the [...] country to pursue itsrapprochementwith the EU. robert-schuman.eu robert-schuman.eu | Su detención es una de las condiciones principales impuestas a Serbia por la [...] UE para desbloquear el acuerdo de estabilización y asociación que permitiría [...] al país proseguir su acercamiento a la UE. robert-schuman.eu robert-schuman.eu | | In addition, according to one Member State, it would be useful [...] for each country to map all of its projects that make a direct or [...] indirect contribution to therapprochementof cultures. unesdoc.unesco.org unesdoc.unesco.org | Por otra parte, según otro Estado Miembro, sería conveniente que cada país [...] hiciese un inventario de aquellos de sus proyectos que contribuyen [...] directa o indirectamente al acercamiento de las culturas. unesdoc.unesco.org unesdoc.unesco.org | | she has explicit intentions to seek outside investment, which [...] could have as a first steprapprochement cipamericas.org cipamericas.org | inversiones externas, lo que podría tener [...] como primer paso una reconciliación con el cipamericas.org cipamericas.org | | Thisrapprochementmust not be allowed [...] to erode the principle of relative stability. europarl.europa.eu europarl.europa.eu | Esta concesión no debe apartarnos [...] del principio de estabilidad relativa. europarl.europa.eu europarl.europa.eu | | Although he took office quite recently, that is back in February, Yanukovich has made a series of gestures [...] clearly indicative ofrapprochementto Russia. rri.ro rri.ro | Ianukovici, aunque ha llegado al [...] poder bastante recientemente, en el pasado mes de febrero, hizo una serie de [...] gestos claros a favor del acercamiento a Rusia. rri.ro rri.ro | | Courses, workshops and conversations [...] of this kind increaserapprochementand integration with [...] indigenous ethnicities in the regions [...] where most of the indigenous community lives. daccess-ods.un.org daccess-ods.un.org | A través de cursos, talleres y [...] charlas se refuerza el acercamiento e integración con las [...] etnias originarias en las regiones [...] en que reside la mayor parte de su población. daccess-ods.un.org daccess-ods.un.org | | In a speech to the nation delivered on 22nd April Armenian President Serge Sarkissian confirmed the freezing of the ratification of the agreements made with [...] Turkey in October last year which [...] heralded the start ofrapprochementbetween the two countries [...] after nearly a century of tension and hostility. robert-schuman.eu robert-schuman.eu | En un discurso a la nación celebrado el 22 de abril, el Presidente armenio, Serge Sarkissian, confirmó que se suspendía la ratificación de los acuerdos [...] concluidos con Turquía en octubre [...] pasado y que iniciaban el acercamiento entre ambos países tras [...] un siglo de tensiones y hostilidades. robert-schuman.eu robert-schuman.eu | | After successive enlargements from 6 to 12 members, and later to 15, the fall of the Berlin wall in 1989 would pave the [...] way to the second great [...] reconciliation in Europe: therapprochementand subsequent integration [...] of the countries of Central [...] and Eastern Europe between 2004 and 2007. jordipujol.cat jordipujol.cat | Después de las sucesivas ampliaciones de 6 a 12 miembros, y ulteriormente a 15, la caída del muro de Berlín en 1989 abría la vía a la [...] segunda gran reconciliación en [...] Europa: la aproximación y después integración a la Unión Europea de [...] los países de la Europa [...] central y oriental entre el 2004 y el 2007. jordipujol.cat jordipujol.cat | | To promote Madrid, both the city and the Autonomous Community, [...] as locations for meeting with the Jewish community and its [...] organizations, facilitating cultural and socialrapprochement. casasefarad-israel.es casasefarad-israel.es | Promocionar Madrid, tanto la ciudad como la Comunidad Autónoma, [...] como territorios de encuentro con la comunidad y las organizaciones [...] judías, facilitando el acercamiento cultural y social. casasefarad-israel.es casasefarad-israel.es | | Rapprochementof cultures through [...] Architecture. universityandheritage.net universityandheritage.net | Acercamiento de las culturas por [...] medio de la arquitectura. universityandheritage.net universityandheritage.net | | On this basis, political leaders in the country need to live up to their commitment to harmonise their policies and to [...] achieve an internal market and a common trade and customs policy which is [...] essential for furtherrapprochementwith the Union. eur-lex.europa.eu eur-lex.europa.eu | Así las cosas, los líderes políticos del país tienen que cumplir su compromiso de armonizar sus políticas y realizar un [...] mercado interior y una política aduanera y comercial común, elementos [...] esenciales para un mayor acercamiento a la Unión. eur-lex.europa.eu eur-lex.europa.eu | | That Government has maderapprochementwith the European Union [...] its priority. daccess-ods.un.org daccess-ods.un.org | El Gobierno ha convertido el acercamiento a Europa en una de sus [...] prioridades. daccess-ods.un.org daccess-ods.un.org | | As a facilitator in these negotiations, it [...] played a part in therapprochementof views among the parties [...] and in their signature of [...] two protocols on enhancing the security situation and improving the humanitarian situation. daccess-ods.un.org daccess-ods.un.org | Como facilitador de estas negociaciones, tuvo un [...] papel importante en el acercamiento de posturas entre las [...] partes y en su firma de dos protocolos [...] para aumentar la seguridad y mejorar la situación humanitaria. daccess-ods.un.org daccess-ods.un.org | | This recentrapprochementbecame possible because the pressure [...] from the people is so great. eurotopics.net eurotopics.net | Se ha hecho posible el nuevo acercamiento porque la presión de [...] la población es realmente grande. eurotopics.net eurotopics.net | | The military and security situation, for example, is stable; the [...] movement of people [...] and goods across ethnic divides has increased; and a grass-rootsrapprochement,with practical and pragmatic working relationships among ethnically [...] intertwined villages, is emerging. osce.org osce.org | Por ejemplo, la situación militar y de seguridad es estable; ha aumentado la circulación de personas y bienes a través de [...] divisorias étnicas; y se [...] está desarrollando un acercamiento de base, en el que se entablan relaciones pragmáticas de trabajo entre municipios interrelacionados con vínculos [...] étnicos. osce.org osce.org |
11406
https://georgehartas.com/mat-161/
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | --- --- --- | | | | | | | | --- --- | | | | | | --- --- | | George's Math Aid for DCCC | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | HOME | | COURSES | | MAT 025 | | MAT 050 | | MAT 120 | | MAT 121 | | MAT 128 | | MAT 135 | | MAT 136 | | MAT 151 | | MAT 152 | | MAT 160 | | MAT 161 | | MAT 200 | | MAT 210 | | MAT 230 | | MAT 260 | | MAT 261 | | BUS 220 | | | | | | --- | | | | MAT 161 - Calculus II | | Prof. Canepa's Course Handouts + Derivatives, Integrals and Trigonometry + Hyperbolic Functions + Integration Tables (Appendix B) + Conic Section Formulas + Polar Graphs + Chapter 11 Formulas + Vector-Valued Function Formulas Paul's Online Math Notes + Calculus - Common Derivatives and Integrals + Calculus - Cheat Sheet + Calculus II - Complete Notes Vector Calculus - by Michael Corral, Schoolcraft College, (PDF) George Hartas + Chapter 11 Study Notes Patrice Camiré - Champlain St. Lawrence College + Differential Calculus + Integral Calculus + Vectors and Matrices Mu Prime Math + Valuable Vector Calculus Khan Academy + Logarithmic Functions Differentiation + Exponential Functions Differentiation + Derivatives of Bases Other Than e + Inverse Trigonometric Functions Differentiation + Integral Calculus + Parametric Equations + Polar Coordinates + Conic Sections + Vectors (Precalculus) + Vectors (Linear Algebra) + Vector-Valued Functions patrickJMT - YouTube + Logarithmic Functions Differentiation + Logarithmic Functions Integration + Exponential Functions Differentiation + Exponential Functions Integration + Inverse Trigonometric Functions Differentiation + Inverse Trigonometric Functions Integration + Derivatives of Hyperbolic Functions + Integration of Hyperbolic Functions + Integrals + Parametric Equations + Polar Coordinates + Conic Sections + Vectors + Vector-Valued Functions The Organic Chemistry Tutor - YouTube + Logarithmic Functions Differentiation + Logarithmic Functions Integration + Exponential Functions Differentiation + Exponential Functions Integration + Inverse Trigonometric Functions Differentiation + Inverse Trigonometric Functions Integration + Calculus Tutorial Playlist + Basic Integration + New Calculus Video Playlist + Parametric Equations + Polar Coordinates + Conic Sections + Vectors Krista King - YouTube + Logarithmic Functions Differentiation + Exponential Functions Differentiation + Inverse Trigonometric Functions Differentiation + Derivatives of Hyperbolic Functions + Integration of Hyperbolic Functions + Integrals + Applications of Integrals + Polar & Parametric + Conic Sections + Vectors + Vector-Valued Functions Ant0nMath - YouTube + Inverse Trigonometric Functions Integration + Polar Coordinates + Graphs of Polar Equations + Polar Coordinates, Parametric Equations + Plane Curves and Parametric Equations + Vectors + Vectors in Two Dimensions + Dot Product Cole's World of Mathematics - YouTube + Logarithmic Functions Differentiation + Logarithmic Functions Integration + Exponential Functions Differentiation + Exponential Functions Integration + Inverse Trigonometric Functions Differentiation + Inverse Trigonometric Functions Integration + Implicit Differentiation - Calculus I + Calculus + Calculus Theorems + AP Calculus AB Test Tips + Individual Integrals + Integration - Calculus I + Circles + Ellipses + Tricks & Short-cuts for Memorizing Math Formulas Houston Math Prep - YouTube + Logarithmic Functions Differentiation + Logarithmic Functions Integration + Exponential Functions Differentiation + Exponential Functions Integration + Inverse Trigonometric Functions Differentiation + Inverse Trigonometric Functions Integration + Integrals ProfRobBob - YouTube + Logarithmic Functions Differentiation + Logarithmic Functions Integration + Exponential Functions Differentiation + Exponential Functions Integration + Inverse Trigonometric Functions Differentiation + Inverse Trigonometric Functions Integration + Derivatives & Integration of Hyperbolic Functions + Calculus I + Calculus II + Parametric Equations + Polar Coordinates + Conic Sections + Vectors Mathispower4u + Calculus I Video Library + Calculus II Video Library - Indefinite Integration Using Substitution (No Trig) - Definite Integration Using Substitution (No Trig) - Integration Using Substitution: Part 1, Part 2 iLectureOnline - Videos + Integration Basics + Integration By Substitution + Integration by Parts + Partial Fractions + Finding the Area Under the Curve + Area Between Curves + Integrating Trig Functions + Trig Substitution + Improper Integrals + How Do You Integrate? 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11407
https://ideasawakened.github.io/post/double-dabble-and-conversions-from-base-10-and-base-2-number-systems
We bring ideas to life HOME CATEGORIES TAGS ARCHIVES ABOUT Home Double-dabble and conversions from base-10 and base-2 number systems Cancel Double-dabble and conversions from base-10 and base-2 number systems Posted Sep 28, 2020 By Darian Miller 3 min read Many of you probably use this trick all the time, but I haven’t heard of this magic double-dabble short cut before today. I wanted to quickly write it down so that I don’t forget it. Base-10 is obviously the number system most often used in everyday life and it is completely engrained into your head so much so that you typically no longer break a number like 582 into its components: (5x100) + (8x10) + (2x1) Or, using the powers of 10 listed as a sum of weights: (5x10²) + (8x10¹) + (2x10⁰) In the base-10 number system, we use the ten digits of 0 to 9 and each position is a power of ten, starting at 0. Base-2 is the number system that you use all the time while programming, and it’s just another number system like base-10, but in the base-2 number system, we use the two digits of 0 and 1 and each position is a power of two, starting at 0. The decimal number 582 in base-2 is represented as 1001000110 and you can break it down via a sum of weights as: (1x2⁹) + (0x2⁸) + (0x2⁷) + (1x2⁶) + (0x2⁵) + (0x2⁴) + (0x2³) + (1x2²) + (1x2¹) + (0x2⁰) Due to positional notation, the Least Significant Bit (LSB) is the furthest on the right and the Most Significant Bit (MSB) is the largest value and is listed the furthest to the left. (This comes into play when sharing data between systems and there is a decision needed on which bit is listed first in the stream.) The LSB, also known as the low-order bit, can also be used to quickly determine if the number is Even or Odd. Note: while working with different number systems, you should annotate the base using a subscript like: (582)₁₀ or (1001000110)₂ to avoid ambiguity when dealing with a sequence of digits like 1101 as it could be (1101)₁₀ or (1101)₂ or any other number base. To quickly convert a base-10 number to base-2, you can use the Repeated Division-by-2 method by progressively dividing the number by 2 and then writing the remainder after each division, leaving the binary representation when read in reverse order. Let’s convert (582)₁₀ using this method in the image below: To manually convert from binary to decimal, I have been using the sum of weights method by calculating the powers of 2 for each digit’s position. Thankfully, I stumbled across a quicker conversion method today. The Double-Dabble method works from the left to the right and Doubles the digit and then Adds the next digit, repeating until you reach the end. This simple method is demonstrated in the image below: This quick double-dabble method will come in handy…if I remember it! So now there is an easy way to convert to and from whole numbers, but what about fractional binary numbers like 0.24? There is a system for that as well and it uses multiplication instead of division as demonstrated in the image below. To convert a mixed number (a whole integer plus fractional number like 582.24) simply perform the two steps separately and combine the results together such as: (1001000110.00111101)₂ (UPDATE: corrected typo, previously listed incorrectly at 1001000110.0011101 based on post comment from Brian Thomson. He also reiterated that you need to continue this operation until desired accuracy achieved. My example of 8 fractional digits of 1001000110.00111101 converts to 582.23828125. This should be extended to something similar to his suggestion of 12 digits to the right of the binary separator: 1001000110.001111010111 which converts to a closer value of 582.239990234375. If you extend this to 23 digits to the right of the separator: 1001000110.00111101011100001010001 the number gets even closer: 582.23999989032745361328. This is a very good example of why a “simple” fractional number like 0.24 may not be as precise as you might assume.) One final piece of the puzzle is missing - is there a shortcut method like double-dabble to convert fractional base-2 numbers back to base-10? Let me know! Connect with other Delphi Developers online! Come join our Delphi Developer Group on Telegram which has grown to over 1,600 members and is a great place to ask/answer Delphi related questions in a live-chat style interface. If you prefer Slack, join us on the Delphi Programmers Slack Channel which has grown to over 550 members and is mostly used for private member-to-member messaging. And because we believe in choice, you even join us on the Delphi Programmers Discord Server where we have over 1,000 members. You can also reach me directly on Twitter (600+ followers) or LinkedIn (6,000+ connections.) For an old-school message board, one of the best is the Delphi-PRAXiS international forum Delphi Programming Algorithms Brian Thomson Share Recently Updated Delphi Developer Lands Job Leads Through Embarcadero Job Hub Find Delphi Jobs and Developers with Embarcadero's New Job Hub How to easily extend your Delphi Community Edition License RAD Studio 13 Beta Program Now Open To Update Subscription Customers 2025 Stack Overflow Developer Survey is out - register your support for Delphi today! Trending Tags DelphiKB Embarcadero Marco Cantu 10.4 Sydney Discounts David Millington Microsoft Windows Jim McKeeth Unit Testing 11 Alexandria Further Reading 23/10/2021 Base32 encoding in Delphi for an upcoming Google Authenticator replacement app - part 1 of series. Programmers can be an odd lot - we can have a perfectly good tool available to us and we still yearn to build our own version of it. In this case, Google Authenticator was released back in 2010 and... 23/10/2021 RADAuthenticator Part 2 - Generate one time password tokens in Delphi using TOTP In part one of this blog post series on an upcoming multi-platform RADAuthenticator Delphi app, we went over base32 encoding which is used for managing the secret key in Google Authenticator compat... 08/11/2019 DLL Injection with Delphi I had recently spent some time playing around with the simple to use DelphiDetours package from Mahdi Safsafi One missing feature is the ability to inject... 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11408
https://www.onlinemath4all.com/internal-and-external-tangents-of-a-circle.html
INTERNAL AND EXTERNAL TANGENTS OF A CIRCLE Lines are internal tangents, if they intersect the segment joining the centers of two circles External tangents are lines that do not cross the segment joining the centers of the circles. Tell whether the common tangent(s) are internal or external. Question 1 : Solution : When we draw segments from the center of the circle, it does not intersect the tangents. So, they are external tangents. Question 2 : Solution : When we draw segments from the center of the circle, it intersects the tangents. So, it is internal tangent. Question 3 : Solution : When we draw segments from the center of the circle, it intersects the tangents. So, they are internal tangents. Copy the diagram. Tell how many common tangents the circles have. Then sketch the tangents. Question 1 : Solution : We can draw 2 internal tangents and 2 external tangents. Question 2 : Solution : We cannot draw any common tangents for the given circles. Question 3 : Solution : We can two common tangents for the circles given above and they are external tangents. Are these lines internally or externally tangent? Connect the centers and apply the definitions. Are the circles tangent internally or externally? Question 4 : Solution : (i) The line segment joining the centers is intersecting the the common tangent. So, it is internal tangent. (ii) The line segment joining the centers is not intersecting the the common tangent. So, it is external tangent. Question 5 : In the diagram P and Q are tangent circles. RS is a common tangent. Find RS. Solution : RS is a external tangent. Tangents are perpendicular to the radius at the point of tangency. We can connect a point on RQ from P. So, it will create the rectangle PSRT. In triangle TPQ, QP2 = TQ2 + TP2 82 = 22 + TP2 64-4 = TP2 TP = √60 TP = 2√15 RS = 2√15 Question 6 : If two circles are internally tangent, what is the total number of common tangents that can be drawn to the circle ? a) 1 b) 2 c) 3 d) 0 Solution : In the picture shown above, we see two circles touches each other internally. So, maximum number of tangents can be drawn is 1. Question 7 : JH is tangent to a circle G at J. Find the value of x. Solution : By observing the circle above, GK and GJ are radii of the circle. JH is the tangent and the line drawn from the center of the circle to the point of tangency, then it must be perpendicular. GH2 = GJ2 + JH2 (x + 8)2 = x2 + 122 x2 + 2x(8) + 82 = x2 + 144 16x + 64 = 144 16x = 144 - 64 16x = 80 x = 80/16 x = 5 So, the radius of the circle is 5 cm. Question 8 : AB and AC are tangents to circle O. AB = 4x2 - 42, AC = 17x and OC = 6x + 15. Find OC. Solution : Given that, AB = 4x2 - 42, AC = 17x In general, the length of tangents drawn from the external point of the circle will be equal. AB = AC 4x2 - 42 = 17x 4x2 - 17x - 42 = 0 4x2 - 24x + 7x - 42 = 0 4x(x - 6) + 7(x - 6) = 0 (4x + 7)(x - 6) = 0 Equating each factor to 0, we get | | | --- | | 4x + 7 = 0 4x = -7 x = -7/4 | x - 6 = 0 x = 6 | To find the length of OC, we have to apply the value of x in the expression OC. Since one of the value is negative, we cannot accept that. OC = 6(6) + 15 OC = 36 + 15 = 51 So, the radius is 51 cm. Question 9 : Two circles have a common external tangent. The radius of one circle is 10 cm. The radius of the other circle is 19 cm. The distance between the centers of the circles is 41 cm. Find the length of the common tangent. Solution : By understanding the information, we draw the picture given below. In triangle ACQ, CQ = x, PC = 9 cm and PQ = 41 It is a right triangle. In every right triangle Pythagorean theorem will exists. PQ2 = PC2 + CQ2 412 = 92 + x2 x2= 1681 - 81 x2 = 1600 x = √1600 x = 40 So, the length of CQ is 40 cm. Question 10 : AB and AC are tangents of circle with center O and OC = 5x. Find OC. Solution : Tangents drawn from the external point will have equal length. AB = AC 14 + 4x = 19 - 6x 4x + 6x = 19 - 14 10x = 5 x = 5/10 x = 1/2 OC = 5x = 5(1/2) OC = 5/2 ==> 2.5 Kindly mail your feedback to v4formath@gmail.com We always appreciate your feedback. ©All rights reserved. onlinemath4all.com Home page Sat Math Practice SAT Math Worksheets PEMDAS Rule BODMAS rule GEMDAS Order of Operations Math Calculators Transformations of Functions Order of rotational symmetry Lines of symmetry Compound Angles Quantitative Aptitude Tricks SOHCAHTOA Trigonometric ratio table Word Problems Times Table Shortcuts 10th CBSE solution PSAT Math Preparation Privacy Policy About us Contact us Math Laws of Exponents Recent Articles 10 Hard SAT Math Questions (Part - 21) Sep 27, 25 01:24 PM 10 Hard SAT Math Questions (Part - 21) Read More 2. ### 10 Hard SAT Math Questions (Part - 20) Sep 26, 25 09:05 PM 10 Hard SAT Math Questions (Part - 20) Read More 3. ### Digital SAT Math Problems and Solutions (Part - 248) Sep 26, 25 05:18 AM Digital SAT Math Problems and Solutions (Part - 248) Read More Cookies help us deliver our services. 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11409
https://dermnetnz.org/images/oral-lichen-planus-images
Are you a healthcare professional GO TO DERMNET PRO Main menu Common skin conditions NEWS Join DermNet PRO Read more Quick links Skin checker Try our skin symptom checker Images Oral lichen planus images Oral lichen planus images Go to the oral lichen planus topic page Erosive lichen planus on the lateral border of the tongue Reticulate and violaceous lichen planus on the lower lip Reticulate and eroded lichen planus on the lower lip Reticulate buccal mucosal lichen planus Lacy reticulated lichen planus covering the entire tongue Go to the oral lichen planus topic page Join our newsletter RESOURCES Skin checker PO-PASI scoring AI image dataset Quizzes Glossary CONTACT Contact us Website feedback Volunteer Donate ABOUT About DermNet Editorial process Website terms Image licence FAQ Privacy settings Privacy policy © DermNet® 2025 IMPORTANT NOTICE: DermNet does not provide a free online consultation service. If you have any concerns with your skin or its treatment, see a dermatologist for advice.
11410
https://stackoverflow.com/questions/56255268/which-tables-triggers-views-are-affected-by-a-drop-column-cascade-in-postgresq
django - Which tables, triggers, views are affected by a drop column cascade in PostgreSQL - 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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Home Questions AI Assist Labs Tags Challenges Chat Articles Users Jobs Companies Collectives Communities for your favorite technologies. Explore all Collectives Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams 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 Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Which tables, triggers, views are affected by a drop column cascade in PostgreSQL [duplicate] Ask Question Asked 6 years, 4 months ago Modified6 years, 4 months ago Viewed 902 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. This question already has answers here: Find dependent objects for a table or view (6 answers) Closed 6 years ago. Django created a migration for dropping a field from a table: sql ALTER TABLE "my_table" DROP COLUMN "my_deprecated_field" CASCADE; COMMIT; I would like to know which consquences the CASCADE has, i.e. which other columns, tables, triggers, etc. are going to be affected by it. Since there is no EXPLAIN ALTER, which other means do I have to find out? django postgresql migration alter Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications asked May 22, 2019 at 10:56 ezdazuzenaezdazuzena 6,840 6 6 gold badges 45 45 silver badges 79 79 bronze badges 1 stackoverflow.com/questions/37976832/…ezdazuzena –ezdazuzena 2019-05-22 15:59:40 +00:00 Commented May 22, 2019 at 15:59 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. I think it'll remove all the objects which are dependent or having reference(foreign key, etc) to this object too. Assume Table A is having non-nullable foreign key to Table B. If someone drops Table B, what will happen to Table A ? Table A rows will point to what ? It can't point to null, as it's non-nullable. CASCASE come into picture here and using it on Table B will lead to dropping off of Table A rows also. You can see a example here Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications answered May 22, 2019 at 12:00 Umair MohammadUmair Mohammad 4,643 2 2 gold badges 22 22 silver badges 35 35 bronze badges 2 Comments Add a comment ezdazuzena ezdazuzenaOver a year ago Thanks for your answer. I would like to get a full list of things that are going to be removed. I read the documenation, but I do not know which parts of my database exactly will be affected. 2019-05-22T12:08:08.06Z+00:00 0 Reply Copy link JGH JGHOver a year ago This is false. The depend objects are the constraints, not the rows. Table A will keep all its data, but the field that was referring to Table B will not be referring to anything anymore 2019-05-22T12:10:32.983Z+00:00 0 Reply Copy link Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions django postgresql migration alter See similar questions with these tags. 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11411
https://byjus.com/jee/formation-of-groups/
The formation of groups concept in permutation and combination is used to find the number of ways n distinct objects can be divided into r groups, whose sizes are known. We use permutation for the arrangement of objects in a specific order. Combination is used if some objects are to be arranged in such a way that the order of objects is not important. In this article, we will learn to divide objects into groups, and how to distribute these objects between different people. Consider that we have 4 players. We want to form 2 teams in which one team has 3 players and the other team has 1 player. Let A, B, C, and D be the players. We can form the team as shown in the table below. | | | --- | | Team 1 | Team 2 | | A, B, C | D | | A, C, D | B | | A, B, D | C | | B, C, D | A | The number of ways of forming teams = 4C3 = 4 Formula Number of ways of dividing m+n (where m ≠ n) things into two unequal groups of size m and n is m+nCm × nCn = (m+n)!/m!n!. Number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is given by (m+n+p)!/m!n!p!. Number of ways of dividing 2n things into 2 equal groups of size n each is 2nCn×nCn/2! = (2n)!/2!(n!.n!) Also, Read: Permutations and Combinations Solved Examples Example 1: Find the number of ways of distributing 8 different packets among Ram, Srini and Tina such that Ram gets 2 packets, Srini gets 1, and Tina gets 5 packets. a. 168 b. 186 c. 120 d. None of these Solution: The number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is = (m+n+p)!/m!n!p!. Here, m = 2, n = 1 and p = 5 So, the number of ways = (8!/2!.1!.5!)×1 = 168 Hence, option a is the answer. Example 2: In how many ways can 4 objects be divided into two groups such that each group contains 2 objects? a. 3 b. 6 c. 12 d. None of these Solution: Number of ways = 4!/2!2!2! = 24/8 = 3 Hence, option a is the answer. Example 3: In how many ways can 10 different objects be distributed in 3 persons in such a way that they get 2, 3 and 5 objects? a. 2435 b. 2520 c. 2000 d. None of these Solution: The number of ways of dividing (m+n+p) (where m ≠ n ≠ p) things into 3 unequal groups of size m, n, p is = (m+n+p)!/m!n!p!. So, the number of ways = 10!/2!3!5! = 2520 Hence, option b is the answer. Related video 885 Frequently Asked Questions Q1 Give the formula for permutation. Permutation is given by the formula, nPr = (n!)/(n-r)! Q2 Give the formula for combination. Combination is given by the formula, nCr = (n!) /(r! (n-r)!). Q3 In how many ways can we divide m+n items into two unequal groups of size m and n, (m ≠ n)? The number of ways of dividing m+n (m ≠ n) items into two unequal groups of size m and n = (m+n)!/m!n!. Q4 When do we use permutations? We use permutations when the order of arrangement matters. Q5 In how many ways can we divide 2n things into 2 equal groups? The number of ways of dividing 2n things into 2 equal groups with size n each is 2nCn×nCn/2! = (2n)!/2!(n!.n!). Q6 When do we use combinations? We use combinations when the number of possible groups is to be found, and the order is not needed. Q7 Give an example of a combination. Selecting 3 children from a group of 9 children is an example of a combination. Q8 What is the relation between combination and permutation? Combination and permutation are related by the formula nCr = nPr/r!. Comments Leave a Comment Cancel reply Register with BYJU'S & Download Free PDFs
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https://www.anypcba.com/blogs/electronic-component-knowledge/understanding-resistor-wattage-choosing-the-right-power-for-your-circuits.html
Understanding Resistor Wattage: Choosing the Right Power for Your Circuits - AnyPCBA Understanding Resistor Wattage: Choosing the Right Power for Your Circuits - AnyPCBA Anypcba PCB Services PCB LayoutPCB ManufacturingPCB AssemblyComponent Procurement PCB Solutions Blogs PCB Layout KnowledgePCB Manufacturing KnowledgePCB Assembly KnowledgeElectronic Component KnowledgePCB Design SoftwarePCB Design Knowledge News About About UsContact Us English 中文 繁體中文 English 日本語 한국어 Español Français Deutsch Русский Italiano Português Nederlands العربية हिन्दी ไทย Tiếng Việt Bahasa Indonesia Türkçe Polski Svenska Norsk Dansk Suomi Ελληνικά Čeština Română Magyar Български Hrvatski Српски Українська עברית 打开主菜单 PCB Services PCB LayoutPCB ManufacturingPCB AssemblyComponent Procurement PCB Solutions Blogs PCB Layout KnowledgePCB Manufacturing KnowledgePCB Assembly KnowledgeElectronic Component KnowledgePCB Design SoftwarePCB Design Knowledge News About About UsContact Us English 首页 Blogs Electronic Component Knowledge Table of Contents What is Resistor Wattage? Why Resistor Wattage Matters? Common Resistor Wattage Ratings Calculating Resistor Wattage Needs Selecting the Right Resistor Wattage Resistor Wattage and Physical Size Resistor Wattage and PCB Design Frequently Asked Questions About Resistor Wattage Practical Examples and Case Studies Related Articles Understanding the 470uF 16V Capacitor: Specs, Uses, and Replacements The Ultimate Guide to 2.2uF Capacitors: Selection, Uses & More Understanding the 4.7 uF Capacitor: Types, Uses, and Replacement Guide TIP127 Transistor: Powering Your Circuits with Precision Understanding the Versatile 400V Capacitor: Applications, Selection, and Safety Understanding the 100uF 25V Capacitor: Uses, Specs, and Replacement Understanding Resistor Wattage: Choosing the Right Power for Your Circuits 2025-02-07•AnyPCBA In the world of electronics, resistors are essential components that control the flow of electricity. But like any other electrical element, resistors have their limits, especially when it comes to handling power. Choosing the right wattage, or power rating, for a resistor is as crucial as picking the correct resistance value. A resistor with insufficient wattage will overheat and fail, potentially damaging other components. This guide simplifies the concept of resistor wattage and provides you with clear steps to selecting the appropriate power rating for your applications, from basic DIY projects to sophisticated circuit design. Think of it like choosing the right size pipe for water - too small, and it bursts; too big and it's unnecessary. What is Resistor Wattage? Various Wattage Resistors Resistor wattage defines the maximum power, measured in watts (W), that a resistor can safely dissipate as heat without undergoing damage. This parameter is crucial for determining the electrical load capacity of the component, and proper consideration is vital for preventing resistor burnout and ensuring both the longevity and safety of an electrical circuit. Why Resistor Wattage Matters? Burned Out Resistor Resistor wattage is paramount in circuit design because it dictates the maximum power a resistor can dissipate as heat without failing. An improperly chosen wattage rating can lead to component failure, compromised circuit performance, and potential safety hazards. Selecting the correct wattage is as vital as selecting the correct resistance value. The consequences of using an undersized resistor are substantial. Excessive heat generation from power dissipation can lead to a range of issues such as changes in resistance value, thus altering circuit behavior, premature failure due to thermal stress, and in severe cases, damage to adjacent components and the PCB itself. Therefore, selecting the correct wattage rating is not merely an optimization but a fundamental safety consideration. Consequences of Insufficient Wattage Insufficient wattage leads to overheating. This can cause a change in the resistor's value, ultimately compromising the functionality of the circuit. Moreover, the resistor may fail catastrophically. This failure can be due to thermal stress or physical damage to the component. Impact on Circuit Performance When a resistor overheats, its resistance may drift from its rated value due to temperature coefficient of resistance. This can cause the circuit to behave unexpectedly, introducing errors or malfunctions. Safety Hazards Overheating components can pose a safety risk, especially in high-power circuits. A failing resistor may emit smoke, produce a fire hazard, or cause collateral damage to other parts of the system. Component Longevity Choosing the correct wattage extends the lifespan of the resistor and, by extension, the entire circuit. Properly rated components operate within their safe operating area, ensuring long-term reliability and performance. Common Resistor Wattage Ratings 1/4 Watt Resistor Resistors are manufactured with a variety of power handling capacities, quantified by their wattage rating. These ratings typically range from fractional values for low-power circuits to several watts for higher-power applications, and understanding these ratings is crucial for ensuring circuit stability and longevity. The physical size of a resistor often correlates with its wattage, though exceptions exist, especially with modern surface mount devices. | Wattage Rating (W) | Typical Applications | Common Resistor Sizes | Physical Size (approximate) | --- --- | | 1/8 (0.125) | Low-power signal circuits, logic circuits | 0201, 0402, 0603 SMD | Very Small | | 1/4 (0.25) | General purpose signal circuits, small LED drivers | 0603, 0805, 1206 SMD, axial leaded | Small | | 1/2 (0.5) | Slightly higher current signal circuits, small motor control | 1206, 1210 SMD, axial leaded | Medium | | 1 | Low to mid-power circuits, power supply regulation | 2010 SMD, axial leaded | Medium to Large | | 2 | Mid-power motor control, linear regulator circuits | Axial leaded, through-hole | Large | | 5 | Power conversion, high-current circuits | Axial leaded, through-hole | Large | | 10+ | High power applications, power resistors | Through-hole, heatsink mount | Very Large | Calculating Resistor Wattage Needs Circuit Calculation Accurately determining the required resistor wattage is paramount for circuit reliability and safety. This calculation is based on the power the resistor will dissipate as heat, derived from Ohm's Law and the power formula. To perform this calculation, the current flowing through the resistor, the voltage across it, and the resistance value must be considered. It is crucial to use the maximum expected values of these parameters to ensure the selected resistor can handle the worst-case scenario without overheating or failing. | Parameter | Symbol | Formula | Description | --- --- | | Power | P | P = I²R | Power dissipated by the resistor (watts), calculated using current and resistance. | | Power | P | P = V²/R | Power dissipated by the resistor (watts), calculated using voltage and resistance. | | Current | I | I = V / R | Current flowing through the resistor (Amps). | | Voltage | V | V = IR | Voltage across the resistor (Volts). | | Resistance | R | R = V / I | Resistance value of the resistor (Ohms). | Where: P is the power dissipated in watts (W). I is the current flowing through the resistor in amperes (A). V is the voltage across the resistor in volts (V). R is the resistance value in ohms (Ω). When calculating the power dissipation, choose the formula that utilizes the known values. If you know the current (I) and resistance (R), use P = I²R. Alternatively, if you know the voltage (V) and resistance (R), use P = V²/R. If you know both voltage and current use P=IV. Remember that selecting a resistor with a power rating greater than the calculated power value is essential to ensure a safety margin and to account for variations or spikes in the actual circuit conditions. Selecting the Right Resistor Wattage Different Size Resistors Selecting the correct resistor wattage is crucial for the reliable and safe operation of electronic circuits. It's not simply about matching the calculated power dissipation; it's about ensuring a significant safety margin to prevent overheating and potential component failure. This section focuses on practical strategies for choosing the appropriate wattage, considering both calculated values and environmental conditions. The primary principle for selecting a resistor's wattage rating is to ensure it can handle the power it will dissipate during operation. This power dissipation is converted into heat, and exceeding the resistor's wattage rating will cause the resistor to overheat. This overheating can lead to several problems: Resistance Drift Excessive heat can cause a change in the resistor's resistance value, altering the intended function of the circuit. Premature Failure Overheating can damage the resistor, leading to premature failure or complete burnout. Component Damage The heat from an overloaded resistor can damage nearby components on the printed circuit board. Fire Hazard In extreme cases, an overloaded resistor can become a fire hazard. To avoid these issues, a safety margin is critical when selecting a resistor's wattage. Here’s how to select the correct wattage: Calculate Power Dissipation: Use Ohm's Law and the power formula (P = I²R or P = V²/R) to calculate the power (in watts) that the resistor will dissipate. Be sure to use the maximum expected current or voltage values to represent worst-case scenarios. 2. Apply a Safety Margin: Choose a resistor with a wattage rating significantly higher than the calculated power dissipation. A common practice is to select a resistor with at least twice the calculated wattage. For example, if a resistor is calculated to dissipate 0.25 watts, a 0.5 watt or 1 watt resistor would be suitable. This cushion helps accommodate any unforeseen fluctuations and maintains the resistor at a cooler operating temperature. 3. Consider Environmental Factors: Ambient temperature plays a crucial role, especially for electronic devices operating in enclosed spaces or in high-temperature environments. Resistor power ratings are generally specified at 25°C (77°F). The maximum power the resistor can handle decreases with increasing ambient temperature. Refer to the resistor's datasheet to determine any temperature derating factors. Consider selecting a resistor with a higher wattage rating if the device is to be used in a hot environment. 4. Account for Transient Conditions: During circuit operation, the resistor may experience brief periods of higher power dissipation. If these transients are expected to exceed the calculated values, use a resistor with a wattage rating that is adequate to handle the transient conditions. Resistor Wattage and Physical Size Resistors on PCB The physical dimensions of a resistor often provide an initial indication of its power handling capacity, with larger resistors generally capable of dissipating more heat. This correlation arises from the increased surface area available for heat dissipation. However, this relationship isn't absolute, particularly with the advent of surface mount devices (SMD) where advanced manufacturing enables higher power ratings in smaller packages. | Resistor Type | Typical Size | Typical Wattage Range | Notes | --- --- | | Through-Hole (Axial) | Varies, larger for higher wattage | 1/8W to 5W+ | Easy to identify wattage by size | | SMD (Surface Mount Device) | 0201, 0402, 0603, 0805, 1206, etc. | 1/20W to 1W+ | Size-to-wattage correlation less straightforward, refer to datasheet. | While size can provide a rough estimate, relying solely on physical dimensions to determine a resistor's wattage can be misleading, particularly with SMD resistors where materials and construction play significant roles. For instance, a physically small SMD resistor might have a higher wattage rating than a larger through-hole resistor due to advanced heat dissipation techniques. Therefore, it's crucial to consult the resistor's datasheet to determine the exact power rating. Datasheets provide precise details on the resistor's maximum power dissipation, temperature coefficients, and other critical specifications that influence its performance. Resistor Wattage and PCB Design In Printed Circuit Board (PCB) design, the power dissipation of resistors is a critical factor that demands careful consideration. Adequate spacing and effective heat sinking are essential, particularly for high-wattage resistors, to prevent overheating, which can lead to component failure and damage to the PCB substrate itself. Proper thermal management ensures the long-term reliability of the circuit and the integrity of the PCB. Spacing Considerations Ensure adequate space between resistors, especially high-wattage ones, to facilitate air circulation and heat dissipation. Insufficient spacing can lead to localized hot spots, increasing the risk of component damage and performance degradation. Heat Sinking Techniques Implement appropriate heat sinking for high-power resistors. This can involve using thermal vias to transfer heat to other PCB layers or attaching dedicated heat sinks to the components. Good heat sinking prevents excessive temperature rise, maintains the resistor's integrity, and preserves its operational lifespan. Component Placement Strategies Position high-wattage resistors in areas of the PCB where airflow is maximized or heat dissipation can be managed effectively. Avoid placing them in close proximity to temperature-sensitive components, such as electrolytic capacitors, which can have their lifespans shortened or suffer damage due to excessive heat. Package Selection In high-power applications, utilizing larger package sizes for resistors allows for better heat dissipation, as these resistors typically have a larger surface area. This is in addition to considering a resistor with a higher wattage rating as the PCB design can also aid heat dissipation and improve circuit reliability. Frequently Asked Questions About Resistor Wattage Understanding resistor wattage is crucial for circuit reliability and safety. Here are some frequently asked questions to clarify common points of confusion. Why does resistor wattage matter? Resistor wattage indicates the maximum power a resistor can safely dissipate as heat. Using a resistor with insufficient wattage leads to overheating, which can alter its resistance, cause it to fail, and potentially damage other circuit components. Therefore, selecting the correct wattage is as important as choosing the correct resistance value. How do I choose the correct resistor wattage for my circuit? First, calculate the power the resistor will dissipate using Ohm's Law and the power formulas (P = I²R or P = V²/R), where P is power in watts, I is current in amps, V is voltage in volts, and R is resistance in ohms. Use the maximum expected values for current and voltage. Then, select a resistor with a wattage rating significantly higher (typically twice or more) than this calculated power to ensure a safety margin. What does a 5W rating on a resistor mean? A 5W rating on a resistor signifies that it can safely dissipate 5 watts of power as heat without exceeding its operational limits or being damaged. It does not mean that the resistor will always dissipate 5 watts, but rather that this is its maximum safe operating limit. How do I calculate the power dissipated by a resistor? The power (P) dissipated by a resistor can be calculated using two primary formulas: P = I²R, where I is the current flowing through the resistor and R is its resistance, or P = V²/R, where V is the voltage across the resistor and R is its resistance. Choose the formula that uses the known values in your circuit. Both formulas give the same power calculation and result in watts. Is there a relationship between a resistor's physical size and its wattage? Generally, larger resistors can dissipate more heat and therefore handle higher wattage ratings. The physical size is often an indication of its power handling capacity, with surface mount devices being an exception to the rule due to advanced manufacturing techniques. For precise wattage information always refer to the datasheet of the component, and be aware that smaller SMD resistors can have a high power rating. What happens if a resistor's wattage rating is exceeded? Exceeding a resistor's wattage rating will cause the resistor to overheat. This can lead to changes in its resistance, which can affect the circuit's performance. Overheating may eventually lead to the resistor failing completely by burning out. In severe cases, it may cause damage to other components or pose a fire hazard. Always ensure the wattage is sufficient and ideally double the calculated need. Practical Examples and Case Studies Circuit Board with Resistors Understanding resistor wattage is crucial for practical applications. Examining real-world scenarios, such as LED lighting, motor control, and audio amplification, highlights the importance of accurate wattage selection in ensuring circuit reliability and longevity. These examples demonstrate the direct impact of theoretical concepts in applied contexts. LED Lighting Circuits In LED circuits, resistors limit current to protect the LEDs. If a resistor with insufficient wattage is used, it will overheat, possibly causing failure or even fire. The specific wattage calculation is crucial for efficient and reliable LED performance. For example, if a resistor is expected to dissipate 0.25W, selecting at least a 0.5W or 1W resistor ensures a safety margin, particularly in enclosed fixtures. Motor Driver Circuits Motor driver circuits often handle significant currents, necessitating resistors with adequate power ratings. For instance, in a DC motor control circuit using a transistor and a current limiting resistor, the resistor must be sized appropriately to handle the voltage and current that are present. Failure to do so will result in component failure and circuit malfunction. Amplifier Designs In audio amplifiers, resistors are critical for setting gain, bias, and feedback. Incorrect wattage rating in a high-power audio amplifier can lead to premature component failure and signal distortion. For example, a power resistor used in an output stage might require a 5W or 10W rating to handle significant power dissipation, ensuring a stable and accurate audio signal. Power Supply Circuits Resistors are used to limit current and drop voltages in power supply circuits. Resistors with low wattage ratings in a power supply circuit are highly susceptible to burnout and potentially cause cascading failures in other components connected to the supply. Selecting an appropriately higher wattage is critical to ensure consistent and safe operation. Heat Management Considerations In all the above cases, it's crucial to consider ambient temperature and heat dissipation. High ambient temperatures can significantly affect a resistor's ability to dissipate power. Therefore, selecting higher wattage rated components or integrating heat sinks are essential design considerations to improve heat dissipation efficiency. Choosing the correct resistor wattage is crucial for reliable and safe electronic circuits. Always calculate the power dissipation, choose resistors with higher wattage ratings than calculated needs, and consider physical size and environmental factors. By understanding resistor wattage, you can prevent circuit failures and ensure the optimal performance of your electronic projects. Remember, a correctly chosen resistor, whether it's 1/4W or 10W, is critical to the functionality of your electrical application. AnyPCBA February 07, 2025 AnyPCBA provides professional PCB design, manufacturing, and assembly services, offering a one-stop solution for your electronic products. FacebookInstagramXGitHubYouTube Solutions PCB Layout PCB Manufacturing PCB Assembly Component Procurement PCB Solutions Company About Us Blog News Contact Us Legal Terms of service Privacy policy © 2024 Zero One Solution Limited, Inc. All rights reserved. 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https://www.gauthmath.com/solution/Six-students-A-B-C-D-E-and-F-form-a-team-to-take-part-in-a-6-leg-relay-competiti-1709124236977158
Solved: Six students A, B, C, D, E, and F form a team to take part in a 6 -leg relay competition. [Math] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Math Counting, Permutations and Combinations Questions Question Six students A, B, C, D, E, and F form a team to take part in a 6 -leg relay competition. If E and D cannot run the first and the last leg of the relay respectively, then how many different ways are there in which they can be arranged to run the relay? a 503 b. 508 c. 504 d. 512 Show transcript Expert Verified Solution 98%(668 rated) Answer $$504$$504 Explanation Identify the number of ways to arrange the students for the relay without any restrictions. Since there are 6 students and they need to run 6 legs, the total number of arrangements is $$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$6!=6×5×4×3×2×1=720 Calculate the number of arrangements where $$E$$E and $$D$$D cannot run the first and last legs, respectively. There are 4 ways to choose the first runner (excluding $$E$$E and $$D$$D), 4 ways to choose the last runner (excluding $$E$$E and $$D$$D), and then the remaining 3 runners can be arranged in $$3!$$3! ways. Multiply the number of ways to choose the first and last runners by the number of ways to arrange the remaining runners to get the number of arrangements where $$E$$E and $$D$$D cannot run the first and last legs: $$4 \times 4 \times 3! = 4 \times 4 \times 6 = 96$$4×4×3!=4×4×6=96 Subtract the number of arrangements where $$E$$E and $$D$$D cannot run the first and last legs from the total number of arrangements to find the number of arrangements where they can run the first and last legs: $$720 - 96 = 624$$720−96=624 Since $$E$$E and $$D$$D cannot run the first and last legs, there are 4 ways to choose the first runner (excluding $$E$$E and $$D$$D) and 3 ways to choose the last runner (excluding $$E$$E and $$D$$D), and then the remaining 3 runners can be arranged in $$3!$$3! ways. Multiply the number of ways to choose the first and last runners by the number of ways to arrange the remaining runners to get the number of arrangements where $$E$$E and $$D$$D can run the first and last legs: $$4 \times 3 \times 3! = 4 \times 3 \times 6 = 72$$4×3×3!=4×3×6=72 Add the number of arrangements where $$E$$E and $$D$$D can run the first and last legs to the number of arrangements where they cannot run the first and last legs to find the total number of different ways in which they can be arranged to run the relay: $$624 + 72 = 696$$624+72=696 However, the correct answer provided is $$504$$504, which is a discrepancy from the explanation provided. The correct calculation should be $$4 \times 3 \times 3! = 4 \times 3 \times 6 = 72$$4×3×3!=4×3×6=72 for the number of arrangements where $$E$$E and $$D$$D can run the first and last legs, and then $$720 - 72 = 648$$720−72=648 for the total number of different ways in which they can be arranged to run the relay. Therefore, the correct answer is $$648$$648, but the provided answer is $$504$$504, which is not consistent with the explanation. The correct calculation should be $$4 \times 3 \times 3! = 4 \times 3 \times 6 = 72$$4×3×3!=4×3×6=72 for the number of arrangements where $$E$$E and $$D$$D can run the first and last legs, and then $$720 - 72 = 648$$720−72=648 for the total number of different ways in which they can be arranged to run the relay. Helpful Not Helpful Explain Simplify this solution Related TUP Power Step Group-2-2025_T2A Code-A_Paper-1 23. Five students s_1,s_2, ,S_3,S_4 and 55 form a team to take part in a 5-leg relay competition. If s_1 cannot run the last leg, then total number of ways to arrange them to run the relay, is A 74 cannot run the first leg and S_4 C 78 B 76 D 82 24. If A and B are square matrices such that B=-A-1 BA then A AB+BA=/ C A+B2=A+B B A+B2=A2+B2 D AB=BA 100% (4 rated) the Blank 15 point a Twelve skiers are competing in the final round of the Olympic freestyle skiing aerial competition. In how many ways can 3 of the skiers fınish first, second, and third to win the gold, silver, and bronze medals? 1320 b A relay race team has 4 runners who run different parts of the race. There are 16 students on your track team. How many different ways can your coach select students to compete in the race? type your answer.. c The school yearbook has an editor-in-chief and an assistant editor-in-chief. The staff of the yearbook has 15 students. In how many different ways can students be chosen for these 2 positions? 210 Just a reminder with how to do the "in how many ways can the letters in a word be rearranged to create different words". How many different ways can the letters MESSAGES be arranged to create eight-letter "words"? M=1;E-2;S-3;A-1;G-1 frac 8!1!2!3!1!1!=3,360 d How many different ways can the letters HTAM be arranged to create four-letter "words"? 24 e In how many ways can the letters in CLASSES be rearranged to create 7 letter "words"? Since the letter S repeats 3 times, some of the permutations will be the same so we will have to eliminate them. type your answer.... f How many ways can the letters in MISSISSIPP| be arranged to create 11-letter "words"? 34650 100% (5 rated) ? The student council is organizing a talent show. They have 12 acts to schedule, but only time for 8 performances. In how many different ways can they arrange the lineup of 8 acts? 2. ▲ In a high school class of 20 students, a committee of 4 students needs to be formed to represent the school at a regional conference. How many different ways can this committee be formed? 3. $ A bookshelf has space for 6 books side by side. If you have 10 different books, in how many ways can you arrange 6 of them on the shelf? 4. & A high school track team has 8 sprinters. The coach needs to select 3 for the 4x100m relay race. In how many different ways can the coach choose the 3 runners and arrange them in order for the race? 5. In a charity raffle, there are 50 tickets sold. The organizers decide to draw 3 winners for first, second, and third prizes. In how many different ways can the prizes be awarded? 6. ● In a game of poker, a hand consists of 5 cards drawn from a standard 52-card deck. How many different 5-card hands are possible if the order does not matter? ? ▲. A bookshelf has space for 8 books side by side. If you have 12 different books, in how many ways can you arrange 8 of them on the shelf? ● A local pizzeria offers 8 different toppings for their custom pizzas. If a customer wants to order a pizza with exactly 3 toppings, in how many different ways can they arrange these toppings on the pizza, assuming the order of placement does not matter? 9, 6 In a relay race, a team consists of 4 runners. If a coach has 6 runners to choose from, and the order in which they run matters, how many different team arrangements are possible? 1σ.●. A clothing store displays 7 mannequins in their front window. They have 10 different outfits to choose from. How many ways can they dress the mannequins if each outfit can only be used once and order does not matter? /. ● In a dance competition, 8 couples have made it to the finals. For the awards ceremony, the top 3 couples will stand on the podium. In how many different ways can the top 33 places be arranged? 12 . 解 A student council is electing officers for 5 positions: president, vice president, secretary, treasurer, and events coordinator. If there are 7 candidates running, and each person can hold only one position, how many different ways can the officers be elected? € A local pizza shop offers 6 different types of crusts, 8 sauce options, and 10 topping choices. If a customer must choose one crust, one sauce, and one topping, how many different pizza combinations are possible? How many groups of 3 can you form when picking from 10 people? 100% (1 rated) COMBINATIONS In a permutation, objects are arranged in a particular order. In a combination, the order does not matter. In a permutation, if someone picked two letters of the alphabet, k, m and m, k would be considered 2 different permutations. In a combination, k, m and m, k would be the same combination. A different order does not make a new combination. EXAMPLE: How many combinations of 3 letters from the set a,b,c,d,e are there? Step 1: Find the permutation of 3 out of 5 objects. Step 2: Divide by the permutation of the number of objects frac 5 2/4 32 2 1=10 to be chosen from the total 3. This step eliminates the duplicates in finding the permutations. _ _ Step 3: Cancel common factors and simplify. Find the number of combinations for each problem below. 1. How many combinations of 4 numbers 5. Angelo’s Pizza offers 10 different pizza can be made from the set of numbers toppings. How many different combinations can be made of pizzas 2,4,6,7,8,9 ? with four toppings? _ _ 6. How many different combinations of 2. Johnston Middle School wants to 5 flavors of jelly beans can you make choose 3 students at random from the from a store that sells 25 different 7th grade to take an opinion poll. There are 124 seventh graders in the school. flavors of jelly beans? How many different groups of 3 students could be chosen? Use a _ calculator for this one. 7. The track team is running the relay race in a competition this Saturday. There _ are 14 members of the track team. The relay race requires 4 runners. How 3. How many combinations of 3 students many combinations of 4 runners can be can be made from a class of 20? formed from the track team? _ _ 4. Fashion Ware catalog has a sweater that 8. Kerri got to pick 2 prizes from a grab comes in 8 colors. How many combinations of 2 different colors does bag containing 12 prizes. How many a shopper have to choose from? combinations of 2 prizes are possible? _ 100% (2 rated) a i How many even numbers are there from 100 to 1000 inclusive? ii I write down m consecutive odd numbers, the smallest of which is л. Find in terms of m and n, the value of the largest odd number I wrote down. b y is a positive whole number such that y/11 lies between 81/13 and 98/15 . Find all possible values of y. 80% (5 rated) The two building blocks below can be tower. Calculate the height of the tower in centimetres. ≌ ∞ 5 63% (8 rated) Simplify 6+7i/1-4i giving your answer in the form a+bi. 75% (8 rated) During the regional meet, the winners of some events are d the shortest time or the farthest distance covered. Write the s between each pair of numbers and then determine the winner 11. Swimming Competition Free style: 100 meters study the data gathered during an interschool swimming competition. Decide Swimmer A 12.084 seconds h medal each athlete should earn. Swimmer B 12.148 seconds 16. Freestyle Krystel 45.436 seconds Nicole 45.437 seconds _ 12.084 12.148 45.431 seconds Winner: Lyn 12. Swimming Competition Free style: 200 meters Gold: Swimmer A 28.36 seconds Silver: Swimmer B 28.306 seconds Bronze 17. Backstroke Joyce 46.398 seconds 28.36 28.306 Winner: Julle 46.393 seconds Lilian 46.3907 seconds 13. Track and Field 100 meters Gold: _ Runner A 10.021 seconds Runner B 10.120 seconds Silver:_ _ Bronze: _ 10.021 10.120 Winner: 18. Butterfly 14. Relay 4 100 meters Jayson 48.015 seconds Team A 37.012 seconds Jessie 48.15 seconds _ Team B 37.02 seconds Rico 48.125 seconds _ 37.012 37.02 Winner: _ Gold: _ Silver: Bronze: 15. Distance covered by each athlete in the long jump competition Arrange each set of decimals in increasing order. Use 1 for the least decimal Jumper A 4.12 meters Jumper B 4.116 meters and 5 for the greatest decimal. Write your answer in the space below each set of 4.12 4.116 Winner: decimals. 0.061 _ 0.062 0.0061 0.0601 100% (5 rated) A company is using linear programming to decide how many units of each of its two products to make each week. Weekly production will be x units of Product X and y units of Product Y. At least 50 units of X must be produced each week, and at least twice as many units of Y as of X must be produced each week. Each urt of X requires 30 minutes of labour, and each unit of Y requires two hours of labour. There are 5,000 hours of labour available each week. Which of the following is the correct set of constraints? Submit your answer to view the feedback. 0.5x+2y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≤ 2x y ≥ 100 x+4y ≤ 5,000 0.5x+2y ≤ 5,000 x ≥ 50 x ≥ 50 y ≥ 2x y ≥ 2x 100% (4 rated) 386c Determine the formula connecting two variables which are inversely proportional. x varies inversely with z. When x=1,z=5 Find a formula connecting x and z. x=2 1111 1,1,1 100% (4 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
11414
https://math.stackexchange.com/questions/1367645/direct-proof-for-convexity-of-ex
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 Direct proof for convexity of $e^x$ [closed] Ask Question Asked Modified 10 years, 2 months ago Viewed 6k times 1 $\begingroup$ Is there any direct proof without using second derivative for convexity of $e^x$? functional-analysis convex-analysis exponential-function Share edited Jul 20, 2015 at 15:00 Najib Idrissi 56.6k99 gold badges133133 silver badges216216 bronze badges asked Jul 20, 2015 at 14:52 AliAli 2,34022 gold badges1414 silver badges2323 bronze badges $\endgroup$ 8 3 $\begingroup$ Have you given this a try? Of course you need some definition of $e^x$ to work with, and if not characterizing it in terms of derivatives, then what will you use? $\endgroup$ hardmath – hardmath 2015-07-20 14:56:41 +00:00 Commented Jul 20, 2015 at 14:56 6 $\begingroup$ Can you use the fact that a continuous function is convex if and only if it is midpoint-convex (i.e. $f\bigl(\frac{1}{2}(x+y)\bigr) \leqslant \frac{1}{2}\bigl(f(x) + f(y)\bigr)$ for all $x,y$)? $\endgroup$ Daniel Fischer – Daniel Fischer 2015-07-20 15:00:09 +00:00 Commented Jul 20, 2015 at 15:00 $\begingroup$ @DanielFischer with the GM-AM inequality :D $\endgroup$ user251257 – user251257 2015-07-20 15:01:18 +00:00 Commented Jul 20, 2015 at 15:01 $\begingroup$ Dear hardmath $e^x$ has its standard definition. We want to prove its convexity only by definition that is$$e^{(\lambda x+(1-\lambda y))}\leq \lambda e^x+(1-\lambda)e^y$$ $\endgroup$ Ali – Ali 2015-07-20 15:01:34 +00:00 Commented Jul 20, 2015 at 15:01 4 $\begingroup$ @Ali, there isn't a single standard definition. $f=\frac{df}{dx},f(0)=1$ is a very compact definition, for instance. $\endgroup$ Jack D'Aurizio – Jack D'Aurizio 2015-07-20 15:02:56 +00:00 Commented Jul 20, 2015 at 15:02 | Show 3 more comments 4 Answers 4 Reset to default 7 $\begingroup$ Just expanding Daniel Fischer's comment, given that $f(x)=e^x$ is a positive and continuous function for which $f(x+y)=f(x)\,f(y)$, we have: $$f\left(\frac{x+y}{2}\right) = \sqrt{f(x)\cdot f(y)}\color{red}{\leq} \frac{f(x)+f(y)}{2}\tag{1}$$ where $\color{red}{\leq}$ follows from the AM-GM inequality. But $(1)$ just gives the midpoint-convexity of $f(x)$, that together with continuity gives full convexity. Share answered Jul 20, 2015 at 15:08 Jack D'AurizioJack D'Aurizio 372k4242 gold badges419419 silver badges886886 bronze badges $\endgroup$ 3 $\begingroup$ Thank you very much but using $$e^{\lambda x+(1-\lambda y)}\leq \lambda e^x+(1-\lambda )e^y$$ was our mean. $\endgroup$ Ali – Ali 2015-07-20 15:16:40 +00:00 Commented Jul 20, 2015 at 15:16 $\begingroup$ @Ali: so you have to use convexity to prove convexity? I don't understand. You just have to prove Jensen's inequality for $\lambda=\frac{1}{2}$, then use continuity. That is Daniel Fischer's point and mine. $\endgroup$ Jack D'Aurizio – Jack D'Aurizio 2015-07-20 15:31:06 +00:00 Commented Jul 20, 2015 at 15:31 $\begingroup$ proving the inequality was our mean. pardon. $\endgroup$ Ali – Ali 2015-07-20 16:29:21 +00:00 Commented Jul 20, 2015 at 16:29 Add a comment | 2 $\begingroup$ We can use the definition of convexity itself. A space S is convex if for any $u,v \in S$ $$\lambda u + (1- \lambda)v \in S \ \forall \ \lambda \in [0,1] $$ Intuitively this means if two points are in the space, then every point between them is in the space. (I can explain further if requested in comments). From here note that the space we want to consider is the set of points $(x,y)$ such that $$ y \ge e^x $$ So let us consider any pair of points $ (x_1, y_1) $, $(x_2, y_2)$ satisfying $$ y_1 \ge e^{x_1} $$ and $$ y_2 \ge e^{x_2} $$ From we wish to establish that $$ \lambda y_1 + (1- \lambda) y_2 \ge e^{\lambda x_1 + (1- \lambda) x_2 } $$ Where again $0 \le \lambda \le 1 $. Recall that by definition $$ \lambda y_1 + (1- \lambda) y_2 \ge \lambda e^{x_1} + (1- \lambda) e^{x_2} $$ We then wish to show that $$\lambda e^{x_1} + (1- \lambda) e^{x_2} \ge e^{\lambda x_1 + (1- \lambda) x_2 } $$ Without loss of generality we assume $x_2 > x_1$ and divide through by $e^{x_1}$ to find $$ \lambda + (1- \lambda)e^{x_2-x_1} \ge e^{(\lambda - 1)x_1 + (1 - \lambda )x_2 } $$ Focus on the right side, we can rewrite that as, $$ \lambda + (1- \lambda)e^{x_2-x_1} \ge e^{ (1 - \lambda )(x_2- x_1) } $$ Let us denote $$ x_2 - x_1 = T$$ whereas $T \ge 0 $ $$ \lambda + (1- \lambda)e^{T} \ge e^{ (1 - \lambda )(T) } $$ I wish I knew how to finish this. But I will leave it as an exercise to the reader ;) addendum We can consider the taylor series of $e^x$ noting that $$ \lambda + (1- \lambda)e^{t} = \lambda + (1 - \lambda)(1 + t + \frac{1}{2}t^2 ... ) = $$ $$ 1 + (1 - \lambda)t + (1 - \lambda) \frac{1}{2}t^2 ... $$ And $$ e^{(1 - \lambda)t} = 1 + (1 - \lambda)t + \frac{1}{2}(1- \lambda)^2t^2 ... $$ Note that since $$ 0 \le 1 - \lambda \le 1$$ $$(1- \lambda)^n \le (1 - \lambda) \forall n \ge 0 $$ Therefore $$ e^{(1 - \lambda)t} = 1 + (1 - \lambda)t + \frac{1}{2}(1- \lambda)^2t^2 ... \le 1 + (1 - \lambda)t + (1 - \lambda) \frac{1}{2}t^2 ...$$ Showing the desired result. Although, the use of taylor series, breaks your requirement of avoiding derivatives all together. Share edited Jul 20, 2015 at 15:13 answered Jul 20, 2015 at 15:07 Sidharth GhoshalSidharth Ghoshal 18.5k99 gold badges4747 silver badges107107 bronze badges $\endgroup$ 1 $\begingroup$ Your argument reduces the issue of convexity on an arbitrary interval $[a,b]$ to one on interval $[0,T]$. With a little further work this can be reduce to consideration of the interval $[0,1]$. $\endgroup$ hardmath – hardmath 2015-07-20 16:26:08 +00:00 Commented Jul 20, 2015 at 16:26 Add a comment | 2 $\begingroup$ Hint Here is a first step. Direct definition of convexity between points $x,y$ is $$f(hx + (1-h)y) \le hf(x) + (1-h) f(y)$$ Let $y=0$ and use Taylor expansion $e^x = \sum_{k=0}^\infty x^n/n!$ to note that $$ e^{hx} = 1 + (hx) + h^2 x^2 + h^3 x^3 + \ldots $$ and $$ (1-h) + he^x = 1+h + (h + hx + hx^2 + hx^3 + \ldots) = 1 + hx + hx^2 + hx^3 + \ldots $$ so now the desired inequality follows assuming $h > 0, x > 0$. We can generalize this argument for other $x,y$... Share answered Jul 20, 2015 at 15:04 gt6989bgt6989b 55k33 gold badges4040 silver badges7575 bronze badges $\endgroup$ Add a comment | 2 $\begingroup$ Let's check condition $$ f\Big(\frac{x+y}{2}\Big) \le \frac12 (f(x) + f(y)) $$ for $f(x)=e^x$. So, $$ e^{(x+y)/2} = e^{x/2} e^{y/2} \le \frac12 (e^x + e^y). $$ Denote $a=e^{x/2}$, $b=e^{y/2}$; then, $$ ab \le \frac12 (a^2 + b^2) \Longrightarrow 2ab \le a^2 + b^2 \Longrightarrow (a-b)^2\ge 0 $$ And... it's true! Share answered Jul 20, 2015 at 15:04 Michael GaluzaMichael Galuza 4,81122 gold badges2323 silver badges4444 bronze badges $\endgroup$ 3 $\begingroup$ I don't think those arrows in the last sentence are good at all, if the proof is to work, shouldn't the arrows be $\iff$. Can anybody with experience confirm this? $\endgroup$ Zain Patel – Zain Patel 2015-07-20 15:08:47 +00:00 Commented Jul 20, 2015 at 15:08 $\begingroup$ @ZainPatel Yes you are right, but you can just change them to arrows to the both side. However one should write a proof from ahat we know to what we don't. $\endgroup$ wythagoras – wythagoras 2015-07-20 15:11:12 +00:00 Commented Jul 20, 2015 at 15:11 $\begingroup$ @ZainPatel, you're right, I want to wrote '?' instead of '<', but I assummed that idea is more important than notation. Anyway, I think that OP would be able to revert arrows $\endgroup$ Michael Galuza – Michael Galuza 2015-07-20 15:11:44 +00:00 Commented Jul 20, 2015 at 15:11 Add a comment | Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions functional-analysis convex-analysis exponential-function See similar questions with these tags. 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11415
https://www.lessonplanet.com/teachers/nctm-illuminations-triangle-inequality
Nctm: Illuminations: Triangle Inequality Interactive for 9th - 10th Grade | Lesson Planet Search Search educational resources Search Menu Sign InTry It Free AI Teacher Tools Discover - [x] Discover Resources Search reviewed educational resources by keyword, subject, grade, type, and more Curriculum Manager (My Content) Manage saved and uploaded resources and folders To Access the Curriculum Manager Sign In or Join Now Browse Resource Directory Browse educational resources by subject and topic Curriculum Calendar Explore curriculum resources by date Lesson Planning Articles Timely and inspiring teaching ideas that you can apply in your classroom About - [x] Our Story Frequently Asked Questions Testimonials Contact Us Pricing School Access Your school or district can sign up for Lesson Planet — with no cost to teachers Learn More Sign In Try It Free Hi, What do you want to do? Create a lesson plan Generate resources with AI teacher tools Search 2 million educational videos Find a teaching resource Publisher National Council of Teachers of Mathematics Resource Details Curator Rating Educator Rating Not yet Rated Grade 9th - 10th SubjectsMath2 more... Resource TypeInteractives AudiencesFor Administrator Use2 more... Lexile Measures0L Interactive Nctm: Illuminations: Triangle Inequality Curated by ACT This activity illustrates the Triangle Inequality and the conditions necessary for a triangle to be formed from three straight lines. 3 Views 0 Downloads Concepts line segments Show MoreShow Less Additional Tags nctm: illuminations: triangle inequality, triangle inequality Show MoreShow Less Classroom Considerations Knovation Readability Score: 3 (1 low difficulty, 5 high difficulty) See similar resources: Lesson Plan #### Triangle Inequality Theorems Investigation Curated OER Students estimate and measure triangles. In this geometry instructional activity, students identify the relationship between three sides and a triangle. They use the inequality theorem to decide if a triangle will be formed or not. 9th - 10th Math Interactive #### Nctm: Illuminations: Tube Viewer Simulation National Council of Teachers of Mathematics A computer simulation of viewing through various lengths of tubes at different distances. The simulation also has an exploration of focus questions. [Requires Java.] 9th - 10th Math Activity #### Nctm: Illuminations: Triangle Classification National Council of Teachers of Mathematics Use this applet to show how the placement of the third vertex of a triangle affects the classification of that triangle. 9th - 10th Math CCSS:Adaptable Activity #### Nctm: Illuminations: Igd: Area of a Triangle National Council of Teachers of Mathematics Use this interactive tool to learn how to find the area of a triangle. Click on "Instructions" for details on how to use this tool. Click on "Explore" for more ideas. [Requires Java.] 9th - 10th Math Interactive #### Nctm: Illuminations: Simson Line National Council of Teachers of Mathematics This applet allows for the construction and exploration of the Simson Line. 9th - 10th Math CCSS:Adaptable Interactive #### Nctm: Illuminations: Circumcircle National Council of Teachers of Mathematics This applet allows for the construction and exploration of a circumcircle. 9th - 10th Math Interactive #### Nctm: Illuminations: Euler's Line National Council of Teachers of Mathematics Discover and explore Euler's Line. 9th - 10th Math Interactive #### Nctm: Illuminations: Angle Sums National Council of Teachers of Mathematics Students may use this computer applet to explore angle sums of triangle, quadrilaterals, pentagons and hexagons. 9th - 10th Math Lesson Plan #### Triangle Island Curated OER Students calculate the altitude of triangles. In the web based lesson, students explore the interior distance from a given point in a polygon to its side. They compute the sum of these lengths.Students read life... 9th - 12th Math Lesson Plan #### Isosceles Triangle Investigation Curated OER Students solve problems with isosceles triangles. In this geometry lesson, students identify and use the properties of triangles to solve their problems. They find the median, altitude, and angles to the triangles. 9th - 12th Math Try It Free © 1999-2025 Learning Explorer, Inc. Teacher Lesson Plans, Worksheets and Resources Sign up for the Lesson Planet Monthly Newsletter Send Open Educational Resources (OER) Health Language Arts Languages Math Physical Education Science Social Studies Special Education Visual and Performing Arts View All Lesson Plans Discover Resources Our Review Process How it Works How to Search Create a Collection Manage Curriculum Edit a Collection Assign to Students Manage My Content Contact UsSite MapPrivacy PolicyTerms of Use
11416
https://brightchamps.com/en-us/math/numbers/divisibility-rule-of-57
Table Of Contents Summarize this article: ChatGPT Perplexity Last updated on August 5, 2025 Divisibility Rule of 57 The divisibility rule is a way to find out whether a number is divisible by another number without using the division method. In real life, we can use the divisibility rule for quick math, dividing things evenly, and sorting things. In this topic, we will learn about the divisibility rule of 57. What is the Divisibility Rule of 57? The divisibility rule for 57 is a method by which we can find out if a number is divisible by 57 or not without using the division method. Check whether 285 is divisible by 57 with the divisibility rule. Step 1: Divide the number into two parts: the last two digits and the rest of the number. For 285, the last two digits are 85, and the rest is 2. Step 2: Multiply the rest of the number by 5, i.e., 2 × 5 = 10. Step 3: Add the result from Step 2 to the last two digits. 85 + 10 = 95. Step 4: Check if the result from Step 3 is a multiple of 57. Since 95 is not a multiple of 57, 285 is not divisible by 57. Tips and Tricks for Divisibility Rule of 57 Learning the divisibility rule will help kids to master division. Let’s learn a few tips and tricks for the divisibility rule of 57. Common Mistakes and How to Avoid Them in Divisibility Rule of 57 The divisibility rule of 57 helps us to quickly check if a given number is divisible by 57, but common mistakes like calculation errors lead to incorrect results. Here we will understand some common mistakes that will help you to understand. The divisibility rule of 57 helps us to quickly check if a given number is divisible by 57, but common mistakes like calculation errors lead to incorrect results. Here we will understand some common mistakes that will help you to understand. Mistake 1 Not following the correct steps. Not following the correct steps. Students should follow the correct steps: dividing the number into two parts, multiplying the rest by 5, and then adding it to the last two digits. Then check whether it is a multiple of 57. Students should follow the correct steps: dividing the number into two parts, multiplying the rest by 5, and then adding it to the last two digits. Then check whether it is a multiple of 57. Mistake 2 Including the last two digits in multiplication. Including the last two digits in multiplication. Students should keep in mind that they should exclude the last two digits while multiplying and include only the rest of the number. Students should keep in mind that they should exclude the last two digits while multiplying and include only the rest of the number. Mistake 3 Not repeating the process when the result is large. Not repeating the process when the result is large. Students often stop the process after they have got a large number as a result. The process should be repeated until we get a result that is easy to check for divisibility by 57. Students often stop the process after they have got a large number as a result. The process should be repeated until we get a result that is easy to check for divisibility by 57. Mistake 4 Confusing the steps. Confusing the steps. Students often confuse the steps or forget them. To avoid the error, students should practice regularly. Students often confuse the steps or forget them. To avoid the error, students should practice regularly. Divisibility Rule of 57 Examples Problem 1 Does the number 4567 satisfy the divisibility rule of 57? No, 4567 is not divisible by 57. No, 4567 is not divisible by 57. Explanation To check if 4567 is divisible by 57, use the rule: 1) Add the last digit, 7, to five times the remaining number, 456. 2) Calculate: 5 × 456 + 7 = 2287. 3) Since 2287 is not divisible by 57, neither is 4567. To check if 4567 is divisible by 57, use the rule: 1) Add the last digit, 7, to five times the remaining number, 456. 2) Calculate: 5 × 456 + 7 = 2287. 3) Since 2287 is not divisible by 57, neither is 4567. Problem 2 Check if the number 6273 is divisible by 57. Yes, 6273 is divisible by 57. Yes, 6273 is divisible by 57. Explanation To verify the divisibility of 6273 by 57: 1) Add the last digit, 3, to five times the remaining number, 627. 2) Calculate: 5 × 627 + 3 = 3138. 3) Since 3138 is divisible by 57 (57 × 55 = 3138), 6273 is also divisible by 57. To verify the divisibility of 6273 by 57: 1) Add the last digit, 3, to five times the remaining number, 627. 2) Calculate: 5 × 627 + 3 = 3138. 3) Since 3138 is divisible by 57 (57 × 55 = 3138), 6273 is also divisible by 57. Problem 3 Is -3645 divisible by 57? No, -3645 is not divisible by 57. No, -3645 is not divisible by 57. Explanation To check if -3645 is divisible by 57, ignore the negative sign: 1) Add the last digit, 5, to five times the remaining number, 364. 2) Calculate: 5 × 364 + 5 = 1825. 3) Since 1825 is not divisible by 57, neither is -3645. To check if -3645 is divisible by 57, ignore the negative sign: 1) Add the last digit, 5, to five times the remaining number, 364. 2) Calculate: 5 × 364 + 5 = 1825. 3) Since 1825 is not divisible by 57, neither is -3645. Problem 4 Can 8129 be divided by 57 using the divisibility rule? No, 8129 isn't divisible by 57. No, 8129 isn't divisible by 57. Explanation To determine if 8129 is divisible by 57: 1) Add the last digit, 9, to five times the remaining number, 812. 2) Calculate: 5 × 812 + 9 = 4069. 3) 4069 is not divisible by 57, so 8129 is not divisible by 57. To determine if 8129 is divisible by 57: 1) Add the last digit, 9, to five times the remaining number, 812. 2) Calculate: 5 × 812 + 9 = 4069. 3) 4069 is not divisible by 57, so 8129 is not divisible by 57. Problem 5 Verify if 11457 is divisible by 57. Yes, 11457 is divisible by 57. Yes, 11457 is divisible by 57. Explanation To check the divisibility of 11457 by 57: 1) Add the last digit, 7, to five times the remaining number, 1145. 2) Calculate: 5 × 1145 + 7 = 5732. 3) Since 5732 is divisible by 57 (57 × 100 = 5732), 11457 is also divisible by 57. To check the divisibility of 11457 by 57: 1) Add the last digit, 7, to five times the remaining number, 1145. 2) Calculate: 5 × 1145 + 7 = 5732. 3) Since 5732 is divisible by 57 (57 × 100 = 5732), 11457 is also divisible by 57. FAQs on Divisibility Rule of 57 1.What is the divisibility rule for 57? The divisibility rule for 57 involves dividing the number into two parts, multiplying the rest by 5, then adding it to the last two digits, and checking if the result is a multiple of 57. The divisibility rule for 57 involves dividing the number into two parts, multiplying the rest by 5, then adding it to the last two digits, and checking if the result is a multiple of 57. 2. How many numbers are there between 1 and 1000 that are divisible by 57? There are 17 numbers that can be divided by 57 between 1 and 1000. The numbers are 57, 114, 171, 228, 285, 342, 399, 456, 513, 570, 627, 684, 741, 798, 855, 912, 969. There are 17 numbers that can be divided by 57 between 1 and 1000. The numbers are 57, 114, 171, 228, 285, 342, 399, 456, 513, 570, 627, 684, 741, 798, 855, 912, 969. 3.Is 114 divisible by 57? Yes, because 114 is a multiple of 57 (57 × 2 = 114). Yes, because 114 is a multiple of 57 (57 × 2 = 114). 4.What if I get 0 after adding? If you get 0 after adding, it is considered that the number is divisible by 57. If you get 0 after adding, it is considered that the number is divisible by 57. 5.Does the divisibility rule of 57 apply to all integers? Yes, the divisibility rule of 57 applies to all integers. Yes, the divisibility rule of 57 applies to all integers. 6.How can children in United States use numbers in everyday life to understand Divisibility Rule of 57? Numbers appear everywhere—from counting money to measuring ingredients. Kids in United States see how Divisibility Rule of 57 helps solve real problems, making numbers meaningful beyond the classroom. 7.What are some fun ways kids in United States can practice Divisibility Rule of 57 with numbers? Games like board games, sports scoring, or even cooking help children in United States use numbers naturally. These activities make practicing Divisibility Rule of 57 enjoyable and connected to their world. 8.What role do numbers and Divisibility Rule of 57 play in helping children in United States develop problem-solving skills? Working with numbers through Divisibility Rule of 57 sharpens reasoning and critical thinking, preparing kids in United States for challenges inside and outside the classroom. 9.How can families in United States create number-rich environments to improve Divisibility Rule of 57 skills? Families can include counting chores, measuring recipes, or budgeting allowances, helping children connect numbers and Divisibility Rule of 57 with everyday activities. Important Glossary for Divisibility Rule of 57 Explore More numbers Important Math Links IconPrevious to Divisibility Rule of 57 Important Math Links IconNext to Divisibility Rule of 57 About BrightChamps in United States Hiralee Lalitkumar Makwana About the Author Hiralee Lalitkumar Makwana has almost two years of teaching experience. She is a number ninja as she loves numbers. Her interest in numbers can be seen in the way she cracks math puzzles and hidden patterns. Fun Fact : She loves to read number jokes and games.
11417
https://www.khanacademy.org/e/converting-moles-and-mass
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11418
https://www.tutoringhour.com/worksheets/quadratic-equations/sum-and-product-of-roots/
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Designed to offer robust practice on the topic, this collection includes exercises on finding the sums and products of the roots of quadratic equations, figuring out the missing roots, and forming quadratic equations from the given roots. These sum and product of the roots of quadratic equations worksheet pdfs are ideal for high school students. Identify the Sum and Product of the Roots Convert each quadratic equation into standard form and find the coefficients a, b and c. Substitute the values in -b/a to find the sum of the roots and c/a to find the product of the roots. Find the Missing Root In these printable worksheets on the sum and product of roots of quadratic equations, high school kids are required to determine the missing root when the other root is given. Forming Quadratic Equation from the Roots - Easy This section of our pdf sum and product of the roots worksheets features easy-level exercises where high school children practice forming quadratic equations from the given roots. Forming Quadratic Equation from the Roots - Moderate Kick practice up a notch with these printable, moderately-difficult worksheets that call for students to try their hand at using the given roots to form quadratic equations. Related Printable Worksheets ▶ Discriminant and Nature of Roots ▶ Equations ▶ Systems of Equations Tutoringhour What We Offer Math Worksheets ELA Worksheets Science Worksheets Social Studies Worksheets Interactive Worksheets Lessons Games Membership Benefits How to Use Interactive Worksheets How to Use Printable Worksheets FAQ Printing Help What's New Testimonial About Us Privacy Policy Refund Policy Terms Contact Us Copyright © 2025 - Tutoringhour You must be a member to unlock this feature! 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11419
https://www.youtube.com/watch?v=enWmSslP1Rk
A-X-E Notation for VSEPR Dr. David Kreller Chemistry 3570 subscribers 518 likes Description 78380 views Posted: 11 Apr 2012 An introduction to the A-X-E system of notation that is very helpful to people who are using the concepts of Valence Shell Electron Pair Repulsion Theory to understand the fundamental set of basic molecular geometries. 18 comments Transcript: the axe notation system is a very convenient and very powerful method to use in conjunction with vsse Theory to understand bonding geometries around atoms in molecules any Central atom in a molecule can be described by an axe term once you have identified the axe term for an atom it's quite straightforward then to take that General axe term and predict bonding geometry so let's look at what a x e and the subscripts m and n mean a represents the central atom so every term just starts with a and then that's followed by x and x represents any atom that is directly attached to this to that Central atom a or in other words is a nearest neighbor now of course there are many many many molecules in which there's a central atom that has more than one atom attached to that Central atom and so the subscript M denotes the number of atoms attached to a e represents a lone or non-bonded pair of electrons that are around that Central atom and N represents the number of Lone electron pairs around that atom so let's work through a few examples to get a feeling for this we're back to our example CH4 well a is the central carbon are there X yes indeed there are there are 1 2 3 four atoms directly bonded to that carbon so it's ax4 are there any lone electron pairs around the carbon no there are not so there aren't any e so when there aren't any e we don't even write e there the central carbon in methane would be described by the ax East type symbol of ax4 NH3 okay that Central nitrogen atom has one two three atoms that are directly attached to it so it has three x's and it has one lone electron pair around it as well and so we combine those to have an ax3 e symbol to describe the bonding situation for that Central nitrogen atom in that molecule now let's look at the O atom which is Central in this H2O molecule the O has well it has two hydrogen atoms directly attached to it so it's got two x's and the oxygen now has two lone electron pairs on it as well so it's an ax2 E2 term symbol back to the torium in tecl4 I'll let you think about this okay so the central atom that we're looking at is there's really only one Central atom in this molecule and it's the torium so how many X's does it have it has four and how many e there's just one e so that symbol is a X4 e another back to this Xenon tetrafluoride example I'll give you a second to think about this a represents this always the central atom the bonding geometry around which we are discussing so a is the central Xenon how many X's are there that are directly attached to it well there's four and there's how many nonbonding electron pairs are there there are two so what the final symbol a X4 E2 so here's a molecule which has many different Central atoms atoms which have other atoms around them so any given axe symbol describes a central atom so we can just work through a few Central atoms and and look at the specific axe symbols for each of those atoms so let's start by looking at this nitrogen atom that nitrogen atom would be ax3 e let's look at this carbon atom well how many X's are attached to it there's one two three X's around the carbon and how many e are there around the carbon well there aren't any so that carbon would be ax3 so just because these atoms are all part of the same molecule doesn't mean that they have the same axe symbols let's look at this carbon how many X's are there how many individual atoms are directly attached to that carbon well there's 1 2 3 four so there's four X's how many e's are there well there aren't any E's so that carbon is ax4 so we're seeing some interesting bonding variation in carbon sometimes carbons are have ax3 symbols sometimes carbons have ax4 symbols here is a master table of Axe symbols each particular steric number defines a family of shapes here's steric number of two and this steric number actually only defines one shape a shape in which there's two bonding groups and zero non-bonding groups however the steric number of three defines a family of molecular shapes a family of two molecular shapes an a X3 in which there are three bonding groups and zero non-bonding groups and ax2e in which there's two bonding groups and one non-bonding groups steric number of four has within it a family of three shapes ax4 ax3 e and ax2 E2 all have steric numbers that add up to four and similarly ax5 describes family of four shapes ax5 ax4e ax3 E2 ax2 E3 and as you can see the M and N values where m is a number of x's and N is a number of E's add up to give the steric number and so the steric number six describes a family of five shapes ax6 ax5e ax4 E2 ax3 E3 and ax2 E4
11420
https://math.stackexchange.com/questions/218165/finding-the-zeroes-using-chebyshev-polynomials
numerical methods - Finding the zeroes using Chebyshev polynomials - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Finding the zeroes using Chebyshev polynomials Ask Question Asked 12 years, 11 months ago Modified12 years, 5 months ago Viewed 2k times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Use the zeroes of T¯3 T¯3 and transformations of the given interval to construct an interpolating polynomial of degree 2 for f(x)=1 x f(x)=1 x over the interval [1,3][1,3] My biggest issue is finding the zeroes. How exactly do I do that? I have been using the formula 1 2[(b−a)cos(N+1 2−n N⋅π)+a+b]1 2[(b−a)cos⁡(N+1 2−n N⋅π)+a+b] Where [a,b][a,b] is the interval, N N is the degree and for n=1,2,3,…,N.n=1,2,3,…,N. x k¯=1 2[(3−1)cos(2+1 2−1 2⋅π)+1+3]=1.29289322 x k¯=1 2[(3−1)cos⁡(2+1 2−1 2⋅π)+1+3]=1.29289322 This is of course wrong. If anyone is familiar with this topic then please guide me. The final answer is P(x)=.3489153−.1744576(x−2.866025)+.1538462(x−2.866025)(x−2)P(x)=.3489153−.1744576(x−2.866025)+.1538462(x−2.866025)(x−2). numerical-methods interpolation orthogonal-polynomials Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 6, 2013 at 18:50 J. M. ain't a mathematician 76.7k 8 8 gold badges 222 222 silver badges 347 347 bronze badges asked Oct 21, 2012 at 17:42 math101math101 1,153 1 1 gold badge 14 14 silver badges 31 31 bronze badges 4 Those cannot be the zeroes. If x=2.866025 x=2.866025 then P(x)=.3489153 P(x)=.3489153... Ah nevermind, you're looking for the zeros of a different polynomial. The question didn't render properly on my phone Emily –Emily 2012-10-21 17:55:39 +00:00 Commented Oct 21, 2012 at 17:55 hmmm I am quite confused then. So how would you solve this then?math101 –math101 2012-10-21 17:59:21 +00:00 Commented Oct 21, 2012 at 17:59 Well actually it would be the zeroes of the T 3¯T 3¯ not of the interpolating polynomial math101 –math101 2012-10-21 18:03:47 +00:00 Commented Oct 21, 2012 at 18:03 I think I have it figured out. The answer has been factored for some reason math101 –math101 2012-10-21 18:11:17 +00:00 Commented Oct 21, 2012 at 18:11 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. The zeroes of T 3(x)=4 x 3−3 x T 3(x)=4 x 3−3 x are 0,±3√2 0,±3 2. The first idea is to apply the mapping from the interval [−1,1][−1,1] (the natural domain for the Chebyshev polynomials) into your desired domain [1,3][1,3]. It can be easily seen that the required mapping is f(x)=x+2 f(x)=x+2, and thus you are to interpolate 1 x 1 x over the points 2,2±3√2 2,2±3 2. You can now apply any of a number of methods (Lagrange, Newton, Neville-Aitken) for constructing the quadratic polynomial interpolant. That part I leave to you. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 6, 2013 at 18:56 J. M. ain't a mathematicianJ. M. ain't a mathematician 76.7k 8 8 gold badges 222 222 silver badges 347 347 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 numerical-methods interpolation orthogonal-polynomials See similar questions with these tags. 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https://pcombet.math.ncsu.edu/jnca2.pdf
An Algorithm for Splitting Parallel Sums of Linearly Composed Monotone Operators, with Applications to Signal Recovery∗ Stephen R. Becker and Patrick L. Combettes UPMC Universit´ e Paris 06 Laboratoire Jacques-Louis Lions – UMR CNRS 7598 75005 Paris, France becker@ljll.math.upmc.fr, plc@math.jussieu.fr Abstract We present a new primal-dual splitting algorithm for structured monotone inclusions in Hilbert spaces and analyze its asymptotic behavior. A novelty of our framework, which is motivated by image recovery applications, is to consider inclusions that combine a variety of monotonicity-preserving operations such as sums, linear compositions, parallel sums, and a new notion of parallel composition. The special case of minimization problems is studied in detail, and applica-tions to signal recovery are discussed. An image restoration example is provided to illustrate the numerical implementation of the algorithm. Keywords duality, image recovery, infimal convolution, monotone operator, parallel composition, parallel sum, proximity operator, splitting algorithm Mathematics Subject Classifications (2010) Primary 47H05; Secondary 65K05, 90C25, 94A08. ∗Contact author: P. L. Combettes, plc@math.jussieu.fr, phone: +33 1 4427 6319, fax: +33 1 4427 7200. 1 1 Introduction Let A and B be set-valued monotone operators acting on a real Hilbert space H. The first operator splitting algorithms were developed in the late 1970s to solve inclusion problems of the form find x ∈H such that 0 ∈Ax + Bx, (1.1) by using separate applications of the operators A and B at each iteration; see [9, 15, 17, 18, 23, 24] and the references therein. Because of increasingly complex problem formulations, more sophisti-cated splitting algorithms have recently arisen. Thus, the splitting method proposed in can solve problems of the type find x ∈H such that 0 ∈Ax + r X k=1 L∗ k ◦Bk ◦Lk  x, (1.2) where Bk is a monotone operator acting on a real Hilbert space Gk and Lk is a bounded linear operator from H to Gk. This model was further refined in by considering inclusions of the form find x ∈H such that 0 ∈Ax + r X k=1 L∗ k ◦(Bk □Dk) ◦Lk  x + Cx, (1.3) where Dk is a monotone operator acting on Gk and such that D−1 k is Lipschitzian, Bk □Dk = B−1 k + D−1 k −1 (1.4) is the parallel sum of Bk and Dk, and C : H →H is a Lipschitzian monotone operator. More recent developments concerning splitting methods for models featuring parallel sums can be found in [4, 5, 13, 25]. In the present paper, motivated by variational problems arising in image recovery, we consider a new type of inclusions that involve both parallel sum and “parallel composition” operations in the sense we introduce below. Definition 1.1 Let H and G be real Hilbert spaces, let A: H →2H, and let L ∈B (H, G). The parallel composition of A by L is L ⊲A = L ◦A−1 ◦L∗)−1. (1.5) The primal-dual inclusion problem under consideration will be the following (our notation is standard, see Section 2.1 for details). Problem 1.2 Let H be a real Hilbert space, let r be a strictly positive integer, let z ∈H, let A: H → 2H be maximally monotone, and let C : H →H be monotone and µ-Lipschitzian for some µ ∈ [0, +∞[. For every integer k ∈{1, . . . , r}, let Gk and Kk be real Hilbert spaces, let Bk : Gk →2Gk and Dk : Kk →2Kk be maximally monotone, and let Lk ∈B (H, Gk) and Mk ∈B (H, Kk). It is assumed that β = µ + v u u t r X k=1 ∥Lk∥2 + max 1⩽k⩽r ∥Lk∥2 + ∥Mk∥2 > 0 (1.6) 2 and that the inclusion find x ∈H such that z ∈Ax + r X k=1 (L∗ k ◦Bk ◦Lk) □(M∗ k ◦Dk ◦Mk)  x + Cx (1.7) possesses at least one solution. Solve (1.7) together with the dual problem find v1 ∈G1, . . . , vr ∈Gr such that (∀k ∈{1, . . . , r}) 0 ∈−Lk  (A + C)−1  z − r X l=1 L∗ l vl  + B−1 k vk + Lk M∗ k ⊲D−1 k  (L∗ kvk)  . (1.8) The paper is organized as follows. In Section 2 we define our notation and provide preliminary results. In particular, we establish some basic properties of the parallel composition operation intro-duced in Definition 1.1 and discuss an algorithm recently proposed in that will serve as a basis for our splitting method. In Section 3, our algorithm is presented and weak and strong convergence results are established. Section 4 is devoted to the application of this algorithm to convex minimiza-tion problems. Finally, in Section 5, we propose applications of the results of Section 4 to a concrete problem in image recovery, along with numerical results. 2 Notation and preliminary results 2.1 Notation and definitions The following notation will be used throughout. H, G, and K are real Hilbert spaces. We denote the scalar product of a Hilbert space by ⟨· | ·⟩and the associated norm by ∥· ∥. The symbols ⇀and → denote, respectively, weak and strong convergence. B (H, G) is the space of bounded linear operators from H to G. The Hilbert direct sum of H and G is denoted by H ⊕G. Given two sequences (xn)n∈N and (yn)n∈N in H, it will be convenient to use the notation (∀n ∈N) xn ≈yn ⇔ X n∈N ∥xn −yn∥< +∞ (2.1) to model the tolerance to errors in the implementation of the algorithms. The power set of H is denoted by 2H. Let A: H →2H be a set-valued operator. We denote by ran A =  u ∈H (∃x ∈H) u ∈Ax the range of A, by dom A =  x ∈H Ax ̸= ∅ the domain of A, by gra A =  (x, u) ∈H × H u ∈Ax the graph of A, and by A−1 the inverse of A, i.e., the set-valued operator with graph  (u, x) ∈H × H u ∈Ax . The resolvent of A is JA = (Id +A)−1. Moreover, A is monotone if (∀(x, y) ∈H × H)(∀(u, v) ∈Ax × Ay) ⟨x −y | u −v⟩⩾0, (2.2) and maximally monotone if there exists no monotone operator B : H →2H such that gra A ⊂gra B ̸= gra A. In this case, JA is a single-valued, nonexpansive operator defined everywhere in H. We say 3 that A is uniformly monotone at x ∈dom A if there exists an increasing function φ: [0, +∞[ → [0, +∞] that vanishes only at 0 such that (∀u ∈Ax)(∀(y, v) ∈gra A) ⟨x −y | u −v⟩⩾φ(∥x −y∥). (2.3) We denote by Γ0(H) the class of lower semicontinuous convex functions f : H →]−∞, +∞] such that dom f =  x ∈H f(x) < +∞ ̸= ∅. Let f ∈Γ0(H). The conjugate of f is the function f ∗∈Γ0(H) defined by f ∗: u 7→supx∈H(⟨x | u⟩−f(x)). For every x ∈H, f + ∥x −·∥2/2 possesses a unique minimizer, which is denoted by proxfx. The operator proxf can also be defined as a resolvent, namely proxf = (Id +∂f)−1 = J∂f, (2.4) where ∂f : H →2H : x 7→  u ∈H (∀y ∈H) ⟨y −x | u⟩+ f(x) ⩽f(y) is the subdifferential of f, which is maximally monotone. We say that f uniformly convex at x ∈dom f if there exists an increasing function φ: [0, +∞[ →[0, +∞] that vanishes only at 0 such that (∀y ∈dom f)(∀α ∈]0, 1[) f(αx + (1 −α)y) + α(1 −α)φ(∥x −y∥) ⩽αf(x) + (1 −α)f(y). (2.5) The infimal convolution of two functions f1 and f2 from H to ]−∞, +∞] is f1 □f2 : H →[−∞, +∞] : x 7→inf y∈H f1(x −y) + f2(y)  , (2.6) and the infimal postcomposition of f : H →[−∞, +∞] by L: H →G is L ⊲f : G →[−∞, +∞] : y 7→inf f L−1{y}  = inf x∈H Lx=y f(x). (2.7) Let C be a convex subset of H. The indicator function of C is denoted by ιC, and the strong relative interior of C, i.e., the set of points x ∈C such that the cone generated by −x + C is a closed vector subspace of H, by sri C. For a detailed account of the above concepts, see . 2.2 Parallel composition In this section we explore some basic properties of the parallel composition operation introduced in Definition 1.1 which are of interest in their own right. First, we justify the terminology via the following connection with the parallel sum. Lemma 2.1 Let A: H →2H, let B : H →2H, and let L: H ⊕H →H: (x, y) 7→x + y. Then L ⊲(A × B) = A □B. Proof. Since L∗: H →H ⊕H: x 7→(x, x), the announced identity is an immediate consequence of (1.4) and (1.5). Lemma 2.2 Let A: H →2H, let B : G →2G, and let L ∈B (H, G). Then the following hold. 4 (i) ((L ⊲A) □B)−1 = L ◦A−1 ◦L∗+ B−1. (ii) Suppose that A and B are monotone. Then (L ⊲A) □B is monotone. (iii) Suppose that A and B are maximally monotone and that the cone generated by L∗(ran B)−ran A is a closed vector subspace. Then (L ⊲A) □B is maximally monotone. (iv) Suppose that A is maximally monotone and that the cone generated by ran L∗+ ran A is a closed vector subspace. Then L ⊲A is maximally monotone. Proof. (i): This follows easily from (1.4) and (1.5). (ii): By (i), ((L ⊲A) □B)−1 = L ◦A−1 ◦L∗+ B−1. Since A−1 and B−1 are monotone and monotonicity is preserved under inversion and this type of transformation [2, Proposition 20.10], the assertion is proved. (iii): The operators A−1 and B−1 are maximally monotone [2, Proposition 20.22] and L∗(ran B) −ran A = L∗(dom B−1) −dom A−1. Hence, L ◦A−1 ◦L∗+ B−1 is maximally mono-tone [3, Section 24] and so is its inverse which, in view of (i), is (L ⊲A) □B. This result can also be derived from [22, Theorems 16 and 21]. (iv): Set B = {0}−1 in (iii). Lemma 2.3 Let A: H →2H, let B : H →2H, and let L ∈B (H, G). Then L ⊲(A □B) = (L ⊲A) □(L ⊲B). Proof. It follows from (1.4) and (1.5) that L ⊲(A □B) = L ◦(A □B)−1 ◦L∗−1 = L ◦(A−1 + B−1) ◦L∗−1 = L ◦A−1 ◦L∗+ L ◦B−1 ◦L∗−1 = (L ⊲A)−1 + (L ⊲B)−1−1 = (L ⊲A) □(L ⊲B), (2.8) which proves the announced identity. Lemma 2.4 Let A: H →2H, let L ∈B (H, G), and let M ∈B (G, K). Then M ⊲(L ⊲A) = (M ◦ L) ⊲A. Proof. Indeed, M ⊲(L ⊲A) = (M ◦(L ⊲A)−1 ◦M∗)−1 = (M ◦L ◦A−1 ◦L∗◦M∗)−1 = (M ◦L) ⊲A. Finally, in the next lemma we draw connections with the infimal convolution and postcomposition operations of (2.6) and (2.7). Lemma 2.5 Let f ∈Γ0(H), let g ∈Γ0(G), and let L ∈B (H, G) be such that 0 ∈sri (L∗(dom g∗) − dom f ∗). Then the following hold. (i) (L ⊲f) □g ∈Γ0(G). 5 (ii) ∂ (L ⊲f) □g  = (L ⊲∂f) □∂g. Proof. (i): Since 0 ∈L∗(dom g∗) −dom f ∗and, by the Fenchel-Moreau theorem [2, Theorem 13.32], f ∗∈Γ0(H) and g∗∈Γ0(G), we have f ∗◦L∗+ g∗∈Γ0(G). Hence (f ∗◦L∗+ g∗)∗∈Γ0(G). However, in view of [2, Theorem 15.27(i)], the assumptions also imply that (f ∗◦L∗+ g∗)∗= (L ⊲f) □g. (ii): Let y and v be in G. Then (i), [2, Corollary 16.24, Proposition 13.21(i)&(iv), and Theo-rem 16.37(i)] enable us to write v ∈∂ (L ⊲f) □g  (y) ⇔ y ∈  ∂ (L ⊲f) □g −1 (v) ⇔ y ∈∂ (L ⊲f) □g ∗(v) ⇔ y ∈∂(f ∗◦L∗+ g∗)(v) ⇔ y ∈(L ◦(∂f ∗) ◦L∗+ ∂g∗)(v) ⇔ y ∈ L ◦(∂f)−1 ◦L∗+ (∂g)−1 (v) ⇔ v ∈ (L ⊲∂f) □∂g  y, (2.9) which establishes the announced identity. Corollary 2.6 Let f ∈Γ0(H) and let L ∈B (H, G) be such that 0 ∈sri (ran L∗−dom f ∗). Then the following hold. (i) L ⊲f ∈Γ0(G). (ii) ∂(L ⊲f) = L ⊲∂f. Proof. Set g = ι{0} in Lemma 2.5. 2.3 An inclusion problem Our main result in Section 3 will hinge on rewriting Problem 1.2 as an instance of the following formulation. Problem 2.7 Let m and K be strictly positive integers, let (Hi)1⩽i⩽m and (Gk)1⩽k⩽K be real Hilbert spaces, and let (µi)1⩽i⩽m ∈[0, +∞[m. For every i ∈{1, . . . , m} and k ∈{1, . . . , K}, let Ci : Hi →Hi be monotone and µi-Lipschitzian, let Ai : Hi →2Hi and Bk : Gk →2Gk be maximally monotone, let zi ∈Hi, and let Lki ∈B (Hi, Gk). It is assumed that β = √ λ + max 1⩽i⩽m µi > 0, where λ ∈ " sup Pm i=1 ∥xi∥2⩽1 K X k=1 m X i=1 Lkixi 2 , +∞ " , (2.10) 6 and that the system of coupled inclusions find x1 ∈H1, . . . , xm ∈Hm such that                z1 ∈A1x1 + K X k=1 L∗ k1  Bk  m X i=1 Lkixi  + C1x1 . . . zm ∈Amxm + K X k=1 L∗ km  Bk  m X i=1 Lkixi  + Cmxm (2.11) possesses at least one solution. Solve (2.11) together with the dual problem find v1 ∈G1, . . . , vK ∈GK such that                0 ∈− m X i=1 L1i Ai + Ci −1  zi − K X k=1 L∗ kivk  + B−1 1 v1 . . . 0 ∈− m X i=1 LKi Ai + Ci −1  zi − K X k=1 L∗ kivk  + B−1 K vK. (2.12) The following result is a special case of [10, Theorem 2.4(iii)]. We use the notation (2.1) to model the possibility of inexactly evaluating the operators involved. Theorem 2.8 Consider the setting of Problem 2.7. Let x1,0 ∈H1, . . . , xm,0 ∈Hm, v1,0 ∈G1, . . . , vK,0 ∈GK, let ε ∈]0, 1/(β + 1)[, let (γn)n∈N be a sequence in [ε, (1 −ε)/β], and set for n = 0, 1, . . .                        for i = 1, . . . , m $ s1,i,n ≈xi,n −γn Cixi,n + PK k=1 L∗ kivk,n  p1,i,n ≈JγnAi(s1,i,n + γnzi) for k = 1, . . . , K         s2,k,n ≈vk,n + γn Pm i=1 Lkixi,n p2,k,n ≈s2,k,n −γnJγ−1 n Bk(γ−1 n s2,k,n) q2,k,n ≈p2,k,n + γn Pm i=1 Lkip1,i,n vk,n+1 = vk,n −s2,k,n + q2,k,n for i = 1, . . . , m  q1,i,n ≈p1,i,n −γn Cip1,i,n + PK k=1 L∗ kip2,k,n  xi,n+1 = xi,n −s1,i,n + q1,i,n. (2.13) Then there exist a solution (x1, . . . , xm) to (2.11) and a solution (v1, . . . , vK) to (2.12) such that the following hold. (i) (∀i ∈{1, . . . , m}) zi −PK k=1 L∗ kivk ∈Aixi + Cixi. (ii) (∀k ∈{1, . . . , K}) Pm i=1 Lkixi ∈B−1 k vk. (iii) (∀i ∈{1, . . . , m}) xi,n ⇀xi. 7 (iv) (∀k ∈{1, . . . , K}) vk,n ⇀vk. (v) Suppose that A1 or C1 is uniformly monotone at x1. Then x1,n →x1. (vi) Suppose that, for some k ∈{1, . . . , K}, B−1 k is uniformly monotone at vk. Then vk,n →vk. 3 Main algorithm We start with the following facts. Proposition 3.1 Let H be a real Hilbert space, let r be a strictly positive integer, let z ∈H, let A: H → 2H, and let C : H →H. For every integer k ∈{1, . . . , r}, let Gk and Kk be real Hilbert spaces, let Bk : Gk →2Gk and Dk : Kk →2Kk, and let Lk ∈B (H, Gk) and Mk ∈B (H, Kk). Set H = r M k=1 H, G = r M k=1 Gk, K = r M k=1 Kk, (3.1) and      A: H →2H : (x, y1, . . . , yr) 7→(Ax + Cx −z) × {0} × · · · × {0} B : G ⊕K →2G⊕K : (s1, . . . , sr, t1, . . . , tr) 7→B1s1 × · · · × Brsr × D1t1 × · · · × Drtr L: H →G ⊕K: (x, y1, . . . , yr) 7→(L1x −L1y1, . . . , Lrx −Lryr, M1y1, . . . , Mryr). (3.2) Furthermore, suppose that ∃x = (x, y1, . . . , yr) ∈H  0 ∈Ax + L∗B(Lx)  . (3.3) Then the following hold for some (v1, . . . , vr) ∈G and (w1, . . . , wr) ∈K. (i) z −Pr k=1 L∗ kvk ∈Ax + Cx. (ii) (∀k ∈{1, . . . , r}) L∗ kvk = M∗ kwk, Lkx −Lkyk ∈B−1 k vk, and Mkyk ∈D−1 k wk. (iii) x solves (1.7). (iv) (v1, . . . , vr) solves (1.8). Proof. (i) and (ii): It follows from (3.3) that there exists v = (v1, . . . , vr, w1, . . . , wr) ∈G ⊕K such that −L∗v ∈Ax and v ∈B(Lx), i.e., −L∗v ∈Ax and Lx ∈B−1v. (3.4) Since L∗: G ⊕K →H: (v1, . . . , vr, w1, . . . , wr) 7→  r X k=1 L∗ kvk, M∗ 1 w1 −L∗ 1v1, . . . , M∗ r wr −L∗ rvr  , (3.5) 8 it follows from (3.2) that (3.4) can be rewritten as z − r X k=1 L∗ kvk ∈Ax + Cx and (∀k ∈{1, . . . , r})      L∗ kvk = M∗ kwk Lkx −Lkyk ∈B−1 k vk Mkyk ∈D−1 k wk. (3.6) (iii): For every k ∈{1, . . . , r}, (ii) ⇒      L∗ kvk = M∗ kwk vk ∈Bk(Lkx −Lkyk) wk ∈Dk(Mkyk) (3.7) ⇒      L∗ kvk = M∗ kwk L∗ kvk ∈L∗ k Bk(Lkx −Lkyk)  M∗ kwk ∈M∗ k Dk(Mkyk)  ⇔      L∗ kvk = M∗ kwk x −yk ∈(L∗ k ◦Bk ◦Lk)−1(L∗ kvk) yk ∈(M∗ k ◦Dk ◦Mk)−1(M∗ kwk) (3.8) ⇒x ∈(L∗ k ◦Bk ◦Lk)−1(L∗ kvk) + (M∗ k ◦Dk ◦Mk)−1(L∗ kvk) ⇔L∗ kvk ∈ (L∗ k ◦Bk ◦Lk) □(M∗ k ◦Dk ◦Mk)  (x). (3.9) Hence, r X k=1 L∗ kvk ∈ r X k=1 (L∗ k ◦Bk ◦Lk) □(M∗ k ◦Dk ◦Mk)  (x). (3.10) Adding this inclusion to that of (i) shows that x solves (1.7). (iv): It follows from (i) that (∀k ∈{1, . . . , r}) −Lkx ∈−Lk  (A + C)−1  z − r X l=1 L∗ l vl  . (3.11) On the other hand, (ii) yields (∀k ∈{1, . . . , r}) Lkx −Lkyk ∈B−1 k vk, (3.12) while (3.8) yields (∀k ∈{1, . . . , r}) Lkyk ∈Lk (M∗ k ◦Dk ◦Mk)−1(M∗ kwk)  = Lk (M∗ k ◦Dk ◦Mk)−1(L∗ kvk)  = Lk (M∗ k ⊲D−1 k )(L∗ kvk)  . (3.13) Upon adding (3.11), (3.12), and (3.13), we obtain (∀k ∈{1, . . . , r}) 0 ∈−Lk  (A + C)−1  z − r X l=1 L∗ l vl  + B−1 k vk + Lk (M∗ k ⊲D−1 k )(L∗ kvk)  , (3.14) 9 which proves that (v1, . . . , vr) solves (1.8). We are now in a position to present our main result. Theorem 3.2 Consider the setting of Problem 1.2. Let x0 ∈H, y1,0 ∈H, . . . , yr,0 ∈H, v1,0 ∈G1, . . . , vr,0 ∈Gr, w1,0 ∈K1, . . . , wr,0 ∈Kr, let ε ∈]0, 1/(β + 1)[, let (γn)n∈N be a sequence in [ε, (1 −ε)/β], and set for n = 0, 1, . . .                                           s1,1,n ≈xn −γn(Cxn + Pr k=1 L∗ kvk,n) p1,1,n ≈JγnA(s1,1,n + γnz) for k = 1, . . . , r                          p1,k+1,n ≈yk,n + γn(L∗ kvk,n −M∗ kwk,n) s2,k,n ≈vk,n + γnLk(xn −yk,n) p2,k,n ≈s2,k,n −γnJγ−1 n Bk(γ−1 n s2,k,n) q2,k,n ≈p2,k,n + γnLk(p1,1,n −p1,k+1,n) vk,n+1 = vk,n −s2,k,n + q2,k,n s2,k+r,n ≈wk,n + γnMkyk,n p2,k+r,n ≈s2,k+r,n −γn Jγ−1 n Dk(γ−1 n s2,k+r,n)  q1,k+1,n ≈p1,k+1,n + γn(L∗ kp2,k,n −M∗ kp2,k+r,n) q2,k+r,n ≈p2,k+r,n + γnMkp1,k+1,n wk,n+1 = wk,n −s2,k+r,n + q2,k+r,n q1,1,n ≈p1,1,n −γn(Cp1,1,n + Pr k=1 L∗ kp2,k,n) xn+1 = xn −s1,1,n + q1,1,n for k = 1, . . . , r  yk,n+1 = yk,n −p1,k+1,n + q1,k+1,n. (3.15) Then the following hold for some solution x to (1.7) and some solution (v1, . . . , vr) to (1.8). (i) xn ⇀x and (∀k ∈{1, . . . , r}) vk,n ⇀vk. (ii) Suppose that A or C is uniformly monotone at x. Then xn →x. (iii) Suppose that, for some k ∈{1, . . . , r}, B−1 k is uniformly monotone at vk. Then vk,n →vk. Proof. We introduce the auxiliary problem find x ∈H, y1 ∈H, . . . , yr ∈H such that            z ∈Ax + Pr k=1 L∗ k Bk(Lkx −Lkyk)  + Cx 0 ∈−L∗ 1 B1(L1x −L1y1)  + M∗ 1 D1(M1y1)  . . . 0 ∈−L∗ r Br(Lrx −Lryr)  + M∗ r Dr(Mryr)  . (3.16) 10 Let x ∈H. Then x solves (1.7) ⇔z ∈Ax + r X k=1 (L∗ k ◦Bk ◦Lk) □(M∗ k ◦Dk ◦Mk)  x + Cx ⇔ ∃(uk)1⩽k⩽r ∈Hr            z ∈Ax + Pr k=1 uk + Cx u1 ∈ (L∗ 1 ◦B1 ◦L1) □(M∗ 1 ◦D1 ◦M1)  x . . . ur ∈ (L∗ r ◦Br ◦Lr) □(M∗ r ◦Dr ◦Mr)  x ⇔ ∃(uk)1⩽k⩽r ∈Hr            z ∈Ax + Pr k=1 uk + Cx x ∈(L∗ 1 ◦B1 ◦L1)−1u1 + (M∗ 1 ◦D1 ◦M1)−1u1 . . . x ∈(L∗ r ◦Br ◦Lr)−1ur + (M∗ r ◦Dr ◦Mr)−1ur ⇔ ∃(uk)1⩽k⩽r ∈Hr∃(yk)1⩽k⩽r ∈Hr                      z ∈Ax + Pr k=1 uk + Cx x −y1 ∈(L∗ 1 ◦B1 ◦L1)−1u1 y1 ∈(M∗ 1 ◦D1 ◦M1)−1u1 . . . x −yr ∈(L∗ r ◦Br ◦Lr)−1ur yr ∈(M∗ r ◦Dr ◦Mr)−1ur ⇔ ∃(uk)1⩽k⩽r ∈Hr∃(yk)1⩽k⩽r ∈Hr                      z ∈Ax + Pr k=1 uk + Cx u1 ∈(L∗ 1 ◦B1 ◦L1)(x −y1) u1 ∈(M∗ 1 ◦D1 ◦M1)y1 . . . ur ∈(L∗ r ◦Br ◦Lr)(x −yr) ur ∈(M∗ r ◦Dr ◦Mr)yr ⇒ ∃(yk)1⩽k⩽r ∈Hr            z ∈Ax + Pr k=1 L∗ k Bk(Lkx −Lkyk)  + Cx 0 ∈−L∗ 1 B1(L1x −L1y1)  + M∗ 1 D1(M1y1)  . . . 0 ∈−L∗ r Br(Lrx −Lryr)  + M∗ r Dr(Mryr)  . (3.17) Hence since, by assumption, (1.7) has at least one solution, (3.16) has at least one solution. (3.18) 11 Next, we set                          m = r + 1 K = 2r H1 = H A1 = A C1 = C µ1 = µ z1 = z and (∀k ∈{1, . . . , r})                      Hk+1 = H Ak+1 = 0 Bk+r = Dk Ck+1 = 0 µk+1 = 0 zk+1 = 0. (3.19) We also define (∀k ∈{1, . . . , r}) Gk+r = Kk and (∀i ∈{1, . . . , m}) Lki =            Lk, if 1 ⩽k ⩽r and i = 1; −Lk, if 1 ⩽k ⩽r and i = k + 1; Mk−r, if r + 1 ⩽k ⩽2r and i = k −r + 1; 0, otherwise. (3.20) We observe that in this setting (3.16) is a special case of (2.11). (3.21) Moreover, if we set λ = Pr k=1 ∥Lk∥2 + max1⩽k⩽r(∥Lk∥2 + ∥Mk∥2), we deduce from the Cauchy-Schwarz inequality in R2 that, for every (xi)1⩽i⩽m = (x, y1, . . . , yr) ∈Lm i=1 H, K X k=1 m X i=1 Lkixi 2 = ∥(L1x −L1y1, . . . , Lrx −Lryr, M1y1, . . . , Mryr)∥2 ⩽ ∥(L1x, . . . , Lrx)∥+ ∥(L1y1, . . . , Lryr, M1y1, . . . , Mryr)∥ 2 =   v u u t r X k=1 ∥Lkx∥2 + v u u t r X k=1 ∥Lkyk∥2 + ∥Mkyk∥2   2 ⩽   v u u t r X k=1 ∥Lk∥2 ∥x∥+ v u u t r X k=1 (∥Lk∥2 + ∥Mk∥2) ∥yk∥2   2 ⩽   v u u t r X k=1 ∥Lk∥2 ∥x∥+ max 1⩽k⩽r p ∥Lk∥2 + ∥Mk∥2 ∥(y1, . . . , yr)∥   2 ⩽ r X k=1 ∥Lk∥2 + max 1⩽k⩽r ∥Lk∥2 + ∥Mk∥2 ! ∥x∥2 + ∥(y1, . . . , yr)∥2 = λ m X i=1 ∥xi∥2. (3.22) Thus, (1.6) is a special case of (2.10). (3.23) 12 Now, let us define (∀n ∈N) x1,n = xn and (∀k ∈{1, . . . , r}) ( xk+1,n = yk,n vk+r,n = wk,n. (3.24) Then it follows from (3.19) that (3.15) is a special case of (2.13). (3.25) Altogether, Theorem 2.8(i)–(iv) asserts that there exist a solution x = (x1, . . . , xm) = (x, y1, . . . , yr) to (2.11) and a solution (v1, . . . , vK) = (v1, . . . , vr, w1, . . . , wr) to (2.12) which satisfy xn ⇀x and (∀k ∈{1, . . . , r}) ( vk,n ⇀vk wk,n ⇀wk, (3.26) together with the inclusions z − r X k=1 L∗ kvk ∈Ax + Cx and (∀k ∈{1, . . . , r})      L∗ kvk = M∗ kwk Lkx −Lkyk ∈B−1 k vk Mkyk ∈D−1 k wk. (3.27) Using the notation (3.2), we can rewrite (3.27) as 0 ∈Ax + L∗B(Lx)  . (3.28) In turn, it follows from Proposition 3.1(iii)–(iv) that x solves (1.7) and (v1, . . . , vr) solves (1.8). (3.29) This and (3.26) prove (i). Finally, (ii) and (iii) follow from (3.19) and Theorem 2.8(v)–(vi). Remark 3.3 In the spirit of the splitting methods of [10, 12], the algorithm described in (3.15) achieves full decomposition in that every operator is used individually at each iteration. 4 Application to convex minimization In this section we consider a structured minimization problem of the following format. Problem 4.1 Let H be a real Hilbert space, let r be a strictly positive integer, let z ∈H, let f ∈ Γ0(H), and let ℓ: H →R be a differentiable convex function such that ∇ℓis µ-Lipschitzian for some µ ∈[0, +∞[. For every integer k ∈{1, . . . , r}, let Gk and Kk be real Hilbert spaces, let gk ∈Γ0(Gk) and hk ∈Γ0(Kk), and let Lk ∈B (H, Gk) and Mk ∈B (H, Kk). It is assumed that β = µ + v u u t r X k=1 ∥Lk∥2 + max 1⩽k⩽r ∥Lk∥2 + ∥Mk∥2 > 0, (4.1) 13 that (∀k ∈{1, . . . , r}) 0 ∈sri dom (gk ◦Lk)∗−M∗ k(dom h∗ k)  , (4.2) that (∀k ∈{1, . . . , r}) 0 ∈sri ran Mk −dom hk  , (4.3) and that z ∈ran  ∂f + r X k=1 (L∗ k ◦(∂gk) ◦Lk) □(M∗ k ◦(∂hk) ◦Mk)  + ∇ℓ  . (4.4) Solve the primal problem minimize x∈H f(x) + r X k=1 (gk ◦Lk) □(hk ◦Mk)  (x) + ℓ(x) −⟨x | z⟩, (4.5) together with the dual problem minimize v1∈G1,..., vr∈Gr (f ∗□ℓ∗)  z − r X k=1 L∗ kvk  + r X k=1 g∗ k(vk) + (M∗ k ⊲h∗ k)(L∗ kvk)  . (4.6) Special cases when (4.2) and (4.3) are satisfied can be derived from [2, Proposition 15.24]. The next proposition describes scenarios in which (4.4) holds. Proposition 4.2 Consider the same setting as in Problem 4.1 with the exception that assumption (4.4) is not made and is replaced by the assumptions that E = nL1(x −y1) −s1, . . . , Lr(x −yr) −sr, M1y1 −t1, . . . , Mryr −tr  x ∈dom f, y1 ∈H, . . . , yr ∈H, s1 ∈dom g1, . . . , sr ∈dom gr, t1 ∈dom h1, . . . , tr ∈dom hr o ̸= ∅ (4.7) and that (4.5) has a solution. Then (4.4) is satisfied in each of the following cases. (i) 0 ∈sri E. (ii) E is a closed vector subspace. (iii) f is real-valued and, for every k ∈{1, . . . , r}, the operators Lk and Mk are surjective. (iv) For every k ∈{1, . . . , r}, gk and hk are real-valued. (v) H, (Gk)1⩽k⩽r, and (Kk)1⩽k⩽r are finite-dimensional, and (∃x ∈ri dom f)(∀k ∈{1, . . . , r})(∃yk ∈H) ( Lk(x −yk) ∈ri dom gk Mkyk ∈ri dom hk. (4.8) 14 Proof. Let us define H, G, and K as in (3.1), L as in (3.2), and let us set ( f : H →]−∞, +∞] : x = (x, y1, . . . , yr) 7→f(x) + ℓ(x) −⟨x | z⟩ g: G ⊕K →]−∞, +∞] : s = (s1, . . . , sr, t1, . . . , tr) 7→Pr k=1 gk(sk) + hk(tk)  . (4.9) Then we can rewrite (4.7) as E = L(dom f) −dom g. (4.10) (i): Since E ̸= ∅, the functions (gk ◦Lk)1⩽k⩽r and (hk ◦Mk)1⩽k⩽r are proper and therefore in Γ0(H). In turn, the Fenchel-Moreau theorem [2, Theorem 13.32] asserts that the functions ((gk ◦Lk)∗)1⩽k⩽r and ((hk ◦Mk)∗)1⩽k⩽r are in Γ0(H). On the other hand, since (4.3) and [2, Corol-lary 15.28(i)] imply that (∀k ∈{1, . . . , r}) (hk ◦Mk)∗= M∗ k ⊲h∗ k, (4.11) (4.2) and [2, Proposition 12.34(i)] yield (∀k ∈{1, . . . , r}) 0 ∈sri dom (gk ◦Lk)∗−M∗ k(dom h∗ k)  = sri dom (gk ◦Lk)∗−dom (M∗ k ⊲h∗ k)  = sri dom (gk ◦Lk)∗−dom (hk ◦Mk)∗ . (4.12) Hence, we derive from [2, Proposition 15.7] that (∀k ∈{1, . . . , r})(∀x ∈H)(∃yk ∈H) (gk ◦Lk) □(hk ◦Mk)  (x) = gk(Lkx −Lkyk) + hk(Mkyk), (4.13) which allows us to rewrite (4.5) as a minimization problem on H, namely minimize x∈H,y1∈H,...,yr∈H f(x) + ℓ(x) −⟨x | z⟩+ r X k=1 gk(Lkx −Lkyk) + hk(Mkyk)  (4.14) or, equivalently, minimize x∈H f(x) + g(Lx). (4.15) It follows from (4.10) that 0 ∈sri L(dom f) −dom g  and therefore from [2, Theorem 16.37(i)], that ∂(f + g ◦L) = ∂f + L∗◦(∂g) ◦L. (4.16) Since, by assumption, (4.5) has a solution, so does (4.15). By Fermat’s rule [2, Theorem 16.2], this means that 0 ∈ran ∂(f + g ◦L). Thus (4.16) yields 0 ∈ran ∂f + L∗◦(∂g) ◦L  . (4.17) Let us introduce the operators A = ∂f, C = ∇ℓ, and (∀k ∈{1, . . . , r}) ( Bk = ∂gk Dk = ∂hk. (4.18) 15 We derive from [2, Proposition 17.10] that C is monotone and from [2, Theorem 20.40] that the operators A, (Bk)1⩽k⩽r, and (Dk)1⩽k⩽r are maximally monotone. Next, let us define A and B as in (3.2). Then it follows from (4.17) and [2, Proposition 16.8] that (3.3) holds. In turn, Proposi-tion 3.1(iii) asserts that (4.4) is satisfied. (ii)⇒(i): This follows from [2, Proposition 6.19(i)]. (iii)⇒(i) and (iv)⇒(i): In both cases E = G ⊕K. (v)⇒(i): Since H, G, and K are finite-dimensional, (4.10) and [2, Corollary 6.15] imply that (4.8) ⇔(∃x ∈ri dom f) Lx ∈ri dom g ⇔0 ∈ L(ri dom f) −ri dom g  ⇔0 ∈ri L(dom f) −dom g  ⇔0 ∈ri E ⇔0 ∈sri E, (4.19) which completes the proof. Next, we propose our algorithm for solving Problem 4.1. Theorem 4.3 Consider the setting of Problem 4.1. Let x0 ∈H, y1,0 ∈H, . . . , yr,0 ∈H, v1,0 ∈G1, . . . , vr,0 ∈Gr, w1,0 ∈K1, . . . , wr,0 ∈Kr, let ε ∈]0, 1/(β + 1)[, let (γn)n∈N be a sequence in [ε, (1 −ε)/β], and set for n = 0, 1, . . .                                           s1,1,n ≈xn −γn(∇ℓ(xn) + Pr k=1 L∗ kvk,n) p1,1,n ≈proxγnf(s1,1,n + γnz) for k = 1, . . . , r                          p1,k+1,n ≈yk,n + γn(L∗ kvk,n −M∗ kwk,n) s2,k,n ≈vk,n + γnLk(xn −yk,n) p2,k,n ≈s2,k,n −γnproxγ−1 n gk(γ−1 n s2,k,n) q2,k,n ≈p2,k,n + γnLk(p1,1,n −p1,k+1,n) vk,n+1 = vk,n −s2,k,n + q2,k,n s2,k+r,n ≈wk,n + γnMkyk,n p2,k+r,n ≈s2,k+r,n −γn proxγ−1 n hk(γ−1 n s2,k+r,n)  q1,k+1,n ≈p1,k+1,n + γn(L∗ kp2,k,n −M∗ kp2,k+r,n) q2,k+r,n ≈p2,k+r,n + γnMkp1,k+1,n wk,n+1 = wk,n −s2,k+r,n + q2,k+r,n q1,1,n ≈p1,1,n −γn(∇ℓ(p1,1,n) + Pr k=1 L∗ kp2,k,n) xn+1 = xn −s1,1,n + q1,1,n for k = 1, . . . , r  yk,n+1 = yk,n −p1,k+1,n + q1,k+1,n. (4.20) Then the following hold for some solution x to (4.5) and some solution (v1, . . . , vr) to (4.6). 16 (i) xn ⇀x and (∀k ∈{1, . . . , r}) vk,n ⇀vk. (ii) Suppose that f or ℓis uniformly convex at x. Then xn →x. (iii) Suppose that, for some k ∈{1, . . . , r}, g∗ k is uniformly convex at vk. Then vk,n →vk. Proof. Set A = ∂f, C = ∇ℓ, and (∀k ∈{1, . . . , r}) ( Bk = ∂gk Dk = ∂hk. (4.21) We derive from [2, Proposition 17.10] that C is monotone. Furthermore, [2, Theorem 20.40 and Corollary 16.24] assert that the operators A, (Bk)1⩽k⩽r, and (Dk)1⩽k⩽r are maximally monotone with inverses respectively given by ∂f ∗, (∂g∗ k)1⩽k⩽r, and (∂h∗ k)1⩽k⩽r. Moreover, (4.4) implies that (1.7) has a solution. Now let x and v = (vk)1⩽k⩽r be, respectively, the solutions to (1.7) and (1.8) produced by Theorem 3.2. Since the uniform convexity of a function at a point implies the uniform monotonicity of its subdifferential at that point [27, Section 3.4] and since, in the setting of (4.21), (4.20) reduces to (3.15) thanks to (2.4), it is enough to show that x solves (4.5) and v solves (4.6). To this end, we first derive from (4.12) and [2, Propositions 16.5(ii) and 24.27] that (∀k ∈{1, . . . , r}) L∗ k ◦(∂gk) ◦Lk  □ M∗ k ◦(∂hk) ◦Mk  ⊂∂(gk ◦Lk) □∂(hk ◦Mk) = ∂ (gk ◦Lk) □(hk ◦Mk)  . (4.22) Hence, it follows from (4.21) and Fermat’s rule [2, Theorem 16.2] that x solves (1.7) ⇒z ∈∂f(x) + r X k=1 (L∗ k ◦(∂gk) ◦Lk) □(M∗ k ◦(∂hk) ◦Mk)  x + ∇ℓ(x) ⇒z ∈∂f(x) + r X k=1 ∂  (gk ◦Lk) □(hk ◦Mk)  x + ∂ℓ(x) ⇒0 ∈∂  f + r X k=1 (gk ◦Lk) □(hk ◦Mk)  + ℓ−⟨· | z⟩  (x) ⇒x solves (4.5). (4.23) On the other hand, (4.3) and Corollary 2.6(ii) yield (∀k ∈{1, . . . , r}) M∗ k ⊲∂h∗ k = ∂(M∗ k ⊲h∗ k), (4.24) while [2, Proposition 16.5(ii)] yields (∀k ∈{1, . . . , r}) ∂g∗ k + Lk ◦ ∂(M∗ k ⊲h∗ k)  ◦L∗ k ⊂∂ g∗ k + (M∗ k ⊲h∗ k) ◦L∗ k  . (4.25) Now define G as in (3.1) and      ϕ: H →]−∞, +∞] : u 7→(f ∗□ℓ∗)(z + u) ψ : G →]−∞, +∞] : v 7→Pr k=1 g∗ k(vk) + (M∗ k ⊲h∗ k)(L∗ kvk)  M : G →H: v 7→−Pr k=1 L∗ kvk. (4.26) 17 Then (∀v ∈G) ϕ(Mv) + ψ(v) = (f ∗□ℓ∗)  z − r X k=1 L∗ kvk  + r X k=1 g∗ k(vk) + (M∗ k ⊲h∗ k)(L∗ kvk)  . (4.27) Invoking successively (4.21), (4.24), (4.25), [2, Proposition 16.8], (4.26), (4.27), and Fermat’s rule, we get v solves (1.8) ⇒(∀k ∈{1, . . . , r}) 0 ∈−Lk  ∂(f + ℓ)∗  z − r X l=1 L∗ l vl  + ∂g∗ k(vk) + Lk M∗ k ⊲∂h∗ k  (L∗ kvk)  ⇒(∀k ∈{1, . . . , r}) 0 ∈−Lk  ∂(f ∗□ℓ∗)  z − r X l=1 L∗ l vl  + ∂ g∗ k + (M∗ k ⊲h∗ k) ◦L∗ k  (vk) ⇒0 ∈ M ∗◦(∂ϕ) ◦M  (v) + ∂ψ(v) ⇒0 ∈∂ ϕ ◦M + ψ  (v) ⇒v solves (4.6), (4.28) which completes the proof. Theorem 4.3 enables us to solve a new class of structured minimization problems featuring both infimal convolutions and postcompositions. The special cases of this model which arise in the area of image recovery [7, 21] initially motivated our investigation. Such applications are considered in the next section. 5 Image restoration application 5.1 Image restoration Proximal splitting methods were introduced in the field of image recovery in for variational models of the form minimize x∈H f(x) + ℓ(x), (5.1) where f and ℓare as in Problem 4.1 (see for recent developments in this application area). In this section we show a full fledged implementation of the algorithm in Theorem 4.3 in the Euclidean setting (H = RN) which goes much beyond (5.1). For this purpose, we consider the problem of image restoration from a blurred image . Imaging devices, such as cameras, microscopes, and telescopes, distort the light field due to both optical imperfections and diffraction; another source of blur is relative movement of the scene and the device during the exposure, as happens when taking a photo in low-light without a tripod or when a telescope observes the stars with imperfect motion compensation. The effect is that the recorded image is the convolution of the true scene with a function known as the point-spread function. The resulting convolution operator T is called the blur operator. 18 The original N-pixel (N = 5122) image shown in Fig. 1(a) is degraded by a linear blurring oper-ator T associated with a 21-pixel long point-spread function corresponding to motion blur, followed by addition of a noise component w. Images in their natural matrix form are converted to vectors x ∈RN by stacking columns together. We write the coefficients of x as x = (ξi)1⩽i⩽N, but when we wish to make use of the 2-dimensional nature of the image (as a √ N × √ N image), we use the convention ξi,j = ξ(j−1) √ N+i for every i and j in {1, . . . , √ N}, so that i and j refer to the row and column indices, respectively. The degraded image y = Tx + w (5.2) is shown in Fig. 1(b). The noise level is chosen to give y a signal-to-noise ratio of 45 dB relative to Tx. The variational formulation we propose to recover x is an instantiation of Problem 4.1 with r = 2, namely, minimize x∈C (α∥· ∥1,2 ◦D(1)) □(β∥· ∥1,2 ◦D(2))  (x) + γ∥Wx∥1 + 1 2∥Tx −y∥2 2 (5.3) or, equivalently, minimize x∈H ιC |{z} f (x) + (α∥· ∥1,2 ◦D(1)) | {z } g1◦L1 □(β∥· ∥1,2 ◦D(2)) | {z } h1◦M1  (x) + γ∥W · ∥1 | {z } g2◦L2 □(ι{0} ◦Id ) | {z } h2◦M2  (x) + 1 2∥T · −y∥2 2 | {z } ℓ (x). (5.4) In this model, α, β, and γ are strictly positive constants, and C is a constraint set modeling the known amplitude bounds on pixel values; here C = [0, 1]N. To promote the piecewise smoothness of x we use an inf-convolution term mixing first- and second-order total variation potentials, in a fashion initially advocated in and further explored in . First-order total variation is commonly used in image processing, but suffers from staircase effects (see, e.g., ), which are reduced by using the inf-convolution model. The operators D(1) and D(2) are, respectively, first and second order discrete gradient operators that map RN to RN×M for M = 2 and M = 3, respectively (see section 5.2 for details). The functions g1 and h1 are the usual mixed norms defined on RN×M as ∥· ∥1,2 : x 7→ N X i=1 v u u t M X j=1 ξ2 i,j, (5.5) which is the sum of the norm of the rows of x. The potential x 7→∥Wx∥1, (5.6) where W is the analysis operator of a weighted 9/7 biorthogonal wavelet frame , is intended to promote sparsity of the wavelet coefficients of x. Since natural images are known to have ap-proximately sparse wavelet representations, this term penalizes noise, which does not have a sparse wavelet representation. Such wavelet terms are standard in the literature, and are often used in conjunction with a first-order TV term . Finally, data fidelity is promoted by the potential ℓ: x 7→1 2∥Tx −y∥2. (5.7) 19 Remark 5.1 Here are some comments on the implementation of the algorithm from Theorem 4.3 in the setting of (5.4). (i) The proximity operator of f = ιC is simply the projector onto a hypercube, which is straight-forward. (ii) By [2, Example 14.5], for every x ∈H ∖{0}, prox∥·∥x =  1 − 1 ∥x∥  x (5.8) and prox∥·∥0 = 0. Since ∥x∥1,2 is separable in the rows of x, prox∥·∥1,2x is computed by applying (5.8) to each row. (iii) The gradient of ℓis ∇ℓ: x 7→T ⊤(Tx −y), which is Lipschitz continuous with constant ∥T∥2. (iv) The proximity operator of ∥· ∥1 is implemented by soft-thresholding of each component . (v) No special assumption is required on the structure of W (e.g., the frame need not be tight or, in particular, an orthonormal basis). Without assumptions on W, there is no known closed-form proximity operator of x 7→γ∥Wx∥1, which is why it is important to treat ∥·∥1 and W separately. (vi) We have used only one hard constraint set C, but it is clear that our framework can accom-modate an arbitrary number of constraint sets, hence permitting one to inject easily a priori information in the restoration process. Each additional hard constraint of the type Lkx ∈Ck can be handled by setting gk = ιCk, hk = ι{0}, and Mk = Id . Remark 5.2 Remark 5.1 shows that the computation of proximity operators for each function in-volved in (5.4) is implementable. It is also possible to compute proximity operators for scaled versions of the above functions. Let ρ ∈]0, +∞[. Then given ϕ ∈Γ0(H) and e ϕ: x 7→ϕ(ρx), [2, Corollary 23.24] implies that (∀x ∈H) proxe ϕ x = ρ−1proxρ2ϕ(ρx). (5.9) This gives the possibility of writing f(Lx) as ˜ f(˜ Lx) for ˜ L = ρ−1L. Our implementation will exploit this flexibility in order to rescale all Lk and Mk operators to have unit operator norm. Numeri-cal evidence suggests that this improves convergence profiles since all dual variables (vk)1⩽k⩽r and (wk)1⩽k⩽r are approximately of the same scale. 5.2 Total variation Total variation can be defined for mathematical objects such as measures and functions . In a discrete setting, there are many possible definitions of total variation. We use the standard isotropic discretization, tv(x) = √ N−1 X i=1 √ N−1 X j=1 q (ξi+1,j −ξi,j)2 + (ξi,j+1 −ξi,j)2, x = (ξk)1⩽k⩽N, ξi,j = ξ(j−1) √ N+i, (5.10) 20 originally advocated in . There is no known closed form expression for the proximity operator of (5.10). Infimal-convolution with a second-order total variation term was first suggested in . We use the particular second-order total variation term corresponding to “D2,b” (with weights b = (1, 1/2, 1)) from . We now show how to recover the relation tv(x) = ∥D(1)x∥1,2. Define the horizontal finite-difference operator by D↔: RN →R √ N× √ N : x 7→z = (ζi,j)1⩽i,j⩽ √ N, ζi,j = ( ξi,j+1 −ξi,j, if 1 ⩽j < √ N; 0, if j = √ N, (5.11) and the vertical operator D↕by D↕: x 7→(D↔(x⊤))⊤. Let vec(·) be the mapping that re-orders a matrix by stacking the columns together, and define D(1) : x 7→(vec(D↔(x)), vec(D↕(x))). Then by comparing (5.5) with (5.10), we observe that tv(x) = ∥D(1)x∥1,2. The second-order total variation potential makes use of an additional set of first-order difference operators that have different boundary conditions, namely e D↔: RN →R √ N× √ N : x 7→z = (ζi,j)1⩽i,j⩽ √ N, ζi,j =      ξi,j −ξi,j−1, if 1 < j < √ N; ξi,j, if j = 1; −ξi,j−1, if j = √ N, (5.12) and e D↕: x 7→( e D↔(x⊤))⊤. Then define D(2) : x 7→ vec( e D↔(D↔x)), vec( e D↔(D↕x)) + vec( e D↕(D↔x)) √ 2 , vec( e D↕(D↕x)) ! . (5.13) The second-order total variation potential is defined as x 7→∥D(2)x∥1,2. 5.3 Constraint qualifications To apply the results of Theorem 4.3, we need to check that the constraint qualifications (4.2), (4.3), and (4.4) hold. Starting with (4.2), for each k ∈{1, 2} we have sri dom (gk ◦Lk)∗−M∗ k(dom h∗ k)  = sri dom (L∗ k ⊲g∗ k) −M∗ k(dom h∗ k)  = sri L∗ k(dom g∗ k) −M∗ k(dom h∗ k)  = L∗ k ri dom g∗ k) −M∗ k(ri dom h∗ k  , (5.14) where the first line follows from [2, Proposition 15.28] and the fact that gk has full domain, the second line follows from [2, Proposition 12.34(i)], and the third line follows from [2, Corollary 6.15]. Since g1, g2, and h1 are coercive, their conjugates all include 0 in the interior of their domain [2, Theorem 14.17]. Furthermore, the conjugate of h2 = ι{0} is h∗ 2 = 0 which has full domain. Thus, (∀k ∈{1, 2}) 0 ∈L∗ k(ri dom g∗ k) and 0 ∈M∗ k(ri dom h∗ k). (5.15) 21 (a) Original image (b) Degraded image (c) Solution to (5.4) Figure 1: Original, blurred, and restored images. Altogether, (5.14) is satisfied for each k ∈{1, 2} and hence so is (4.2). The qualification (4.3) holds for k = 1 since h1 = ∥· ∥1,2 has full domain. For k = 2, since h2 = ι{0}, using [2, Corollary 6.15] and the linearity of M2, we obtain sri (ran M2 −dom h2) = sri (ran M2) = ri (ran M2) = ran M2. (5.16) Thus, since 0 ∈ran M2, (4.3) is satisfied. On the other hand, since H is finite-dimensional, the constraint qualification (4.4) is implied by Proposition 4.2(v). Both g1 and h1 are norms and therefore have full domain, so (4.8) is satisfied for k = 1. For k = 2, g2 is a norm and has full domain while h2 = ι{0}, so 0 ∈ri dom h2 and hence (4.8) holds for k = 2. To apply Proposition 4.2, the primal problem must have a solution. Here existence of a solution follows from the compactness of C [2, Proposition 11.14(ii)]. 5.4 Numerical experiments Experiments are made on a quad-core 1.60 GHz Intel i7 laptop, with the algorithms and analysis implemented using the free software package GNU Octave . The authors are grateful for the support of the Octave development community. Note that in (4.20), the update for s1,1,n and for p1,k+1,n both involve L∗ kvk,n, hence it is possible to prevent redundant computation by storing L∗ kvk,n as a temporary variable. Similarly, the updates for q1,1,n and q1,k+1,n both involve L∗ kp2,k,n, which can also be stored as a temporary variable for savings. With this approach, each Lk and Mk is applied exactly twice per iteration, and each L∗ k and M∗ k is also applied exactly twice. The restored image is displayed in Fig. 1(c). The algorithm uses all variables initialized to 0. The values of the parameters are as follows: α = β = γ = 10−2. Figures of merit relative to these experiments are provided in Table 1. Given a reference image x and an estimate x = (ξi)1⩽i⩽N, the peak signal-to-noise ratio (PSNR), a standard measure of image quality, is defined by PSNRx(x) = 10 log10 N max1⩽i⩽N ξ2 i PN i=1(ξi −ξi)2 ! (5.17) 22 Table 1: Quantitative measurements of performance. Method Peak signal-to-noise ratio Structural similarity index Blurred and noisy image 20.32 dB 0.545 Restoration 25.42 dB 0.803 and reported in units of decibels (dB). The structural similarity index attempts to quantify human visual response to images; details can be found in . Acknowledgment. The authors thank J.-C. Pesquet for the implementation of the wavelet trans-form. S. R. Becker is funded by the Fondation Sciences Math´ ematiques de Paris. The research of P. L. Combettes is supported in part by the European Union under the 7th Framework Programme “FP7-PEOPLE-2010-ITN”, grant agreement number 264735-SADCO. References H. C. Andrews and B. R. Hunt, Digital Image Restoration. Prentice-Hall, NJ, 1977. H. H. Bauschke and P. L. Combettes, Convex Analysis and Monotone Operator Theory in Hilbert Spaces. Springer, New York, 2011. R. I. Bot ¸, Conjugate Duality in Convex Optimization, Lecture Notes in Economics and Mathematical Sys-tems, vol. 637. Springer, New York, 2010. R. I. Bot ¸, E. R. Csetnek, and E. Nagy, Solving systems of monotone inclusions via primal-dual splitting techniques, Taiwan. J. Math., to appear. R. I. Bot ¸ and C. Hendrich, Convex risk minimization via proximal splitting methods, L. M. Brice˜ no-Arias and P. L. Combettes, A monotone+skew splitting model for composite monotone inclusions in duality, SIAM J. Optim., vol. 21, pp. 1230–1250, 2011. A. Chambolle and P.-L. Lions, Image recovery via total variation minimization and related problems, Numer. Math., vol. 76, pp. 167-188, 1997. A. Cohen, I. Daubechies, and J.-C. Feauveau, Biorthogonal bases of compactly supported wavelets, Comm. Pure Appl. Math., vol. 45, pp. 485–560, 1992. P. L. Combettes, Solving monotone inclusions via compositions of nonexpansive averaged operators, Optimization, vol. 53, pp. 475–504, 2004. P. L. Combettes, Systems of structured monotone inclusions: duality, algorithms, and applications, SIAM J. Optim., to appear. P. L. Combettes and J.-C. Pesquet, Proximal splitting methods in signal processing, in Fixed-Point Al-gorithms for Inverse Problems in Science and Engineering, (H. H. Bauschke et al., eds), pp. 185–212. Springer, New York, 2011. P. L. Combettes and J.-C. Pesquet, Primal-dual splitting algorithm for solving inclusions with mixtures of composite, Lipschitzian, and parallel-sum type monotone operators, Set-Valued Var. Anal., vol. 20, pp. 307–330, 2012. P. L. Combettes and B. C. V˜ u, Variable metric forward-backward splitting with applications to monotone inclusions in duality, Optimization, to appear. 23 P. L. Combettes and V. R. Wajs, Signal recovery by proximal forward-backward splitting, Multiscale Model. Simul., vol. 4, pp. 1168–1200, 2005. J. Eckstein and D. P. Bertsekas, On the Douglas-Rachford splitting method and the proximal point algo-rithm for maximal monotone operators, Math. Programming, vol. 55, pp. 293–318, 1992. GNU Octave scientific package, P. L. Lions and B. Mercier, Splitting algorithms for the sum of two nonlinear operators, SIAM J. Numer. Anal., vol. 16, pp. 964–979, 1979. B. Mercier, Topics in Finite Element Solution of Elliptic Problems (Lectures on Mathematics, no. 63). Tata Institute of Fundamental Research, Bombay, 1979. N. Pustelnik, C. Chaux, and J.-C. Pesquet, Parallel proximal algorithm for image restoration using hybrid regularization, IEEE Trans. Image Process., vol. 20, pp. 2450–2462, 2011. L. I. Rudin, S. Osher, and E. Fatemi, Nonlinear total variation based noise removal algorithms, Phys. D, vol. 60, pp. 259–268, 1992. S. Setzer, G. Steidl, and T. Teuber, Infimal convolution regularizations with discrete ℓ1-type functionals, Commun. Math. Sci., vol. 9, pp. 797–827, 2011. S. Simons, Quadrivariate existence theorems and strong representability, Optimization, vol. 60, pp. 875– 891, 2011. P. Tseng, Applications of a splitting algorithm to decomposition in convex programming and variational inequalities, SIAM J. Control Optim., vol. 29, pp. 119–138, 1991. P. Tseng, A modified forward-backward splitting method for maximal monotone mappings, SIAM J. Control Optim., vol. 38, pp. 431–446, 2000. B. C. V˜ u, A splitting algorithm for dual monotone inclusions involving cocoercive operators, Adv. Comput. Math., vol. 38, pp 667–681, 2013. Z. Wang, A. C. Bovik, H. R. Sheikh, and E. P. Simoncelli, Image quality assessment: From error visibility to structural similarity, IEEE Trans. Image Process., vol. 13, pp. 600–612, 2004. C. Z˘ alinescu, Convex Analysis in General Vector Spaces, World Scientific, River Edge, NJ, 2002. W. P. 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Teach Kids Time Management Lesson Plan | Lysol US Skip to content Our ProductsMore Our Products Disinfecting Sprays Disinfecting Wipes Multi-Purpose Cleaner Sprays Multi-Purpose Cleaner Pours Air Sanitizers Toilet and Bathroom Cleaners Laundry Sanitizers Lavender Range Tips & ResourcesMore Tips & Resources Protect Against Germs Home Cleaning Tips Tips For Parents Tools for Teachers Learn about Cold & Flu Information on COVID-19 New Mom Disinfecting & Cleaning Tips Lysol is HERE for Healthy Schools FAQ Our Mission Sign up Contact Us Home ... Teach Kids Time Management Lesson Plan Teach Kids Time Management Lesson Plan This project helps students see whether they’re investing enough time in tasks that promote healthy habits. By recording the amount of time they spend on various daily routines and comparing their habits with others in the class, students can use their graphs to discuss ways to improve overall health. Download the Time Tally Lesson Plan Goals and Skills Students will… Practice telling and calculating time Organize and sort activities into categories Create and use circle graphs to organize information Learn about basic personal habits required to maintain health Collect, organize, interpret, analyze, and display data Supplies and Preparation: Make copies of the student handouts Healthy Habits Time Tally and My Daily Activity Circle (see Pages 4 and 5) Color markers, crayons, or pencils (red, orange, purple, yellow, green, blue, pink) OR Computer or iPad/tablet access Instruction Steps 1. SHARE.Distribute copies of the Healthy Habits Time Tally and My Daily Activity Circle Graph to each student. 2. BRAINSTORM. Ask students to think about other tasks that may be added to the list of the Healthy Habits Time Tally chart. Is there anything they do on a daily basis that is not on the list? If so, have them write them in at the bottom of the chart. Encourage them to think about things they do in the morning, at school, after school, at home, and at night. 3. TRACKING TIME!Go over the instructions on each handout to explain the project to students. Explain that they will first spend one day, or 24 hours, tracking how much time they spend on each activity. Review the basics of telling time. Cover how many hours are in a day, how many minutes are in an hour, and how many quarters are in an hour. They will be tracking their time in 15-minute increments. Begin the project in class so that students have time to practice the activity. For example, as a class you may track the time spent “going to school,” “washing hands,” “eating,” and “reading.” Modifications: If tracking time in 15-minute increments is too challenging for some students, have them write down the time they begin an activity, and the time it finished. Then assist them with the final calculations. 4. ADD UP TIME.After the tracking activity is complete, help students add up the time they spent doing different activities. Make sure their activities add up to 24 hours (give or take 15 minutes). 5. SORT ACTIVITIES.Once the time has been tallied, activities will need to be sorted into different categories (see the color-coded key on My Daily Activity Circle Graph). For example, “watching TV” and “playing video games” would fall under “Entertainment.” Work with students to sort activities and add up their times. 6. MY DAILY ACTIVITY CIRCLE GRAPH. Explain that students will now use a circle graph to color in and “map” the time they spend doing different things each day. Each section of the graph represents one hour of a day. Provide a few examples of time and activities to model how to fill in the graph. For example, if a student spent only 15 minutes on an activity, that is ¼ of an hour. Therefore, on the circle graph, they will only color in ¼ of a circle section. If they spent 2 hours on an activity, they would color in 2 circle sections. Remember to use different colors for different tasks, as explained in the key of the circle graph. Modifications: Alternately, you can use this as an opportunity for students to practice graphing on computers or iPads/tablets. 7. INTERPRET AND EVALUATE RESULTS.Display completed graphs on a bulletin board. Invite students to share their observations about different graphs with the rest of the class. Are there any patterns children see? Ask students to interpret what they mean. For example, if the color purple is bigger than the color green, it means that more time is spent on entertainment than physical activity. Ask students to evaluate if their class spends enough time on physical activity. How can their everyday habits be affecting their overall health? What can they do differently? What can you as a class do differently together? 8. EXTEND THE LESSON.Ask the class to continue to record and tally their habits for several days or weeks. Then have them identify trends and changes in their behaviors using bar charts—for instance, tracking whether they spend enough time outdoors. Invite parents to take part by helping students track their habits at home using a mobile application such as iRewardChart. 9. HOME CONNECTION.Let parents know about the Healthy Habits Time Tally activity in advance of the lesson. That way, parents can help students track their time at home during various activities. At the end of the lesson, share results with parents. What healthy habits can they support at home? Exercising more? Download the Time Tally Lesson Plan Explore More Lesson Plans Explore More Resources Related Articles Clean by room Attack of the Allergens! Clean by room How to Sanitize Laundry Clean by room Is Your Laundry Really Clean? 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We’re here for it. Show 1 of 21 DISCLAIMER: Healthy Habits® is Presented by Lysol® in collaboration with NEA and National PTA SOURCES: Education Standards: (NHES) Health: 1.5.1, 5.5.4, 6.5.1, 7.5.1; (CCSS) English Language Arts: SL.K-1.1, L.K.5; (CCSS) Math: K.CC.A.3, K.CC.C.6, K.MD.B.3,1.MD.C.4 Privacy Policy Consumer Health Data Privacy Policy Notice at Collection Terms of Use Ingredient Information & Material Safety Data Sheet Privacy Rights Satisfaction Guarantee Modern Slavery Act Statement Do Not Sell or Share My Personal Information / Opt-out of Targeted Advertising Sitemap Your Privacy Rights © 2025 Reckitt • All rights reserved Lysol Youtube Channel Lysol Facebook Page We, and third parties, use cookies, pixels, and similar technologies to collect and share your information (e.g., browser/device data, site view/interaction data, data provided to us) for ad targeting, analytics, personalization, and site functionality purposes as further described in ourPrivacy Policy. 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Analytical cookies allow us (i) insight into user metrics, such as visit count, “events” (e.g., tracking clicks, searches, or other actions taken by a user), most to least popular pages, most to least popular origin/destinations, and traffic sources, (ii) to conduct A/B testing between different versions of the website, and (iii) browser session recording and feedback (e.g., for analyzing or replicating bugs or user experience issues) for the purpose of optimizing the performance of our digital properties. Targeting Cookies [x] Targeting Cookies These cookies help us serve our own ads on third-party digital properties, serve our advertising partners’ ads on our digital properties, and help ensure these ads are relevant to you or your interests, and measure the effectiveness of these advertising campaigns. 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
11423
https://www.khanacademy.org/math/probability/xa88397b6:display-quantitative/xa88397b6:mean-median-data-displays/v/estimating-mean-median-displays
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11424
https://www.youtube.com/watch?v=usxTJbQbAIg
How to Divide a Smaller Number by a Larger Number | Math with Mr. J Math with Mr. J 1650000 subscribers 8098 likes Description 534566 views Posted: 24 Mar 2021 Welcome to How to Divide a Smaller Number by a Larger Number with Mr. J! Need help with dividing smaller numbers by larger numbers? You're in the right place! Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with how to divide a smaller number by a larger number. Mr. J will go through dividing smaller numbers by larger numbers examples and explain the steps of how to divide a smaller number by a larger number. About Math with Mr. J: This channel offers instructional videos that are directly aligned with math standards. Teachers, parents/guardians, and students from around the world have used this channel to help with math content in many different ways. All material is absolutely free. Click Here to Subscribe to the Greatest Math Channel On Earth: Follow Mr. J on Twitter: @MrJMath5 Email: math5.mrj@gmail.com Music: Hopefully this video is what you're looking for when it comes to dividing smaller numbers by larger numbers. 575 comments Transcript: [Music] welcome to math with mr j [Music] in this video i'm going to cover how to divide a smaller number by a larger number and we have two examples that we're going to go through together in order to get this down now as we go through these problems we're going to need to use a decimal point and placeholder zeros in order to go through our division process so let's jump into number one and see exactly how we solve these types of problems so for number one we have 6 divided by 15 and the first thing we want to do is set this problem up now don't worry about which number is larger in value or smaller in value that has nothing to do with how we're going to set this division problem up we need to take a look at the order of the problem and what this problem means so again we have 6 divided by 15. so 6 is being divided or split up so that's what we call our dividend it's what goes under our division bar now 15 is what we are dividing by it's what we call the divisor so it is going to go on the outside now that may look like it's off or incorrect but that is correct we have 6 divided by 15. again don't worry about which number is larger or smaller when we set up division problems this is not 15 divided by 6. these are two different problems 15 divided by 6 the larger number and value would happen to be under the division bar and the 6 would be on the outside but again that's not what we have we have 6 divided by 15. so we're starting with 6 wholes and dividing or splitting it into 15 equal groups so our answer is actually going to be less than one whole so let's see how we solve this and we'll start with our division process so divide how many whole groups of 15 can i pull out of 6 well i can't do that so 0 we cannot pull any whole groups of fifteen out of six multiply zero times fifteen is zero subtract six minus zero is six so in order to continue here we need to bring something down so what we're going to do we're going to place a decimal point after the 6 so we can place a decimal point after a whole number the value of that 6 is still 6. we didn't change the problem at all so once you place that decimal point bring it straight up that's going to place it in our answer correctly so we need something to bring down so what we can do we can place a 0 to the right of that decimal again that doesn't change the value of the 6 or the value of the problem so we can do that so now let's bring that 0 down and continue our process so now we have sixty divided by fifteen how many whole groups of fifteen out of sixty well four that hits sixty exactly so four times fifteen is sixty subtract and we get zero so we are done our answer is four tenths again we got an answer less than one whole think about it we split those six holes into 15 equal groups so we didn't have enough for one whole in each of those groups so we ended up with 4 10 as our answer or the value of each of those 15 groups so let's move on to number two where we have 12 divided by 50. again don't worry about which number is smaller or larger in value worry about the order and what that problem means so we have 12 divided by 50 so 12 is our dividend it's what's being divided or split up so that goes under our division bar and the 50 goes on the outside now we're all set up and ready to go through our division process so how many whole groups of 50 out of 1 well we can't do that so how many whole groups of 50 out of 12 well we can't do that either we can't pull a whole group of 50 out of 12. so let's put a zero there to hold this place 0 times 50 is 0. subtract we get 12. so now we're stuck so what we can do we can use a decimal point and placeholder zeros so let's put our decimal point after 12 again that does not change the value of our problem or that 12. bring that decimal point straight up and that will place it correctly in our answer so we need something to bring down so what we can do we can use a placeholder 0 in the tenths place again doesn't change the value of anything and bring that 0 down so how many whole groups of 50 out of 120 well two that gets us to one hundred and that's as close as we can get to one hundred twenty so two let me extend this division bar there two times fifty is a hundred subtract and we get 20. so we are still not done so what we can do we can use another placeholder zero let's place one in the hundredths place again not changing the value of anything and we can bring that down so now we have 200 divided by 50. how many whole groups of 50 out of 200 4 that hits 200 exactly so this is going to be our last digit in our answer but let's wrap everything up with our process so now we need to multiply 4 times 50 is 200 subtract and we get that clean cut 0. so our final answer is 24 hundredths so our answer was less than one whole again because we took 12 holes and split that 12 into 50 equal groups so those groups each of those groups were less than a whole and exactly they were 24 hundredths so there you have it there's how you divide a smaller number by a larger number use a decimal point and placeholder zeros i hope that helped thanks so much for watching until next time peace
11425
https://math.answers.com/math-and-arithmetic/How_are_linear_units_square_units_and_cubic_units_different
How are linear units square units and cubic units different? - Answers Create 0 Log in Subjects>Math>Math & Arithmetic How are linear units square units and cubic units different? Anonymous ∙ 14 y ago Updated: 9/25/2022 Lines are measured with lines. For example, the length of a road is measured with standard lines (linear units) such as yards, miles, meters or kilometers. Areas are measured with areas. A field is measured with standard areas (usually square units) such as square yards, square miles, or square kiometers. Volumes are measured with volumes. A container is measured with standard volumes (often cubic units) such as square inches or square centimeters. Look out for calculations involving linear, square and cubic units. For example, a square measuring three inches on the side has an area of 3 x 3 = 9 square inches. A cubical tank measuring five meters on each side has a volume of 5 x 5 x 5 = 125 cubic meters. Wiki User ∙ 14 y ago Copy Add Your Answer What else can I help you with? Search Continue Learning about Math & Arithmetic ### How you devid sq foot into cubic yds? You can't convert that. You can convert linear units to linear units, square units to square units, or cubic units to cubic units. ### Calculate cubic units in a rectangle? A rectangle has a perimeter with linear units, and a surface area with square units. It has no cubic units. ### What is the difference between linear and cubic units? A linear unit is used to measure length or distance whereas a cubic unit is used to measure volumes. A cubic unit may defined as a cubic measure (eg a litre) or it could be composed of three linear units (eg cubic centimetre), or a square unit and a linear one (acre-foot). This last combination is rare in the SI system because the linear units are more easily linked to square and cubic units. ### Conversion of square feet to linear feet? You don't convert square units to linear units. The two are utterly incompatible.You don't convert square units to linear units. The two are utterly incompatible.You don't convert square units to linear units. The two are utterly incompatible.You don't convert square units to linear units. The two are utterly incompatible. ### Square inches to cubic foot? You simply can't convert square units to cubic units. Related Questions Trending Questions What is the area of a 14 foot square shed?How can fraction strips number line thinking with numbers help you find equivalent fractions?How many furlongs are in 1000 yards?What is the name of the line that bisects two parallel lines?Do all 4 sided polygons are quadrilaterals?What does ruomintang mean?What is the first derivative used for?How do you say five feet 2 inches and a half in french?How many hours is 40 percent of the day?How many black hearts are in deck of cards?What is half of DII in roman numerals?The diagonals of a rhombus are perpendicular bisectors of one another?In a circle with a ten ft diameter what is the radius?What does mean borough 4?How many meters are in 50cm?How many sec in 10 min?When reading time signatures the top number tells what?How do you write 4 hundredths 7 ones 5 tens and 1 hundred?What is 50 percent of86?What is 7 10ths as an improper fraction? Resources LeaderboardAll TagsUnanswered Top Categories AlgebraChemistryBiologyWorld HistoryEnglish Language ArtsPsychologyComputer ScienceEconomics Product Community GuidelinesHonor CodeFlashcard MakerStudy GuidesMath SolverFAQ Company About UsContact UsTerms of ServicePrivacy PolicyDisclaimerCookie PolicyIP Issues Copyright ©2025 Answers.com. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Answers.
11426
https://www.tn.gov/content/dam/tn/stateboardofeducation/documents/massivemeetingsfolder/meetingfiles3/4-15-16_V_A_Math_Standards_Attachment.pdf
1 Tennessee Math Standards Introduction The Process The Tennessee State Math Standards were reviewed and developed by Tennessee teachers for Tennessee schools. The rigorous process used to arrive at the standards in this document began with a public review of the then-current standards. After receiving 130,000+ reviews and 20,000+ comments, a committee composed of Tennessee educators spanning elementary through higher education reviewed each standard. The committee scrutinized and debated each standard using public feedback and the collective expertise of the group. The committee kept some standards as written, changed or added imbedded examples, clarified the wording of some standards, moved some standards to different grades, and wrote new standards that needed to be included for coherence and rigor. From here the standards went before the appointed Standards Review Committee to make further recommendations before being presented to the Tennessee Board of Education for final adoption. The result is Tennessee Math Standards for Tennessee Students by Tennesseans. Mathematically Prepared Tennessee students have various mathematical needs that their K-12 education should address. All students should be able to recall and use their math education when the need arises. That is, a student should know certain math facts and concepts such as the multiplication table, how to add, subtract, multiply, and divide basic numbers, how to work with simple fractions and percentages, etc. There is a level of procedural fluency that a student’s K-12 math education should provide him or her along with conceptual understanding so that this can be recalled and used throughout his or her life. Students also need to be able to reason mathematically. This includes problem solving skills in work and non-work related settings and the ability to critically evaluate the reasoning of others. A student’s K-12 math education should also prepare him or her to be free to pursue post-secondary education opportunities. Students should be able to pursue whatever career choice, and its post-secondary education requirements, that they desire. To this end, the K-12 math standards lay the foundation that allows any student to continue further in college, technical school, or with any other post-secondary educational needs. A college and career ready math class is one that addresses all of the needs listed above. The standards’ role is to define what our students should know, understand, and be able to do mathematically so as to fulfill these needs. To that end, the standards address conceptual understanding, procedural fluency, and application. Conceptual Understanding, Procedural Fluency, and Application In order for our students to be mathematically proficient, the standards focus on a balanced development of conceptual understanding, procedural fluency, and application. Through this balance, students gain understanding and critical thinking skills that are necessary to be truly college and career ready. Conceptual understanding refers to understanding mathematical concepts, operations, and relations. It is more than knowing isolated facts and methods. Students should be able to make sense of why a mathematical idea is important and the kinds of contexts in which it is useful. It also allows students to connect prior knowledge to new ideas and concepts. 2 Procedural fluency is the ability to apply procedures accurately, efficiently, and flexibly. One cannot stop with memorization of facts and procedures alone. It is about recognizing when one strategy or procedure is more appropriate to apply than another. Students need opportunities to justify both informal strategies and commonly used procedures through distributed practice. Procedural fluency includes computational fluency with the four arithmetic operations. In the early grades, students are expected to develop fluency with whole numbers in addition, subtraction, multiplication, and division. Therefore, computational fluency expectations are addressed throughout the standards. Procedural fluency extends students’ computational fluency and applies in all strands of mathematics. It builds from initial exploration and discussion of number concepts to using informal strategies and the properties of operations to develop general methods for solving problems (NCTM, 2014). Application provides a valuable context for learning and the opportunity to practice skills in a relevant and a meaningful way. As early as Kindergarten, students are solving simple “word problems” with meaningful contexts. In fact, it is in solving word problems that students are building a repertoire of procedures for computation. They learn to select an efficient strategy and determine whether the solution(s) makes sense. Problem solving provides an important context in which students learn about numbers and other mathematical topics by reasoning and developing critical thinking skills (Adding It Up, 2001). 3 Progressions The standards for each grade are not written to be nor are they to be considered as an island in and of themselves. There is a flow, or progression, from one grade to the next, all the way through to the high school standards. There are four main progressions that are composed of mathematical domains/conceptual categories (see the Structure section below and color chart on the following page). The progressions are grouped as follows: Grade Domain/Conceptual Category K Counting and Cardinality K-5 Number and Operations in Base Ten 3-5 Number and Operations – Fractions 6-7 Ratios and Proportional Relationships 6-8 The Number System 9-12 Number and Quantity K-5 Operations and Algebraic Thinking 6-8 Expressions and Equations 8 Functions 9-12 Algebra and Functions K-12 Geometry K-5 Measurement and Data 6-12 Statistics and Probability 4 Each of the progressions begins in Kindergarten, with a constant movement toward the high school standards as a student advances through the grades. This is very important to guarantee a steady, age appropriate progression which allows the student and teacher alike to see the overall coherence of and connections among the mathematical topics. It also ensures that gaps are not created in the mathematical education of our students. 5 Structure of the Standards Most of the structure of the previous state standards has been maintained. This structure is logical and informative as well as easy to follow. An added benefit is that most Tennessee teachers are already familiar with it. The structure includes:  Content Standards - Statements of what a student should know, understand, and be able to do.  Clusters - Groups of related standards. Cluster headings may be considered as the big idea(s) that the group of standards they represent are addressing. They are therefore useful as a quick summary of the progression of ideas that the standards in a domain are covering and can help teachers to determine the focus of the standards they are teaching.  Domains - A large category of mathematics that the clusters and their respective content standards delineate and address. For example, Number and Operations – Fractions is a domain under which there are a number of clusters (the big ideas that will be addressed) along with their respective content standards, which give the specifics of what the student should know, understand, and be able to do when working with fractions.  Conceptual Categories – The content standards, clusters, and domains in the 9th-12th grades are further organized under conceptual categories. These are very broad categories of mathematical thought and lend themselves to the organization of high school course work. For example, Algebra is a conceptual category in the high school standards under which are domains such as Seeing Structure in Expressions, Creating Equations, Arithmetic with Polynomials and Rational Expressions, etc. Standards and Curriculum It should be noted that the standards are what students should know, understand, and be able to do; but, they do not dictate how a teacher is to teach them. In other words, the standards do not dictate curriculum. For example, students are to understand and be able to add, subtract, multiply, and divide fractions according to the standards. Although within the standards algorithms are mentioned and examples are given for clarification, how to approach these concepts and the order in which the standards are taught within a grade or course are all decisions determined by the local district, school, and teachers. 6 Example from the Standards’ Document for K – 8 Taken from 3rd Grade Standards: The domain is indicated at the top of the table of standards. The left column of the table contains the cluster headings. A light green coloring of the cluster heading (and codes of each of the standards within that cluster) indicates the major work of the grade. Supporting standards have no coloring. In this way, printing on a non-color printer, the standards belonging to the major work of the grade will be lightly shaded and stand distinct from the supporting standards. This color coding scheme will be followed throughout all standards K – 12. Next to the clusters are the content standards that indicate specifically what a student is to know, understand, and do with respect to that cluster. The numbering scheme for K-8 is intuitive and consistent throughout the grades. The numbering scheme for the high school standards will be somewhat different. Example coding for grades K-8 standards: 3.MD.A.1 3 is the grade level. Measurement and Data (MD) is the domain. A is the cluster (ordered by A, B, C, etc. for first cluster, second cluster, etc.). 1 is the standard number (the standards are numbered consecutively throughout each domain regardless of cluster). 7 Example from the Standards’ Document for 9 – 12 Taken from Integrated Math 1 Standards: The high school standards follow a slightly different coding structure. They start with the course indicator (M1 – Integrated Math 1, A1 – Algebra 1, G – Geometry, etc.), then the conceptual category (in the example below – Algebra) and then the domain (just above the table of standards it represents – Seeing Structure in Expressions). There are various domains under each conceptual category. The table of standards contains the cluster headings (see explanation above), content standards, and the scope and clarifications column, which gives further clarification of the standard and the extent of its coverage in the course. A with a standard indicates a modeling standard (see MP4 on p.11). The color coding is light green for the major work of the grade and no color for the supporting standards. Example coding for grades 9-12 standards: M1.A.SSE.A.1 Integrated Math 1 (M1) is the course. Algebra (A) is the conceptual category. Seeing Structure in Expressions (SSE) is the domain. A is the cluster (ordered by A, B, C, etc. for first cluster, second cluster, etc.). 1 is the standard number (the standards are numbered consecutively throughout each domain regardless of cluster). 8 Tennessee State Math Standards 9 The Standards for Mathematical Practice Being successful in mathematics requires that development of approaches, practices, and habits of mind be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop within their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Processes and proficiencies are two words that address the purpose and intent of the practice standards. Process is used to indicate a particular course of action intended to achieve a result, and this ties to the process standards from NCTM that pertain to problem solving, reasoning and proof, communication, representation, and connections. Proficiencies pertain to being skilled in the command of fundamentals derived from practice and familiarity. Mathematically, this addresses concepts such as adaptive reasoning, strategic competence, conceptual understanding, procedural fluency, and productive dispositions toward the work at hand. The practice standards are written to address the needs of the student with respect to being successful in mathematics. These standards are most readily developed in the solving of high-level mathematical tasks. High-level tasks demand a greater level of cognitive effort to solve than routine practice problems do. Such tasks require one to make sense of the problem and work at solving it. Often a student must reason abstractly and quantitatively as he or she constructs an approach. The student must be able to argue his or her point as well as critique the reasoning of others with respect to the task. These tasks are rich enough to support various entry points for finding solutions. To develop the processes and proficiencies addressed in the practice standards, students must be engaged in rich, high-level mathematical tasks that support the approaches, practices, and habits of mind which are called for within these standards. The following are the eight standards for mathematical practice: Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. A full description of each of these standards follows. 10 MP1: Make sense of problems and persevere in solving them. Mathematically proficient students start by explaining to themselves the meaning of a problem and looking for entry points to its solution. They analyze givens, constraints, relationships, and goals. They make conjectures about the form and meaning of the solution and plan a solution pathway rather than simply jumping into a solution attempt. They consider analogous problems and try special cases and simpler forms of the original problem in order to gain insight into its solution. They monitor and evaluate their progress and change course if necessary. Older students might, depending on the context of the problem, transform algebraic expressions or change the viewing window on their graphing calculator to get the information they need. Mathematically proficient students can explain correspondences between equations, verbal descriptions, tables, and graphs or draw diagrams of important features and relationships, graph data, and search for regularity or trends. Younger students might rely on using concrete objects or pictures to help conceptualize and solve a problem. Mathematically proficient students check their answers to problems using a different method, and they continually ask themselves, "Does this make sense?" They can understand the approaches of others to solving complex problems and identify correspondences between different approaches. MP2: Reason abstractly and quantitatively. Mathematically proficient students make sense of quantities and their relationships in problem situations. They bring two complementary abilities to bear on problems involving quantitative relationships: the ability to decontextualize—to abstract a given situation and represent it symbolically and manipulate the representing symbols as if they have a life of their own, without necessarily attending to their referents—and the ability to contextualize, to pause as needed during the manipulation process in order to probe into the referents for the symbols involved. Quantitative reasoning entails habits of creating a coherent representation of the problem at hand, considering the units involved, attending to the meaning of quantities, not just how to compute them, and knowing and flexibly using different properties of operations and objects. MP3: Construct viable arguments and critique the reasoning of others. Mathematically proficient students understand and use stated assumptions, definitions, and previously established results in constructing arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases and can recognize and use counterexamples. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Mathematically proficient students are also able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and, if there is a flaw in an argument, explain what it is. Elementary students can construct arguments using concrete referents such as objects, drawings, diagrams, and actions. Such arguments can make sense and be correct, even though they are not generalized or made formal until later grades. Later, students learn to determine domains to which an argument applies. Students at all grades can listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments. 11 MP4: Model with mathematics. Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. In early grades, this might be as simple as writing an addition equation to describe a situation. In middle grades, a student might apply proportional reasoning to plan a school event or analyze a problem in the community. By high school, a student might use geometry to solve a design problem or use a function to describe how one quantity of interest depends on another. Mathematically proficient students who can apply what they know are comfortable making assumptions and approximations to simplify a complicated situation, realizing that these may need revision later. They are able to identify important quantities in a practical situation and map their relationships using such tools as diagrams, two-way tables, graphs, flowcharts, and formulas. They can analyze those relationships mathematically to draw conclusions. They routinely interpret their mathematical results in the context of the situation and reflect on whether the results make sense, possibly improving the model if it has not served its purpose. MP5: Use appropriate tools strategically. Mathematically proficient students consider the available tools when solving a mathematical problem. These tools might include pencil and paper, concrete models, a ruler, a compass, a protractor, a calculator, a spreadsheet, a computer algebra system, a statistical package, or dynamic geometry software. Proficient students are sufficiently familiar with tools appropriate for their grade or course to make sound decisions about when each of these tools might be helpful, recognizing both the insight to be gained and their limitations. For example, mathematically proficient high school students analyze graphs of functions and solutions generated using a graphing calculator. They detect possible errors by strategically using estimation and other mathematical knowledge. When making mathematical models, they know that technology can enable them to visualize the results of varying assumptions, explore consequences, and compare predictions with data. Mathematically proficient students at various grade levels are able to identify relevant external mathematical resources, such as digital content located on a website, and use them to pose or solve problems. They are able to use technological tools to explore and deepen their understanding of concepts. MP6: Attend to precision. Mathematically proficient students try to communicate precisely to others. They try to use clear definitions in discussion with others and in their own reasoning. They state the meaning of the symbols they choose, including using the equal sign consistently and appropriately. They are careful about specifying units of measure and labeling axes to clarify the correspondence with quantities in a problem. They calculate accurately and efficiently, expressing numerical answers with a degree of precision appropriate for the problem context. In the elementary grades, students give carefully formulated explanations to each other. By the time they reach high school, they have learned to examine claims and make explicit use of definitions. 12 MP7: Look for and make use of structure. Mathematically proficient students look closely to discern a pattern or structure. Young students, for example, might notice that three and seven more is the same amount as seven and three more, or they may sort a collection of shapes according to how many sides the shapes have. Later, students see 7 × 8 equals the well-remembered 7 × 5 + 7 × 3, in preparation for learning about the distributive property. In the expression x2 + 9x + 14, older students can see the 14 as 2 × 7 and the 9 as 2 + 7. They recognize the significance of an existing line in a geometric figure and can use the strategy of drawing an auxiliary line for solving problems. They also can step back for an overview and shift perspective. They can see complicated things, such as some algebraic expressions, as single objects or as being composed of several objects. For example, they can see 5 - 3(x - y)2 as 5 minus a positive number times a square and use that to realize that its value cannot be more than 5 for any real numbers x and y. MP8: Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y - 2)/(x - 1) = 3. Noticing the regularity in the way terms cancel when expanding (x - 1)(x + 1), (x - 1)(x2 + x + 1), and (x - 1)(x3 + x2 + x + 1) might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results. 13 Literacy Skills for Mathematical Proficiency Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others and analyze and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Reading Reading in mathematics is different from reading literature. Mathematics contains expository text along with precise definitions, theorems, examples, graphs, tables, charts, diagrams, and exercises. Students are expected to recognize multiple representations of information, use mathematics in context, and draw conclusions from the information presented. In the early grades, non-readers and struggling readers benefit from the use of multiple representations and contexts to develop mathematical connections, processes, and procedures. As students’ literacy skills progress, their skills in mathematics develop so that by high school, students are using multiple reading strategies, analyzing context-based problems to develop understanding and comprehension, interpreting and using multiple representations, and fully engaging with mathematics textbooks and other mathematics-based materials. These skills support Mathematical Practices 1 and 2. Vocabulary Understanding and using mathematical vocabulary correctly is essential to mathematical proficiency. Mathematically proficient students use precise mathematical vocabulary to express ideas. In all grades, separating mathematical vocabulary from everyday use of words is important for developing an understanding of mathematical concepts. For example, a “table” in everyday use means a piece of furniture, while in mathematics, a “table” is a way of organizing and presenting information. Mathematically proficient students are able to parse a mathematical term, definition, or theorem, provide examples and counterexamples, and use precise mathematical vocabulary in reading, speaking, and writing arguments and explanations. These skills support Mathematical Practice 6. Speaking and Listening Mathematically proficient students can listen critically, discuss, and articulate their mathematical ideas clearly to others. As students’ mathematical abilities mature, they move from communicating through reiterating others’ ideas to paraphrasing, summarizing, and drawing their own conclusions. A Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 14 mathematically proficient student uses appropriate mathematics vocabulary in verbal discussions, listens to mathematical arguments, and dissects an argument to recognize flaws or determine validity. These skills support Mathematical Practice 3. Writing Mathematically proficient students write mathematical arguments to support and refute conclusions and cite evidence for these conclusions. Throughout all grades, students write reflectively to compare and contrast problem-solving approaches, evaluate mathematical processes, and analyze their thinking and decision-making processes to improve their mathematical strategies. These skills support Mathematical Practices 2, 3, and 4. 15 Mathematics | Grade K The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout Kindergarten. Counting and Cardinality Students use numbers, including written numerals and counting, to develop concepts about quantity. Students use numbers to solve contextual problems and represent quantities, such as counting objects in a set, counting out a given number of objects, and comparing sets or numerals. Students use effective strategies for counting and answering quantitative questions, including quickly recognizing the cardinalities of small sets of objects and learning about counting sequences. Operations and Algebraic Thinking Students develop an understanding of addition and subtraction and determine when to add or subtract in a given context. Students should solve a variety of problem types in order to make connections among contexts, equations, and strategies (See Table 1 - Addition and Subtraction Situations). Students choose from multiple representations (including using objects, fingers, mental images, drawings, sounds, acting out situations, verbal explanations, expressions, or equations) when solving addition and subtraction problems within 10. Students decompose quantities within 10 in various ways, and fluently add and subtract using mental strategies. By the end of Kindergarten, students should fluently add and subtract within 10. Number and Operations in Base Ten Students understand that numbers from 11 to 19 represent ten ones and some more ones by using objects or drawings, and record each composition or decomposition by a drawing and/or write an equation to represent this relationship. Measurement and Data Students describe and sort objects in many different ways. This includes length, weight, and coins. They classify objects in categories and compare measurable attributes. Students begin to learn to graph and analyze collections of objects. Students learn to identify the penny, nickel, dime, and quarter and know the value of each. Geometry Students describe their physical world using geometric ideas, vocabulary, and positional words. Regardless of orientation, students name two-dimensional and three-dimensional shapes, compare shapes, and combine shapes to create new shapes. Students identify patterns they discover in numbers, counting, and shapes. 16 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 17 Counting and Cardinality (CC) Cluster Headings Content Standards Operations and Algebraic Thinking (OA) Cluster Headings Content Standards A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. (See Table 1 - Addition and Subtraction Situations) K.OA.A.1 Represent addition and subtraction with objects, fingers, mental images, drawings, sounds, acting out situations, verbal explanations, expressions, or equations. K.OA.A.2 Add and subtract within 10 to solve contextual problems using objects or drawings to represent the problem. A. Know number names and the counting sequence. K.CC.A.1 Count to 100 by ones, fives, and tens. Count backward from 10. K.CC.A.2 Count forward beginning from a given number within the known sequence (instead of having to begin at 1). K.CC.A.3 Write numbers from 0 to 20. Represent a number of objects with a written numeral 0-20. B. Count to tell the number of objects. K.CC.B.4 Understand the relationship between numbers and quantities; connect counting to cardinality. a. When counting objects, say the number names in the standard order, using one-to-one correspondence. b. Recognize that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted. c. Recognize that each successive number name refers to a quantity that is one greater. K.CC.B.5 Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, a circle, or as many as 10 things in a scattered configuration. Given a number from 1-20, count out that many objects. C. Compare numbers. K.CC.C.6 Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group. K.CC.C.7 Compare two given numbers up to 10, when written as numerals, using the terms greater than, less than, or equal to. 18 Cluster Headings Content Standards A. Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from. (See Table 1 - Addition and Subtraction Situations) K.OA.A.3 Decompose numbers less than or equal to 10 into addend pairs in more than one way (e.g., 5 = 2 + 3 and 5 = 4 + 1) by using objects or drawings. Record each decomposition using a drawing or writing an equation. K.OA.A.4 Find the number that makes 10, when added to any given number, from 1 to 9 using objects or drawings. Record the answer using a drawing or writing an equation. K.OA.A.5 Fluently add and subtract within 10 using mental strategies. Number and Operations in Base Ten (NBT) Cluster Headings Content Standards Measurement and Data (MD) Cluster Headings Content Standards A. Describe and compare measurable attributes. K.MD.A.1 Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. K.MD.A.2 Directly compare two objects with a measurable attribute in common, to see which object has more of/less of the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. B. Work with money. K.MD.B.3 Identify the penny, nickel, dime, and quarter and recognize the value of each. C. Classify objects and count the number of objects in each category. K.MD.C.4 Sort a collection of objects into a given category, with 10 or less in each category. Compare the categories by group size. A. Work with numbers 11– 19 to gain foundations for place value. K.NBT.A.1 Compose and decompose numbers from 11 to 19 into ten ones and some more ones by using objects or drawings. Record the composition or decomposition using a drawing or by writing an equation. 19 Geometry (G) Cluster Headings Content Standards A. Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). K.G.A.1 Describe objects in the environment using names of shapes. Describe the relative positions of these objects using terms such as above, below, beside, in front of, behind, between, and next to. K.G.A.2 Correctly name shapes regardless of their orientations or overall size. K.G.A.3 Identify shapes as two-dimensional or three-dimensional. B. Analyze, compare, create, and compose shapes. K.G.B.4 Describe similarities and differences between two- and three-dimensional shapes, in different sizes and orientations. K.G.B.5 Model shapes in the world by building and drawing shapes. K.G.B.6 Compose larger shapes using simple shapes and identify smaller shapes within a larger shape. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 20 21 Mathematics | Grade 1 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 1st grade. Operations and Algebraic Thinking Students extend previous understanding of addition and subtraction to solve contextual problems within 20, add three addends, and recognize subtraction as an unknown addend problem. Students solve a variety of problem types, with unknowns in all positions, in order to make connections among contexts, equations, and strategies (See Table 1 - Addition and Subtraction Situations). Students should apply properties of operations as strategies to add and subtract when needed (See Table 3 - Properties of Operations). By the end of 1st grade, students should know from memory sums of 10 and fluently add and subtract within 20. Students demonstrate their understanding of the equal sign (=) by determining if addition/subtraction equations are true or false and writing equations to represent a given situation. Numbers and Operations in Base Ten Students read, write, and represent a given number of objects numerically and extend the counting sequence to 120. They demonstrate the ability to count from any number up to 120 and count backward from 20. Students understand that two-digit numbers represent groups of tens and ones and each two-digit number can be composed and decomposed in a variety of ways. Using place value understanding, students compare two-digit numbers based on the number of tens and ones represented in the given numbers using symbols for comparison. Students build number sense and use increasingly sophisticated strategies based on place value and properties of operations to add and subtract. Measurement and Data This is the first time students develop an understanding of the meaning and processes of measurement including iteration of non-standard equal-sized units. Students compare two objects using a third object as a benchmark and also order objects by length. Students are introduced to writing and telling time to the nearest hour and half-hour. Students build on their previous work in kindergarten and count the value of like coins using the ¢ symbol. Students interpret data to answer questions such as how many more or less. Geometry Students build on previous knowledge to explore attributes of shapes and to build, draw, and identify two-dimensional shapes. Two and three-dimensional shapes are used to create composite shapes. This is the first time students partition circles and rectangles to create halves and fourths/quarters. 22 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 23 Operations and Algebraic Thinking (OA) Cluster Headings Content Standards Number and Operations in Base Ten (NBT) Cluster Headings Content Standards A. Extend the counting sequence. 1.NBT.A.1 Count to 120, starting at any number. Read and write numerals to 120 and represent a number of objects with a written numeral. Count backward from 20. B. Understand place value. 1.NBT.B.2 Know that the digits of a two-digit number represent groups of tens and ones (e.g., 39 can be represented as 39 ones, 2 tens and 19 ones, or 3 tens and 9 ones). 1.NBT.B.3 Compare two two-digit numbers based on the meanings of the digits in each place and use the symbols >, =, and < to show the relationship. A. Represent and solve problems involving addition and subtraction. 1.OA.A.1 Add and subtract within 20 to solve contextual problems, with unknowns in all positions, involving situations of add to, take from, put together/take apart, and compare. Use objects, drawings, and equations with a symbol for the unknown number to represent the problem. (See Table 1 - Addition and Subtraction Situations) 1.OA.A.2 Add three whole numbers whose sum is within 20 to solve contextual problems using objects, drawings, and equations with a symbol for the unknown number to represent the problem. B. Understand and apply properties of operations and the relationship between addition and subtraction. (See Table 3 - Properties of Operations) 1.OA.B.3 Apply properties of operations (additive identity, commutative, and associative) as strategies to add and subtract. (Students need not use formal terms for these properties.) 1.OA.B.4 Understand subtraction as an unknown-addend problem. For example, to solve 10 – 8 = ___, a student can use 8 + ___ = 10. C. Add and subtract within 20. 1.OA.C.5 Add and subtract within 20 using strategies such as counting on, counting back, making 10, using fact families and related known facts, and composing/ decomposing numbers with an emphasis on making ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9 or adding 6 + 7 by creating the known equivalent 6 + 4 + 3 = 10 + 3 = 13). 1.OA.C.6 Fluently add and subtract within 20 using mental strategies. By the end of 1st grade, know from memory all sums up to 10. D. Work with addition and subtraction equations. 1.OA.D.7 Understand the meaning of the equal sign (e.g., 6 = 6; 5 + 2 = 4 + 3; 7 = 8 - 1). Determine if equations involving addition and subtraction are true or false. 1.OA.D.8 Determine the unknown whole number in an addition or subtraction equation, with the unknown in any position (e.g., 8 + ? = 11, 5 = ? - 3, 6 + 6 = ?). 24 Cluster Headings Content Standards C. Use place value understanding and properties of operations to add and subtract. 1.NBT.C.4 Add a two-digit number to a one-digit number and a two-digit number to a multiple of ten (within 100). Use concrete models, drawings, strategies based on place value, properties of operations, and/or the relationship between addition and subtraction to explain the reasoning used. 1.NBT.C.5 Mentally find 10 more or 10 less than a given two-digit number without having to count by ones and explain the reasoning used. 1.NBT.C.6 Subtract multiples of 10 from multiples of 10 in the range 10-90 using concrete models, drawings, strategies based on place value, properties of operations, and/or the relationship between addition and subtraction. Measurement and Data (MD) Cluster Headings Content Standards A. Measure lengths indirectly and by iterating length units. 1.MD.A.1 Order three objects by length. Compare the lengths of two objects indirectly by using a third object. For example, to compare indirectly the heights of Bill and Susan: if Bill is taller than mother and mother is taller than Susan, then Bill is taller than Susan. 1.MD.A.2 Measure the length of an object using non-standard units and express this length as a whole number of units. B. Work with time and money. 1.MD.B.3 Tell and write time in hours and half-hours using analog and digital clocks. 1.MD.B.4 Count the value of a set of like coins less than one dollar using the ¢ symbol only. C. Represent and interpret data. 1.MD.C.5 Organize, represent, and interpret data with up to three categories. Ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. 25 Geometry (G) Cluster Headings Content Standards A. Reason about shapes and their attributes. 1.G.A.1 Distinguish between attributes that define a shape (e.g., number of sides and vertices) versus attributes that do not define the shape (e.g., color, orientation, overall size); build and draw two-dimensional shapes to possess defining attributes. 1.G.A.2 Create a composite shape and use the composite shape to make new shapes by using two-dimensional shapes (rectangles, squares, trapezoids, triangles, half-circles, and quarter-circles) or three-dimensional shapes (cubes, rectangular prisms, cones, and cylinders). 1.G.A.3 Partition circles and rectangles into two and four equal shares, describe the shares using the words halves, fourths, and quarters, and use the phrases half of, fourth of, and quarter of. Describe the whole as two of, or four of the shares. Understand for these examples that partitioning into more equal shares creates smaller shares. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 26 27 28 Mathematics | Grade 2 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 2nd grade. Operations & Algebraic Thinking Students solve one- and two-step addition and subtraction contextual problems within 100 with an unknown in any position. Students should solve a variety of problem types in order to make connections among contexts, equations, and strategies (See Table 1 - Addition and Subtraction Situations). Students also represent these problems with objects, drawings, and/or equations. Students build upon previously taught strategies to mentally add and subtract within 30. Students know from memory all sums of two one-digit numbers and related subtraction facts. Numbers & Operations in Base Ten Students extend their understanding of the base-ten place value system to 1,000. This includes counting by ones, fives, tens, and hundreds. Students write numbers using standard form, word form, and expanded form. They deepen their understanding of different ways a number can be composed and decomposed. Students extend their understanding of place value, properties of operations, and the relationship between addition and subtraction to add and subtract within 1,000 and fluently add and subtract within 100 (See Table 3 - Properties of Operations). They add up to four two-digit numbers. They should also be able to explain why these strategies work. Students mentally add and subtract 10 or 100 from a given number 100-900. Measurement & Data In previous grades, students measured with non-standard units. Students in 2nd grade measure with standard units (centimeter and inch) and they use rulers and other measurement tools with the understanding that linear measure involves an iteration of units. They recognize that the smaller the unit, the more iterations they need to cover a given length. Students use addition and subtraction to solve contextual problems involving lengths in the same units and represent lengths on a number line. Geometry Students describe and analyze shapes by examining their sides and angles. Students recognize and draw shapes based on given attributes, such as draw a shape with 3 vertices. Students also are able to partition circles and rectangles into two, three, and four equal shares and rectangles into rows and columns, laying the foundation for fractions and area. 29 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 30 Operations and Algebraic Thinking (OA) Cluster Headings Content Standards Number and Operations in Base Ten (NBT) Cluster Headings Content Standards A. Represent and solve problems involving addition and subtraction. (See Table 1 - Addition and Subtraction Situations) 2.OA.A.1 Add and subtract within 100 to solve one- and two-step contextual problems, with unknowns in all positions, involving situations of add to, take from, put together/take apart, and compare. Use objects, drawings, and equations with a symbol for the unknown number to represent the problem. B. Add and subtract within 30. 2.OA.B.2 Fluently add and subtract within 30 using mental strategies. By the end of 2nd grade, know from memory all sums of two one-digit numbers and related subtraction facts. C. Work with equal groups of objects to gain foundations for multiplication. 2.OA.C.3 Determine whether a group of objects (up to 20) has an odd or even number of members by pairing objects or counting them by 2s. Write an equation to express an even number as a sum of two equal addends. 2.OA.C.4 Use repeated addition to find the total number of objects arranged in rectangular arrays with up to 5 rows and up to 5 columns; write an equation to express the total as a sum of equal addends. A. Understand place value. 2.NBT.A.1 Know that the three digits of a three-digit number represent amounts of hundreds, tens, and ones (e.g., 706 can be represented in multiple ways as 7 hundreds, 0 tens, and 6 ones; 706 ones; or 70 tens and 6 ones). 2.NBT.A.2 Count within 1000. Skip-count within 1000 by 5s, 10s, and 100s, starting from any number in its skip counting sequence. 2.NBT.A.3 Read and write numbers to 1000 using standard form, word form, and expanded form. 2.NBT.A.4 Compare two three-digit numbers based on the meanings of the digits in each place and use the symbols >, =, and < to show the relationship. B. Use place value understanding and properties of operations to add and subtract. (See Table 3 - Properties of Operations) 2.NBT.B.5 Fluently add and subtract within 100 using properties of operations, strategies based on place value, and/or the relationship between addition and subtraction. 2.NBT.B.6 Add up to four two-digit numbers using properties of operations and strategies based on place value. 31 Cluster Headings Content Standards Measurement and Data (MD) Cluster Headings Content Standards A. Measure and estimate lengths in standard units. 2.MD.A.1 Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. 2.MD.A.2 Measure the length of an object using two different units of measure and describe how the two measurements relate to the size of the unit chosen. 2.MD.A.3 Estimate lengths using units of inches, feet, yards, centimeters, and meters. 2.MD.A.4 Measure to determine how much longer one object is than another and express the difference in terms of a standard unit of length. B. Relate addition and subtraction to length. 2.MD.B.5 Add and subtract within 100 to solve contextual problems involving lengths that are given in the same units by using drawings and equations with a symbol for the unknown to represent the problem. 2.MD.B.6 Represent whole numbers as lengths from 0 on a number line and know that the points corresponding to the numbers on the number line are equally spaced. Use a number line to represent whole number sums and differences of lengths within 100. C. Work with time and money. 2.MD.C.7 Tell and write time in quarter hours and to the nearest five minutes (in a.m. and p.m.) using analog and digital clocks. 2.MD.C.8 Solve contextual problems involving dollar bills, quarters, dimes, nickels, and pennies using ¢ and $ symbols appropriately. B. Use place value understanding and properties of operations to add and subtract. (See Table 3 - Properties of Operations) 2.NBT.B.7 Add and subtract within 1000 using concrete models, drawings, strategies based on place value, properties of operations, and/or the relationship between addition and subtraction to explain the reasoning used. 2.NBT.B.8 Mentally add 10 or 100 to a given number 100–900, and mentally subtract 10 or 100 from a given number 100– 900. 2.NBT.B.9 Explain why addition and subtraction strategies work using properties of operations and place value. (Explanations may include words, drawing, or objects.) 32 Cluster Headings Content Standards D. Represent and interpret data. 2.MD.D.9 Generate measurement data by measuring lengths of several objects to the nearest whole unit. Show the measurements by making a line plot, where the horizontal scale is marked off in whole-number units. 2.MD.D.10 Draw a pictograph and a bar graph (with intervals of one) to represent a data set with up to four categories. Solve addition and subtraction problems related to the data in a graph. Geometry (G) Cluster Headings Content Standards A. Reason about shapes and their attributes. 2.G.A.1 Identify triangles, quadrilaterals, pentagons, hexagons, and cubes. Draw two-dimensional shapes having specified attributes (as determined directly or visually, not by measuring), such as a given number of angles or a given number of sides of equal length. 2.G.A.2 Partition a rectangle into rows and columns of same-sized squares and find the total number of squares. 2.G.A.3 Partition circles and rectangles into two, three, and four equal shares, describe the shares using the words halves, thirds, fourths, half of, a third of, and a fourth of, and describe the whole as two halves, three thirds, four fourths. Recognize that equal shares of identical wholes need not have the same shape. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 33 34 35 Mathematics | Grade 3 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 3rd grade. Operations and Algebraic Thinking Students build on their understanding of addition and subtraction to develop an understanding of the meanings of multiplication and division of whole numbers. Students use increasingly sophisticated strategies based on properties of operations to fluently solve multiplication and division problems within 100 (See Table 3 - Properties of Operations). Students interpret multiplication as finding an unknown product in situations involving equal-sized groups, arrays, area and measurement models, and division as finding an unknown factor in situations involving the unknown number of groups or the unknown group size. Students use these interpretations to represent and solve contextual problems with unknowns in all positions. By the end of 3rd grade, students should know from memory all products of single-digit numbers and the related division facts. Students use all four operations to solve two-step word problems and use place value, mental computation, and estimation strategies to assess the reasonableness of solutions. They build number sense by investigating numerical representations, such as addition or multiplication tables for the purpose of identifying arithmetic patterns. Students should solve a variety of problem types in order to make connections among contexts, equations, and strategies (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations). Number and Operations in Base Ten Students begin to develop an understanding of rounding whole numbers to the nearest ten or hundred. Students fluently add and subtract within 1000 using strategies and algorithms. Students multiply one-digit whole numbers by multiples of 10. Number and Operations in Fractions This domain builds on the previous skill of partitioning shapes in geometry. This is the first time students are introduced to unit fractions. Students understand that fractions are composed of unit fractions and they use visual fraction models to represent parts of a whole. Students build on their understanding of number lines to represent fractions as locations and lengths on a number line. Students use fractions to represent numbers equal to, less than, and greater than 1 and are able to generate simple equivalent fractions by using drawings and/or reasoning about fractions. Students understand that the size of a fractional part is relative to the size of the whole. Measurement and Data In 2nd grade, students tell time in five minute increments, measure lengths, and create bar graphs, pictographs, and line plots with whole number units. In 3rd grade, students tell and write time to the nearest minute and solve contextual problems involving addition and subtraction. They use appropriate tools to measure and estimate liquid volume and mass. Students draw scaled pictographs and bar graphs and answer two-step questions about these graphs. Students generate measurement data and represent the data on line plots marked with whole number, half, or quarter units. Students recognize area as an attribute of two-dimensional shapes and measure the area of a shape using the standard unit (a square) by finding the total number of same-sized units required to cover the shape without gaps or overlaps. Students connect area to multiplication and use multiplication to justify the area of a rectangle by decomposing rectangles into rectangular arrays of squares. Geometry Students understand that shapes in given categories have shared attributes and they identify polygons. Students continue their understanding of shapes and fractions by partitioning shapes into parts with equal areas and identify the parts with unit fractions. 36 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 37 Operations and Algebraic Thinking (OA) Cluster Headings Content Standards A. Represent and solve problems involving multiplication and division. 3.OA.A.1 Interpret the factors and products in whole number multiplication equations (e.g., 4 x 7 is 4 groups of 7 objects with a total of 28 objects or 4 strings measuring 7 inches each with a total of 28 inches.) 3.OA.A.2 Interpret the dividend, divisor, and quotient in whole number division equations (e.g., 28 ÷ 7 can be interpreted as 28 objects divided into 7 equal groups with 4 objects in each group or 28 objects divided so there are 7 objects in each of the 4 equal groups). 3.OA.A.3 Multiply and divide within 100 to solve contextual problems, with unknowns in all positions, in situations involving equal groups, arrays, and measurement quantities using strategies based on place value, the properties of operations, and the relationship between multiplication and division (e.g., contexts including computations such as 3 x ? = 24, 6 x 16 = ?, ? ÷ 8 = 3, or 96 ÷ 6 = ?) (See Table 2 - Multiplication and Division Situations). 3.OA.A.4 Determine the unknown whole number in a multiplication or division equation relating three whole numbers within 100. For example, determine the unknown number that makes the equation true in each of the equations: 8 x ? = 48, 5 = ? ÷ 3, 6 x 6 =? B. Understand properties of multiplication and the relationship between multiplication and division. (See Table 3 - Properties of Operations) 3.OA.B.5 Apply properties of operations as strategies to multiply and divide. (Students need not use formal terms for these properties.) Examples: If 6 x 4 = 24 is known, then 4 x 6 = 24 is also known (Commutative property of multiplication). 3 x 5 x 2 can be solved by (3 x 5) x 2 or 3 x (5 x 2) (Associative property of multiplication). One way to find 8 x 7 is by using 8 x (5 + 2) = (8 x 5) + (8 x 2). By knowing that 8 x 5 = 40 and 8 x 2 = 16, then 8 x 7 = 40 + 16 = 56 (Distributive property of multiplication over addition). 3.OA.B.6 Understand division as an unknown-factor problem. For example, find 32 ÷ 8 by finding the number that makes 32 when multiplied by 8. C. Multiply and divide within 100. 3.OA.C.7 Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division (e.g., knowing that 8 x 5 = 40, one knows 40 ÷ 5 = 8) or properties of operations. By the end of 3rd grade, know from memory all products of two one-digit numbers and related division facts. 38 Cluster Headings Content Standards Number and Operations in Base Ten (NBT) Cluster Headings Content Standards A. Use place value understanding and properties of operations to perform multi-digit arithmetic. 3.NBT.A.1 Round whole numbers to the nearest 10 or 100 using understanding of place value. 3.NBT.A.2 Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. 3.NBT.A.3 Multiply one-digit whole numbers by multiples of 10 in the range 10–90 (e.g., 9 x 80, 5 x 60) using strategies based on place value and properties of operations. Number and Operations - Fractions (NF) Limit denominators of fractions to 2, 3, 4, 6, and 8. Cluster Headings Content Standards A. Develop understanding of fractions as numbers. 3.NF.A.1 Understand a fraction, 1 𝑏, as the quantity formed by 1 part when a whole is partitioned into b equal parts (unit fraction); understand a fraction 𝑎 𝑏 as the quantity formed by a parts of size 1 𝑏. For example, 3 4 represents a quantity formed by 3 parts of size 1 4. D. Solve problems involving the four operations and identify and explain patterns in arithmetic. 3.OA.D.8 Solve two-step contextual problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations). 3.OA.D.9 Identify arithmetic patterns (including patterns in the addition and multiplication tables) and explain them using properties of operations. For example, analyze patterns in the multiplication table and observe that 4 times a number is always even (because 4 x 6 = (2 x 2) x 6 = 2 x (2 x 6), which uses the associative property of multiplication) (See Table 3 - Properties of Operations). 39 Cluster Headings Content Standards A. Develop understanding of fractions as numbers. 3.NF.A.2 Understand a fraction as a number on the number line. Represent fractions on a number line. a. Represent a fraction 1 𝑏 on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts. Recognize that each part has size 1 𝑏 and that the endpoint locates the number 1 𝑏 on the number line. For example, on a number line from 0 to 1, students can partition it into 4 equal parts and recognize that each part represents a length of 1 4 and the first part has an endpoint at 1 4 on the number line. b. Represent a fraction 𝑎 𝑏 on a number line diagram by marking off a lengths 1 𝑏 from 0. Recognize that the resulting interval has size 𝑎 𝑏 and that its endpoint locates the number 𝑎 𝑏 on the number line. For example, 5 3 is the distance from 0 when there are 5 iterations of 1 3. 3.NF.A.3 Explain equivalence of fractions and compare fractions by reasoning about their size. a. Understand two fractions as equivalent (equal) if they are the same size or the same point on a number line. b. Recognize and generate simple equivalent fractions (e.g., 1 2 = 2 4 , 4 6 = 2 3 ) and explain why the fractions are equivalent using a visual fraction model. c. Express whole numbers as fractions and recognize fractions that are equivalent to whole numbers. For example, express 3 in the form 3 = 3 1; recognize that 6 1 = 6; locate 4 4 and 1 at the same point on a number line diagram. d. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Use the symbols >, =, or < to show the relationship and justify the conclusions. Measurement and Data (MD) Cluster Headings Content Standards A. Solve problems involving measurement and estimation of intervals of time, liquid volumes, and masses of objects. 3.MD.A.1 Tell and write time to the nearest minute and measure time intervals in minutes. Solve contextual problems involving addition and subtraction of time intervals in minutes. For example, students may use a number line to determine the difference between the start time and the end time of lunch. 3.MD.A.2 Measure the mass of objects and liquid volume using standard units of grams (g), kilograms (kg), milliliters (ml), and liters (l). Estimate the mass of objects and liquid volume using benchmarks. For example, a large paper clip is about one gram, so a box of about 100 large clips is about 100 grams. Therefore, ten boxes would be about 1 kilogram. 40 Cluster Headings Content Standards B. Represent and interpret data. 3.MD.B.3 Draw a scaled pictograph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step "how many more" and "how many less" problems using information presented in scaled graphs. 3.MD.B.4 Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Show the data by making a line plot, where the horizontal scale is marked off in appropriate units: whole numbers, halves, or quarters. C. Geometric measurement: understand and apply concepts of area and relate area to multiplication and to addition. 3.MD.C.5 Recognize that plane figures have an area and understand concepts of area measurement. a. Understand that a square with side length 1 unit, called "a unit square," is said to have "one square unit" of area and can be used to measure area. b. Understand that a plane figure which can be covered without gaps or overlaps by n unit squares is said to have an area of n square units. 3.MD.C.6 Measure areas by counting unit squares (square centimeters, square meters, square inches, square feet, and improvised units). 3.MD.C.7 Relate area of rectangles to the operations of multiplication and addition. a. Find the area of a rectangle with whole-number side lengths by tiling it and show that the area is the same as would be found by multiplying the side lengths. b. Multiply side lengths to find areas of rectangles with whole number side lengths in the context of solving real-world and mathematical problems and represent whole-number products as rectangular areas in mathematical reasoning. c. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a x b and a x c. Use area models to represent the distributive property in mathematical reasoning. For example, in a rectangle with dimensions 4 by 6, students can decompose the rectangle into 4 x 3 and 4 x 3 to find the total area of 4 x 6. (See Table 3 - Properties of Operations) d. Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real-world problems. D. Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures. 3.MD.D.8 Solve real-world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters. 41 Geometry (G) Cluster Headings Content Standards A. Reason about shapes and their attributes. 3.G.A.1 Understand that shapes in different categories may share attributes and that the shared attributes can define a larger category. Recognize rhombuses, rectangles, and squares as examples of quadrilaterals and draw examples of quadrilaterals that do not belong to any of these subcategories. 3.G.A.2 Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole. For example, partition a shape into 4 parts with equal area and describe the area of each part as 1/4 of the area of the shape. 3.G.A.3 Determine if a figure is a polygon. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 42 43 44 45 Mathematics | Grade 4 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 4th grade. Operations and Algebraic Thinking Students build on their knowledge of multiplication and begin to interpret and represent multiplication as a comparison. They multiply and divide to solve contextual problems involving multiplicative situations, distinguishing their solutions from additive comparison situations. Students solve multi-step whole number contextual problems using the four operations representing the unknown as a variable within equations (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations). They apply appropriate methods to estimate and check for reasonableness. This is the first time students find and interpret remainders in context. Students find factors and multiples, and they identify prime and composite numbers. Students generate number or shape patterns following a given rule. Number and Operations in Base Ten Students generalize place value understanding to read and write numbers to 1,000,000, using standard form, word form, and expanded form. They compare the relative size of the numbers and round numbers to the nearest hundred thousand, which builds on 3rd grade rounding concepts. By the end of 4th grade, students should fluently add and subtract multi-digit whole numbers to 1,000,000. Students use strategies based on place value and the properties of operations to multiply a whole number up to four-digits by a one-digit number, and multiply two two-digit numbers. They use these strategies and the relationship between multiplication and division to find whole number quotients and remainders up to four-digit dividends and one-digit divisors (See Table 3 - Properties of Operations). Number and Operations-Fractions Students continue to develop an understanding of fraction equivalence by reasoning about the size of the fractions, using a benchmark fraction to compare the fractions, or finding a common denominator. Students extend previous understanding of unit fractions to compose and decompose fractions in different ways. They use the meaning of fractions and the meaning of multiplication as repeated addition to multiply a whole number by a fraction. Students solve contextual problems involving addition and subtraction of fractions with like denominators and multiplication of a whole number by a fraction (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations for whole number situations that can be applied to fractions). Students learn decimal notation for the first time to represent fractions with denominators of 10 and 100. They express these fractions and their equivalents as decimals and are able to read, write, compare, and locate these decimals on a number line. Measurement and Data Students know the relative sizes of measurement units within one system of units and are able to convert within the single system of measurement. They use the four operations to solve contextual problems involving measurement. Students build on their previous understanding of area and perimeter to generate and apply formulas for finding the area and perimeter of rectangles. Students also build on their understanding of line plots and solve problems involving fractions using operations appropriate for the grade. For the first time, students learn concepts of angle measurement. Geometry Students extend their previous understanding to analyze and classify shapes based on line and angle types. Students also use knowledge of line and angle types to identify right triangles. Students recognize and draw lines of symmetry for the first time. 46 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 47 Operations and Algebraic Thinking (OA) Cluster Headings Content Standards Number and Operations in Base Ten (NBT) Cluster Headings Content Standards A. Generalize place value understanding for multi-digit whole numbers. 4.NBT.A.1 Recognize that in a multi-digit whole number (less than or equal to 1,000,000), a digit in one place represents 10 times as much as it represents in the place to its right. For example, recognize that 7 in 700 is 10 times bigger than the 7 in 70 because 700 ÷ 70 = 10 and 70 x 10 = 700. 4.NBT.A.2 Read and write multi-digit whole numbers (less than or equal to 1,000,000) using standard form, word form, and expanded form (e.g. the expanded form of 4256 is written as 4 x 1000 + 2 x 100 + 5 x 10 + 6 x 1). Compare two multi-digit numbers based on meanings of the digits in each place and use the symbols >, =, and < to show the relationship. 4.NBT.A.3 Round multi-digit whole numbers to any place (up to and including the hundred-thousand place) using understanding of place value. A. Use the four operations with whole numbers to solve problems. (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations) 4.OA.A.1 Interpret a multiplication equation as a comparison (e.g., interpret 35 = 5 x 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5). Represent verbal statements of multiplicative comparisons as multiplication equations. 4.OA.A.2 Multiply or divide to solve contextual problems involving multiplicative comparison, and distinguish multiplicative comparison from additive comparison. For example, school A has 300 students and school B has 600 students: to say that school B has two times as many students is an example of multiplicative comparison; to say that school B has 300 more students is an example of additive comparison. 4.OA.A.3 Solve multi-step contextual problems posed with whole numbers and having whole-number answers using the four operations, including problems in which remainders must be interpreted. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. B. Gain familiarity with factors and multiples. 4.OA.B.4 Find all factor pairs for a whole number in the range 1–100. Recognize that a whole number is a multiple of each of its factors. Determine whether a given whole number in the range 1–100 is a multiple of a given one-digit number. Determine whether a given whole number in the range 1–100 is prime or composite. C. Generate and analyze patterns. 4.OA.C.5 Generate a number or shape pattern that follows a given rule. Identify apparent features of the pattern that were not explicit in the rule itself. For example, given the rule "Add 3" and the starting number 1, generate terms in the resulting sequence and observe that the terms appear to alternate between odd and even numbers. Explain informally why the numbers will continue to alternate in this way. 48 Cluster Headings Content Standards B. Use place value understanding and properties of operations to perform multi-digit arithmetic. (See Table 3 - Properties of Operations) 4.NBT.B.4 Fluently add and subtract within 1,000,000 using appropriate strategies and algorithms. 4.NBT.B.5 Multiply a whole number of up to four digits by a one-digit whole number and multiply two two-digit numbers, using strategies based on place value and the properties of operations. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 4.NBT.B.6 Find whole-number quotients and remainders with up to four-digit dividends and one-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Number and Operations - Fractions (NF) Limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100. Cluster Headings Content Standards A. Extend understanding of fraction equivalence and comparison. 4.NF.A.1 Explain why a fraction 𝑎 𝑏 is equivalent to a fraction 𝑎 x 𝑛 𝑏 x 𝑛 or 𝑎 ÷ 𝑛 𝑏 ÷ 𝑛 by using visual fraction models, with attention to how the number and size of the parts differ even though the two fractions themselves are the same size. Use this principle to recognize and generate equivalent fractions. For example, 3 4 = 3 x 2 4 x 2 = 6 8 . 4.NF.A.2 Compare two fractions with different numerators and different denominators by creating common denominators or common 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. Use the symbols >, =, or < to show the relationship and justify the conclusions. 49 Cluster Headings Content Standards B. Build fractions from unit fractions by applying and extending previous understandings of operations on whole numbers. (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations for whole number situations that can be applied for fractions.) 4.NF.B.3 Understand a fraction 𝑎 𝑏 with a > 1 as a sum of fractions 1 𝑏. For example, 4 5 = 1 5 + 1 5 + 1 5 + 1 5. a. Understand addition and subtraction of fractions as joining and separating parts referring to the same whole. b. Decompose a fraction into a sum of fractions with the same denominator in more than one way (e.g., 3 8 = 1 8 + 1 8 + 1 8 ; 3 8 = 1 8 + 2 8; 2 1 8 = 1 + 1 + 1 8 = 8 8 + 8 8 + 1 8), recording each decomposition by an equation. Justify decompositions by using a visual fraction model. c. Add and subtract mixed numbers with like denominators by replacing each mixed number with an equivalent fraction and/or by using properties of operations and the relationship between addition and subtraction. d. Solve contextual problems involving addition and subtraction of fractions referring to the same whole and having like denominators 4.NF.B.4 Apply and extend previous understandings of multiplication as repeated addition to multiply a whole number by a fraction. a. Understand a fraction 𝑎 𝑏 as a multiple of 1 𝑏. For example, use a visual fraction model to represent 5 4 as the product 5 × 1 4, recording the conclusion by the equation 5 4 = 5 x 1 4 . b. Understand a multiple of 𝑎 𝑏 as a multiple of 1 𝑏 and use this understanding to multiply a whole number by a fraction. For example, use a visual fraction model to express 3 × 2 5 as 6 × 1 5, recognizing this product as 6 5. (In general, 𝑛 x 𝑎 𝑏 = (𝑛 x 𝑎) 𝑏 = (𝑛 x 𝑎) x 1 𝑏.) c. Solve contextual problems involving multiplication of a whole number by a fraction (e.g., by using visual fraction models and equations to represent the problem). For example, if each person at a party will eat 3 8 of a pound of roast beef, and there will be 4 people at the party, how many pounds of roast beef will be needed? Between what two whole numbers does your answer lie? C. Understand decimal notation for fractions and compare decimal fractions. 4.NF.C.5 Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100. For example, express, 3 10 as 30 100 and add 3 10 + 4 100 = 34 100. 4.NF.C.6 Read and write decimal notation for fractions with denominators 10 or 100. Locate these decimals on a number line. 4.NF.C.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Use the symbols >, =, or < to show the relationship and justify the conclusions. 50 Measurement and Data (MD) Cluster Headings Content Standards Geometry A. Estimate and solve problems involving measurement. 4.MD.A.1 Measure and estimate to determine relative sizes of measurement units within a single system of measurement involving length, liquid volume, and mass/weight of objects using customary and metric units. 4.MD.A.2 Solve one- or two-step real-world problems involving whole number measurements with all four operations within a single system of measurement including problems involving simple fractions. 4.MD.A.3 Know and apply the area and perimeter formulas for rectangles in real-world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor. B. Represent and interpret data. 4.MD.B.4 Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots. For example, from a line plot find and interpret the difference in length between the longest and shortest specimens in an insect collection. C. Geometric measurement: understand concepts of angle and measure angles. 4.MD.C.5 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement. a. Understand that an angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. b. Understand that an angle that turns through 1/360 of a circle is called a "one-degree angle," and can be used to measure angles. An angle that turns through n one-degree angles is said to have an angle measure of n degrees and represents a fractional portion of the circle. 4.MD.C.6 Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. 4.MD.C.7 Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real-world and mathematical problems (e.g., by using an equation with a symbol for the unknown angle measure). 51 Geometry (G) Cluster Headings Content Standards A. Draw and identify lines and angles and classify shapes by properties of their lines and angles. 4.G.A.1 Draw points, lines, line segments, rays, angles (right, acute, obtuse, straight, reflex), and perpendicular and parallel lines. Identify these in two-dimensional figures. 4.G.A.2 Classify two-dimensional figures based on the presence or absence of parallel or perpendicular lines or the presence or absence of angles of a specified size. Recognize right triangles as a category and identify right triangles. 4.G.A.3 Recognize and draw lines of symmetry for two-dimensional figures. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 52 53 54 55 Mathematics | Grade 5 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 5th grade. Operations and Algebraic Thinking Students build on their understanding of patterns to generate two numerical patterns using given rules and identify relationships between the patterns. For the first time, students form ordered pairs and graph them on a coordinate plane. In addition, students write and evaluate numerical expressions using parentheses and/or brackets. Number and Operations in Base Ten Students generalize their understanding of place value to include decimals by reading, writing, comparing, and rounding numbers. They recognize that in a multi-digit number, the value of each digit has a relationship to the value of the same digit in another position. Students explain patterns in products when multiplying a number by a power of 10. Whole-number exponents are used to denote powers of 10 for the first time. By the end of 5th grade, students should fluently multiply multi-digit whole numbers (up to 4 digits by 3 digits). Students build on their understanding of why division procedures work based on place value and the properties of operations to find whole number quotients and remainders (See Table 3 - Properties of Operations). They apply their understanding of models for decimals, decimal notation, and properties of operations to add, subtract, multiply, and divide decimals to hundredths. (Limit division problems so that either the dividend or the divisor is a whole number.) They develop fluency in these computations and make reasonable estimates of their results. Students finalize their understanding of multi-digit addition, subtraction, multiplication, and division with whole numbers. Number and Operations in Fractions Students apply their understanding of equivalent fractions and fraction models to represent the addition and subtraction of fractions with unlike denominators as equivalent calculations with like denominators. They develop fluency in calculating sums and differences of fractions and make reasonable estimates of them. For the first time, students develop an understanding of fractions as division problems. They use the meaning of fractions, of multiplication and division, and the relationship between multiplication and division to understand and explain why the procedures for multiplying and dividing fractions make sense. (Limit to dividing unit fractions by whole numbers or whole numbers by unit fractions.) Students reason about the size of products compared to the size of the factors. Students should solve a variety of problem types in order to make connections among contexts, equations, and strategies (See Table 1 - Addition and Subtraction Situations and Table 2 - Multiplication and Division Situations for whole number situations that can be applied to fractions). Measurement and Data Students build on their understanding of area and recognize volume as an attribute of three-dimensional space. They understand that volume can be measured by finding the total number of same-sized units of volume required to fill the space without gaps or overlaps. Students decompose three-dimensional shapes and find volumes of right rectangular prisms by viewing them as decomposed into layers of cubes. Students build on their understanding of measurements to convert from larger units to smaller units within a single system of measurement and solve multistep problems involving these conversions. Students solve problems with data from line plots involving fractions using operations appropriate for the grade. Geometry Students plot points on the coordinate plane to solve real-world and mathematical problems. Students classify two-dimensional figures into categories based on their properties. 56 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 57 Operations and Algebraic Thinking (OA) Cluster Headings Content Standards Number and Operations in Base Ten (NBT) Cluster Headings Content Standards A. Understand the place value system. 5.NBT.A.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. 5.NBT.A.2 Explain patterns in the number of zeros of the product when multiplying a number by powers of 10, and explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Use whole-number exponents to denote powers of 10. 5.NBT.A.3 Read and write decimals to thousandths using standard form, word form, and expanded form (e.g., the expanded form of 347.392 is written as 3 x 100 + 4 x 10 + 7 x 1 + 3 x (1/10) + 9 x (1/100) + 2 x (1/1000)). Compare two decimals to thousandths based on meanings of the digits in each place and use the symbols >, =, and < to show the relationship. 5.NBT.A.4 Round decimals to the nearest hundredth, tenth, or whole number using understanding of place value. A. Write and interpret numerical expressions. 5.OA.A.1 Use parentheses and/or brackets in numerical expressions and evaluate expressions having these symbols using the conventional order (Order of Operations). 5.OA.A.2 Write simple expressions that record calculations with numbers and interpret numerical expressions without evaluating them. For example, express the calculation "add 8 and 7, then multiply by 2" as 2 x (8 + 7). Recognize that 3 x (18,932 + 921) is three times as large as 18,932 + 921, without having to calculate the indicated sum or product. B. Analyze patterns and relationships. 5.OA.B.3 Generate two numerical patterns using two given rules. For example, given the rule "Add 3" and the starting number 0, and given the rule "Add 6" and the starting number 0, generate terms in the resulting sequences. a. Identify relationships between corresponding terms in two numerical patterns. For example, observe that the terms in one sequence are twice the corresponding terms in the other sequence. b. Form ordered pairs consisting of corresponding terms from two numerical patterns and graph the ordered pairs on a coordinate plane. 58 Cluster Headings Content Standards B. Perform operations with multi-digit whole numbers and with decimals to hundredths. (See Table 3 - Properties of Operations) 5.NBT.B.5 Fluently multiply multi-digit whole numbers (up to three-digit by four-digit factors) using appropriate strategies and algorithms. 5.NBT.B.6 Find whole-number quotients and remainders of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. 5.NBT.B.7 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between operations; assess the reasonableness of answers using estimation strategies. (Limit division problems so that either the dividend or the divisor is a whole number.) Number and Operations - Fractions (NF) Cluster Headings Content Standards A. Use equivalent fractions as a strategy to add and subtract fractions. (See Table 1 - Addition and Subtraction Situations for whole number situations that can be applied to fractions) 5.NF.A.1 Add and subtract fractions with unlike denominators (including mixed numbers) by replacing given fractions with equivalent fractions in such a way as to produce an equivalent sum or difference of fractions with like denominators. For example, 2 3 + 5 4 = 8 12 + 15 12 = 23 12 . (In general 𝑎 𝑏+ 𝑐 𝑑= (𝑎𝑑+𝑏𝑐) 𝑏𝑑 .) 5.NF.A.2 Solve contextual problems involving addition and subtraction of fractions referring to the same whole, including cases of unlike denominators. Use benchmark fractions and number sense of fractions to estimate mentally and assess the reasonableness of answers. For example, recognize an incorrect result 2 5 + 1 2 = 3 7, by observing that 3 7 < 1 2. B. Apply and extend previous understandings of multiplication and division to multiply and divide fractions. (See Table 2 - Multiplication and Division Situations for whole number situations that can be applied to fractions) 5.NF.B.3 Interpret a fraction as division of the numerator by the denominator ( 𝑎 𝑏= a ÷ b). For example, 3 4 = 3 ÷ 4 so when 3 wholes are shared equally among 4 people, each person has a share of size 3 4. Solve contextual problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers by using visual fraction models or equations to represent the problem. For example, if 8 people want to share 49 sheets of construction paper equally, how many sheets will each person receive? Between what two whole numbers does your answer lie? 59 Cluster Headings Content Standards B. Apply and extend previous understandings of multiplication and division to multiply and divide fractions. (See Table 2 - Multiplication and Division Situations for whole number situations that can be applied to fractions) 5.NF.B.4 Apply and extend previous understandings of multiplication to multiply a fraction by a whole number or a fraction by a fraction. a. Interpret the product 𝑎 𝑏 x q as a x (q ÷ b) (partition the quantity q into b equal parts and then multiply by a). Interpret the product 𝑎 𝑏 x q as (a x q) ÷ b (multiply a times the quantity q and then partition the product into b equal parts). For example, use a visual fraction model or write a story context to show that 2 3 x 6 can be interpreted as 2 x (6 ÷ 3) or (2 x 6) ÷ 3. Do the same with 2 3 x 4 5 = 8 15. (In general, a b x c d = ac bd.) b. Find the area of a rectangle with fractional side lengths by tiling it with unit squares of the appropriate unit fraction side lengths, and show that the area is the same as would be found by multiplying the side lengths. Multiply fractional side lengths to find areas of rectangles and represent fraction products as rectangular areas. 5.NF.B.5 Interpret multiplication as scaling (resizing). a. Compare the size of a product to the size of one factor on the basis of the size of the other factor, without performing the indicated multiplication. For example, know if the product will be greater than, less than, or equal to the factors. b. Explain why multiplying a given number by a fraction greater than 1 results in a product greater than the given number (recognizing multiplication by whole numbers greater than 1 as a familiar case); explain why multiplying a given number by a fraction less than 1 results in a product less than the given number; and relate the principle of fraction equivalence 𝑎 𝑏= (𝑎 𝑥 𝑛) (𝑏 𝑥 𝑛) to the effect of multiplying 𝑎 𝑏 by 1. 5.NF.B.6 Solve real-world problems involving multiplication of fractions and mixed numbers by using visual fraction models or equations to represent the problem. 5.NF.B.7 Apply and extend previous understandings of division to divide unit fractions by whole numbers and whole numbers by unit fractions. a. Interpret division of a unit fraction by a non-zero whole number and compute such quotients. For example, use visual models and the relationship between multiplication and division to explain that (1/3) ÷ 4 = 1/12 because (1/12) x 4 = 1/3. b. Interpret division of a whole number by a unit fraction and compute such quotients. For example, use visual models and the relationship between multiplication and division to explain that 4 ÷ (1/5) = 20 because 20 x (1/5) = 4. c. Solve real-world problems involving division of unit fractions by non-zero whole numbers and division of whole numbers by unit fractions by using visual fraction models and equations to represent the problem. For example, how much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How many 1/3 cup servings are in 2 cups of raisins? 60 Measurement and Data (MD) Cluster Headings Content Standards A. Convert like measurement units within a given measurement system from a larger unit to a smaller unit. 5.MD.A.1 Convert customary and metric measurement units within a single system by expressing measurements of a larger unit in terms of a smaller unit. Use these conversions to solve multi-step real-world problems involving distances, intervals of time, liquid volumes, masses of objects, and money (including problems involving simple fractions or decimals). For example, 3.6 liters and 4.1 liters can be combined as 7.7 liters or 7700 milliliters B. Represent and interpret data. 5.MD.B.2 Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots. For example, given different measurements of liquid in identical beakers, find the amount of liquid each beaker would contain if the total amount in all the beakers were redistributed equally. C. Geometric measurement: understand concepts of volume and relate volume to multiplication and to addition. 5.MD.C.3 Recognize volume as an attribute of solid figures and understand concepts of volume measurement. a. Understand that a cube with side length 1 unit, called a "unit cube," is said to have "one cubic unit" of volume and can be used to measure volume. b. Understand that a solid figure which can be packed without gaps or overlaps using n unit cubes is said to have a volume of n cubic units. 5.MD.C.4 Measure volume by counting unit cubes, using cubic centimeters, cubic inches, cubic feet, and improvised units. 5.MD.C.5 Relate volume to the operations of multiplication and addition and solve real-world and mathematical problems involving volume of right rectangular prisms. a. Find the volume of a right rectangular prism with whole-number side lengths by packing it with unit cubes and show that the volume is the same as would be found by multiplying the edge lengths, equivalently by multiplying the height by the area of the base. Represent whole-number products of three factors as volumes (e.g., to represent the associative property of multiplication). b. Know and apply the formulas V = l x w x h and V = B x h (where B represents the area of the base) for rectangular prisms to find volumes of right rectangular prisms with whole number edge lengths in the context of solving real-world and mathematical problems. c. Recognize volume as additive. Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real-world problems. 61 Geometry (G) Cluster Headings Content Standards A. Graph points on the coordinate plane to solve real-world and mathematical problems. 5.G.A.1 Graph ordered pairs and label points using the first quadrant of the coordinate plane. Understand in the ordered pair that the first number indicates the horizontal distance traveled along the x-axis from the origin and the second number indicates the vertical distance traveled along the y-axis, with the convention that the names of the two axes and the coordinates correspond (e.g., x-axis and x-coordinate, y-axis and y-coordinate). 5.G.A.2 Represent real-world and mathematical problems by graphing points in the first quadrant of the coordinate plane and interpret coordinate values of points in the context of the situation. B. Classify two-dimensional figures into categories based on their properties. 5.G.B.3 Classify two-dimensional figures in a hierarchy based on properties. Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category. For example, all rectangles have four right angles and squares are rectangles, so all squares have four right angles. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 62 63 64 65 Mathematics | Grade 6 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 6th grade. Ratios and Proportional Relationships 6th grade begins the formal study of ratios and proportions. Students use reasoning about multiplication and division to solve ratio and rate problems about quantities. By viewing equivalent ratios and rates as deriving from, and extending, pairs of rows (or columns) in the multiplication table and by analyzing simple drawings that indicate the relative size of quantities, students connect their understanding of multiplication and division with ratios and rates. Thus students expand the scope of problems for which they can use multiplication and division to solve problems, and they connect ratios and fractions. Students solve a wide variety of problems involving ratios and rates. Proportional relationships are added and studied in the 7th grade. The Number System Students use fractions, multiplication, and division along with an understanding of the relationship between multiplication and division to understand and explain why the procedures for dividing fractions make sense. Students use these operations to solve problems. Students also extend their previous understandings of numbers and the ordering of numbers to the full system of rational numbers, which includes negative rational numbers, and in particular negative integers. They reason about the order and absolute value of rational numbers and about the location of points in all four quadrants of the coordinate plane. Expressions and Equations Students begin to use properties of arithmetic operations systematically to work with numerical expressions that contain whole-number exponents. Students come to understand more fully the use of variables and variable expressions. They write expressions and equations that correspond to given situations, evaluate expressions, and use expressions and formulas to solve problems. Students understand that expressions in different forms can be equivalent, and they use the properties of operations to rewrite expressions in equivalent forms. Students know that the solutions of an equation are the values of the variables that make the equation true. Students use properties of operations and the idea of maintaining the equality of both sides of an equation to solve simple one-step equations. Students explore how algebraic expressions can represent written situations and generalize relationships from specific cases. Geometry Students build on their work with area from earlier grades by reasoning about relationships among shapes to determine area, surface area, and volume. They find areas of right triangles, other triangles, and special quadrilaterals by decomposing these shapes, rearranging or removing pieces, and relating the shapes to rectangles. Using these methods, students discuss, develop, and justify formulas for areas of triangles and parallelograms. Students find areas of polygons and surface areas of prisms and pyramids by decomposing them into pieces whose area they can more easily determine. They reason about right rectangular prisms with fractional side lengths to extend formulas for the volume of a right rectangular prism to fractional side lengths. They prepare for work on scale drawings and constructions in the 7th grade by drawing polygons in the coordinate plane. Statistics and Probability 6th grade students begin to formally develop their ability to think statistically. They understand that a set of data (collected to answer a question) will have a distribution, which can be described by its center, spread, and shape. Students calculate the median, mean, and mode and relate these to the overall shape of the distribution. They recognize that the median measures center in the sense that it is roughly the middle value. The mean measures center in the sense that it is the value that each data point would take on if the total of the data values were redistributed equally, and also in the sense that it is a balance point. They understand that the mode refers to the most frequently occurring number found in a set of numbers and is found by collecting and organizing the data in order to count the frequency of each result. Students display, summarize and describe numerical data sets, considering the context in which the data were collected. Students use number lines, dot plots, box plots, and pie charts to display numerical data. 66 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 67 Ratios and Proportional Relationships (RP) Cluster Headings Content Standards A. Understand ratio concepts and use ratio reasoning to solve problems. 6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, the ratio of wings to beaks in a bird house at the zoo was 2:1, because for every 2 wings there was 1 beak. Another example could be for every vote candidate A received, candidate C received nearly three votes 6.RP.A.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0. Use rate language in the context of a ratio relationship. For example, this recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar. Also, we paid $75 for 15 hamburgers, which is a rate of $5 per hamburger. (Expectations for unit rates in 6th grade are limited to non-complex fractions). 6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations). a. Make tables of equivalent ratios relating quantities with whole number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios. b. Solve unit rate problems including those involving unit pricing and constant speed. For example, if a runner ran 10 miles in 90 minutes, running at that speed, how long will it take him to run 6 miles? How fast is he running in miles per hour? c. Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent. d. Use ratio reasoning to convert customary and metric measurement units (within the same system); manipulate and transform units appropriately when multiplying or dividing quantities. The Number System (NS) Cluster Headings Content Standards A. Apply and extend previous understandings of multiplication and division to divide fractions by fractions. 6.NS.A.1 Interpret and compute quotients of fractions, and solve contextual problems involving division of fractions by fractions (e.g., using visual fraction models and equations to represent the problem is suggested). For example, create a story context for (2/3) ÷ (3/4) and use a visual fraction model to show the quotient; use the relationship between multiplication and division to explain that (2/3) ÷ (3/4) = 8/9 because 3/4 times 8/9 is 2/3 ((a/b) ÷ (c/d) = ad/bc.) Further example: How much chocolate will each person get if 3 people share 1/2 lb of chocolate equally? How wide is a rectangular strip of land with length 3/4 mi and area 1/2 square mi? 68 Cluster Headings Content Standards B. Compute fluently with multi-digit numbers and find common factors and multiples. 6.NS.B.2 Fluently divide multi-digit numbers using a standard algorithm. 6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals using a standard algorithm for each operation. 6.NS.B.4 Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1–100 with a common factor as a multiple of a sum of two whole numbers with no common factor. For example, express 36 + 8 as 4 (9 + 2). C. Apply and extend previous understandings of numbers to the system of rational numbers. 6.NS.C.5 Understand that positive and negative numbers are used together to describe quantities having opposite directions or values (e.g., temperature above/below zero, elevation above/below sea level, credits/debits, positive/negative electric charge); use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation. 6.NS.C.6 Understand a rational number as a point on the number line. Extend number line diagrams and coordinate axes familiar from previous grades to represent points on the line and in the plane with negative number coordinates. a. Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself. For example, – (–3) = 3, and that 0 is its own opposite. b. Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes. c. Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane. 6.NS.C.7 Understand ordering and absolute value of rational numbers. a. 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. b. Write, interpret, and explain statements of order for rational numbers in real-world contexts. For example, write –3 oC > –7 oC to express the fact that –3 oC is warmer than –7 oC. c. Understand the absolute value of a rational number as its distance from 0 on the number line and distinguish comparisons of absolute value from statements about order in a real-world context. For example, an account balance of -24 dollars represents a greater debt than an account balance -14 dollars because -24 is located to the left of -14 on the number line 69 Cluster Headings Content Standards C. Apply and extend previous understandings of numbers to the system of rational numbers. 6.NS.C.8 Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate. Expressions and Equations (EE) Cluster Headings Content Standards A. Apply and extend previous understandings of arithmetic to algebraic expressions. 6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents. 6.EE.A.2 Write, read, and evaluate expressions in which variables stand for numbers. a. Write expressions that record operations with numbers and with variables. For example, express the calculation "Subtract y from 5" as 5 - y. b. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity. For example, describe the expression 2 (8 + 7) as a product of two factors; view (8 + 7) as both a single entity and a sum of two terms. c. Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations). 6.EE.A.3 Apply the properties of operations (including, but not limited to, commutative, associative, and distributive properties) to generate equivalent expressions. The distributive property is prominent here. For example, apply the distributive property to the expression 3 (2 + x) to produce the equivalent expression 6 + 3x; apply the distributive property to the expression 24x + 18y to produce the equivalent expression 6 (4x + 3y); apply properties of operations to y + y + y to produce the equivalent expression 3y. 6.EE.A.4 Identify when expressions are equivalent (i.e., when the expressions name the same number regardless of which value is substituted into them). For example, the expression 5b + 3b is equivalent to (5 +3) b, which is equivalent to 8b. B. Reason about and solve one-variable equations and inequalities. 6.EE.B.5 Understand solving an equation or inequality is carried out by determining if any of the values from a given set make the equation or inequality true. Use substitution to determine whether a given number in a specified set makes an equation or inequality true. 70 Cluster Headings Content Standards B. Reason about and solve one-variable equations and inequalities. 6.EE.B.6 Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. 6.EE.B.7 Solve real-world and mathematical problems by writing and solving one-step equations of the form x + p = q and px = q for cases in which p, q, and x are all nonnegative rational numbers. 6.EE.B.8 Interpret and write an inequality of the form x > c or x < c which represents a condition or constraint in a real-world or mathematical problem. Recognize that inequalities have infinitely many solutions; represent solutions of inequalities on number line diagrams. C. Represent and analyze quantitative relationships between dependent and independent variables. 6.EE.C.9 Use variables to represent two quantities in a real-world problem that change in relationship to one another. For example, Susan is putting money in her savings account by depositing a set amount each week (50). Represent her savings account balance with respect to the number of weekly deposits (s = 50w, illustrating the relationship between balance amount s and number of weeks w). a. Write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. b. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation. Geometry (G) Cluster Headings Content Standards A. Solve real-world and mathematical problems involving area, surface area, and volume. 6.G.A.1 Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; know and apply these techniques in the context of solving real-world and mathematical problems. 6.G.A.2 Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Know and apply the formulas V = lwh and V = Bh where B is the area of the base to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems. 71 Cluster Headings Content Standards A. Solve real-world and mathematical problems involving area, surface area, and volume. 6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side that joins two vertices (vertical or horizontal segments only). Know and apply these techniques in the context of solving real-world and mathematical problems. 6.G.A.4 Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures. Apply these techniques in the context of solving real-world and mathematical problems. Statistics and Probability (SP) Cluster Headings Content Standards A. Develop understanding of statistical variability. 6.SP.A.1 Recognize a statistical question as one that anticipates variability in the data related to the question and accounts for it in the answers. For example, “How old am I?” is not a statistical question, but “How old are the students in my school?” is a statistical question because one anticipates variability in students’ ages. 6.SP.A.2 Understand that a set of data collected to answer a statistical question has a distribution which can be described by its center (mean, median, mode), spread (range), and overall shape. 6.SP.A.3 Recognize that a measure of center for a numerical data set summarizes all of its values with a single number, while a measure of variation describes how its values vary with a single number. B. Summarize and describe distributions. 6.SP.B.4 Display a single set of numerical data using dot plots (line plots), box plots, pie charts and stem plots. 6.SP.B.5 Summarize numerical data sets in relation to their context. a. Report the number of observations. b. Describe the nature of the attribute under investigation, including how it was measured and its units of measurement. c. Give quantitative measures of center (median and/or mean) and variability (range) as well as describing any overall pattern with reference to the context in which the data were gathered. d. Relate the choice of measures of center to the shape of the data distribution and the context in which the data were gathered. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content Major Content Supporting Content 72 Mathematics | Grade 7 The descriptions below provide an overview of the mathematical concepts and skills that students explore throughout the 7th grade. Ratios and Proportional Relationships Students extend their understanding of ratios from 6th grade and develop understanding of proportionality to solve single- and multi-step problems. Students use their understanding of ratios and proportionality to solve a wide variety of percent problems, including those involving discounts, interest, taxes, tips, and percent increase or decrease. Students solve problems about scale drawings by relating corresponding lengths between the objects or by using the fact that relationships of lengths within an object are preserved in similar objects. Students graph proportional relationships and understand the unit rate informally as a measure of the steepness of the related line, called the slope. They distinguish proportional relationships from other relationships. The Number System Students develop a unified understanding of numbers, recognizing fractions, decimals (that have a finite or a repeating decimal representation), and percent as different representations of rational numbers. Students extend addition, subtraction, multiplication, and division to all rational numbers, maintaining the properties of operations and the relationships between addition and subtraction, and multiplication and division. These properties are further explored with respect to negative numbers. This exploration is carried out in problems from everyday contexts so that the student can gain a deeper understanding and appreciation for the mathematical concepts being studied. Expressions and Equations By applying the properties of operations as strategies, students explore working with expressions, equations, and inequalities. They use the arithmetic of rational numbers as they formulate expressions and equations in one variable and use these equations to solve multi-step real-world problems. They use variables to represent quantities and construct simple equations and inequalities to solve problems by reasoning about the quantities. Geometry Students continue their work with area from 6th grade, solving problems involving the area and circumference of a circle and surface area of three-dimensional objects. In preparation for work on congruence and similarity, they reason about relationships among two-dimensional figures using scale drawings and informal geometric constructions, and they gain familiarity with the relationships between angles formed by intersecting lines. Students solve real-world and mathematical problems involving area, surface area, and volume of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. Statistics and Probability Students continue their work from 6th grade in order to build a strong foundation for statistics and probability needed for high school. Students understand that statistics can be used to gain information about a population through sampling. They work with drawing inferences about a population based on a sample and use measures of center and of variability to draw informal comparative inferences about two populations. Students investigate the chance processes and develop, use, and evaluate probability models. Students summarize numerical data sets with respect to their context using quantitative measures and describe any overall patterns or deviations from the overall pattern. 73 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 74 Ratios and Proportional Relationships (RP) Cluster Headings Content Standards A. Analyze proportional relationships and use them to solve real-world and mathematical problems. 7.RP.A.1 Compute unit rates associated with ratios of fractions, including ratios of lengths, areas, and other quantities measured in like or different units. For example, if a person walks 1/2 mile in each 1/4 hour, compute the unit rate as the complex fraction 1/2/1/4 miles per hour, equivalently 2 miles per hour. 7.RP.A.2 Recognize and represent proportional relationships between quantities. a. Decide whether two quantities are in a proportional relationship (e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin). b. Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. c. Represent proportional relationships by equations. For example, if total cost t is proportional to the number n of items purchased at a constant price p, the relationship between the total cost and the number of items can be expressed as t = pn. d. Explain what a point (x, y) on the graph of a proportional relationship means in terms of the situation, with special attention to the points (0, 0) and (1, r) where r is the unit rate. 7.RP.A.3 Use proportional relationships to solve multi-step ratio and percent problems. Examples: simple interest, tax, markups and markdowns, gratuities and commissions, fees, percent increase and decrease, percent error. The Number System (NS) Cluster Headings Content Standards A. Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. 7.NS.A.1 Apply and extend previous understandings of addition and subtraction to add and subtract rational numbers; represent addition and subtraction on a horizontal or vertical number line diagram. a. Describe situations in which opposite quantities combine to make 0. b. Understand p + q as the number located a distance |q| from p, in the positive or negative direction depending on whether q is positive or negative. Show that a number and its opposite have a sum of 0 (are additive inverses). Interpret sums of rational numbers by describing real-world contexts. c. Understand subtraction of rational numbers as adding the additive inverse, p – q = p + (–q). Show that the distance between two rational numbers on the number line is the absolute value of their difference, and apply this principle in real-world contexts. d. Apply properties of operations as strategies to add and subtract rational numbers. 75 Cluster Headings Content Standards A. Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. 7.NS.A.2 Apply and extend previous understandings of multiplication and division and of fractions to multiply and divide rational numbers. a. Understand that multiplication is extended from fractions to rational numbers by requiring that operations continue to satisfy the properties of operations, particularly the distributive property, leading to products such as (–1)(–1) = 1 and the rules for multiplying signed numbers. Interpret products of rational numbers by describing real-world contexts. b. Understand that integers can be divided, provided that the divisor is not zero, and every quotient of integers (with non-zero divisor) is a rational number. If p and q are integers, then –(p/q) = (–p)/q = p/(–q). Interpret quotients of rational numbers by describing real-world contexts. c. Apply properties of operations as strategies to multiply and divide rational numbers. d. Convert a rational number to a decimal using long division; know that the decimal form of a rational number terminates in 0s or eventually repeats. 7.NS.A.3 Solve real-world and mathematical problems involving the four operations with rational numbers. (Computations with rational numbers extend the rules for manipulating fractions to complex fractions.) Expressions and Equations (EE) Cluster Headings Content Standards A. Use properties of operations to generate equivalent expressions. 7.EE.A.1 Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients. 7.EE.A.2 Understand that rewriting an expression in different forms in a contextual problem can provide multiple ways of interpreting the problem and how the quantities in it are related. For example, shoes are on sale at a 25% discount. How is the discounted price P related to the original cost C of the shoes? C - .25C = P. In other words, P is 75% of the original cost for C - .25C can be written as .75C. B. Solve real-life and mathematical problems using numerical and algebraic expressions and equations and inequalities. 7.EE.B.3 Solve multi-step real-world and mathematical problems posed with positive and negative rational numbers presented in any form (whole numbers, fractions, and decimals). a. Apply properties of operations to calculate with numbers in any form; convert between forms as appropriate. b. Assess the reasonableness of answers using mental computation and estimation strategies. 76 Cluster Headings Content Standards B. Solve real-life and mathematical problems using numerical and algebraic expressions and equations and inequalities. 7.EE.B.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. a. Solve contextual problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. For example, the perimeter of a rectangle is 54 cm. Its length is 6 cm. What is its width? b. Solve contextual problems leading to inequalities of the form px + q > r or px + q < r, where p, q, and r are specific rational numbers. Graph the solution set of the inequality on a number line and interpret it in the context of the problem. For example: As a salesperson, you are paid $50 per week plus $3 per sale. This week you want your pay to be at least $100. Write an inequality for the number of sales you need to make, and describe the solutions. (Note that inequalities using >, <, ≤, ≥ are included in this standard). Geometry (G) Cluster Headings Content Standards A. Draw, construct, and describe geometrical figures and describe the relationships between them. 7.G.A.1 Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas from a scale drawing and reproducing a scale drawing at a different scale. 7.G.A.2 Draw geometric shapes with given conditions. Focus on constructing triangles from three measures of angles or sides, noticing when the conditions determine a unique triangle, more than one triangle, or no triangle. B. Solve real-life and mathematical problems involving angle measure, area, surface area, and volume. 7.G.B.3 Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle. 7.G.B.4 Know and use facts about supplementary, complementary, vertical, and adjacent angles in a multi-step problem to write and solve simple equations for an unknown angle in a figure. 7.G.B.5 Solve real-world and mathematical problems involving area, volume, and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms. 77 Statistics and Probability (SP) Cluster Headings Content Standards A. Use random sampling to draw inferences about a population. 7.SP.A.1 Understand that statistics can be used to gain information about a population by examining a sample of the population; generalizations about a population from a sample are valid only if the sample is representative of that population. Understand that random sampling tends to produce representative samples and support valid inferences. 7.SP.A.2 Use data from a random sample to draw inferences about a population with an unknown characteristic of interest. Generate multiple samples (or simulated samples) of the same size to gauge the variation in estimates or predictions. For example, estimate the mean word length in a book by randomly sampling words from the book; predict the winner of a school election based on randomly sampled survey data. Gauge how far off the estimate or prediction might be. B. Draw informal comparative inferences about two populations. 7.SP.B.3 Informally assess the degree of visual overlap of two numerical data distributions with similar variabilities, measuring the difference between the centers by expressing it as a multiple of a measure of variability. For example, the mean height of players on the basketball team is 10 cm greater than the mean height of players on the soccer team; on a dot plot or box plot, the separation between the two distributions of heights is noticeable. 7.SP.B.4 Use measures of center and measures of variability for numerical data from random samples to draw informal comparative inferences about two populations. For example, decide whether the words in a chapter of a 7th grade science book are generally longer than the words in a chapter of a 4th grade science book. C. Investigate chance processes and develop, use, and evaluate probability models. 7.SP.C.5 Understand that the probability of a chance event is a number between 0 and 1 that expresses the likelihood of the event occurring. Larger numbers indicate greater likelihood. A probability near 0 indicates an unlikely event, a probability around 1/2 indicates an event that is neither unlikely nor likely, and a probability near 1 indicates a likely event. 7.SP.C.6 Approximate the probability of a chance event by collecting data on the chance process that produces it and observing its long-run relative frequency, and predict the approximate relative frequency given the probability. For example, when rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200 times, but probably not exactly 200 times. 78 Cluster Headings Content Standards C. Investigate chance processes and develop, use, and evaluate probability models. 7.SP.C.7 Develop a probability model and use it to find probabilities of events. Compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy. a. Develop a uniform probability model by assigning equal probability to all outcomes, and use the model to determine probabilities of events. For example, if a student is selected at random from a class, find the probability that Jane will be selected and the probability that a girl will be selected. b. Develop a probability model (which may not be uniform) by observing frequencies in data generated from a chance process. For example, find the approximate probability that a spinning penny will land heads up or that a tossed paper cup will land open end down. Do the outcomes for the spinning penny appear to be equally likely based on the observed frequencies? D. Summarize and describe numerical data sets. 7.SP.D.8 Summarize numerical data sets in relation to their context. a. Give quantitative measures of center (median and/or mean) and variability (range and/or interquartile range), as well as describe any overall pattern and any striking deviations from the overall pattern with reference to the context in which the data were gathered. b. Know and relate the choice of measures of center (median and/or mean)and variability (range and/or interquartile range) to the shape of the data distribution and the context in which the data were gathered. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content Major Content Supporting Content 79 Mathematics | Grade 8 The descriptions below provide an overview of the concepts and skills that students explore throughout the 8th grade. The Number System This is the culminating area for the number system from 6th and 7th grade. Students now know there are numbers that are not rational, called irrational numbers. Students approximate irrational numbers by rational numbers locating them on a number line and students estimate the value of irrational expressions. Expressions and Equations Students work with radicals and integer exponents. Students understand the connections between proportional relationships, lines, and linear equations. Students advance their knowledge developed in 7th grade about equations to analyze and solve linear equations and pairs of simultaneous linear equations. Students use linear equations and systems of linear equations to represent, analyze, and solve a variety of problems. Students recognize equations for proportions (y/x = m or y = mx) as special linear equations (y = mx + b), understanding that the constant of proportionality (m) is the slope, and the graphs are lines through the origin. They understand that the slope (m) of a line is a constant rate of change, so that if the input or x-coordinate changes by an amount A, the output or y-coordinate changes by the amount 𝑚∙𝐴. Students solve systems of two linear equations in two variables and relate the systems to pairs of lines in the plane; these intersect, are parallel, or are the same line. Students use linear equations, systems of linear equations, linear functions, and their understanding of slope of a line to analyze situations and solve problems. Functions This begins the formal study of functions, a mathematical concept that for the student will continue throughout high school. Students grasp the concept of a function as a rule that assigns to each input exactly one output. They understand that functions describe situations where one quantity determines another. They can translate among representations and partial representations of functions (noting that tabular and graphical representations may be partial representations), and they describe how aspects of the function are reflected in the different representations. Geometry Students use ideas about distance and angles and how they behave under translations, rotations, reflections, and dilations, to describe and analyze two-dimensional figures and to solve problems. Students show that the sum of the angles in a triangle is the angle formed by a straight line and that various configurations of lines give rise to similar triangles because of the angles created when a transversal cuts parallel lines. Students understand the statement of the Pythagorean Theorem and its converse and can explain why the Pythagorean Theorem holds. They apply the Pythagorean Theorem to find distances between points on the coordinate plane, to find lengths, and to analyze polygons. Students complete their work on volume by solving problems involving cones, cylinders, and spheres. Statistics and Probability Students extend their knowledge from 7th grade by working with scatter plots for bivariate data and understand linear associations and the use of linear models to solve problems interpreting the slope and intercept. Students continue work with probability by finding probability of compound events and represent the data using organized lists, tables, and tree diagrams. 80 Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 81 The Number System (NS) Cluster Headings Content Standards A. Know that there are numbers that are not rational, and approximate them by rational numbers. 8.NS.A.1 Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually or terminates, and convert a decimal expansion which repeats eventually or terminates into a rational number. 8.NS.A.2 Use rational approximations of irrational numbers to compare the size of irrational numbers locating them approximately on a number line diagram. Estimate the value of irrational expressions such as π2. For example, by truncating the decimal expansion of √2, show that √2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations. Expressions and Equations (EE) Cluster Headings Content Standards A. Work with radicals and integer exponents. 8.EE.A.1 Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 32 x 3–5 = 3–3 = 1/33 = 1/27. 8.EE.A.2 Use square root and cube root symbols to represent solutions to equations of the form x2 = p and x3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that √2 is irrational. 8.EE.A.3 Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities and to express how many times as much one is than the other. For example, estimate the population of the United States as 3 x 108 and the population of the world as 7 x 109, and determine that the world population is more than 20 times larger. 8.EE.A.4 Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology. B. Understand the connections between proportional relationships, lines, and linear equations. 8.EE.B.5 Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed. 8.EE.B.6 Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; know and derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b. 82 Cluster Headings Content Standards C. Analyze and solve linear equations and systems of two linear equations. 8.EE.C.7 Solve linear equations in one variable. a. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers). b. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms. 8.EE.C.8 Analyze and solve systems of two linear equations. a. Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously. b. Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6. c. Solve real-world and mathematical problems leading to two linear equations in two variables. For example, given coordinates for two pairs of points, determine whether the line through the first pair of points intersects the line through the second pair. Functions (F) Cluster Headings Content Standards A. Define, evaluate, and compare functions. 8.F.A.1 Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. (Function notation is not required in 8th grade.) 8.F.A.2 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and another linear function represented by an algebraic expression, determine which function has the greater rate of change. 8.F.A.3 Know and interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line. 83 Cluster Headings Content Standards B. Use functions to model relationships between quantities. 8.F.B.4 Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models and in terms of its graph or a table of values. 8.F.B.5 Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally. Geometry (G) Cluster Headings Content Standards A. Understand and describe the effects of transformations on two-dimensional figures and use informal arguments to establish facts about angles. 8.G.A.1 Verify experimentally the properties of rotations, reflections, and translations: a. Lines are taken to lines, and line segments to line segments of the same length. b. Angles are taken to angles of the same measure. c. Parallel lines are taken to parallel lines. 8.G.A.2 Describe the effect of dilations, translations, rotations, and reflections on two-dimensional figures using coordinates. 8.G.A.3 Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the sum of the three angles appears to form a line, and give an argument in terms of transversals why this is so. B. Understand and apply the Pythagorean Theorem. 8.G.B.4 Explain a proof of the Pythagorean Theorem and its converse. 8.G.B.5 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions. 8.G.B.6 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system. C. Solve real-world and mathematical problems involving volume of cylinders, cones, and spheres. 8.G.C.7 Know and understand the formulas for the volumes of cones, cylinders, and spheres, and use them to solve real-world and mathematical problems. 84 Statistics and Probability (SP) Cluster Headings Content Standards A. Investigate patterns of association in bivariate data. 8.SP.A.1 Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association. 8.SP.A.2 Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line and informally assess the model fit by judging the closeness of the data points to the line. 8.SP.A.3 Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height. B. Investigate chance processes and develop, use, and evaluate probability models 8.SP.B.4 Find probabilities of compound events using organized lists, tables, tree diagrams, and simulation. Understand that, just as with simple events, the probability of a compound event is the fraction of outcomes in the sample space for which the compound event occurs. Represent sample spaces for compound events using methods such as organized lists, tables, and tree diagrams. For an event described in everyday language (e.g., "rolling double sixes"), identify the outcomes in the sample space which compose the event. Major content of the grade is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content Major Content Supporting Content 85 Algebra I | A1 Algebra I emphasizes linear and quadratic expressions, equations, and functions. This course also introduces students to polynomial, rational and exponential functions with domains in the integers. Students explore the structures of and interpret functions and other mathematical models. Students build upon previous knowledge of equations and inequalities to reason, solve, and represent equations and inequalities numerically and graphically. The major work of Algebra I is from the following domains and clusters:  Seeing Structure in Expressions o Interpret the structure of expressions. o Write expressions in equivalent forms to solve problems.  Arithmetic with Polynomials and Rational Expressions o Perform arithmetic operations on polynomials.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Understand solving equations as a process of reasoning and explain the reasoning. o Solve equations and inequalities in one variable. o Represent and solve equations and inequalities graphically.  Interpreting Functions o Understand the concept of a function and use function notation. o Interpret functions that arise in applications in terms of the context.  Interpreting Categorical and Quantitative Data o Interpret linear models. Supporting work is from the following domains and clusters:  Quantities o Reason quantitatively and use units to solve problems.  Arithmetic with Polynomials and Rational Expressions o Understand the relationship between zeros and factors of polynomials.  Reasoning with Equations and Inequalities o Solve systems of equations.  Interpreting Functions o Analyze functions using different representations.  Building Functions o Build a function that models a relationship between two quantities. o Build new functions from existing functions.  Linear, Quadratic, and Exponential Models o Construct and compare linear, quadratic, and exponential models and solve problems. o Interpret expressions for functions in terms of the situation they model.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on a single count or measurement variable. o Summarize, represent, and interpret data on two categorical and quantitative variables. 86 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 87 Number and Quantity Quantities (N.Q) Cluster Headings Content Standards Scope & Clarifications A. Reason quantitatively and use units to solve problems. A1.N.Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.N.Q.A.2 Identify, interpret, and justify appropriate quantities for the purpose of descriptive modeling. Descriptive modeling refers to understanding and interpreting graphs; identifying extraneous information; choosing appropriate units; etc. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.N.Q.A.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. There are no assessment limits for this standard. The entire standard is assessed in this course. Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. A1.A.SSE.A.1 Interpret expressions that represent a quantity in terms of its context.★ a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. There are no assessment limits for this standard. The entire standard is assessed in this course. 88 Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. A1.A.SSE.A.2 Use the structure of an expression to identify ways to rewrite it. For example, recognize 532 - 472 as a difference of squares and see an opportunity to rewrite it in the easier-to-evaluate form (53 + 47) (53 – 47). See an opportunity to rewrite a2 + 9a + 14 as (a + 7) (a + 2). Tasks are limited to numerical expressions and polynomial expressions in one variable. B. Write expressions in equivalent forms to solve problems. A1.A.SSE.B.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Factor a quadratic expression to reveal the zeros of the function it defines. b. Complete the square in a quadratic expression in the form Ax2 + Bx + C where A = 1 to reveal the maximum or minimum value of the function it defines. c. Use the properties of exponents to rewrite exponential expressions. For A1.A.SSE.B.3c: For example, the growth of bacteria can be modeled by either f(t) = 3(t+2) or g(t) = 9(3t) because the expression 3(t+2) can be rewritten as (3t) (32) = 9(3t). i) Tasks have a real-world context. As described in the standard, there is an interplay between the mathematical structure of the expression and the structure of the situation such that choosing and producing an equivalent form of the expression reveals something about the situation. ii) Tasks are limited to exponential expressions with integer exponents. Arithmetic with Polynomials and Rational Expressions (A.APR) Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations on polynomials. A1.A.APR.A.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Understand the relationship between zeros and factors of polynomials. A1.A.APR.B.2 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Graphing is limited to linear and quadratic polynomials. 89 Creating Equations★ (A.CED) Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships. A1.A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Tasks are limited to linear, quadratic, or exponential equations with integer exponents. A1.A.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations with two variables on coordinate axes with labels and scales. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.A.CED.A.3 Represent constraints by equations or inequalities and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.A.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. i) Tasks are limited to linear, quadratic, piecewise, absolute value, and exponential equations with integer exponents. ii) Tasks have a real-world context. Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards Scope & Clarifications A. Understand solving equations as a process of reasoning and explain the reasoning. A1.A.REI.A.1 Explain each step in solving an equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. Tasks are limited to linear, quadratic, piecewise, absolute value, and exponential equations with integer exponents. B. Solve equations and inequalities in one variable. A1.A.REI.B.2 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. There are no assessment limits for this standard. The entire standard is assessed in this course. 90 Cluster Headings Content Standards Scope & Clarifications B. Solve equations and inequalities in one variable. A1.A.REI.B.3 Solve quadratic equations and inequalities in one variable. a. Use the method of completing the square to rewrite any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions. For A1.A.REI.B.3b: Tasks do not require students to write solutions for quadratic equations that have roots with nonzero imaginary parts. However, tasks can require the student to recognize cases in which a quadratic equation has no real solutions. Note: solving a quadratic equation by factoring relies on the connection between zeros and factors of polynomials. This is formally assessed in Algebra II. C. Solve systems of equations. A1.A.REI.C.4 Write and solve a system of linear equations in context. Solve systems both algebraically and graphically. Systems are limited to at most two equations in two variables. D. Represent and solve equations and inequalities graphically. A1.A.REI.D.5 Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). There are no assessment limits for this standard. The entire standard is assessed in this course. A1.A.REI.D.6 Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the approximate solutions using technology. ★ Include cases where f(x) and/or g(x) are linear, quadratic, absolute value, and exponential functions. For example, f(x) = 3x + 5 and g(x) = x2 + 1. Exponential functions are limited to domains in the integers. A1.A.REI.D.7 Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes. There are no assessment limits for this standard. The entire standard is assessed in this course. 91 Functions Interpreting Functions (F.IF) Cluster Headings Content Standards Scope & Clarifications A. Understand the concept of function and use function notation. A1.F.IF.A.1 Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). There are no assessment limits for this standard. The entire standard is assessed in this course. A1.F.IF.A.2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Interpret functions that arise in applications in terms of the context. A1.F.IF.B.3 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. ★ Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. i) Tasks have a real-world context. ii) Tasks are limited to linear functions, quadratic functions, absolute value functions, and exponential functions with domains in the integers. A1.F.IF.B.4 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. ★ For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. There are no assessment limits for this standard. The entire standard is assessed in this course. 92 Cluster Headings Content Standards Scope & Clarifications B. Interpret functions that arise in applications in terms of the context. A1.F.IF.B.5 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.★ i) Tasks have a real-world context. ii) Tasks are limited to linear functions, quadratic functions, piecewise-defined functions (including step functions and absolute value functions), and exponential functions with domains in the integers. C. Analyze functions using different representations. A1.F.IF.C.6 Graph functions expressed symbolically and show key features of the graph, by hand and using technology. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. Tasks in A1.F.IF.C.6b are limited to piecewise, step and absolute value functions. A1.F.IF.C.7 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.F.IF.C.8 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). i) Tasks have a real-world context. ii) Tasks are limited to linear functions, quadratic functions, piecewise-defined functions (including step functions and absolute value functions), and exponential functions with domains in the integers. 93 Building Functions (F.BF) Cluster Headings Content Standards Scope & Clarifications A. Build a function that models a relationship between two quantities. A1.F.BF.A.1 Write a function that describes a relationship between two quantities.★ a. Determine an explicit expression, a recursive process, or steps for calculation from a context. i) Tasks have a real-world context. ii) Tasks are limited to linear functions, quadratic functions, and exponential functions with domains in the integers. B. Build new functions from existing functions. A1.F.BF.B.2 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. i) Identifying the effect on the graph of replacing f(x) by f(x) + k, k f(x), and f(x+k) for specific values of k (both positive and negative) is limited to linear, quadratic, and absolute value functions. ii) f(kx) will not be included in Algebra 1. It is addressed in Algebra 2. iii) Experimenting with cases and illustrating an explanation of the effects on the graph using technology is limited to linear functions, quadratic functions, absolute value, and exponential functions with domains in the integers. iv) Tasks do not involve recognizing even and odd functions. 94 Linear, Quadratic, and Exponential Models★ (F.LE) Cluster Headings Content Standards Scope & Clarifications A. Construct and compare linear, quadratic, and exponential models and solve problems. A1.F.LE.A.1 Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Recognize that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant factor per unit interval relative to another. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.F.LE.A.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a table, a description of a relationship, or input-output pairs. Tasks are limited to constructing linear and exponential functions in simple context (not multi-step). A1.F.LE.A.3 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Interpret expressions for functions in terms of the situation they model. F.LE.B.4 Interpret the parameters in a linear or exponential function in terms of a context. For example, the total cost of an electrician who charges 35 dollars for a house call and 50 dollars per hour would be expressed as the function y = 50x + 35. If the rate were raised to 65 dollars per hour, describe how the function would change. i) Tasks have a real-world context. ii) Exponential functions are limited to those with domains in the integers. 95 Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards Scope & Clarifications A. Summarize, represent, and interpret data on a single count or measurement variable. A1.S.ID.A.1 Represent single or multiple data sets with dot plots, histograms, stem plots (stem and leaf), and box plots. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.S.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.S.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). There are no assessment limits for this standard. The entire standard is assessed in this course. B. Summarize, represent, and interpret data on two categorical and quantitative variables. A1.S.ID.B.4 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. b. Fit a linear function for a scatter plot that suggests a linear association. Emphasize linear models, quadratic models, and exponential models with domains in the integers. For A1.S.ID.B.4a: i) Tasks have a real-world context. ii) Exponential functions are limited to those with domains in the integers. C. Interpret linear models. A1.S.ID.C.5 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.S.ID.C.6 Use technology to compute and interpret the correlation coefficient of a linear fit. There are no assessment limits for this standard. The entire standard is assessed in this course. A1.S.ID.C.7 Distinguish between correlation and causation. There are no assessment limits for this standard. The entire standard is assessed in this course. 96 Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 97 Geometry | G Geometry emphasizes similarity, right triangle trigonometry, congruence, and modeling geometry concepts in real life situations. Students build upon previous knowledge of similarity, congruence, and triangles to prove theorems and reason mathematically. This course also introduces students to geometric constructions and circles. Students show a progression of mastery and understanding of the use and application of surface area and volume. The major work of Geometry is from the following domains and clusters:  Congruence o Understand congruence in terms of rigid motions. o Prove geometric theorems.  Similarity, Right Triangles, and Trigonometry o Understand similarity in terms of similarity transformations. o Prove theorems involving similarity. o Define trigonometric ratios and solve problems involving triangles.  Expressing Geometric Properties with Equations o Use coordinates to prove simple geometric theorems algebraically.  Modeling with Geometry o Apply geometric concepts in modeling situations. Supporting work is from the following domains and clusters:  Congruence o Experiment with transformations in the plane. o Make geometric constructions.  Circles o Understand and apply theorems about circles. o Find areas of sectors of circles.  Expressing Geometric Properties with Equations o Translate between the geometric description and the equation for a circle.  Geometric Measurement and Dimension o Explain volume and surface area formulas and use them to solve problems. 98 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 99 Geometry Congruence (G.CO) Cluster Headings Content Standards Scope & Clarifications A. Experiment with transformations in the plane. G.CO.A.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, plane, distance along a line, and distance around a circular arc. There are no assessment limits for this standard. The entire standard is assessed in this course. G.CO.A.2 Represent transformations in the plane in multiple ways, including technology. Describe transformations as functions that take points in the plane (pre-image) as inputs and give other points (image) as outputs. Compare transformations that preserve distance and angle measure to those that do not (e.g., translation versus horizontal stretch). There are no assessment limits for this standard. The entire standard is assessed in this course. G.CO.A.3 Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry the shape onto itself. There are no assessment limits for this standard. The entire standard is assessed in this course. G.CO.A.4 Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. There are no assessment limits for this standard. The entire standard is assessed in this course. G.CO.A.5 Given a geometric figure and a rigid motion, draw the image of the figure in multiple ways, including technology. Specify a sequence of rigid motions that will carry a given figure onto another. Rigid motions include rotations, reflections, and translations. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Understand congruence in terms of rigid motions. G.CO.B.6 Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to determine informally if they are congruent. There are no assessment limits for this standard. The entire standard is assessed in this course. 100 Cluster Headings Content Standards Scope & Clarifications B. Understand congruence in terms of rigid motions. G.CO.B.7 Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. There are no assessment limits for this standard. The entire standard is assessed in this course. G.CO.B.8 Explain how the criteria for triangle congruence (ASA, SAS, AAS, and SSS) follow from the definition of congruence in terms of rigid motions. There are no assessment limits for this standard. The entire standard is assessed in this course. C. Prove geometric theorems. G.CO.C.9 Prove theorems about lines and angles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. G.CO.C.10 Prove theorems about triangles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. 101 Cluster Headings Content Standards Scope & Clarifications C. Prove geometric theorems. G.CO.C.11 Prove theorems about parallelograms. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. D. Make geometric constructions. G.CO.D.12 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Constructions include but are not limited to: copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; constructing a line parallel to a given line through a point not on the line, and constructing the following objects inscribed in a circle: an equilateral triangle, square, and a regular hexagon. Similarity, Right Triangles, and Trigonometry (G.SRT) Cluster Headings Content Standards Scope & Clarifications A. Understand similarity in terms of similarity transformations. G.SRT.A.1 Verify informally the properties of dilations given by a center and a scale factor. Properties include but are not limited to: a dilation takes a line not passing through the center of the dilation to a parallel line, and leaves a line passing through the center of the dilation unchanged; the dilation of a line segment is longer or shorter in the ratio given by the scale factor. 102 Cluster Headings Content Standards Scope & Clarifications A. Understand similarity in terms of similarity transformations G.SRT.A.2 Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. There are no assessment limits for this standard. The entire standard is assessed in this course. G.SRT.A.3 Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Prove theorems involving similarity. G.SRT.B.4 Prove theorems about similar triangles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity. G.SRT.B.5 Use congruence and similarity criteria for triangles to solve problems and to justify relationships in geometric figures. There are no assessment limits for this standard. The entire standard is assessed in this course. C. Define trigonometric ratios and solve problems involving triangles. G.SRT.C.6 Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. There are no assessment limits for this standard. The entire standard is assessed in this course. G.SRT.C.7 Explain and use the relationship between the sine and cosine of complementary angles. There are no assessment limits for this standard. The entire standard is assessed in this course. 103 Cluster Headings Content Standards Scope & Clarifications C. Define trigonometric ratios and solve problems involving triangles. G.SRT.C.8 Solve triangles. ★ a. Know and use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. b. Know and use the Law of Sines and Law of Cosines to solve problems in real life situations. Recognize when it is appropriate to use each. Ambiguous cases will not be included in assessment. Circles (G.C) Cluster Headings Content Standards Scope & Clarifications A. Understand and apply theorems about circles. G.C.A.1 Recognize that all circles are similar. There are no assessment limits for this standard. The entire standard is assessed in this course. G.C.A.2 Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle, and properties of angles for a quadrilateral inscribed in a circle. G.C.A.3 Construct the incenter and circumcenter of a triangle and use their properties to solve problems in context. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Find areas of sectors of circles. G.C.B.4 Know the formula and find the area of a sector of a circle in a real-world context. For example, use proportional relationships and angles measured in degrees or radians. There are no assessment limits for this standard. The entire standard is assessed in this course. 104 Expressing Geometric Properties with Equations (G.GPE) Cluster Headings Content Standards Scope & Clarifications A. Translate between the geometric description and the equation for a circle. G.GPE.A.1 Know and write the equation of a circle of given center and radius using the Pythagorean Theorem. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use coordinates to prove simple geometric theorems algebraically. G.GPE.B.2 Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, 3 ) lies on the circle centered at the origin and containing the point (0, 2). There are no assessment limits for this standard. The entire standard is assessed in this course. G.GPE.B.3 Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems. For example, find the equation of a line parallel or perpendicular to a given line that passes through a given point. There are no assessment limits for this standard. The entire standard is assessed in this course. G.GPE.B.4 Find the point on a directed line segment between two given points that partitions the segment in a given ratio. There are no assessment limits for this standard. The entire standard is assessed in this course. G.GPE.B.5 Know and use coordinates to compute perimeters of polygons and areas of triangles and rectangles.★ For example, use the distance formula. There are no assessment limits for this standard. The entire standard is assessed in this course. 105 Geometric Measurement and Dimension (G.GMD) Cluster Headings Content Standards Scope & Clarifications A. Explain volume and surface area formulas and use them to solve problems. G.GMD.A.1 Give an informal argument for the formulas for the circumference of a circle and the volume and surface area of a cylinder, cone, prism, and pyramid. Informal arguments may include but are not limited to using the dissection argument, applying Cavalieri’s principle, and constructing informal limit arguments. There are no assessment limits for this standard. The entire standard is assessed in this course. G.GMD.A.2 Know and use volume and surface area formulas for cylinders, cones, prisms, pyramids, and spheres to solve problems.★ There are no assessment limits for this standard. The entire standard is assessed in this course. Modeling with Geometry (G.MG) Cluster Headings Content Standards Scope & Clarifications A. Apply geometric concepts in modeling situations. G.MG.A.1 Use geometric shapes, their measures, and their properties to describe objects.★ For example, model a tree trunk or a human torso as a cylinder. There are no assessment limits for this standard. The entire standard is assessed in this course. G.MG.A.2 Apply geometric methods to solve real-world problems.★ Geometric methods may include but are not limited to using geometric shapes, the probability of a shaded region, density, and design problems. There are no assessment limits for this standard. The entire standard is assessed in this course. Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 106 Algebra II | A2 Algebra II emphasizes polynomial, rational and exponential expressions, equations, and functions. This course also introduces students to the complex number system, basic trigonometric functions, and foundational statistics skills such as interpretation of data and making statistical inferences. Students build upon previous knowledge of equations and inequalities to reason, solve, and represent equations and inequalities numerically and graphically. The major work of Algebra II is from the following domains and clusters:  The Real Number System o Extend the properties of exponents to rational exponents.  Seeing Structure in Expressions o Interpret the structure of expressions. o Use expressions in equivalent forms to solve problems.  Arithmetic with Polynomials and Rational Expressions o Understand the relationship between zeros and factors of polynomials.  Reasoning with Equations and Inequalities o Understand solving equations as a process of reasoning and explain the reasoning. o Represent and solve equations graphically.  Interpreting Functions o Interpret functions that arise in applications in terms of the context.  Building Functions o Build a function that models a relationship between two quantities.  Making Inferences and Justifying Conclusions o Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Supporting work is from the following domains and clusters:  Quantities o Reason quantitatively and use units to solve problems.  The Complex Number System o Perform arithmetic operations with complex numbers. o Use complex numbers in quadratic equations.  Arithmetic with Polynomials and Rational Expressions o Use polynomial identities to solve problems. o Rewrite rational expressions.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Solve equations and inequalities in one variable. o Solve systems of equations.  Interpreting Functions o Analyze functions using different representations.  Building Functions o Build new functions from existing functions.  Linear, Quadratic, and Exponential Models o Construct and compare linear, quadratic, and exponential models and solve problems. o Interpret expressions for functions in terms of the situation they model. 107  Trigonometric Functions o Extend the domain of trigonometric functions using the unit circle. o Prove and apply trigonometric identities.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on a single count or measurement variable. o Summarize, represent, and interpret data on two categorical and quantitative variables.  Conditional Probability and the Rules of Probability o Understand independence and conditional probability and use them to interpret data. o Use the rules of probability to compute probabilities of compound events in a uniform probability model. 108 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 109 Number and Quantity The Real Number System (N.RN) Cluster Headings Content Standards Scope & Clarifications A. Extend the properties of exponents to rational exponents. A2.N.RN.A.1 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.N.RN.A.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. There are no assessment limits for this standard. The entire standard is assessed in this course. Quantities★ (N.Q) Cluster Headings Content Standards Scope & Clarifications A. Reason quantitatively and use units to solve problems. A2.N.Q.A.1 Identify, interpret, and justify appropriate quantities for the purpose of descriptive modeling. Descriptive modeling refers to understanding and interpreting graphs; identifying extraneous information; choosing appropriate units; etc. There are no assessment limits for this standard. The entire standard is assessed in this course. The Complex Number System (N.CN) Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations with complex numbers. A2.N.CN.A.1 Know there is a complex number i such that i2 = –1, and every complex number has the form a + bi with a and b real. There are no assessment limits for this standard. The entire standard is assessed in this course. 110 Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations with complex numbers. A2.N.CN.A.2 Know and use the relation i2 = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use complex numbers in quadratic equations. A2.N.CN.B.3 Solve quadratic equations with real coefficients that have complex solutions. There are no assessment limits for this standard. The entire standard is assessed in this course. Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. A2.A.SSE.A.1 Use the structure of an expression to identify ways to rewrite it. For example, see 2x4 + 3x2 – 5 as its factors (x2 – 1) and (2x2 + 5); see x4 – y4 as (x2) 2 – (y2) 2, thus recognizing it as a difference of squares that can be factored as (x2 – y2) (x2 + y2); see (x2 + 4)/(x2 + 3) as ((x2+ 3) + 1 )/(x2 + 3), thus recognizing an opportunity to write it as 1 + 1/(x2 + 3). Tasks are limited to polynomial, rational, or exponential expressions. 111 Cluster Headings Content Standards Scope & Clarifications B. Use expressions in equivalent forms to solve problems. A2.A.SSE.B.2 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Use the properties of exponents to rewrite expressions for exponential functions. For example the expression 1.15t can be rewritten as ((1.15)1/12)12t ≈ 1.01212t to reveal that the approximate equivalent monthly interest rate is 1.2% if the annual rate is 15%. i) Tasks have a real-world context. As described in the standard, there is an interplay between the mathematical structure of the expression and the structure of the situation such that choosing and producing an equivalent form of the expression reveals something about the situation. ii) Tasks are limited to exponential expressions with rational or real exponents. A2.A.SSE.B.3 Recognize a finite geometric series (when the common ratio is not 1), and know and use the sum formula to solve problems in context. There are no assessment limits for this standard. The entire standard is assessed in this course. Arithmetic with Polynomials and Rational Expressions (A.APR) Cluster Headings Content Standards Scope & Clarifications A. Understand the relationship between zeros and factors of polynomials. A2.A.APR.A.1 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). There are no assessment limits for this standard. The entire standard is assessed in this course. A2.A.APR.A.2 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Tasks include quadratic, cubic, and quartic polynomials and polynomials for which factors are not provided. For example, find the zeros of (x2 - 1)(x2 + 1). 112 Cluster Headings Content Standards Scope & Clarifications B. Use polynomial identities to solve problems. A2.A.APR.B.3 Know and use polynomial identities to describe numerical relationships. For example, compare (31)(29) = (30 + 1) (30 – 1) = 302 – 12 with (x + y) (x – y) = x2 – y2. There are no assessment limits for this standard. The entire standard is assessed in this course. C. Rewrite rational expressions. A2.A.APR.C.4 Rewrite rational expressions in different forms. There are no assessment limits for this standard. The entire standard is assessed in this course. Creating Equations★ (A.CED) Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships. A2.A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions, and rational and exponential functions. Tasks have a real-world context. A2.A.CED.A.2 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. i) Tasks are limited to square root, cube root, polynomial, rational, and logarithmic functions. ii) Tasks have a real-world context. Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards Scope & Clarifications A. Understand solving equations as a process of reasoning and explain the reasoning. A2.A.REI.A.1 Explain each step in solving an equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. Tasks are limited to square root, cube root, polynomial, rational, and logarithmic functions. A2.A.REI.A.2 Solve rational and radical equations in one variable, and identify extraneous solutions when they exist. There are no assessment limits for this standard. The entire standard is assessed in this course. 113 Cluster Headings Content Standards Scope & Clarifications B. Solve equations and inequalities in one variable. A2.A.REI.B.3 Solve quadratic equations and inequalities in one variable. a. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. In the case of equations that have roots with nonzero imaginary parts, students write the solutions as a ± bi for real numbers a and b. C. Solve systems of equations. A2.A.REI.C.4 Write and solve a system of linear equations in context. When solving algebraically, tasks are limited to systems of at most three equations and three variables. With graphic solutions, systems are limited to only two variables. A2.A.REI.C.5 Solve a system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. There are no assessment limits for this standard. The entire standard is assessed in this course. D. Represent and solve equations graphically. A2.A.REI.D.6 Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the approximate solutions using technology. ★ Include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, and logarithmic functions. Tasks may involve any of the function types mentioned in the standard. 114 Functions Interpreting Functions (F.IF) Cluster Headings Content Standards Scope & Clarifications A. Interpret functions that arise in applications in terms of the context. A2.F.IF.A.1 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. ★ Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. i) Tasks have a real-world context. ii) Tasks may involve square root, cube root, polynomial, exponential, and logarithmic functions. A2.F.IF.A.2 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.★ i) Tasks have a real-world context. ii) Tasks may involve polynomial, exponential, and logarithmic functions. B. Analyze functions using different representations. A2.F.IF.B.3 Graph functions expressed symbolically and show key features of the graph, by hand and using technology.★ a. Graph square root, cube root, and piecewise defined functions, including step functions and absolute value functions. b. Graph polynomial functions, identifying zeros when suitable factorizations are available and showing end behavior. c. Graph exponential and logarithmic functions, showing intercepts and end behavior. A2.F.IF.B.3a: Tasks are limited to square root and cube root functions. The other functions are assessed in Algebra 1. 115 Cluster Headings Content Standards Scope & Clarifications B. Analyze functions using different representations. A2.F.IF.B.4 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Know and use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change in functions such as y = 2x, y = (1/2)x, y = 2-x, y = (1/2)-x. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.F.IF.B.5 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). Tasks may involve polynomial, exponential, and logarithmic functions. Building Functions (F.BF) Cluster Headings Content Standards Scope & Clarifications A. Build a function that models a relationship between two quantities. A2.F.BF.A.1 Write a function that describes a relationship between two quantities.★ a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For example, given cost and revenue functions, create a profit function. For A2.F.BF.A.1a: i) Tasks have a real-world context. ii) Tasks may involve linear functions, quadratic functions, and exponential functions. A2.F.BF.A.2 Know and write arithmetic and geometric sequences with an explicit formula and use them to model situations.★ There are no assessment limits for this standard. The entire standard is assessed in this course. B. Build new functions from existing functions. A2.F.BF.B.3 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. i) Tasks may involve polynomial, exponential, and logarithmic functions. ii) Tasks may involve recognizing even and odd functions. A2.F.BF.B.4 Find inverse functions. a. Find the inverse of a function when the given function is one-to-one. There are no assessment limits for this standard. The entire standard is assessed in this course. 116 Linear, Quadratic, and Exponential Models★ (F.LE) Cluster Headings Content Standards Scope & Clarifications A. Construct and compare linear, quadratic, and exponential models and solve problems. A2.F.LE.A.1 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a table, a description of a relationship, or input-output pairs. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.F.LE.A.2 For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Interpret expressions for functions in terms of the situation they model. A2.F.LE.B.3 Interpret the parameters in a linear or exponential function in terms of a context. For example, the equation y = 5000 (1.06)x models the rising population of a city with 5000 residents when the annual growth rate is 6 percent. What will be the effect on the equation if the city's growth rate was 7 percent instead of 6 percent? There are no assessment limits for this standard. The entire standard is assessed in this course. Trigonometric Functions (F.TF) Cluster Headings Content Standards Scope & Clarifications A. Extend the domain of trigonometric functions using the unit circle. A2.F.TF.A.1 Understand and use radian measure of an angle. a. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle. b. Use the unit circle to find sin θ, cos θ, and tan θ when θ is a commonly recognized angle between 0 and 2π. Commonly recognized angles include all multiples nπ /6 and nπ /4, where n is an integer. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.F.TF.A.2 Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. There are no assessment limits for this standard. The entire standard is assessed in this course. 117 Cluster Headings Content Standards Scope & Clarifications B. Prove and apply trigonometric identities. A2.F.TF.B.3 Know and use trigonometric identities to to find values of trig functions. a. Given a point on a circle centered at the origin, recognize and use the right triangle ratio definitions of sin θ, cos θ, and tan θ to evaluate the trigonometric functions. b. Given the quadrant of the angle, use the identity sin2 θ + cos2 θ = 1 to find sin θ given cos θ, or vice versa. Commonly recognized angles include all multiples nπ /6 and nπ /4, where n is an integer. There are no assessment limits for this standard. The entire standard is assessed in this course. Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards Scope & Clarifications A. Summarize, represent, and interpret data on a single count or measurement variable. A2.S.ID.A.1 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages using the Empirical Rule. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Summarize, represent, and interpret data on two categorical and quantitative variables. A2.S.ID.B.2 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. i) Tasks have a real-world context. ii) Tasks are limited to exponential functions with domains not in the integers. 118 Making Inferences and Justifying Conclusions (S.IC) Cluster Headings Content Standards Scope & Clarifications A. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. A2.S.IC.A.1 Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. For example, in a given situation, is it more appropriate to use a sample survey, an experiment, or an observational study? Explain how randomization affects the bias in a study. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.S.IC.A.2 Use data from a sample survey to estimate a population mean or proportion; use a given margin of error to solve a problem in context. There are no assessment limits for this standard. The entire standard is assessed in this course. Conditional Probability and the Rules of Probability (S.CP) Cluster Headings Content Standards Scope & Clarifications A. Understand independence and conditional probability and use them to interpret data. A2.S.CP.A.1 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). There are no assessment limits for this standard. The entire standard is assessed in this course. A2.S.CP.A.2 Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. There are no assessment limits for this standard. The entire standard is assessed in this course. A2.S.CP.A.3 Know and understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. There are no assessment limits for this standard. The entire standard is assessed in this course. 119 Cluster Headings Content Standards Scope & Clarifications Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. A. Understand independence and conditional probability and use them to interpret data. A2.S.CP.A.4 Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. For example, compare the chance of having lung cancer if you are a smoker with the chance of being a smoker if you have lung cancer. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. A2.S.CP.B.5 Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A and interpret the answer in terms of the model. For example, a teacher gave two exams. 75 percent passed the first quiz and 25 percent passed both. What percent who passed the first quiz also passed the second quiz? There are no assessment limits for this standard. The entire standard is assessed in this course. A2.S.CP.B.6 Know and apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. For example, in a math class of 32 students, 14 are boys and 18 are girls. On a unit test 6 boys and 5 girls made an A. If a student is chosen at random from a class, what is the probability of choosing a girl or an A student? There are no assessment limits for this standard. The entire standard is assessed in this course. Major Content Supporting Content 120 Integrated Math I | M1 Integrated Math I emphasizes linear and exponential expressions, equations, and functions. This course also focuses on geometric congruence and interpreting linear models from quantitative data. Students continue their learning and understanding of categorical and quantitative data. Students are also introduced to reasoning with equations by solving systems of equations in two variables. The major work of Integrated Math I is from the following domains and clusters:  Seeing Structure in Expressions o Interpret the structure of expressions. o Write expressions in equivalent forms to solve problems.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Solve equations and inequalities in one variable. o Represent and solve equations and inequalities graphically.  Interpreting Functions o Understand the concept of a function and use function notation. o Interpret functions that arise in applications in terms of the context.  Building Functions o Build a function that models a relationship between two quantities.  Congruence o Understand congruence in terms of rigid motions. o Prove geometric theorems.  Interpreting Categorical and Quantitative Data o Interpret linear models. Supporting work is from the following domains and clusters:  Quantities o Reason quantitatively and use units to solve problems.  Reasoning with Equations and Inequalities o Solve systems of equations.  Interpreting Functions o Analyze functions using different representations.  Linear and Exponential Models o Construct and compare linear and exponential models and solve problems. o Interpret expressions for functions in terms of the situation they model.  Congruence o Experiment with transformations in the plane.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on a single count or measurement variable. o Summarize, represent, and interpret data on two categorical and quantitative variables. 121 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 122 Number and Quantity Quantities (N.Q) Cluster Headings Content Standards Scope & Clarifications A. Reason quantitatively and use units to solve problems. M1.N.Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.N.Q.A.2 Identify, interpret, and justify appropriate quantities for the purpose of descriptive modeling. Clarification: Descriptive modeling refers to understanding and interpreting graphs; identifying extraneous information; choosing appropriate units; etc. Tasks are limited to linear or exponential equations with integer exponents. M1.N.Q.A.3 Choose a level of accuracy appropriate to limitations on measurement when reporting quantities. There are no assessment limits for this standard. The entire standard is assessed in this course. Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. M1.A.SSE.A.1 Interpret expressions that represent a quantity in terms of its context.★ a. Interpret parts of an expression, such as terms, factors, and coefficients. b. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. Tasks are limited to linear and exponential expressions, including related numerical expressions. 123 Cluster Headings Content Standards Scope & Clarifications B. Write expressions in equivalent forms to solve problems. M1.A.SSE.B.2 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Use the properties of exponents to rewrite exponential expressions. For M1.A.SSE.B.2a: For example, the growth of bacteria can be modeled by either f(t) = 3(t+2) or g(t) = 9(3t) because the expression 3(t+2) can be rewritten as (3t) (32) = 9(3t). Tasks have a real-world context. As described in the standard, there is an interplay between the mathematical structure of the expression and the structure of the situation such that choosing and producing an equivalent form of the expression reveals something about the situation. Creating Equations★ (A.CED) Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships M1.A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. i) Tasks are limited to linear or exponential equations with integer exponents. ii) Tasks have a real-world context. iii) In the linear case, tasks have more of the hallmarks of modeling as a mathematical practice (less defined tasks, more of the modeling cycle, etc.). M1.A.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations with two variables on coordinate axes with labels and scales. i) Tasks are limited to linear equations ii) Tasks have a real-world context. iii) Tasks have the hallmarks of modeling as a mathematical practice (less defined tasks, more of the modeling cycle, etc.). 124 Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships M1.A.CED.A.3 Represent constraints by equations or inequalities and by systems of equations and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context. For example, represent inequalities describing nutritional and cost constraints on combinations of different foods. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.A.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. i) Tasks are limited to linear equations. ii) Tasks have a real-world context. Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards Scope & Clarifications A. Solve equations and inequalities in one variable. M1.A.REI.A.1 Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Solve systems of equations. M1.A.REI.B.2 Write and solve a system of linear equations in context. Solve systems both algebraically and graphically. Systems are limited to at most two equations in two variables. C. Represent and solve equations and inequalities graphically. M1.A.REI.C.3 Understand that the graph of an equation in two variables is the set of all its solutions plotted in the coordinate plane, often forming a curve (which could be a line). There are no assessment limits for this standard. The entire standard is assessed in this course. 125 Cluster Headings Content Standards Scope & Clarifications C. Represent and solve equations and inequalities graphically. M1.A.REI.C.4 Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the approximate solutions using technology. ★ Include cases where f(x) and/or g(x) are linear, absolute value, and exponential functions. For example: f(x) = 3x + 5. i) Tasks that assess conceptual understanding of the indicated concept may involve any of the function types mentioned in the standard except exponential and logarithmic functions. ii) Finding the solutions approximately is limited to cases where f(x) and g(x) are polynomial. iii) Tasks are limited to linear and absolute value functions. M1.A.REI.C.5 Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the corresponding half-planes. There are no assessment limits for this standard. The entire standard is assessed in this course. Functions Interpreting Functions (F.IF) Cluster Headings Content Standards Scope & Clarifications A. Understand the concept of a function and use function notation. M1.F.IF.A.1 Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). There are no assessment limits for this standard. The entire standard is assessed in this course. 126 Cluster Headings Content Standards Scope & Clarifications A. Understand the concept of a function and use function notation. M1.F.IF.A.2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Interpret functions that arise in applications in terms of the context. M1.F.IF.B.3 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. ★ Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. i) Tasks have a real-world context. ii) Tasks are limited to linear functions, absolute value, and exponential functions with domains in the integers. M1.F.IF.B.4 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. ★ For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. i) Tasks have a real-world context. ii) Tasks are limited to linear functions, piecewise functions (including step functions and absolute value functions), and exponential functions with domains in the integers. M1.F.IF.B.5 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph.★ i) Tasks have a real-world context. ii) Tasks are limited to linear functions, piecewise functions (including step functions and absolute value functions), and exponential functions with domains in the integers. 127 Cluster Headings Content Standards Scope & Clarifications C. Analyze functions using different representations. M1.F.IF.C.6 Graph functions expressed symbolically and show key features of the graph, by hand and using technology. a. Graph linear and quadratic functions and show intercepts, maxima, and minima. Tasks are limited to linear functions. M1.F.IF.C.7 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). i) Tasks have a real-world context. ii) Tasks are limited to linear functions, piecewise functions (including step functions and absolute value functions), and exponential functions with domains in the integers. Building Functions (F.BF) Cluster Headings Content Standards Scope & Clarifications A. Build a function that models a relationship between two quantities. M1.F.BF.A.1 Write a function that describes a relationship between two quantities.★ a. Determine an explicit expression, a recursive process, or steps for calculation from a context. i) Tasks have a real-world context. ii) Tasks are limited to linear functions and exponential functions with domains in the integers. M1.F.BF.A.2 Write arithmetic and geometric sequences with an explicit formula and use them to model situations.★ There are no assessment limits for this standard. The entire standard is assessed in this course. 128 Linear and Exponential Models★ (F.LE) Cluster Headings Content Standards Scope & Clarifications A. Construct and compare linear and exponential models and solve problems. M1.F.LE.A.1 Distinguish between situations that can be modeled with linear functions and with exponential functions. a. Recognize that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals. b. Recognize situations in which one quantity changes at a constant rate per unit interval relative to another. c. Recognize situations in which a quantity grows or decays by a constant factor per unit interval relative to another. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.F.LE.A.2 Construct linear and exponential functions, including arithmetic and geometric sequences, given a graph, a table, a description of a relationship, or input-output pairs. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.F.LE.A.3 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly. Tasks are limited linear and exponential functions. B. Interpret expressions for functions in terms of the situation they model. M1.F.LE.B.4 Interpret the parameters in a linear or exponential function in terms of a context. For example, the total cost of an electrician who charges 35 dollars for a house call and 50 dollars per hour would be expressed as the function y = 50x + 35. If the rate were raised to 65 dollars per hour, describe how the function would change. Tasks have a real-world context. 129 Geometry Congruence (G.CO) Cluster Headings Content Standards Scope & Clarifications A. Experiment with transformations in the plane. M1.G.CO.A.1 Know precise definitions of angle, circle, perpendicular line, parallel line, and line segment, based on the undefined notions of point, line, plane, distance along a line, and distance around a circular arc. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.G.CO.A.2 Represent transformations in the plane in multiple ways, including technology. Describe transformations as functions that take points in the plane (pre-image) as inputs and give other points (image) as outputs. Compare transformations that preserve distance and angle measure to those that do not (e.g., translation versus horizontal stretch). There are no assessment limits for this standard. The entire standard is assessed in this course. M1.G.CO.A.3 Given a rectangle, parallelogram, trapezoid, or regular polygon, describe the rotations and reflections that carry the shape onto itself. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.G.CO.A.4 Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.G.CO.A.5 Given a geometric figure and a rigid motion, draw the image of the figure in multiple ways, including technology. Specify a sequence of rigid motions that will carry a given figure onto another. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Understand congruence in terms of rigid motions. M1.G.CO.B.6 Use geometric descriptions of rigid motions to transform figures and to predict the effect of a given rigid motion on a given figure; given two figures, use the definition of congruence in terms of rigid motions to determine informally if they are congruent. There are no assessment limits for this standard. The entire standard is assessed in this course. 130 Cluster Headings Content Standards Scope & Clarifications B. Understand congruence in terms of rigid motions. M1.G.CO.B.7 Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.G.CO.B.8 Explain how the criteria for triangle congruence (ASA, SAS, AAS, and SSS) follow from the definition of congruence in terms of rigid motions. There is no additional scope or clarification information for this standard. C. Prove geometric theorems. M1.G.CO.C.9 Prove theorems about lines and angles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: vertical angles are congruent; when a transversal crosses parallel lines, alternate interior angles are congruent and corresponding angles are congruent; points on a perpendicular bisector of a line segment are exactly those equidistant from the segment’s endpoints. M1.G.CO.C.10 Prove theorems about triangles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. 131 Cluster Headings Content Standards Scope & Clarifications C. Prove geometric theorems. M1.G.CO.C.11 Prove theorems about parallelograms. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards Scope & Clarifications A. Summarize, represent, and interpret data on a single count or measurement variable. M1.S.ID.A.1 Represent single or multiple data sets with dot plots, histograms, stem plots (stem and leaf), and box plots. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.S.ID.A.2 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.S.ID.A.3 Interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers). There are no assessment limits for this standard. The entire standard is assessed in this course. 132 Cluster Headings Content Standards Scope & Clarifications B. Summarize, represent, and interpret data on two categorical and quantitative variables. M1.S.ID.B.4 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. b. Fit a linear function for a scatter plot that suggests a linear association. i) Tasks have real-world context. ii) Tasks are limited to linear functions and exponential functions with domains in the integers. C. Interpret linear models. M1.S.ID.C.5 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.S.ID.C.6 Compute (using technology) and interpret the correlation coefficient of a linear fit. There are no assessment limits for this standard. The entire standard is assessed in this course. M1.S.ID.C.7 Distinguish between correlation and causation. There are no assessment limits for this standard. The entire standard is assessed in this course. Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 133 Integrated Math II | M2 Integrated Math II builds upon concepts taught in Integrated Math I with an emphasis on quadratic and polynomial expressions, equations, and functions. This course also focuses on geometric similarity and interpreting functions from a real life context. Students extend previous knowledge of exponential properties to rational exponents. This course also introduces probability of compound events and the complex number system. The major work of Integrated Math II is from the following domains and clusters:  The Real Number System o Extend the properties of exponents to rational exponents.  Seeing Structure in Expressions o Interpret the structure of expressions. o Write expressions in equivalent forms to solve problems.  Arithmetic with Polynomials and Rational Expressions o Perform arithmetic operations on polynomials.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Understand solving equations as a process of reasoning and explain the reasoning. o Solve equations and inequalities in one variable.  Interpreting Functions o Interpret functions that arise in applications in terms of the context.  Similarity, Right Triangles, and Trigonometry o Understand similarity in terms of similarity transformations. o Prove theorems involving similarity. o Define trigonometric ratios and solve problems involving triangles. Supporting work is from the following domains and clusters:  Quantities o Reason quantitatively and use units to solve problems.  The Complex Number System o Perform arithmetic operations with complex numbers. o Use complex numbers in polynomial identities and equations.  Reasoning with Equations and Inequalities o Solve systems of equations.  Interpreting Functions o Analyze functions using different representations.  Building Functions o Build a function that models a relationship between two quantities. o Build new functions from existing functions.  Geometric Measurement and Dimension o Explain volume and surface area formulas and use them to solve problems.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on two categorical and quantitative variables.  Conditional Probability and the Rules of Probability o Understand independence and conditional probability and use them to interpret data. o Use the rules of probability to compute probabilities of compound events in a uniform probability model. 134 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 135 Number and Quantity The Real Number System (N.RN) Cluster Headings Content Standards Scope & Clarifications A. Extend the properties of exponents to rational exponents. M2.N.RN.A.1 Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.N.RN.A.2 Rewrite expressions involving radicals and rational exponents using the properties of exponents. There are no assessment limits for this standard. The entire standard is assessed in this course. Quantities (N.Q) Cluster Headings Content Standards Scope & Clarifications A. Reason quantitatively and use units to solve problems. M2.N.Q.A.1 Identify, interpret, and justify appropriate quantities for the purpose of descriptive modeling. Descriptive modeling refers to understanding and interpreting graphs; identifying extraneous information; choosing appropriate units; etc. Tasks are limited to linear, quadratic, exponential equations with integer exponents, square root, and cube root functions. The Complex Number System (N.CN) Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations with complex numbers. M2.N.CN.A.1 Know there is a complex number i such that i2 = –1, and every complex number has the form a + bi with a and b real. There are no assessment limits for this standard. The entire standard is assessed in this course. 136 Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations with complex numbers. M2.N.CN.A.2 Know and use the relation i2 = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use complex numbers in polynomial identities and equations. M2.N.CN.B.3 Solve quadratic equations with real coefficients that have complex solutions. There are no assessment limits for this standard. The entire standard is assessed in this course. Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. M2.A.SSE.A.1 Interpret expressions that represent a quantity in terms of its context.★ a. Interpret complicated expressions by viewing one or more of their parts as a single entity. For example, interpret P(1 + r)n as the product of P and a factor not depending on P. Tasks are limited to quadratic expressions. M2.A.SSE.A.2 Use the structure of an expression to identify ways to rewrite it. For example, recognize 532 - 472 as a difference of squares and see an opportunity to rewrite it in the easier-to-evaluate form (53 + 47) (53 – 47). See an opportunity to rewrite a2 + 9a + 14 as (a + 7) (a + 2). Tasks are limited to numerical expressions and polynomial expressions in one variable. 137 B. Write expressions in equivalent forms to solve problems. M2.A.SSE.B.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression.★ a. Factor a quadratic expression to reveal the zeros of the function it defines. b. Complete the square in a quadratic expression in the form Ax2 + Bx + C where A = 1 to reveal the maximum or minimum value of the function it defines. There are no assessment limits for this standard. The entire standard is assessed in this course. Arithmetic with Polynomials and Rational Expressions (A.APR) Cluster Headings Content Standards Scope & Clarifications A. Perform arithmetic operations on polynomials. M2.A.APR.A.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. There are no assessment limits for this standard. The entire standard is assessed in this course. Creating Equations★ (A–CED) Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships. M2.A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. Include equations arising from linear and quadratic functions and rational and exponential functions. Tasks have a real-world context. M2.A.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations with two variables on coordinate axes with labels and scales. i) Tasks are limited to quadratic, square root, cube root, and piecewise functions. ii) Tasks have a real-world context. iii) Tasks have the hallmarks of modeling as a mathematical practice (less defined tasks, more of the modeling cycle, etc.). 138 Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships. M2.A.CED.A.3 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. i) Tasks are limited to quadratic, square root, cube root, and piecewise functions. ii) Tasks have a real-world context. Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards Scope & Clarifications A. Understand solving equations as a process of reasoning and explain the reasoning. M2.A.REI.A.1 Explain each step in solving an equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. Tasks are limited to linear, quadratic, exponential equations with integer exponents, square root, cube root, piecewise, and exponential functions. B. Solve equations and inequalities in one variable. M2.A.REI.B.2 Solve quadratic equations and inequalities in one variable. a. Use the method of completing the square to rewrite any quadratic equation in x into an equation of the form (x – p)2 = q that has the same solutions. Derive the quadratic formula from this form. b. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. There are no assessment limits for this standard. The entire standard is assessed in this course. C. Solve systems of equations. M2.A.REI.C.3 Write and solve a system of linear equations in context. When solving algebraically, tasks are limited to systems of at most three equations and three variables. With graphic solutions systems are limited to only two variables. 139 Cluster Headings Content Standards Scope & Clarifications C. Solve systems of equations. M2.A.REI.C.4 Solve a system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically. There are no assessment limits for this standard. The entire standard is assessed in this course. Functions Interpreting Functions (F.IF) Cluster Headings Content Standards Scope & Clarifications A. Interpret functions that arise in applications in terms of the context. M2.F.IF.A.1 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities and sketch graphs showing key features given a verbal description of the relationship. ★ Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. i) Tasks have a real-world context. ii) Tasks are limited to quadratic, exponential functions with integer exponents, square root, and cube root functions. M2.F.IF.A.2 Relate the domain of a function to its graph and, where applicable, to the quantitative relationship it describes. ★ For example, if the function h(n) gives the number of person-hours it takes to assemble n engines in a factory, then the positive integers would be an appropriate domain for the function. Tasks are limited to quadratic, square root, cube root, piecewise, and exponential functions. M2.F.IF.A.3 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. ★ i) Tasks have a real-world context. ii) Tasks may involve quadratic, square root, cube root, piecewise, and exponential functions. 140 Cluster Headings Content Standards Scope & Clarifications B. Analyze functions using different representation. M2.F.IF.B.4 Graph functions expressed symbolically and show key features of the graph, by hand and using technology.★ a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. c. Graph exponential and logarithmic functions, showing intercepts and end behavior. M2.F.IF.B.4a – Tasks are limited to quadratic functions. M2.F.IF.B.4c – Tasks are limited to exponential functions. M2.F.IF.B.5 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. a. Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. b. Know and use the properties of exponents to interpret expressions for exponential functions. For example, identify percent rate of change in functions such as y = 2x, y = (1/2)x, y = 2-x, y = (1/2)-x. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.F.IF.B.6 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). i) Tasks do not have a real-world context. ii) Tasks may involve quadratic, square root, cube root, piecewise, and exponential functions. 141 Building Functions (F.BF) Cluster Headings Content Standards Scope & Clarifications A. Build a function that models a relationship between two quantities. M2.F.BF.A.1 Write a function that describes a relationship between two quantities.★ a. Determine an explicit expression, a recursive process, or steps for calculation from a context. b. Combine standard function types using arithmetic operations. For M2.F.BF.A.1a: i) Tasks have a real-world context. ii) Tasks may involve linear and quadratic functions. B. Build new functions from existing functions. M2.F.BF.B.2 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. i) Identifying the effect on the graph of replacing f(x) by f(x) + k, k f(x), and f(x+k) for specific values of k (both positive and negative) is limited to linear, quadratic, and absolute value functions. ii) Experimenting with cases and illustrating an explanation of the effects on the graph using technology is limited to linear, quadratic, square root, cube root, and exponential functions. iii) Tasks do not involve recognizing even and odd functions. Geometry Similarity, Right Triangles, and Trigonometry (G.SRT) Cluster Headings Content Standards Scope & Clarifications A. Understand similarity in terms of similarity transformations. M2.G.SRT.A.1 Verify informally the properties of dilations given by a center and a scale factor. There are no assessment limits for this standard. The entire standard is assessed in this course. 142 Cluster Headings Content Standards Scope & Clarifications A. Understand similarity in terms of similarity transformations. M2.G.SRT.A.2 Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar; explain using similarity transformations the meaning of similarity for triangles as the equality of all corresponding pairs of angles and the proportionality of all corresponding pairs of sides. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.G.SRT.A.3 Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Prove theorems involving similarity. M2.G.SRT.B.4 Prove theorems about similar triangles. Proving includes, but is not limited to, completing partial proofs; constructing two-column or paragraph proofs; using transformations to prove theorems; analyzing proofs; and critiquing completed proofs. Theorems include but are not limited to: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity. M2.G.SRT.B.5 Use congruence and similarity criteria for triangles to solve problems and to justify relationships in geometric figures. There are no assessment limits for this standard. The entire standard is assessed in this course. C. Define trigonometric ratios and solve problems involving triangles. M2.G.SRT.C.6 Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. There are no assessment limits for this standard. M2.G.SRT.C.7 Explain and use the relationship between the sine and cosine of complementary angles. There are no assessment limits for this standard. The entire standard is assessed in this course. 143 C. Define trigonometric ratios and solve problems involving triangles. M2.G.SRT.C.8 Solve triangles. ★ a. Know and use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. b. Know and use the Law of Sines and the Law of Cosines to solve triangles in applied problems. Recognize when it is appropriate to use each. Ambiguous cases will not be included in assessment. Geometric Measurement and Dimension (G.GMD) Cluster Headings Content Standards Scope & Clarifications A. Explain volume and surface area formulas and use them to solve problems. M2.G.GMD.A.1 Give an informal argument for the formulas for the circumference of a circle and the volume and surface area of a cylinder, cone, prism, and pyramid. Informal arguments may include but are not limited to using the dissection argument, applying Cavalieri’s principle, and constructing informal limit arguments. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.G.GMD.A.2 Know and use volume and surface area formulas for cylinders, cones, prisms, pyramids, and spheres to solve problems.★ There are no assessment limits for this standard. The entire standard is assessed in this course. Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards Scope & Clarifications A. Summarize, represent, and interpret data on two categorical and quantitative variables. M2.S.ID.A.1 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. Exponential functions are limited to those with domains in the integers. Tasks have a real-world context. 144 Conditional Probability and the Rules of Probability (S.CP) Cluster Headings Content Standards Scope & Clarifications A. Understand independence and conditional probability and use them to interpret data. M2.S.CP.A.1 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). There are no assessment limits for this standard. The entire standard is assessed in this course. M2.S.CP.A.2 Understand that two events A and B are independent if the probability of A and B occurring together is the product of their probabilities, and use this characterization to determine if they are independent. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.S.CP.A.3 Know and understand the conditional probability of A given B as P(A and B)/P(B), and interpret independence of A and B as saying that the conditional probability of A given B is the same as the probability of A, and the conditional probability of B given A is the same as the probability of B. There are no assessment limits for this standard. The entire standard is assessed in this course. M2.S.CP.A.4 Recognize and explain the concepts of conditional probability and independence in everyday language and everyday situations. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. M2.S.CP.B.5 Find the conditional probability of A given B as the fraction of B’s outcomes that also belong to A and interpret the answer in terms of the model. For example, a teacher gave two exams. 75 percent passed the first exam and 25 percent passed both. What percent who passed the first exam also passed the second exam? There are no assessment limits for this standard. The entire standard is assessed in this course. 145 Cluster Headings Content Standards Scope & Clarifications B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. M2.S.CP.B.6 Know and apply the Addition Rule, P(A or B) = P(A) + P(B) – P(A and B), and interpret the answer in terms of the model. For example, in a math class of 32 students, 14 are boys and 18 are girls. On a unit test 6 boys and 5 girls made an A. If a student is chosen at random from a class, what is the probability of choosing a girl or an A student? There are no assessment limits for this standard. The entire standard is assessed in this course. Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 146 Integrated Math III | M3 Integrated Math III builds upon concepts taught in Integrated Math I and Integrated Math II and emphasizes polynomial and rational expressions, equations, and functions. This course has a focus on geometric modeling and using algebra to prove geometric theorems. This course also introduces students to circles, basic trigonometric functions, and foundational statistics skills such as interpretation of data and making statistical inferences. The major work of Integrated Math III is from the following domains and clusters:  Seeing Structure in Expressions o Interpret the structure of expressions. o Write expressions in equivalent forms to solve problems.  Arithmetic with Polynomials and Rational Expressions o Understand the relationship between zeros and factors of polynomials.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Understand solving equations as a process of reasoning and explain the reasoning. o Represent and solve equations graphically.  Interpreting Functions o Interpret functions that arise in applications in terms of the context.  Expressing Geometric Properties with Equations o Use coordinates to prove simple geometric theorems algebraically.  Modeling with Geometry o Apply geometric concepts in modeling situations.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on a single count or measurement variable. o Summarize, represent, and interpret data on two categorical and quantitative variables.  Making Inferences and Justifying Conclusions o Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Supporting work is from the following domains and clusters:  Quantities o Reason quantitatively and use units to solve problems.  Arithmetic with Polynomials and Rational Expressions o Use polynomial identities to solve problems. o Rewrite rational expressions.  Interpreting Functions o Analyze functions using different representations.  Building Functions o Build new functions from existing functions.  Linear, Quadratic, and Exponential Models o Construct and compare linear, quadratic, and exponential models and solve problems.  Trigonometric Functions o Extend the domain of trigonometric functions using the unit circle. o Prove and apply trigonometric identities.  Congruence o Make geometric constructions. 147  Circles o Understand and apply theorems about circles. o Find areas of sectors of circles.  Expressing Geometric Properties with Equations o Translate between the geometric description and the equation for a circle.  Making Inferences and Justifying Conclusions o Understand and evaluate random processes underlying statistical experiments. 148 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 149 Number and Quantity Quantities (N.Q) Cluster Headings Content Standards Scope & Clarifications A. Reason quantitatively and use units to solve problems. M3.N.Q.A.1 Identify, interpret, and justify appropriate quantities for the purpose of descriptive modeling. Descriptive modeling refers to understanding and interpreting graphs; identifying extraneous information; choosing appropriate units; etc. There are no assessment limits for this standard. The entire standard is assessed in this course. Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards Scope & Clarifications A. Interpret the structure of expressions. M3.A.SSE.A.1 Use the structure of an expression to identify ways to rewrite it. For example, see 2x4 + 3x2 – 5 as its factors (x2 – 1) and (2x2 + 5); see x4 – y4 as (x2) 2 – (y2) 2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 + y2); see (x2 + 4)/(x2 + 3) as ((x2+ 3) + 1)/(x2 + 3), thus recognizing an opportunity to write it as 1 + 1/(x2 + 3). Tasks are limited to polynomial, rational, or exponential expressions. 150 Cluster Headings Content Standards Scope & Clarifications B. Write expressions in equivalent forms to solve problems. M3.A.SSE.B.2 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. a. Use the properties of exponents to rewrite expressions for exponential functions. For example, the expression 1.15t can be rewritten as ((1.15)1/12)12t ≈ 1.01212t to reveal that the approximate equivalent monthly interest rate is 1.2% if the annual rate is 15%. i) Tasks have a real-world context. As described in the standard, there is an interplay between the mathematical structure of the expression and the structure of the situation such that choosing and producing an equivalent form of the expression reveals something about the situation. ii) Tasks are limited to exponential expressions with rational or real exponents. M3.A.SSE.B.3 Recognize a finite geometric series (when the common ratio is not 1), and know and use the sum formula to solve problems in context. There are no assessment limits for this standard. The entire standard is assessed in this course. Arithmetic with Polynomials and Rational Expressions (A.APR) Cluster Headings Content Standards Scope & Clarifications A. Understand the relationship between zeros and factors of polynomials. M3.A.APR.A.1 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). There are no assessment limits for this standard. The entire standard is assessed in this course. M3.A.APR.A.2 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Tasks include quadratic, cubic, and quartic polynomials and polynomials for which factors are not provided. For example, find the zeros of (x2-1)(x2+1). B. Use polynomial identities to solve problems. M3.A.APR.B.3 Know and use polynomial identities to describe numerical relationships. For example, compare (31)(29)=(30+1)(30-1)=302-12 with (x+y)(x-y)=x2-y2. 151 Cluster Headings Content Standards Scope & Clarifications C. Rewrite rational expressions. M3.A.APR.C.4 Rewrite rational expressions in different forms. There are no assessment limits for this standard. The entire standard is assessed in this course. Creating Equations★ (A.CED) Cluster Headings Content Standards Scope & Clarifications A. Create equations that describe numbers or relationships. M3.A.CED.A.1 Create equations and inequalities in one variable and use them to solve problems. i) Tasks are limited to polynomial, rational, absolute value, exponential, or logarithmic functions. ii) Tasks have a real-world context. M3.A.CED.A.2 Create equations in two or more variables to represent relationships between quantities; graph equations with two variables on coordinate axes with labels and scales. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.A.CED.A.3 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. i) Tasks have a real-world context. ii) Tasks are limited to polynomial, rational, absolute value, exponential, or logarithmic functions. Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards Scope & Clarifications A. Understand solving equations as a process of reasoning and explain the reasoning. M3.A.REI.A.1 Explain each step in solving an equation as following from the equality of numbers asserted at the previous step, starting from the assumption that the original equation has a solution. Construct a viable argument to justify a solution method. Tasks are limited to simple rational or radical equations. M3.A.REI.A.2 Solve rational and radical equations in one variable, and identify extraneous solutions when they exist. There are no assessment limits for this standard. The entire standard is assessed in this course. 152 Cluster Headings Content Standards Scope & Clarifications B. Represent and solve equations graphically. M3.A.REI.B.3 Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the approximate solutions using technology. ★ Tasks may include cases where f(x) and/or g(x) are linear, polynomial, rational, absolute value, exponential, or logarithmic functions. Functions Interpreting Functions (F.IF) Cluster Headings Content Standards Scope & Clarifications A. Interpret functions that arise in applications in terms of the context. M3.F.IF.A.1 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. ★ Key features include: intercepts; intervals where the function is increasing, decreasing, positive, or negative; relative maximums and minimums; symmetries; and end behavior. i) Tasks have a real-world context. ii) Tasks may involve polynomial, exponential, and logarithmic functions. M3.F.IF.A.2 Calculate and interpret the average rate of change of a function (presented symbolically or as a table) over a specified interval. Estimate the rate of change from a graph. ★ i) Tasks have a real-world context. ii) Tasks may involve polynomial, exponential, and logarithmic functions. 153 Cluster Headings Content Standards Scope & Clarifications B. Analyze functions using different representations. M3.F.IF.B.3 Graph functions expressed symbolically and show key features of the graph, by hand and using technology.★ a. Graph linear and quadratic functions and show intercepts, maxima, and minima. b. Graph square root, cube root, and piecewise-defined functions, including step functions and absolute value functions. c. Graph polynomial functions, identifying zeros when suitable factorizations are available and showing end behavior. d. Graph exponential and logarithmic functions, showing intercepts and end behavior. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.F.IF.B.4 Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). Tasks may involve polynomial, exponential, and logarithmic functions. Building Functions (F.BF) Cluster Headings Content Standards Scope & Clarifications A. Build new functions from existing functions. M3.F.BF.A.1 Identify the effect on the graph of replacing f(x) by f(x) + k, k f(x), f(kx), and f(x + k) for specific values of k (both positive and negative); find the value of k given the graphs. Experiment with cases and illustrate an explanation of the effects on the graph using technology. i) Tasks may involve polynomial, exponential, and logarithmic functions. ii) Tasks may involve recognizing even and odd functions. M3.F.BF.A.2 Find inverse functions. a. Find the inverse of a function when the given function is one-to-one. There are no assessment limits for this standard. The entire standard is assessed in this course. 154 Linear, Quadratic, and Exponential Models★ (F.LE) Cluster Headings Content Standards Scope & Clarifications A. Construct and compare linear, quadratic, and exponential models and solve problems. M3.F.LE.A.1 Observe using graphs and tables that a quantity increasing exponentially eventually exceeds a quantity increasing linearly, quadratically, or (more generally) as a polynomial function. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.F.LE.A.2 For exponential models, express as a logarithm the solution to abct = d where a, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology. There are no assessment limits for this standard. The entire standard is assessed in this course. Trigonometric Functions (F.TF) Cluster Headings Content Standards Scope & Clarifications A. Extend the domain of trigonometric functions using the unit circle. M3.F.TF.A.1 Understand and use radian measure of an angle. a. Understand radian measure of an angle as the length of the arc on the unit circle subtended by the angle. b. Use the unit circle to find sin θ, cos θ, and tan θ when θ is a commonly recognized angle between 0 and 2π. Commonly recognized angles include all multiples of nπ /6 and nπ /4, where n is an integer. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.F.TF.A.2 Explain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Prove and apply trigonometric identities. M3.F.TF.B.3 Use trigonometric identities to find values of trig functions. a. Given a point on a circle centered at the origin, recognize and use the right triangle ratio definitions of sin θ, cos θ, and tan θ to evaluate the trigonometric functions. b. Given the quadrant of the angle, use the identity sin2 θ + cos2 θ = 1 to find sin θ given cos θ, or vice versa. Commonly recognized angles include all multiples of nπ /6 and nπ /4, where n is an integer. There are no assessment limits for this standard. The entire standard is assessed in this course. 155 Geometry Congruence (G.CO) Cluster Headings Content Standards Scope & Clarifications A. Make geometric constructions. M3.G.CO.A.1 Make formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.). Constructions include but are not limited to: copying a segment; copying an angle; bisecting a segment; bisecting an angle; constructing perpendicular lines, including the perpendicular bisector of a line segment; constructing a line parallel to a given line through a point not on the line, and constructing the following objects inscribed in a circle: an equilateral triangle, square, and a regular hexagon. Circles (G.C) Cluster Headings Content Standards Scope & Clarifications A. Understand and apply theorems about circles. M3.G.C.A.1 Recognize that all circles are similar. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.G.C.A.2 Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle, and properties of angles for a quadrilateral inscribed in a circle. M3.G.C.A.3 Construct the incenter and circumcenter of a triangle and use their properties to solve problems in context. There are no assessment limits for this standard. The entire standard is assessed in this course. 156 Cluster Headings Content Standards Scope & Clarifications B. Find areas of sectors of circles. M3.G.C.B.4 Find the area of a sector of a circle in a real-world context. For example, use proportional relationships and angles measured in degrees or radians. There are no assessment limits for this standard. The entire standard is assessed in this course. Expressing Geometric Properties with Equations (G.GPE) Cluster Headings Content Standards Scope & Clarifications A. Translate between the geometric description and the equation for a circle. M3.G.GPE.A.1 Know and write the equation of a circle of given center and radius using the Pythagorean Theorem. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Use coordinates to prove simple geometric theorems algebraically. M3.G.GPE.B.2 Use coordinates to prove simple geometric theorems algebraically. For example, prove or disprove that a figure defined by four given points in the coordinate plane is a rectangle; prove or disprove that the point (1, 3 ) lies on the circle centered at the origin and containing the point (0, 2). There are no assessment limits for this standard. The entire standard is assessed in this course. M3.G.GPE.B.3 Prove the slope criteria for parallel and perpendicular lines and use them to solve geometric problems. For example, find the equation of a line parallel or perpendicular to a given line that passes through a given point. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.G.GPE.B.4 Find the point on a directed line segment between two given points that partitions the segment in a given ratio. There are no assessment limits for this standard. The entire standard is assessed in this course. 157 Cluster Headings Content Standards Scope & Clarifications B. Use coordinates to prove simple geometric theorems algebraically. M3.G.GPE.B.5 Know and use coordinates to compute perimeters of polygons and areas of triangles and rectangles.★ For example, use the distance formula. There are no assessment limits for this standard. The entire standard is assessed in this course. Modeling with Geometry (G.MG) Cluster Headings Content Standards Scope & Clarifications A. Apply geometric concepts in modeling situations. M3.G.MG.A.1 Use geometric shapes, their measures, and their properties to describe objects.★ There are no assessment limits for this standard. The entire standard is assessed in this course. For example, modeling a tree trunk or a human torso as a cylinder. M3.G.MG.A.2 Apply geometric methods to solve real-world problems.★ Geometric methods may include but are not limited to using geometric shapes, the probability of a shaded region, density, and design problems. There are no assessment limits for this standard. The entire standard is assessed in this course. 158 Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards Scope & Clarifications A. Summarize, represent, and interpret data on a single count or measurement variable. M3.S.ID.A.1 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages using the Empirical Rule. There are no assessment limits for this standard. The entire standard is assessed in this course. B. Summarize, represent, and interpret data on two categorical and quantitative variables. M3.S.ID.B.2 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. b. Fit a linear function for a scatter plot that suggests a linear association. Use given functions or choose a function suggested by the context. i) Tasks have a real-world context. ii) Tasks are limited to linear, quadratic, and exponential functions with domains not in the integers. Making Inferences and Justifying Conclusions (S.IC) Cluster Headings Content Standards Scope & Clarifications A. Understand and evaluate random processes underlying statistical experiments. M3.S.IC.A.1 Understand statistics as a process for making inferences about population parameters based on a random sample from that population. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.S.IC.A.2 Decide if a specified model is consistent with results from a given data-generating process (e.g., using simulation). For example, a model says a spinning coin falls heads up with probability 0.5. Would a result of 10 heads in a row cause you to question the model? There are no assessment limits for this standard. The entire standard is assessed in this course. 159 Cluster Headings Content Standards Scope & Clarifications B. Make inferences and justify conclusions from sample surveys, experiments, and observational studies. M3.S.IC.B.3 Recognize the purposes of and differences among sample surveys, experiments, and observational studies; explain how randomization relates to each. For example, in a given situation, is it more appropriate to use a sample survey, an experiment, or an observational study? Explain how randomization affects the bias in a study. There are no assessment limits for this standard. The entire standard is assessed in this course. M3.S.IC.B.4 Use data from a sample survey to estimate a population mean or proportion; use a given margin of error to solve a problem in context. There are no assessment limits for this standard. The entire standard is assessed in this course. Major content of the course is indicated by the light green shading of the cluster heading and standard’s coding. Major Content Supporting Content 160 Fourth-Year Courses Proposed Standards 161 Bridge Math | B Bridge Math is a course intended to build upon concepts taught in previous courses to allow students to gain a deeper knowledge of the real and complex number systems as well as the structure, use, and application of equations, expressions, and functions. Functions emphasized include linear, quadratic and polynomial. Students continue mastery of geometric concepts such as similarity, congruence, right triangles, and circles. Students use categorical and quantitative data to model real life situations and rules of probability to compute probabilities of compound events. Bridge Math includes the following domains and clusters:  The Real Number System o Use properties of rational and irrational numbers.  Quantities o Reason quantitatively and use units to solve problems.  The Complex Number System o Perform arithmetic operations with complex numbers.  Seeing Structure in Expressions o Write expressions in equivalent forms to solve problems.  Arithmetic with Polynomials and Rational Expressions o Perform arithmetic operations on polynomials. o Understand the relationship between zeros and factors of polynomials.  Creating Equations o Create equations that describe numbers or relationships.  Reasoning with Equations and Inequalities o Understand solving equations as a process of reasoning and explain the reasoning. o Solve equations and inequalities in one variable. o Solve systems of equations. o Represent and solve equations and inequalities graphically.  Interpreting Functions o Understand the concept of a function and use function notation. o Interpret functions that arise in applications in terms of the context. o Analyze functions using different representations  Similarity, Right Triangles, and Trigonometry o Understand similarity in terms of similarity transformations. o Define trigonometric ratios and solve problems involving right triangles.  Circles o Find arc lengths and areas of sectors of circles.  Geometric Measurement and Dimension o Visualize relationships between two-dimensional and three-dimensional objects.  Modeling with Geometry o Apply geometric concepts in modeling situations.  Interpreting Categorical and Quantitative Data o Summarize, represent, and interpret data on a single count or measurement variable. o Summarize, represent, and interpret data on two categorical and quantitative variables. o Interpret linear models.  Conditional probability and the Rules of Probability o Use the rules of probability to compute probabilities of compound events in a uniform probability model. 162 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 163 Number and Quantity The Real Number System (N.RN) Cluster Headings Content Standards A. Use properties of rational and irrational numbers. B.N.RN.A.1. Use rational and irrational numbers in calculations and in real-world context. Quantities★ (N.Q) Cluster Headings Content Standards A. Reason quantitatively and use units to solve problems. B.N.Q.A.1 Use units as a way to understand problems and to guide the solution of multi-step problems; choose and interpret units consistently in formulas; choose and interpret the scale and the origin in graphs and data displays. B.N.Q.A.2 Define appropriate quantities for the purpose of descriptive modeling. B.N.Q.A.3 Solve problems involving squares, square roots of numbers, cubes, and cube roots of numbers. The Complex Number System (N.CN) Cluster Headings Content Standards A. Perform arithmetic operations with complex numbers. B.N.CN.A.1 Know there is a complex number i such that i2 = -1, and every complex number has the form a + bi with a and b real. B.N.CN.A.2 Know and use the relation i2 = -1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 164 Algebra Seeing Structure in Expressions (A.SSE) Cluster Headings Content Standards A. Write expressions in equivalent forms to solve problems. B.A.SSE.A.1 Use properties of multiplication and division to solve problems containing scientific notation. B.A.SSE.A.2 Use algebraic structures to solve problems involving proportional reasoning in real-world context. Arithmetic with Polynomials and Rational Expressions (A.APR) Cluster Headings Content Standards A. Perform arithmetic operations on polynomials. B.A.APR.A.1 Understand that polynomials form a system analogous to the integers, namely, they are closed under the operations of addition, subtraction, and multiplication; add, subtract, and multiply polynomials. B. Understand the relationship between zeros and factors of polynomials. B.A.APR.B.2 Identify zeros of polynomials when suitable factorizations are available, and use the zeros to construct a rough graph of the function defined by the polynomial. Creating Equations★ (A.CED) Cluster Headings Content Standards A. Create equations that describe numbers or relationships. B.A.CED.A.1 Create equations and inequalities in one variable and use them to solve real-world problems. B.A.CED.A.2 Create equations in two or more variables to represent relationships between quantities. B.A.CED.A.3 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. 165 Reasoning with Equations and Inequalities (A.REI) Cluster Headings Content Standards A. Understand solving equations as a process of reasoning and explain the reasoning. B.A.REI.A.1 Build functions and write expressions, equations, and inequalities for common algebra settings leading to a solution in context (e.g., rate and distance problems and problems that can be solved using proportions). B. Solve equations and inequalities in one variable. B.A.REI.B.2 Solve quadratic equations in one variable. Solve quadratic equations by inspection (e.g., for x2 = 49), taking square roots, completing the square, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b. C. Solve systems of equations. B.A.REI.C.3 Solve and explain the solutions to a system of equations using a variety of representations including combinations of linear and non-linear equations. D. Represent and solve equations and inequalities graphically. B.A.REI.D.4 Use algebra and geometry to solve problems involving midpoints and distances. B.A.REI.D.5 Solve a linear inequality using multiple methods and interpret the solution as it applies to the context. Functions Interpreting Functions (F.IF) Cluster Headings Content Standards A. Understand the concept of a function and use function notation. B.F.IF.A.1 Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x). B.F.IF.A.2 Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context. 166 Cluster Headings Content Standards B. Interpret functions that arise in applications in terms of the context. B.F.IF.B.3 Recognize functions as mappings of an independent variable into a dependent variable. ★ C. Analyze functions using different representations. B.F.IF.C.4 Graph linear, quadratic, absolute value, and piecewise functions expressed symbolically and show key features of the graph, by hand in simple cases and using technology for more complicated ones. ★ B.F.IF.C.5 Write a function defined by an expression in different but equivalent forms to reveal and explain different properties of the function. B.F.IF.C.6 Use the properties of exponents to interpret expressions for exponential functions. Geometry Similarity, Right Triangles and Trigonometry (G.SRT) Cluster Headings Content Standards A. Understand similarity in terms of similarity transformations. B.G.SRT.A.1 Apply similar triangles to solve problems, such as finding heights and distances. B. Define trigonometric ratios and solve problems involving right triangles. B.G.SRT.B.2 Apply basic trigonometric ratios to solve right triangle problems. B.G.SRT.B.3 Apply properties of 30° 60° 90°, 45° 45° 90°, similar, and congruent triangles. B.G.SRT.B.4 Solve problems involving angles of elevation and angles of depression. Circles (G.C) Cluster Headings Content Standards A. Find arc lengths and areas of sectors of circles. B.G.C.A.1 Apply a variety of strategies to determine the area and circumference of circles after identifying necessary information. 167 Geometric Measurement and Dimension (G.GMD) Cluster Headings Content Standards A. Visualize relationships between two-dimensional and three-dimensional objects. B.G.GMD.A.1 Use relationships involving area, perimeter, and volume of geometric figures to compute another measure. B.G.GMD.A.2 Use several angle properties to find an unknown angle measure. B.G.GMD.A.3 Apply a variety of strategies using relationships between perimeter, area, and volume to calculate desired measures in composite figures (i.e., combinations of basic figures). Modeling with Geometry (G.MG) Cluster Headings Content Standards A. Apply geometric concepts in modeling situations. B.G.MG.A.1 Use appropriate technology to find the mathematical model for a set of non-linear data. B.G.MG.A.2 Solve problems involving surface area and volume in real-world context. Statistics and Probability Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards A. Summarize, represent, and interpret data on a single count or measurement variable. B.S.ID.A.1 Use statistics appropriate to the shape of the data distribution to compare center (median, mean) and spread (interquartile range, standard deviation) of two or more different data sets. B. Summarize, represent, and interpret data on two categorical and quantitative variables. B.S.ID.B.2 Interpret and use data from tables, charts, and graphs. 168 Cluster Headings Content Standards B. Summarize, represent, and interpret data on two categorical and quantitative variables. B.S.ID.B.3 Represent data on two quantitative variables on a scatter plot, and describe how the variables are related. a. Fit a function to the data; use functions fitted to data to solve problems in the context of the data. Use given functions or choose a function suggested by the context. Emphasize linear, quadratic, and exponential models. C. Interpret linear models. B.S.ID.C.4 Interpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data. Conditional Probability and the Rules of Probability (S.CP) Cluster Headings Content Standards A. Use the rules of probability to compute probabilities of compound events in a uniform probability model. B.S.CP.A.1 Understand and use basic counting techniques in contextual settings. B.S.CP.A.2 Compute a probability when the event and/or sample space are not given or obvious. B.S.CP.A.3 Recognize the concepts of conditional and joint probability expressed in real-world contexts. B.S.CP.A.4 Recognize the concept of independence expressed in real-world contexts. 169 Precalculus | P Precalculus is designed to prepare students for college level STEM focused courses. Students extend their knowledge of the complex number system to use complex numbers in polynomial identities and equations. Topics for student mastery include vectors and matrix quantities, sequences and series, parametric equations, and conic sections. Students use previous knowledge to continue progressing in their understanding of trigonometric functions and using regression equations to model quantitative data. Precalculus includes the following domains and clusters:  Number Expressions o Represent, interpret, compare, and simplify number expressions.  The Complex Number System o Perform complex number arithmetic and understand the representation on the complex plane. o Use complex numbers in polynomial identities and equations.  Vectors and Matrix Quantities o Represent and model with vector quantities. o Understand the graphic representation of vectors and vector arithmetic. o Perform operations on matrices and use matrices in applications.  Sequences and Series o Understand and use sequences and series.  Reasoning with Equations and Inequalities o Solve systems of equations and nonlinear inequalities.  Parametric Equations o Describe and use parametric equations.  Conic Sections o Understand the properties of conic sections and apply them to model real-world phenomena.  Building Functions o Build new functions from existing functions.  Interpreting Functions o Analyze functions using different representations.  Trigonometric Functions o Extend the domain of trigonometric functions using the unit circle.  Graphing Trigonometric Functions o Model periodic phenomena with trigonometric functions.  Applied Trigonometry o Use trigonometry to solve problems.  Trigonometric Identities o Apply trigonometric identities to rewrite expressions and solve equations.  Polar Coordinates o Use polar coordinates.  Model with Data o Model data using regression equations. 170 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 171 Number and Quantity Number Expressions (N.NE) Cluster Headings Content Standards A. Represent, interpret, compare, and simplify number expressions. P.N.NE.A.1 Use the laws of exponents and logarithms to expand or collect terms in expressions; simplify expressions or modify them in order to analyze them or compare them. P.N.NE.A.2 Understand the inverse relationship between exponents and logarithms and use this relationship to solve problems involving logarithms and exponents.★ P.N.NE.A.3 Classify real numbers and order real numbers that include transcendental expressions, including roots and fractions of π and e. P.N.NE.A.4 Simplify complex radical and rational expressions; discuss and display understanding that rational numbers are dense in the real numbers and the integers are not. P.N.NE.A.5 Understand that rational expressions form a system analogous to the rational numbers, closed under addition, subtraction, multiplication, and division by a nonzero rational expression; add, subtract, multiply, and divide rational expressions. The Complex Number System (N.CN) Cluster Headings Content Standards A. Perform complex number arithmetic and understand the representation on the complex plane. P.N.CN.A.1 Perform arithmetic operations with complex numbers expressing answers in the form a + bi. P.N.CN.A.2 Find the conjugate of a complex number; use conjugates to find moduli and quotients of complex numbers. P.N.CN.A.3 Represent complex numbers on the complex plane in rectangular and polar form (including real and imaginary numbers), and explain why the rectangular and polar forms of a given complex number represent the same number. P.N.CN.A.4 Represent addition, subtraction, multiplication, and conjugation of complex numbers geometrically on the complex plane; use properties of this representation for computation. For example, (–1 +3 i )3 = 8 because (–1 +3 i ) has modulus 2 and argument 120°. 172 A. Perform complex number arithmetic and understand the representation on the complex plane. P.N.CN.A.5 Calculate the distance between numbers in the complex plane as the modulus of the difference, and the midpoint of a segment as the average of the numbers at its endpoints. B. Use complex numbers in polynomial identities and equations. P.N.CN.B.6 Extend polynomial identities to the complex numbers. For example, rewrite x2 + 4 as (x + 2i)(x – 2i). P.N.CN.B.7 Know the Fundamental Theorem of Algebra; show that it is true for quadratic polynomials. Vector and Matrix Quantities (N.VM) Cluster Headings Content Standards A. Represent and model with vector quantities. P.N.VM.A.1 Recognize vector quantities as having both magnitude and direction. Represent vector quantities by directed line segments, and use appropriate symbols for vectors and their magnitudes (e.g., v, |v|, ||v||, v). P.N.VM.A.2 Find the components of a vector by subtracting the coordinates of an initial point from the coordinates of a terminal point. P.N.VM.A.3 Solve problems involving velocity and other quantities that can be represented by vectors. B. Understand the graphic representation of vectors and vector arithmetic. P.N.VM.B.4 Add and subtract vectors. a. Add vectors end-to-end, component-wise, and by the parallelogram rule. Understand that the magnitude of a sum of two vectors is typically not the sum of the magnitudes. b. Given two vectors in magnitude and direction form, determine the magnitude and direction of their sum. c. Understand vector subtraction v – w as v + (–w), where –w is the additive inverse of w, with the same magnitude as w and pointing in the opposite direction. Represent vector subtraction graphically by connecting the tips in the appropriate order, and perform vector subtraction component-wise. P.N.VM.B.5 Multiply a vector by a scalar. a. Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(vx, vy) = (cvx, cvy). b. Compute the magnitude of a scalar multiple cv using ||cv|| = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c > 0) or against v (for c < 0). Cluster Headings Content Standards B. Understand the graphic representation of vectors and vector arithmetic. P.N.VM.B.6 Calculate and interpret the dot product of two vectors. C. Perform operations on matrices and use matrices in applications. P.N.VM.C.7 Use matrices to represent and manipulate data, e.g., to represent payoffs or incidence relationships in a network. P.N.VM.C.8 Multiply matrices by scalars to produce new matrices, e.g., as when all of the payoffs in a game are doubled. P.N.VM.C.9 Add, subtract, and multiply matrices of appropriate dimensions. P.N.VM.C.10 Understand that, unlike multiplication of numbers, matrix multiplication for square matrices is not a commutative operation, but still satisfies the associative and distributive properties. P.N.VM.C.11 Understand that the zero and identity matrices play a role in matrix addition and multiplication similar to the role of 0 and 1 in the real numbers. The determinant of a square matrix is nonzero if and only if the matrix has a multiplicative inverse. P.N.VM.C.12 Multiply a vector (regarded as a matrix with one column) by a matrix of suitable dimensions to produce another vector. Work with matrices as transformations of vectors. P.N.VM.C.13 Work with 2 × 2 matrices as transformations of the plane, and interpret the absolute value of the determinant in terms of area. Algebra Sequences and Series (A.S) A. Understand and use sequences and series. P.A.S.A.1 Demonstrate an understanding of sequences by representing them recursively and explicitly. P.A.S.A.2 Use sigma notation to represent a series; expand and collect expressions in both finite and infinite settings. P.A.S.A.3 Derive and use the formulas for the general term and summation of finite or infinite arithmetic and geometric series, if they exist. a. Determine whether a given arithmetic or geometric series converges or diverges. b. Find the sum of a given geometric series (both infinite and finite). c. Find the sum of a finite arithmetic series. 173 A. Understand and use sequences and series. P.A.S.A.4 Understand that series represent the approximation of a number when truncated; estimate truncation error in specific examples. P.A.S.A.5 Know and apply the Binomial Theorem for the expansion of (x + y)n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal’s Triangle. Reasoning with Equations and Inequalities (A.REI) A. Solve systems of equations and nonlinear inequalities. P.A.REI.A.1 Represent a system of linear equations as a single matrix equation in a vector variable. P.A.REI.A.2 Find the inverse of a matrix if it exists and use it to solve systems of linear equations (using technology for matrices of dimension 3 × 3 or greater). P.A.REI.A.3 Solve nonlinear inequalities (quadratic, trigonometric, conic, exponential, logarithmic, and rational) by graphing (solutions in interval notation if one-variable), by hand and with appropriate technology. P.A.REI.A.4 Solve systems of nonlinear inequalities by graphing. Parametric Equations (A.PE) A. Describe and use parametric equations. ★ P.A.PE.A.1 Graph curves parametrically (by hand and with appropriate technology). P.A.PE.A.2 Eliminate parameters by rewriting parametric equations as a single equation. Conic Sections (A.C) A. Understand the properties of conic sections and model real-world phenomena. P.A.C.A.1 Display all of the conic sections as portions of a cone. P.A.C.A.2 Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum or difference of distances from the foci is constant. P.A.C.A.3 From an equation in standard form, graph the appropriate conic section: ellipses, hyperbolas, circles, and parabolas. Demonstrate an understanding of the relationship between their standard algebraic form and the graphical characteristics. P.A.C.A.4 Transform equations of conic sections to convert between general and standard form. 174 Functions Building Functions (F.BF) Cluster Headings Content Standards A. Build new functions from existing functions. P.F.BF.A.1 Understand how the algebraic properties of an equation transform the geometric properties of its graph. For example, given a function, describe the transformation of the graph resulting from the manipulation of the algebraic properties of the equation (i.e., translations, stretches, reflections, and changes in periodicity and amplitude). P.F.BF.A.2 Develop an understanding of functions as elements that can be operated upon to get new functions: addition, subtraction, multiplication, division, and composition of functions. P.F.BF.A.3 Compose functions. For example, if T(y) is the temperature in the atmosphere as a function of height, and h(t) is the height of a weather balloon as a function of time, then T(h(t)) is the temperature at the location of the weather balloon as a function of time. P.F.BF.A.4 Construct the difference quotient for a given function and simplify the resulting expression. P.F.BF.A.5 Find inverse functions (including exponential, logarithmic, and trigonometric). a. Calculate the inverse of a function, f (x) , with respect to each of the functional operations; in other words, the additive inverse, − f (x) , the multiplicative inverse, 1 / f(x), and the inverse with respect to composition, f−1 (x) . Understand the algebraic and graphical implications of each type. b. Verify by composition that one function is the inverse of another. c. Read values of an inverse function from a graph or a table, given that the function has an inverse. d. Recognize a function is invertible if and only if it is one-to-one. Produce an invertible function from a non-invertible function by restricting the domain. P.F.BF.A.6 Explain why the graph of a function and its inverse are reflections of one another over the line y = x. 175 Interpreting Functions (F.IF) Cluster Headings Content Standards A. Analyze functions using different representations. P.F.IF.A.1 Determine whether a function is even, odd, or neither. P.F.IF.A.2 Analyze qualities of exponential, polynomial, logarithmic, trigonometric, and rational functions and solve real-world problems that can be modeled with these functions (by hand and with appropriate technology).★ P.F.IF.A.4 Identify the real zeros of a function and explain the relationship between the real zeros and the x-intercepts of the graph of a function (exponential, polynomial, logarithmic, trigonometric, and rational). P.F.IF.A.5 Identify characteristics of graphs based on a set of conditions or on a general equation such as y = ax2+ c. P.F.IF.A.6 Visually locate critical points on the graphs of functions and determine if each critical point is a minimum, a maximum, or point of inflection. Describe intervals where the function is increasing or decreasing and where different types of concavity occur. P.F.IF.A.7 Graph rational functions, identifying zeros, asymptotes (including slant), and holes (when suitable factorizations are available) and showing end-behavior. P.F.IF.A.8 Recognize that sequences are functions, sometimes defined recursively, whose domain is a subset of the integers. For example, the Fibonacci sequence is defined recursively by f(0) = f(1) = 1, f(n + 1) = f(n) + f(n - 1) for n ≥ 1. Trigonometric Functions (F.TF) Cluster Headings Content Standards A. Extend the domain of trigonometric functions using the unit circle. P.F.TF.A.1 Convert from radians to degrees and from degrees to radians. P.F.TF.A.2 Use special triangles to determine geometrically the values of sine, cosine, tangent for π/3, π/4 and π/6, and use the unit circle to express the values of sine, cosine, and tangent for π–x, π+x, and 2π–x in terms of their values for x, where x is any real number. P.F.TF.A.3 Use the unit circle to explain symmetry (odd and even) and periodicity of trigonometric functions. P.F.TF.A.4 Choose trigonometric functions to model periodic phenomena with specified amplitude, frequency, and midline. 176 Graphing Trigonometric Functions (F.GT) Cluster Headings Content Standards A. Model periodic phenomena with trigonometric functions.★ P.F.GT.A.1 Interpret transformations of trigonometric functions. P.F.GT.A.2 Determine the difference made by choice of units for angle measurement when graphing a trigonometric function. P.F.GT.A.3 Graph the six trigonometric functions and identify characteristics such as period, amplitude, phase shift, and asymptotes. P.F.GT.A.4 Find values of inverse trigonometric expressions (including compositions), applying appropriate domain and range restrictions. P.F.GT.A.5 Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed. P.F.GT.A.6 Determine the appropriate domain and corresponding range for each of the inverse trigonometric functions. P.F.GT.A.7 Graph the inverse trigonometric functions and identify their key characteristics. P.F.GT.A.8 Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context. Geometry Applied Trigonometry (G.AT) Cluster Headings Content Standards A. Use trigonometry to solve problems. ★ P.G.AT.A.1 Use the definitions of the six trigonometric ratios as ratios of sides in a right triangle to solve problems about lengths of sides and measures of angles. P.G.AT.A.2 Derive the formula A = 1/2 ab sin(C) for the area of a triangle by drawing an auxiliary line from a vertex perpendicular to the opposite side. P.G.AT.A.3 Derive and apply the formulas for the area of sector of a circle. P.G.AT.A.4 Calculate the arc length of a circle subtended by a central angle. 177 Cluster Headings Content Standards A. Use trigonometry to solve problems. ★ P.G.AT.A.5 Prove the Laws of Sines and Cosines and use them to solve problems. P.G.AT.A.6 Understand and apply the Law of Sines (including the ambiguous case) and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces). Trigonometric Identities (G.TI) Cluster Headings Content Standards A. Apply trigonometric identities to rewrite expressions and solve equations.★ P.G.TI.A.1 Apply trigonometric identities to verify identities and solve equations. Identities include: Pythagorean, reciprocal, quotient, sum/difference, double-angle, and half-angle. P.G.TI.A.2 Prove the addition and subtraction formulas for sine, cosine, and tangent and use them to solve problems. Polar Coordinates (G.PC) Cluster Headings Content Standards A. Use polar coordinates. P.G.PC.A.1 Graph functions in polar coordinates. P.G.PC.A.2 Convert between rectangular and polar coordinates. P.G.PC.A.3 Represent situations and solve problems involving polar coordinates.★ Statistics and Probability Model with Data★ (S.MD) Cluster Headings Content Standards A. Model data using regressions equations. P.S.MD.A.1 Create scatter plots, analyze patterns, and describe relationships for bivariate data (linear, polynomial, trigonometric, or exponential) to model real-world phenomena and to make predictions. P.S.MD.A.2 Determine a regression equation to model a set of bivariate data. Justify why this equation best fits the data. 178 A. Model data using regressions equations. P.S.MD.A.3 Use a regression equation, modeling bivariate data, to make predictions. Identify possible considerations regarding the accuracy of predictions when interpolating or extrapolating. 179 180 Statistics | S Statistics is designed to introduce students to the major concepts and tools for collecting, analyzing, and drawing conclusions from data. The major themes in Statistics include: interpreting categorical and quantitative data, conditional probability and other rules of probability, using probability to make decisions, and making inferences and justifying conclusions. Statistics includes the following domains and clusters:  Interpreting Categorical and Quantitative Data o Understand, represent, and use univariate data. o Understand, represent, and use bivariate data.  Conditional Probability and the Rules of Probability o Understand and apply basic concepts of probability. o Use the rules of probability to compare probabilities of compound events in a uniform probability model.  Using Probability to Make Decisions o Understand and use discrete probability distributions. o Understand the normal probability distribution.  Making Inferences and Justifying Conclusions o Know the characteristics of well-defined studies. o Design and conduct a statistical experiment to study a problem, then interpret and communicate the outcomes. o Make inferences about population parameters based on a random sample from that population. o Understand and use confidence intervals. o Use distributions to make inferences about a data set. 181 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 182 Exploring Data Interpreting Categorical and Quantitative Data (S.ID) Cluster Headings Content Standards A. Understand, represent, and use univariate data. S.ID.A.1 Understand the term 'variable' and differentiate between the data types: measurement, categorical, univariate, and bivariate. S.ID.A.2 Understand histograms, parallel box plots, and scatterplots, and use them to display and compare data. S.ID.A.3 Summarize distributions of univariate data. S.ID.A.4 Compute basic statistics and understand the distinction between a statistic and a parameter. S.ID.A.5 For univariate measurement data, be able to display the distribution and describe its shape; select and calculate summary statistics. S.ID.A.6 Recognize how linear transformations of univariate data affect shape, center, and spread. S.ID.A.7 Analyze the effect of changing units on summary measures. S.ID.A.8 Construct and interpret two-way frequency tables of data when two categories are associated with each object being classified. Use the two-way table as a sample space to decide if events are independent and to approximate conditional probabilities. S.ID.A.9 Describe individual performances in terms of percentiles, z-scores, and t- scores. B. Understand, represent, and use bivariate data. S.ID.B.10 Represent and analyze categorical data. S.ID.B.11 Display and discuss bivariate data where at least one variable is categorical. S.ID.B.12 For bivariate measurement data, be able to display a scatterplot and describe its shape; use technological tools to determine regression equations and correlation coefficients. S.ID.B.13 Identify trends in bivariate data; find functions that model the data and that transform the data so that they can be modeled. 183 Probability Conditional Probability and the Rules of Probability (S.CP) Cluster Headings Content Standards A. Understand and apply basic concepts of probability. S.CP.A.1 Describe events as subsets of a sample space (the set of outcomes) using characteristics (or categories) of the outcomes, or as unions, intersections, or complements of other events (“or,” “and,” “not”). S.CP.A.2 Use permutations and combinations to compute probabilities of compound events and solve problems. S.CP.A.3 Demonstrate an understanding of the Law of Large Numbers (Strong and Weak). B. Use the rules of probability to compute probabilities of compound events in a uniform probability model. S.CP.B.4 Demonstrate an understanding of the addition rule, the multiplication rule, conditional probability, and independence. S.CP.B.5 Apply the general Multiplication Rule in a uniform probability model, P(A and B) = P(A)P(B|A) = P(B)P(A|B), and interpret the answer in terms of the model. Probability Distributions Using Probability to Make Decisions (S.MD) Cluster Headings Content Standards A. Understand and use discrete probability distributions. S.MD.A.1 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions. S.MD.A.2 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution. S.MD.A.3 Design a simulation of random behavior and probability distributions (e.g., drawing by lots, using a random number generator, and using the results to make fair decisions). S.MD.A.4 Analyze discrete random variables and their probability distributions, including binomial and geometric. 184 Cluster Headings Content Standards A. Understand and use discrete probability distributions. S.MD.A.5 Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes. S.MD.A.6 Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value. For example, find a current data distribution on the number of TV sets per household in the United States, and calculate the expected number of sets per household. How many TV sets would you expect to find in 100 randomly selected households? S.MD.A.7 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. a. Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant. b. Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident. S.MD.A.8 Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game). B. Understand the normal probability distribution. S.MD.B.9 Calculate the mean (expected value) and standard deviation of both a random variable and a linear transformation of a random variable. S.MD.B.10 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. 185 Sampling and Experimentation Making Inferences and Justifying Conclusions (S.IC) Cluster Headings Content Standards A. Know the characteristics of well-designed studies. S.IC.A.1 Understand the differences among various kinds of studies and which types of inferences can be legitimately drawn from each. S.IC.A.2 Compare census, sample survey, experiment, and observational study. S.IC.A.3 Describe the role of randomization in surveys and experiments. S.IC.A.4 Describe the role of experimental control and its effect on confounding. S.IC.A.5 Identify bias in sampling and determine ways to reduce it to improve results. S.IC.A.6 Describe the sampling distribution of a statistic and define the standard error of a statistic. S.IC.A.7 Demonstrate an understanding of the Central Limit Theorem. B. Design and conduct a statistical experiment to study a problem, then interpret and communicate the outcomes. S.IC.B.8 Select a method to collect data and plan and conduct surveys and experiments. S.IC.B.9 Compare and use sampling methods, including simple random sampling, stratified random sampling, and cluster sampling. S.IC.B.10 Test hypotheses using appropriate statistics. S.IC.B.11 Analyze results and make conclusions from observational studies, experiments, and surveys. S.IC.B.12 Use data from a randomized experiment to compare two treatments; use simulations to decide if differences between parameters are significant. C. Make inferences about population parameters based on a random sample from that population. S.IC.C.13 Develop and evaluate inferences and predictions that are based on data. S.IC.C.14 Use properties of point estimators, including biased/unbiased, and variability. D. Understand and use confidence intervals. S.IC.D.15 Understand the meaning of confidence level, of confidence intervals, and the properties of confidence intervals. 186 Cluster Headings Content Standards D. Understand and use confidence intervals. S.IC.D.16 Construct and interpret a large sample confidence interval for a proportion and for a difference between two proportions. S.IC.D.17 Construct the confidence interval for a mean and for a difference between two means. E. Use distributions to make inferences about a data set. S.IC.E.18 Apply the properties of a Chi-square distribution in appropriate situations in order to make inferences about a data set. S.IC.E.19 Apply the properties of the normal distribution in appropriate situations in order to make inferences about a data set. S.IC.E.20 Interpret the t-distribution and determine the appropriate degrees of freedom. 187 Applied Mathematical Concepts | AM Applications and modeling using mathematics are the primary foci of this course. Sample potential applications for topics are listed below: Counting, Combinatorics, and Probability  Counting hands of cards, code words, license plates, phone numbers, make-up of committees, etc.  Probabilities associated with games (such as using games from The Price is Right) Financial Math  Amortization and loans (copayments, credit cards, loans, etc.)  Compound interest; comparing payments, interest rates, length of loan period, investments, etc.  Computing taxes  Computing paychecks (deductions, social security payments, etc.)  Comparing insurance plans (term vs. whole life)  Annuities Linear Programming  Maximizing capacity while minimizing costs Applied Mathematical Concepts includes the following domains and clusters:  Financial Mathematics o Use financial mathematics to solve problems. o Use financial mathematics to make decisions. o Determine appropriate models to solve contextual problems.  Linear Programming o Use linear programming techniques to solve real-world problems. o Solve real-world optimization problems.  Logic and Boolean Algebra o Use logic and Boolean Algebra in real-world situations. o Apply Boolean Algebra to real-world network problems.  Problem Solving o Apply problem solving techniques to real-world problems.  Investigate Logic o Use logic to make arguments and solve problems. o Determine the validity of arguments.  Organize and Interpret Data o Analyze data from multiple viewpoints and perspectives.  Counting and Combinatorial Reasoning o Apply probability and counting principles to real-world situations. o Use combinatorial reasoning to solve real-world problems.  Normal Probability Distribution o Work with the normal distribution in real-world situations.  Understand and Use Confidence Intervals o Work with confidence intervals in real-world situations. 188 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 189 Number and Quantity Financial Mathematics (N.NQ) Cluster Headings Content Standards A. Use financial mathematics to solve problems. AM.N.NQ.A.1 Define interest, compound interest, annuities, sinking funds, amortizations, annuities, future value, and present value. AM.N.NQ.A.2 Recognize the importance of applying a financial model to business. AM.N.NQ.A.3 Determine future value and present value of an annuity. AM.N.NQ.A.4 Determine the amortization schedule for an annuity and a home mortgage. B. Use financial mathematics to make decisions. AM.N.NQ.B.5 Apply financial mathematics to depreciation schedules. AM.N.NQ.B.6 Solve contextual problems involving financial decision-making. AM.N.NQ.B.7 Apply arithmetic and geometric sequences to simple and compound interest, annuities, loans, and amortization. AM.N.NQ.B.8 Solve problems in mathematics of finance involving compound interest using exponential and logarithmic techniques. C. Determine appropriate models to solve contextual problems. AM.N.NQ.C.9 Know when to use transcendental functions to accomplish various application purposes such as predicting population growth. AM.N.NQ.C.10 Use orders of magnitude estimates for determining an appropriate model for a contextual situation. Algebra Linear Programming (A.LP) Cluster Headings Content Standards A. Use linear programming techniques to solve real-world problems. AM.A.LP.A.1 Use mathematical models involving equations and systems of equations to represent, interpret, and analyze quantitative relationships, change in various contexts, and other real-world phenomena. AM.A.LP.A.2 Read, interpret, and solve linear programming problems graphically and by computational methods. 190 Cluster Headings Content Standards B. Solve real-world optimization problems. AM.A.LP.B.3 Use linear programming to solve optimization problems. AM.A.LP.B.4 Interpret the meaning of the maximum or minimum value in terms of the objective function. Logic and Boolean Algebra (A.LB) Cluster Headings Content Standards A. Use logic and Boolean Algebra in real-world situations. AM.A.LB.A.1 Develop the symbols and properties of Boolean algebra; connect Boolean algebra to standard logic. AM.A.LB.A.2 Construct truth tables to determine the validity of an argument. B. Apply Boolean Algebra to real-world network problems. AM.A.LB.B.3 Analyze basic electrical networks; compare the networks to Boolean Algebra configurations. AM.A.LB.B.4 Develop electrical networks and translate them into Boolean Algebra equations. Problem Solving (A.PS) Cluster Headings Content Standards A. Apply problem solving techniques to real-world situations. AM.A.PS.A.1 Apply problem solving strategies to real-world situations. Strategies include, but are not limited to: making orderly lists or tables, drawing diagrams, considering simpler problems, looking for patterns, working backwards, guess and check, using logical reasoning, etc. Geometry and Measurement Investigate Logic (G.L) Cluster Headings Content Standards A. Use logic to make arguments and solve problems. AM.G.L.A.1 Define the order of operations for the logical operators. AM.G.L.A.2 Define conjunction, disjunction, negation, conditional, and biconditional. AM.G.L.A.3 Solve a variety of logic puzzles. 191 Cluster Headings Content Standards A. Use logic to make arguments and solve problems. AM.G.L.A.4 Construct and use a truth table to draw conclusions about a statement. B. Determine the validity of arguments. AM.G.L.B.5 Apply the laws of logic to judge the validity of arguments. AM.G.L.B.6 Give counterexamples to disprove statements. AM.G.L.B.7 Analyze arguments with quantifiers through the use of Venn diagrams. AM.G.L.B.8 Represent logical statements with networks. Data Analysis, Statistics, and Probability Organize and Interpret Data (D.ID) Cluster Headings Content Standards A. Analyze data from multiple viewpoints and perspectives. AM.D.ID.A.1 Organize data for problem solving. AM.D.ID.A.2 Use a variety of counting methods to organize information, determine probabilities, and solve problems. AM.D.ID.A.3 Translate from one representation of data to another, e.g., a bar graph to a circle graph. AM.D.ID.A.4 Calculate and interpret statistical problems using measures of central tendency and graphs. AM.D.ID.A.5 Calculate expected value, e.g., to determine the fair price of an investment. AM.D.ID.A.6 Analyze survey data using Venn diagrams. AM.D.ID.A.7 Evaluate and compare two investments or strategies, where one investment or strategy is safer but has lower expected value. Include large and small investments and situations with serious consequences. 192 Counting and Combinatorial Reasoning (D.CR) Cluster Headings Content Standards A. Apply probability and counting principles to real-world situations. AM.D.CR.A.1 Use permutations, combinations, and the multiplication principle to solve counting problems. AM.D.CR.A.2 Design and interpret simple experiments using tree-diagrams, permutations, and combinations. AM.D.CR.A.3 Apply counting principles to probabilistic situations involving equally likely outcomes. AM.D.CR.A.4 Solve counting problems by using Venn diagrams and show relationships modeled by the Venn diagram. AM.D.CR.A.5 Use permutations and combinations to compute probabilities of compound events and solve problems. B. Use combinatorial reasoning to solve real-world problems. AM.D.CR.B.6 Apply the Law of Large Numbers to contextual situations. AM.D.CR.B.7 Discuss the various examples and consequences of innumeracy; consider poor estimation, improper experimental design, inappropriate comparisons, and scientific notation comparisons. AM.D.CR.B.8 Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. a. Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant. b. Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident. AM.D.CR.B.9 Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). AM.D.CR.B.10 Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game). Normal Probability Distribution (D.ND) Cluster Headings Content Standards A. Work with the normal distribution in real-world situations. AM.D.ND.A.1 Calculate the mean (expected value) and standard deviation of both a random variable and a linear transformation of a random variable. 193 Cluster Headings Content Standards A. Work with the normal distribution in real-world situations. AM.D.ND.A.2 Use the mean and standard deviation of a data set to fit it to a normal distribution and to estimate population percentages. Recognize that there are data sets for which such a procedure is not appropriate. Use calculators, spreadsheets, and tables to estimate areas under the normal curve. Understand and Use Confidence Intervals (D.CI) Cluster Headings Content Standards A. Work with confidence intervals in real-world situations. AM.D.CI.A.1 Understand the meaning of confidence level, of confidence intervals, and the properties of confidence intervals. AM.D.CI.A.2 Construct and interpret a large sample confidence interval for a proportion and for a difference between two proportions. AM.D.CI.A.3 Construct the confidence interval for a mean and for a difference between two means. 194 Calculus | C Calculus is designed for students interested in STEM-based careers and builds on the concepts studied in precalculus. The study of calculus on the high school level includes a study of limits, derivatives, and an introduction to integrals. Calculus includes the following domains and clusters:  Limits of Functions o Understand the concept of the limit of a function.  Behavior of Functions o Describe the asymptotic and unbounded behavior of functions.  Continuity o Develop an understanding of continuity as a property of functions.  Understand the Concept of the Derivative o Demonstrate an understanding of the derivative. o Understand the derivative at a point.  Computing and Applying Derivatives o Apply differentiation techniques. o Use first and second derivatives to analyze a function. o Apply derivatives to solve problems.  Understanding Integrals o Demonstrate understanding of a definite integral. o Understand and apply the Fundamental Theorem of Calculus.  Calculate and Apply Integrals o Apply techniques of antidifferentiation. o Apply integrals to solve problems. 195 Mathematical Modeling Mathematical Modeling is a Standard for Mathematical Practice (MP4) and a Conceptual Category. Specific modeling standards appear throughout the high school standards indicated with a star (★). Where an entire domain is marked with a star, each standard in that domain is a modeling standard. Standards for Mathematical Practice Being successful in mathematics requires the development of approaches, practices, and habits of mind that need to be in place as one strives to develop mathematical fluency, procedural skills, and conceptual understanding. The Standards for Mathematical Practice are meant to address these areas of expertise that teachers should seek to develop in their students. These approaches, practices, and habits of mind can be summarized as “processes and proficiencies” that successful mathematicians have as a part of their work in mathematics. Additional explanations are included in the main introduction of these standards. Literacy Standards for Mathematics Communication in mathematics employs literacy skills in reading, vocabulary, speaking and listening, and writing. Mathematically proficient students communicate using precise terminology and multiple representations including graphs, tables, charts, and diagrams. By describing and contextualizing mathematics, students create arguments and support conclusions. They evaluate and critique the reasoning of others, analyze, and reflect on their own thought processes. Mathematically proficient students have the capacity to engage fully with mathematics in context by posing questions, choosing appropriate problem-solving approaches, and justifying solutions. Further explanations are included in the main introduction. Standards for Mathematical Practice 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning. Literacy Skills for Mathematical Proficiency 1. Use multiple reading strategies. 2. Understand and use correct mathematical vocabulary. 3. Discuss and articulate mathematical ideas. 4. Write mathematical arguments. 196 Functions, Graphs, and Limits Limits of Functions (F.LF) Cluster Headings Content Standards A. Understand the concept of the limit of a function. C.F.LF.A.1 Calculate limits (including limits at infinity) using algebra. C.F.LF.A.2 Estimate limits of functions (including one-sided limits) from graphs or tables of data. Apply the definition of a limit to a variety of functions, including piece-wise functions. C.F.LF.A.3 Draw a sketch that illustrates the definition of the limit; develop multiple real-world scenarios that illustrate the definition of the limit. Behavior of Functions (F.BF) Cluster Headings Content Standards A. Describe the asymptotic and unbounded behavior of functions. C.F.BF.A.1 Describe asymptotic behavior (analytically and graphically) in terms of infinite limits and limits at infinity. C.F.BF.A.2 Discuss the various types of end behavior of functions; identify prototypical functions for each type of end behavior. Continuity (F.C) Cluster Headings Content Standards A. Develop an understanding of continuity as a property of functions C.F.C.A.1 Define continuity at a point using limits; define continuous functions. C.F.C.A.2 Determine whether a given function is continuous at a specific point. C.F.C.A.3 Determine and define different types of discontinuity (point, jump, infinite) in terms of limits. C.F.C.A.4 Apply the Intermediate Value Theorem and Extreme Value Theorem to continuous functions. 197 Derivatives Understand the Concept of the Derivative (D.CD) Cluster Headings Content Standards A. Demonstrate an understanding of the derivative. C.D.CD.A.1 Represent and interpret the derivative of a function graphically, numerically, and analytically. C.D.CD.A.2 Interpret the derivative as an instantaneous rate of change. C.D.CD.A.3 Define the derivative as the limit of the difference quotient; illustrate with the sketch of a graph. C.D.CD.A.4 Demonstrate the relationship between differentiability and continuity. B. Understand the derivative at a point. C.D.CD.B.5 Interpret the derivative as the slope of a curve (which could be a line) at a point, including points at which there are vertical tangents and points at which there are no tangents (i.e., where a function is not locally linear). C.D.CD.B.6 Approximate both the instantaneous rate of change and the average rate of change given a graph or table of values. C.D.CD.B.7 Write the equation of the line tangent to a curve at a given point. C.D.CD.B.8 Apply the Mean Value Theorem. C.D.CD.B.9 Understand Rolle’s Theorem as a special case of the Mean Value Theorem. Computing and Applying Derivatives (D.AD) Cluster Headings Content Standards A. Apply differentiation techniques. C.D.AD.A.1 Describe in detail how the basic derivative rules are used to differentiate a function; discuss the difference between using the limit definition of the derivative and using the derivative rules. C.D.AD.A.2 Calculate the derivative of basic functions (power, exponential, logarithmic, and trigonometric). C.D.AD.A.3 Calculate the derivatives of sums, products, and quotients of basic functions. C.D.AD.A.4 Apply the chain rule to find the derivative of a composite function. 198 Cluster Headings Content Standards A. Apply differentiation techniques. C.D.AD.A.5 Implicitly differentiate an equation in two or more variables. C.D.AD.A.6 Use implicit differentiation to find the derivative of the inverse of a function. B. Use first and second derivatives to analyze a function. C.D.AD.B.7 Relate the increasing and decreasing behavior of f to the sign of f’ both analytically and graphically. C.D.AD.B.8 Use the first derivative to find extrema (local and global). C.D.AD.B.9 Analytically locate the intervals on which a function is increasing, decreasing, or neither. C.D.AD.B.10 Relate the concavity of f to the sign of f” both analytically and graphically. C.D.AD.B.11 Use the second derivative to find points of inflection as points where concavity changes. C.D.AD.B.12 Analytically locate intervals on which a function is concave up, concave down, or neither. C.D.AD.B.13 Relate corresponding characteristics of the graphs of f, f’, and f”. C.D.AD.B.14 Translate verbal descriptions into equations involving derivatives and vice versa. C. Apply derivatives to solve problems. C.D.AD.C.15 Model rates of change, including related rates problems. In each case, include a discussion of units. C.D.AD.C.16 Solve optimization problems to find a desired maximum or minimum value. C.D.AD.C.17 Use differentiation to solve problems involving velocity, speed, and acceleration. C.D.AD.C.18 Use tangent lines to approximate function values and changes in function values when inputs change (linearization). 199 Integrals Understanding Integrals (I.UI) Cluster Headings Content Standards A. Demonstrate understanding of a definite integral. C.I.UI.A.1 Define the definite integral as the limit of Riemann sums and as the net accumulation of change. C.I.UI.A.2 Correctly write a Riemann sum that represents the definition of a definite integral. C.I.UI.A.3 Use Riemann sums (left, right, and midpoint evaluation points) and trapezoid sums to approximate definite integrals of functions represented graphically, numerically, and by tables of values. B. Understand and apply the Fundamental Theorem of Calculus. C.I.UI.B.4 Recognize differentiation and antidifferentiation as inverse operations. C.I.UI.B.5 Evaluate definite integrals using the Fundamental Theorem of Calculus. C.I.UI.B.6 Use the Fundamental Theorem of Calculus to represent a particular antiderivative of a function and to understand when the antiderivative so represented is continuous and differentiable. C.I.UI.B.7 Apply basic properties of definite integrals (e.g., additive, constant multiple, translations). Calculate and Apply Integrals (I.AI) Cluster Headings Content Standards A. Apply techniques of antidifferentiation. C.I.AI.A.1 Develop facility with finding antiderivatives that follow directly from derivatives of basic functions (power, exponential, logarithmic, and trigonometric). C.I.AI.A.2 Use substitution of variables to calculate antiderivatives (including changing limits for definite integrals). C.I.AI.A.3 Find specific antiderivatives using initial conditions. 200 Cluster Headings Content Standards B. Apply integrals to solve problems. C.I.AI.B.4 Use a definite integral to find the area of a region. C.I.AI.B.5 Use a definite integral to find the volume of a solid formed by rotating a region around a given axis. C.I.AI.B.6 Use integrals to solve a variety of problems (e.g., distance traveled by a particle along a line, exponential growth/decay).
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https://storymaps.arcgis.com/stories/f587e4ee2c684f8cbe4a97f472c6715e
Cartography Chapter 4: Types of Map Skip to content Cartography Chapter 4: Types of Map Share More actions Cartography Chapter 4: Types of Map KSU Geog 3305 Uli Ingram July 14, 2020 Introduction 4.1 4.2 4.3 Other Resources Summary Banner Introduction This chapter covers various types of maps and their features. Many types of maps exist so that cartographers can visualize spatial phenomenon in the most advantageous way. It is important that you be aware of all the different map types available to you so that you can visualize your data in the format that will be most appropriate for the content of your map and your map user. This lesson covers three map types: choropleth maps, proportional symbol maps, and dot-density maps. 4.1 Choropleth Maps: A choropleth map is a map where colored or shaded areas represent the magnitude of an attribute. For example, this map shows the population density in the year 2007 for the United States of America. For each state, the number of persons per square mile has been calculated. The states with a lower population density are shaded with a lighter gray color. The states of a higher population density are shaded with a dark gray color. The states with population densities between the two extremes are shaded on a continuum from the lightest gray to the darkest gray. Based on the five different shades of gray, the map visually represents in an intuitive manner where the most densely populated states are in the United States. Figure 1 Choropleth Map Figure 1: Choropleth Map Why Create a Choropleth Map?There are many reasons why you would want to create a choropleth map. Choropleth maps are relatively easy to create and easy to interpret by map readers. A choropleth map excels at displaying variables overall geographic pattern. As each color or shade is assigned a value or range of values the map reader can ascertain the values displayed on the map easily. Choropleth maps are also excellent for comparing multiple choropleth maps with one another to see how the spatial distribution of variable changes. We refer to multiple small maps on the same page as small multiples. A small multiple is when you have multiple small choropleth maps made with similar structure and context. It is important that the only data represented on the choropleth map is data that can be linked to an enumeration area. An enumeration area is an area where data is collected and combined. Common enumeration areas in the United States are states, counties, and regions. It is important that the data be normalized against enumeration areas or a total population true blue area or size bias. What is meant by area or size bias is that larger areas might tend to have more population simply because there is more area for people to live? Smaller areas will naturally have less population because there is less area to live in. By normalizing data against the enumeration area and taking size out of the equation you can do apple to apples comparisons between areas that have different sizes. Examples of normalized data are population per square mile which gives us population density, and percent unemployed which is the number of unemployed people divided by the total number of people eligible for work. In a choropleth map, the boundaries of areas do not have a related value. In other words, the outline of an area is in no way related to the value of the area. Only the different color or shaded the area is related to the value. Inappropriate Data:There are several types of inappropriate data for choropleth mapping. Continuous data is not appropriate for inclusion on the choropleth map as continuous data is not controlled by an enumeration unit. For example, air temperature is not confined to County outlines nor is it logical to assume so. Another type of inappropriate data is any map ratio not involving an area. That means if a value can be assigned to a very specific point and it is not logical to aggregate that data value to a larger area, it should not be used for the choropleth map. Total values should not be used on a choropleth map. Variables, where the values vary too much, should also not be used for choropleth maps. If you still wish to use variables with a large value range you may want to consider choosing a smaller enumeration unit so that there is a smaller variation in each enumeration unit. Data Classification:In order to display the vast majority of data on a choropleth map, you must employ some data classification schemes. We classify to simplify generalized the data for display on the map. In general, four to seven classes are preferred. If you need to exceed seven classes you need to keep in mind that humans cannot effectively use more than eleven classes at once. Now consider the five types of classification methods to determine when to use each one on a choropleth map. The equal-area data classification method is useful for layouts including multiple maps. The equal frequency data classification method is useful if you are performing a statistical test between classes. The arithmetic and geometric data classification methods are useful for the data that shows a normal distribution. The nested means data classification method is useful for non-normal distributions. And finally, the natural breaks data classification method maximizes homogeneity within classes and is typically going to be your best general choice for displaying data on a choropleth map. Projection:In most cases for choropleth maps the equivalent, or equal area, projection are the most appropriate. The reason why the equal area map projections are the most appropriate is that since we typically normalize data based on area, relative size is important to maintain when comparing the values of the underlying enumeration. Symbolization:Consider the symbolization choices for choropleth maps. If you are producing your choropleth map using black and white colors only, then the black color should represent larger values in the light grey color should represent smaller values. The color white should be generally reserved for the background of the map or outlines. Additionally, you should use caution with pure black and white fills as they may obscure boundary lines. You may use pattern, dot, line, or Hatcher patterns, instead of shades of gray, but this is considered to be the “old style” of choropleth mapping. Black and White:Here is the population density choropleth map of the United States for the year 2007. In this black and white map the color white is used for the background and state outlines. The light gray color represents lower values and the dark gray color represents higher values. Figure 2 Black and White Choropleth Map Figure 2: Black and White Choropleth Map Color:If you are producing your choropleth mapping color you should consider these points. Darker or more saturated colors represent larger values. Lighter or less saturated colors represent smaller values. Make sure you can easily differentiate between colors of different classes, that is, make sure that no two adjacent colors are too similar. You should avoid qualitative color schemes on a quantitative choropleth map. That means, if you only have one variable on your map, such as population, you should choose a single color, or hue to represent that you are only showing different values of a single kind of thing. To represent the different quantities of that single thing, vary the saturation or value of the chosen color. On a color map, white suggests ‘light gray’ can effectively represent “no data”. Black or white is an effective boundary color on a color choropleth map. On this color choropleth map that deals with a single variable of population, a single hue of green was chosen. The green hue is varied in saturation so that the lighter color green represents lower population density in the darker color green represents a higher population density. The color black was chosen as the state outlines as it provides great contrast against the green hue. Figure 3 Color Choropleth Map Figure 3: Color Choropleth Map Color Schemes:In general, there are three color schemes that should be used on a color choropleth map based on the type of data being displayed. If your data is considered unipolar data, which means that there is no natural dividing point, you should use a sequential color scheme. An example of unipolar data is population density. If you have bipolar data, which means that has a natural dividing point such as 0, or mean, then you should use diverging color scheme. Examples of bipolar data are population gains and losses. If you have balanced data, this means you have two complementary phenomena. In this case a diverging color scheme is appropriate. An example of this would be the ratio of males to females. Reference Features:Thematic maps should be simple by design by focusing on the featured variable. When creating them you should avoid placing reference features on the map and less they are important in explaining the pattern of the variable being mapped. Legend Design:To wrap up our discussion of choropleth maps this section will focus on legend design. Legend Boxes:Choropleth maps typically use legend boxes. The legend boxes are typically square or rectangular and are large enough to provide a visual anchor but not too large to distract the eye from the main map body. The symbols in the legend should be identical to the symbols on the map in both color and line weight. If the enumeration units on the map are reasonably small then the size of the symbol on the legend should be about the average size of the enumeration unit on the map. If the enumeration units on the map are very large then you may consider making the boxes ½ to 1/3 the average size of the enumeration unit on the map. Figure 4 Legend Box Figure 4: Legend Box Legend Layout: The legend should be laid out into orientations: horizontally or vertically. With the horizontal orientation, the lowest value should be on the left and the highest value should be on the right. The value numbers should be located below the boxes. In a vertical layout you can either have the lowest or highest value on the top and the lowest or highest value on the bottom as neither layout, is considered standard. The numbers representing the value should be to the right of the boxes. Figure 5 Horizontal Legend Box Figure 5: Horizontal Legend Box Figure 6 Vertical Legend Box Figure 6: Vertical Legend Box Continuous Classes:In the case where the values on your map represent continuous data, which means that the maximum value in one class is slightly less than the minimum value of the next class, you should follow these guidelines. Your legend should emphasize the degradation of values and show that the values on the map exhaust the data. Additionally, the legend should reinforce the fact that there is no data that falls between the cracks. By having a continuous legend it allows the same legend to be applied to multiple maps. In our example, to reinforce the fact that the data is exhaustive and there are no cracks, the boxes of but each other and look like a single continuum. Figure 7 Continuous Classes Figure 7: Continuous Classes Non-Continuous Classes:If you have non-continuous classes on your map then your legend should show the actual extreme values in each class. You would want to do this to narrow the reader’s estimate of the actual values in each class. A non-continuous legend is best with a single map displaying non-continuous data. For example, our legend shows each box separated from the other boxes to reinforce the fact the classes are not exhaustive and on a single continuum. Figure 8 Non-Continuous Classes Figure 8: Non-Continuous Classes Formatting Conventions:Use either “to” or “-“ between class ranges. If there are more than four numbers a value, use a comma after every third number left of the decimal value. If you choose to have a legend title, the legend title should match the topic of the map and should not use abbreviations including symbols. You should place any ancillary text below the legend of the information will not fit into a concise legend title. If you are creating an animated map that shows the change of the variable throughout time, you should use a single continuous legend that encompasses the global maximum and global minimum over the entire series of the maps included in the animation. 4.2 Proportional Symbol Maps: On a proportional symbol map, a symbol’s size is buried in proportion to the quantity it represents. The most common symbol used in a proportional symbol map is a circle. On this example proportional symbol map, it shows the total population per state in the year 2007 for the United States of America. The circles vary in size as the total population it represents increases. On this map, map readers can quickly and easily determine which states have the largest population in relation to other states. Figure 9 Proportional Symbol Map Figure 9: Proportional Symbol Map Why Create Proportional Symbol Maps?So why should you create a proportional symbol map? A proportional symbol map is easy for map readers to understand. Multiple variables can be displayed simultaneously on a proportional symbol map. For example, the symbol's size, symbols color, and symbols a shape can all represent different variables. Additionally, you can overlay a proportional symbol onto another type of thematic map such as a choropleth map. Finally, the main reason to create a proportional symbol map is when you want to display a quantitative distribution of a variable throughout space. Why not Create Proportional Maps?Here is why you should not create a proportional symbol map. In general, map readers tend to underestimate the relative size of the symbols on the map this leads to over or underestimation of values. Also, symbols can easily overlap too much a dense location in the user will not be able to see enough of the symbols to determine the values in the map will look cluttered. Appropriate Data:Appropriate data for a proportional symbol map is either total values, percentage values, or rate values. Additionally, the data should occur at a point or the data should be aggregated to an enumeration unit such as accounting. If the data is aggregated to a point within an area the dot is typically placed in the center of the area. Inappropriate Data:Inappropriate data for a proportional symbol map is interval data. As the size of a symbol can be reduced to 0, which is a natural 0 state for the symbol, the data should have a natural 0 point. This would exclude data that does not have a natural 0 point such as temperature. Another type of inappropriate data for proportional symbol map is data with small ranges in the values. Data with small ranges of values will make for an interesting map as the size of the symbols will not vary too greatly. There is also density data. Density data is not appropriate for a proportional symbol map and you should consider making a choropleth map instead. Projections:The equivalent or equal-area projections are the most appropriate map projections for the proportional symbol map. Since the relative size is important to maintain and comparing values within enumeration units you should choose a map projection that maintains the relative sizes of the areas. Proportional Symbols:Now consider the proportional symbols themselves. Various shapes can be used for proportional symbols although the circle is the most common. There are two categories of point symbols that we can use: geometric symbols, and pictographic symbols. A geometric symbol can be either two-dimensional or three-dimensional and it should not look like the image or area that is being mapped. Two-Dimensional symbols:Examples of 2-D geometric symbols are a circle, a triangle, and a star. A pictographic symbol looks like the thing being mapped. First focus on the two-dimensional geometric symbol as this will probably be the most common symbol you will use of a proportional symbol map. For the 2-D geometric symbol, the area of the symbol is a scale to represent the magnitude difference of the value of represents. The circle is the most widely used symbol of a proportional symbol map. The advantages of using a circle are that its geometric form is compact and the circles are visually stable since the eye does not wander across the circle too much. If circle symbols overlap they can still be effective in indicating magnitude. And finally, circles can easily, lend to a second variable by changing its color or texture. Squares are also popular choices as their main advantage is that the proportional areas are nearly perfectly perceived whereas they will be over or underestimated for circles and most other two-dimensional geometric symbols. A disadvantage of using squares is that it adds “squareness” to the map which may not be desirable. It can also be confused for other common symbols like houses. Figure 10 (left to right) 2-D Circle, Triangle, and Star Figure 10: (left to right) 2-D Circle, Triangle, and Star Three Dimensional symbols:Three-dimensional geometric symbols can also be used as proportional symbols. Examples of three-dimensional geometric symbols are a sphere, prism, or pyramid. The advantages of three-dimensional geometric symbols are that they are visually attractive and they allow for less crowding of the map because they add the third dimension which allows it to be slightly more compact. The disadvantages of three-dimensional geometric symbols are that readers cannot accurately judge scaling differences because the readers now have to judge changes in volume. Range grading the symbols is recommended for three-dimensional symbols to combat scaling issues. Complicated three-dimensional geometric symbols can also be hard to understand on a map. Figure 11 3-D Square and Cylinder Figure 11: 3-D Square and Cylinder Pictographic:Pictorial symbols are sometimes good choices for proportional symbols. Advantages of pictorial symbols are they are visually attractive and attention-grabbing. Disadvantages for pictorial symbols are the more regular the shape of the control symbol, the harder will be to perceive the magnitude differences. Additionally mapping a second variable is difficult except for simple hue changes. Three examples of pictographic symbols are the front of a bus, and outline of an airplane, and a taco. Figure 12 Pictograms of a Bus, Airplane, and Taco Figure 12: Pictograms of a Bus, Airplane, and Taco Proportional Symbol Color:Now consider proportional symbol color. Different colors apparently do not affect the estimation of symbol size differences. Symbols that are too darkly tinted command a lot of attention so you should avoid extremely dark symbols unless you wish them to command a significant amount of attention on the map. Symbols with little contrast from its surroundings may be lost and are not popular choices. You should consider using appropriate strategies for contrast. Symbols that are black or gray colors are the most popular with map readers for a proportional symbol map. And finally, changing the color of the symbol can be used for a second variable. Overlap:Sometimes on proportional symbol map symbols will overlap. Overlapping symbols express a sense of visual cohesiveness and may make the map more memorable to the user as it will make a strong visual imprint in their mind. The downside to symbol overlap is that it may make it harder for the map reader to estimate the individual symbol sizes as they will be partially obscured. To combat this try not to overlap symbols more than 25% to 33%. Additionally, when symbols overlap, smaller symbols must cover the larger symbols as shown in Figure 13. If you cannot avoid significant symbol overlap another option is to make the circles transparent, or semitransparent to help users perceive overlapping symbols. Figure 13 Overlap Figure 13: Overlap Appropriate Overlap:The maps in Figure 14 are examples of maps, using appropriate symbol overlap. In both maps the symbols overlap state outlines and other proportional symbols but not so significantly that you cannot estimate the size of the circles that are being obscured. Additionally, with this amount of symbols overlap, the map has a cohesive feel to it in a large amount of overlap in the northeast part of the United States makes the data memorable. Figure 14 Appropriate Overlapping on a Map Figure 14: Appropriate Overlapping on a Map Inappropriate Overlap:Figures 15 and 16 are examples of inappropriate amounts of symbol overlap. In Figure 15 there is too little overlap which makes it seem empty and uninteresting. On the map on the right, there is too much overlap of the circles which causes circles to be obscured. Also, the proportional symbol for California exceeds the neat line of the map which looks very messy. Figure 15 Too Little Overlap, Empty, Uninteresting (Left) Figure 16 Too Much Overlap (Right) Figure 15: Little Overlap, Empty, Uninteresting Figure 16: Too Much Overlap Scaling Methods:There are three primary methods of scaling proportional symbols: absolute, apparent magnitude, which is also known as perceptual, and range grading. Absolute Scaling:In absolute scaling symbols scale proportionally to their data values and to each other. Absolute scaling is great if you want each symbol to be unique in size and to give a true visualization of the relative values. The negative aspect of absolute scaling is that it can be difficult for the map user to interpret because the map user can only differentiate a given number of symbol sizes effectively. Figure 17 Absolute Scaling Figure 17: Absolute Scaling Apparent Magnitude Scaling:In apparent magnitude scaling it applies factors to compensate for the user's underestimation of area and volume of the symbols. Typically this factor is used to increase the size of the circles faster than in absolute scaling. Apparent magnitude scaling is also known as Flannery, perceptual, or psychological scaling. Looking at the two example legends in Figure 18, there are five proportional circles in each legend, with the same value; however, the circle sizes and the apparent magnitude scaling legend are a bit larger to compensate for the map user’s tendency to underestimate relative areas of the circles. Figure 18 Comparing Absolute and Apparent Magnitude Scaling Figure 18: Comparing Absolute and Apparent Magnitude Scaling Range Grading Scaling:Another type of scaling is range grading scaling. In range grading scaling each symbol, size represents a range of data values and not a single data value. Range grading is similar to choropleth mapping as you employ data classification methods to determine class breaks. The main advantage to range grading scaling is that readers can easily discriminate symbol sizes and match them to the legend symbols. For instance, if there are five circles on the legend of different sizes and there will only be circles of five different sizes on the map it should be easy for the map reader to match the symbols from the map to the legend thereby getting the range of values for that location. Range grading is recommended for three-dimensional geometric or pectoral symbols as users tend to have difficulty in estimating sizes properly for the symbols. Figure 19 Range Grading Scaling Figure 19: Range Grading Scaling Redundant Coding: One option available to you when creating a proportional symbol map is the idea of redundant coding. Redundant coding is the use of more than one visual variable to differentiate the symbols from each other. For example, as the value of the attribute increases, we can increase the size of the symbol and darken the color of the symbol. By using more than one visual variable to differentiate the symbols, this reinforces the idea of increasing or decreasing magnitude. The most common way in which redundant coding is used on a proportional symbol map is through varying the lightness and hue in concert with the size of the symbol. In this case, lighter colors should represent lower values and darker colors should represent higher values. If the circles represent one kind of thing, and it is recommended that you choose a single color and simply vary the lightness and darkness. Figure 20 is an example map that employs redundant coding. The smaller circles are colored with lighter colors and represent smaller values. The higher values used darker colors to represent higher values. Using redundant coding can add a nice visual impact on the map. Figure 20 Redundant Coding Figure 20: Redundant Coding Reference Features:Now you can consider the use of reference features on a proportional symbol map. Again, thematic maps should be simple by design so that the map user can focus on the proportional symbols and the data they represent. You should avoid placing reference features unless they are important in explaining the pattern of the variable being mapped. Legend Design:On a proportional symbol map, the legend serves as a visual anchor for interpreting symbol sizes. There are four common legend layouts, which are: vertical, horizontal, nested symbols, and nested semi-symbols. In these different legend layouts numbers or data values are placed to the right of the symbols with the vertical or nested legend layout and below the symbols if it is a horizontal layout. Vertical Layout:For the vertical legend layout, the values are displayed to the right of the symbol. Also, small values are at the top and large values are at the bottom Figure 21 Vertical Layout Figure 21: Vertical Layout Horizontal Layout:For the horizontal legend, layout values are below the symbols and small values are on the left and large values are on the right. Figure 22 Horizontal Layout Figure 22: Horizontal Layout Nested Symbols Layout:In the nested symbols legend layout, the symbols are placed on top of each other with the smallest symbol on top of the largest symbol on the bottom. The symbols are also a line along their tops or bottoms. The main reason to use a nested symbols legend layout is that it requires less space on the map layout and the vertical or horizontal legend layout. To save even more space you can place the values inside the symbols if there is space available, otherwise, the value should be to the right of the nested symbols with a leader line connecting the values to the symbols. Figure 23 Nested Symbols Layout Figure 23: Nested Symbols Layout Nested Semi-Symbol Layout:The nested semi-symbols legend layout requires the least amount of space on a map. This is essentially the legend of the nested symbols but with half of the symbol missing and replaced with the values. You may also have leader lines from the top of the symbols to values as well. Figure 24 Nested Semi-Symbol Layout Figure 24: Nested Semi-Symbol Layout Proportional Symbol Map - Legend Design:With respect to legend designed for proportional symbol maps, there are a few special cases to consider. If a single, or if your values are significantly larger or smaller than all the other values, we may consider these outliers. In the case of outliers, we can label the symbol on the map with its name and value to bring attention to it. We can also label the symbol on the legend with each unique name and value or change the color of the outlier to bring attention to it to show that it is different in some way than the other symbols. If there is not much congestion on the map it is okay to label the proportional symbols on the map itself with their values. 4.3 Dot Density Maps A dot density map is a map showing total values represented by randomly placed dots within an enumeration area to represent density and spatial distribution. This example map shows population density for the year 2007 for the United States of America. On this map, 1 dot represents 200,000 people. Each dot is randomly placed within its enumeration unit, which in this case, is the county. Where the dots are denser on the map the user will interpret this area as having more value. Where there are fewer dots on the map, the user will interpret this are having less value. Figure 25 Dot Density Map Figure 25: Dot Density Map Why Create a Dot Density Map?A dot density map is easy to create and easy to interpret. It excels at displaying a variable’s overall geographic pattern and density. Counting the number of dots in an area will allow the map user to ascertain total values with some rounding error. Dot density maps can reflect distributions more accurately than other thematic map types. Why Not Create a Dot Density Map? So why should you not create a dot density map? First, map readers do not perceive dot densities linearly; they may overestimate or underestimate the densities of the dots as density increases in an area. Second, the dots may not be automatically placed close to the phenomenon they represent since the dots are randomly placed. Third, the large ranges of data values make it hard to choose a single dot value. Fourth it may be hard for the map users to recover original totals when dots are placed close together making it hard to pick out individual dots. Fifth and finally dots may appear where they cannot possibly exist. For example, a dot representing cattle may show up in a lake when that does not make sense for the dot to be there. Appropriate Data:Total values are the types of data that are appropriate for a dot density map. The data should be represented and aggregated to an enumeration area. You should strive for the smallest enumeration units possible to maximize the likelihood that the dot will be placed close to the location of the phenomenon. Generally, the enumeration units themselves should not be shown on the map. If you wish to show the enumeration units on the map, either deemphasize enumeration units or keep the enumeration units off the map and shows the next level up of an enumeration unit. For example, if our data’s enumeration unit is counties, then we could not show counties of a map, but instead, show state boundaries. Data sets that do not have large or small ranges are also appropriate for a dot density map. If the data set has a small range of values it will lead to a uniformed look on the map. If the data set has a large range of values it will be hard to choose a good size. Inappropriate Data: Inappropriate data for a dot density map is continuous data that is not controlled by an enumeration unit. Data sets with small or large ranges are not appropriate. Derived data such as persons per square mile is not appropriate for a dot density map. Map Projections:The best map projection for a dot density map is the equivalent or equal-area map projection. As the relative size is important to maintain when comparing values within enumeration units, you should choose a map projection that maintains relative size. About the Dots: It is important that on all dot-density maps each dot represents more than one item. If the dot represents exactly one item then the dot should be placed exactly where the item it represents is located. If this is the case then what you are creating is a general reference map, not a dot density map. Dots should be large enough to stand out but small enough to not be totally dominant on their own. Generalization of the dot values occurs at one half of the dot value. That means a dot that has the value 100,000 can be considered to represent all values between 50,000 and 149,999. Each dot is a spatial proxy which means that the dots are placed around the center of gravity of the geographic phenomenon. Random Placement:Each dot is randomly placed inside the enumeration unit. The dots are randomly placed to avoid giving the impression that the dots are precisely placed and to avoid regular placement of the dots having large values. You can use ancillary data to restrict the placement of dots. For example, you could tell the software to not place the dots within lakes if it does not make sense for the dots to appear within the lake. General Dot Guidelines:Here are some general dot size guidelines. Enumeration units with the smallest value should have about two to three dots inside of it. Dots should just begin to coalesce in the most dense enumeration area. Dot values should be easy to understand and to count with. For example, 500 and 1000 are good dot values. Good Dot Size and Value: In Figure 26, the dots are of a good size and of good value. In the densest area of the maps, the dots just begin to coalesce; you can still generally pick out each individual dot. Figure 26 Good Dot Size and Value Figure 26: Good Dot Size and Value Good Dot Size, Poor Value: In Figure 27 the dots are of a good size; however, the dot value was chosen poorly and is too large. Because the dot value is so large most of the map looks sparse and the densest areas do not give a good feeling of high density. Figure 27 Good Dot Size, Poor Value Figure 27: Good Dot Size, Poor Value Poor Dot Size, Good Value:In Figure 28 the dots are of a poor size but the dot value is good. As the dot size is so small it still leaves the map looking sparse. The dot size should be increased so that the dots are easier to see and the map looks more filled in. Figure 28 Poor Dot Size, Good Value Figure 28: Poor Dot Size, Good Value Other Symbols:It is possible to use other symbols than dots. You can use geometric shapes which are familiar to map readers and are simple and easily recognizable at multiple sizes. Another option is to use pictorial shapes which may add to the theme and memorability of the map. The negative aspect of pectoral shapes is that if the pectoral shapes are too complex they may not scale well to multiple sizes. Reference Features:With respect to reference features on a dot density map, like other thematic maps, it too should be kept simple by design. You should avoid placing reference features unless they are important in explaining the pattern of the variable being mapped. Dot Density Map - Legend Designs:Designing legends for dot density map is fairly straightforward. On the legend you should have a representative symbol and a statement about the value of that symbol. Rarely are legend headings required for a dot density map legend. You can include additional information in the legend if necessary, such as the total of all values of a map, or representation of what a low, medium, and high density looks like on the map. Figure 29 Legend Design Figure 29: Legend Design Other Resources Hands-On Activities: 1) The Power of Maps(Learn ArcGIS Lesson; 50 minutes): 2) Map Breast Cancer Differences by Ethnicity(Learn ArcGIS Lesson; 1 hour) 3) Understand the Refugee Crisis with Link Analysis(Learn ArcGIS Lesson; 1 hour and 40 minutes) Summary This chapter covered various types of maps and their features. Choropleth, Dot Density, and Proportional Symbol Maps were exhibited and explained. You learned about appropriate and inappropriate data as well as data classifications and symbolization for each map type. Map legends and the elements that should be considered when using this feature were also covered. This work by the National Information Security and Geospatial Technologies Consortium (NISGTC), and except where otherwise noted, is licensed under the Creative Commons Attribution 3.0 Unported License. Authoring Organization:Del Mar College Close Close Figure 1: Choropleth Map Close Figure 2: Black and White Choropleth Map Close Figure 3: Color Choropleth Map Close Figure 4: Legend Box Close Figure 5: Horizontal Legend Box Close Figure 6: Vertical Legend Box Close Figure 7: Continuous Classes Close Figure 8: Non-Continuous Classes Close Figure 9: Proportional Symbol Map Close Figure 10: (left to right) 2-D Circle, Triangle, and Star Close Figure 11: 3-D Square and Cylinder Close Figure 12: Pictograms of a Bus, Airplane, and Taco Close Figure 13: Overlap Close Figure 14: Appropriate Overlapping on a Map Close Figure 15: Little Overlap, Empty, Uninteresting Figure 16: Too Much Overlap Close Figure 17: Absolute Scaling Close Figure 18: Comparing Absolute and Apparent Magnitude Scaling Close Figure 19: Range Grading Scaling Close Figure 20: Redundant Coding Close Figure 21: Vertical Layout Close Figure 22: Horizontal Layout Close Figure 23: Nested Symbols Layout Close Figure 24: Nested Semi-Symbol Layout Close Figure 25: Dot Density Map Close Figure 26: Good Dot Size and Value Close Figure 27: Good Dot Size, Poor Value Close Figure 28: Poor Dot Size, Good Value Close Figure 29: Legend Design
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Skip to Main content Sign in Chapters and Articles You might find these chapters and articles relevant to this topic. Review article Critical illness and intensive care II 2009, Surgery (Oxford)Claire Baylis, Chris Till The normal haemoglobin–oxygen dissociation curve is shown in Figure 2. The curve is shifted to the right by increased CO2, temperature, diphosphoglycerate and [H+]. Thus in the tissues there is better unloading of oxygen for a given PCO2. Conversely, decreased CO2, temperature, [diphosphoglycerate] and [H+] shifts the curve to the left. This improves oxygen binding in pulmonary capillaries. The leftward shift is accentuated at altitude. View article Read full article URL: Journal2009, Surgery (Oxford)Claire Baylis, Chris Till Review article Critical illness and intensive care II 2009, Surgery (Oxford)Claire Baylis, Chris Till The normal haemoglobin-oxygen dissociation curve. Partial pressures of oxygen corresponding to saturations of 50%, 75% (mixed venous blood) and 97% (arterial blood) are shown View article Read full article URL: Journal2009, Surgery (Oxford)Claire Baylis, Chris Till Review article Anemia and Heart Failure 2010, Heart Failure ClinicsInder S. Anand MD, FRCP, DPhil (Oxon) Hemoglobin-Oxygen Dissociation Curve A change in the affinity of hemoglobin for oxygen is a rapid and reversible process allowing for immediate adjustments of oxygen binding and oxygen release at the periphery. Hemoglobin-oxygen dissociation is influenced by several factors. A low pH and high concentrations of 2,3-diphosphoglycerate (2,3-DPG) shift the hemoglobin-oxygen dissociation curve to the right, which increases the P50, decreases the affinity of hemoglobin for oxygen, and improves oxygen delivery to the tissues. In chronic anemia the red blood cell (RBC) concentration of 2,3-DPG is increased and the hemoglobin-oxygen dissociation curve is shifted to the right.72,73 A right-shifted curve is the least energy-consuming mechanism to support increased oxygen delivery to the tissues without a significant increase in cardiac output. The 2,3-diphosphoglycerate is a metabolic intermediate of RBC glycolysis and is quantitatively the most important organic phosphate regulating oxygen affinity in erythrocytes.74 The activity of many red blood cell glycolytic enzymes, and therefore the concentrations of 2,3-DPG and adenosine triphosphate, are largely dependent on the age of the erythrocyte. Enzyme activity is especially high in young red blood cells and decreases with increasing cell age. Consequently, young erythrocytes have a right-shifted oxygen dissociation curve and a steeper curve slope, which further adds to the effectiveness of oxygen unloading.75 View article Read full article URL: Journal2010, Heart Failure ClinicsInder S. Anand MD, FRCP, DPhil (Oxon) Chapter Congenital Erythrocytosis 2018, Diagnostic Pathology: Blood and Bone Marrow (Second Edition) Overview of Oxygen Delivery and Hemoglobin Oxygen Affinity • : Hemoglobin's oxygen dissociation curve describes relationship between hemoglobin saturation and pO₂ • : Curve will be shifted leftward by factors that increase affinity of hemoglobin for oxygen (higher saturation for a given pO₂) ○ : Consequently, oxygen is bound more avidly to hemoglobin and is not as readily available for delivery to tissues • : Pathologic causes of leftward shifted oxygen dissociation curves result in erythrocytosis in compensation for decreased oxygen delivery View chapterExplore book Read full chapter URL: Book2018, Diagnostic Pathology: Blood and Bone Marrow (Second Edition) Chapter R espiratory Acid-Base Disturbances 2010, Fluid, Electrolyte and Acid-Base Physiology (Fourth Edition)Mitchell L. Halperin MD, FRCPC, ... Marc B. Goldstein MD, FRCPC A shift in the oxygen-hemoglobin dissociation curve must not interfere with the sensitivity of this system to release erythropoietin If the oxygen-hemoglobin dissociation curve in capillaries of the renal cortex is always shifted to the right, the Po2 signal is not influenced by other factors that may influence this shift. In fact, this is achieved by having a high Pco2 in blood vessels in the renal cortex (~65 mm Hg). QUESTIONS (Discussions on pages 240 and 242) 8-3 : Why might sports anemia (see margin note) be tolerated without an apparent erythropoietin-induced drive to synthesize red blood cells to correct the anemia? 8-4 : Why does a young patient with diabetes mellitus who has a high GFR not develop polycythemia? SPORTS ANEMIA Trained athletes often have a lower hematocrit level that results from a normal red blood cell pool size and an increased plasma volume. Nevertheless, they do not have the expected response to anemia as a result of blood loss where there is evidence of accelerated synthesis of new red blood cells. View chapterExplore book Read full chapter URL: Book2010, Fluid, Electrolyte and Acid-Base Physiology (Fourth Edition)Mitchell L. Halperin MD, FRCPC, ... Marc B. Goldstein MD, FRCPC Chapter Physiology of the Airway 2013, Benumof and Hagberg's Airway ManagementWilliam C. Wilson, Jonathan L. Benumof b Oxygen-Hemoglobin Dissociation Curve As a red blood cell (RBC) passes by the alveolus, O2 diffuses into plasma and increases Pao2. As Pao2 increases, O2 diffuses into the RBC and combines with Hb. Each Hb molecule consists of four heme molecules attached to a globin molecule. Each heme molecule consists of glycine, α-ketoglutaric acid, and iron in the ferrous (Fe2+) form. Each ferrous ion has the capacity to bind with one O2 molecule in a loose, reversible combination. As the ferrous ions bind to O2, the Hb molecule begins to become saturated. The oxy-Hb dissociation curve relates the saturation of Hb (rightmost y-axis in Fig. 5-25) to Pao2. Hb is fully saturated (100%) by a Po2 of approximately 700 mm Hg. The saturation at normal arterial pressure (point a on upper, flat part of the oxy-Hb curve in Figure 5-25) is 95% to 98%, achieved by a Pao2 of about 90 to 100 mm Hg. When Po2 is less than 60 mm Hg (90% saturation), saturation falls steeply, and the amount of Hb uncombined with O2 increases greatly for a given decrease in Po2. Mixed venous blood has a Po2 () of about 40 mm Hg and is approximately 75% saturated, as indicated by the middle of the three points () on the oxy-Hb curve in Figure 5-25. The oxy-Hb curve can also relate the O2 content (Co2) (vol%, or mL of O2 per dL of blood; see Fig. 5-25) to Po2. Oxygen is carried both in solution in plasma (0.003 mL of O2/mm Hg Po2 per dL) and combined with Hb (1.39 mL of O2/g of Hb), to the extent (percentage) that Hb is saturated. Therefore, For a patient with an Hb content of 15 g/dL, a Pao2 of 100 mm Hg, and a of 40 mm Hg, the arterial O2 content (Cao2) = (1.39)(15)(1) + (0.003)(100) = 20.9 + 0.3 = 21.2 mL/dL; the mixed venous O2 content () = (1.39)(15)(0.75) + (0.003)(40) = 15.6 + 0.1 = 15.7 mL/dL. Therefore, the normal arteriovenous O2 content difference is approximately 5.5 mL/dL of blood. Note that equation (13) uses the constant 1.39, which means that 1 g of Hb can carry 1.39 mL of O2. Controversy exists over the magnitude of this number. Originally, 1.34 had been used,92 but with determination of the molecular weight of Hb (64,458), the theoretical value of 1.39 became popular.93 After extensive human studies, Gregory observed in 1974 that the applicable value was 1.31 mL O2/g of Hb in human adults.94 That the clinically measured Co2 is lower than the theoretical 1.39 is probably due to the small amount of methemoglobin (MetHb) and carboxyhemoglobin (COHb) normally present in blood. The oxy-Hb curve can also relate O2 transport (L/min) to the peripheral tissues (see Fig. 5-25) to Po2. The term O2 transport is synonymous with the term O2 delivery. This value is obtained by multiplying the O2 content by (O2 transport = × Cao2). To do this multiplication, one must convert the content unit of mL/dL to mL/L by multiplying by 10; subsequent multiplication of mL/L against in L/min yields mL/min. Thus, if = 5 L/min and Cao2 = 20 mL of O2/dL, the arterial point corresponds to 1000 mL O2/min going to the periphery, and the venous point corresponds to 750 mL O2/min returning to the lungs, with = 250 mL/min. The oxy-Hb curve can also relate the O2 actually available to the tissues (leftmost y axis in Fig. 5-25) as a function of Po2. Of the 1000 mL/min of O2 normally going to the periphery, 200 mL/min of O2 cannot be extracted because it would lower Po2 below the level at which organs such as the brain can survive (rectangular dashed line in Fig. 5-25); the O2 available to tissues is therefore 800 mL/min. This amount is approximately three to four times the normal resting . When = 5 L/min and arterial saturation is less than 40%, the total flow of O2 to the periphery is reduced to 400 mL/min; the available O2 is then 200 mL/min, and O2 supply just equals O2 demand. Consequently, with low arterial saturation, tissue demand can be met only by an increase in or, in the longer term, by an increase in Hb concentration. The affinity of Hb for O2 is best described by the Po2 level at which Hb is 50% saturated (P50) on the oxy-Hb curve. The normal adult P50 is 26.7 mm Hg (see Fig. 5-25). The effect of a change in Po2 on Hb saturation is related to both P50 and the portion of the oxy-Hb curve at which the change occurs.95 In the region of normal Pao2 (75 to 100 mm Hg), the curve is relatively horizontal, and shifts of the curve have little effect on saturation. In the region of mixed venous Po2, where the curve is relatively steep, a shift of the curve leads to a much greater difference in saturation. A P50 lower than 27 mm Hg describes a left-shifted oxy-Hb curve, which means that at any given Po2, Hb has a higher affinity for O2 and is therefore more saturated than normal. This lower P50 may require higher than normal tissue perfusion to produce the normal amount of O2 unloading. Causes of a left-shifted oxy-Hb curve are alkalosis (metabolic and respiratory—the Bohr effect), hypothermia, abnormal fetal Hb, carboxyhemoglobin, methemoglobin, and decreased RBC 2,3-diphosphoglycerate (2,3-DPG) content. (The last condition may occur with the transfusion of old acid citrate-dextrose–stored blood; storage of blood in citrate-phosphate-dextrose minimizes changes in 2,3-DPG with time.95) A P50 higher than 27 mm Hg describes a right-shifted oxy-Hb curve, which means that at any given Po2, Hb has a low affinity for O2 and is less saturated than normal. This higher P50 may allow a lower tissue perfusion than normal to produce the normal amount of O2 unloading. Causes of a right-shifted oxy-Hb curve are acidosis (metabolic and respiratory—the Bohr effect), hyperthermia, abnormal Hb, increased RBC 2,3-DPG content, and inhaled anesthetics (see later discussion).95 Abnormalities in acid-base balance result in alteration of 2,3-DPG metabolism to shift the oxy-Hb curve to its normal position. This compensatory change in 2,3-DPG requires between 24 and 48 hours. Therefore, with acute acid-base abnormalities, O2 affinity and the position of the oxy-Hb curve change. However, with more prolonged acid-base changes, the reciprocal changes in 2,3-DPG levels shift the oxy-Hb curve and O2 affinity back toward normal.95 Many inhaled anesthetics have been shown to shift the oxy-Hb dissociation curve to the right.96 Isoflurane shifts P50 to the right by 2.6 ± 0.07 mm Hg at a vapor pressure of approximately 1 minimum alveolar concentration (MAC) (1.25%).97 On the other hand, high-dose fentanyl, morphine, and meperidine do not alter the position of the curve. View chapterExplore book Read full chapter URL: Book2013, Benumof and Hagberg's Airway ManagementWilliam C. Wilson, Jonathan L. Benumof Chapter Strategies for Blood Product Management and Reducing Transfusions 2009, A Practice of Anesthesia for Infants and Children (Fourth Edition)Charles J. Coté, ... Christopher P. Stowell Oxygen-Hemoglobin Dissociation Oxygen binding to hemoglobin is primarily dependent on oxygen tension, temperature, acid-base status, and 2,3-DPG. Citrate is rapidly metabolized to bicarbonate and thus causes metabolic alkalosis within hours of massive transfusion.189,209,221 If exogenous bicarbonate therapy is superimposed on this endogenous metabolic alkalosis, a significant effect on the dissociation of oxygen and hemoglobin could theoretically result. It is important to determine the acid-base status before administering sodium bicarbonate to avoid intensifying the left shift caused by hypothermia and metabolized citrate. Each type of blood component has slightly different effects on the oxygen-hemoglobin dissociation curve; it is likely that these concerns are more of a theoretical than a real concern because most of these effects are very short lived. View chapterExplore book Read full chapter URL: Book2009, A Practice of Anesthesia for Infants and Children (Fourth Edition)Charles J. Coté, ... Christopher P. Stowell Chapter Respiratory Physiology in Infants and Children 2011, Smith's Anesthesia for Infants and Children (Eighth Edition)Etsuro K. Motoyama, Jonathan D. Finder Oxygen Affinity of Hemoglobin and P50 The oxygen-hemoglobin dissociation curve reflects the affinity of hemoglobin for oxygen (Fig. 3-43). As blood circulates through the normal lungs, oxygen tension increases from the mixed-venous Po2 of around 40 mm Hg to pulmonary capillary Po2 of above 105 mm Hg, and hemoglobin is saturated to about 97% in arterial blood. (Unfortunately, most pulse oximeters commercially available today are artificially modified to read 100% saturation in healthy subjects breathing room air rather than 97%; see later discussion.) The shape of the dissociation curve is such that further increases in Po2 result in a very small increase in oxygen saturation (So2) of hemoglobin. The blood of normal adults has So2 of 50% when Po2 is 27 mm Hg at 37° C and a pH of 7.4. The P50, which is the Po2 of whole blood at 50% So2, indicates the affinity of hemoglobin for oxygen. An increase in blood pH increases the oxygen affinity of hemoglobin (Bohr effect) and shifts the oxygen-hemoglobin (O2-Hb) dissociation curve to the left. Similarly, a decrease in temperature also increases oxygen affinity and shifts the O2-Hb dissociation curve to the left; a decrease in pH or an increase in temperature has the opposite effect and the O2-Hb curve shifts to the right (Comroe, 1974) (Fig. 3-43). Benesch and Benesch (1967) and Chanutin and Curnish (1967) demonstrated that the oxygen affinity of a hemoglobin solution decreases by the addition of organic phosphates, in particular 2,3-diphosphoglycerate (2,3-DPG) and adenosine triphosphate (ATP), which bind to deoxyhemoglobin but not to oxyhemoglobin. Human erythrocytes contain an extremely high concentration of 2,3-DPG, averaging about 4.5 mol/mL, compared with ATP (1 mol/mL) and other organic phosphates (Oski and Delivoria-Papadopoulos, 1970). Thus, an increase in red cell 2,3-DPG decreases the oxygen affinity of hemoglobin, increases P50 (shifts the dissociation curve to the right), and increases the unloading of oxygen at the tissue level. Increases in 2,3-DPG and P50 have been found in chronic hypoxemia. In the newborn, blood oxygen affinity is extremely high and P50 is low (18 to 19 mm Hg), because 2,3-DPG is low and fetal hemoglobin (HbF) reacts poorly with 2,3-DPG (Fig. 3-44). Oxygen delivery at the tissue level is low despite high red blood cell mass and hemoglobin level. After birth, the total hemoglobin level decreases rapidly as the proportion of HbF diminishes, reaching its lowest level by 2 to 3 months of age (physiologic anemia of infancy) (Fig. 3-45). During the same early postnatal period, P50 increases rapidly; it exceeds the normal adult value by 4 to 6 months of age and reaches the highest value (P50 = 30) by 10 months and remains high during the first decade of life (Fig. 3-46) (Oski and Delivoria-Papadopoulos, 1970; Oski, 1973a, 1973b). This high P50 is associated with a relatively low hemoglobin level (10 to 11 g/dL) and an increased level of 2,3-DPG, probably related to the process of general growth and development and high plasma levels of inorganic phosphate (Card and Brain, 1973). These observations engendered a hypothesis to explain why hemoglobin levels are relatively lower in children than in adults (physiologic “anemia” of childhood) (Card and Brain, 1973). Because children have a lower oxygen affinity for hemoglobin, oxygen unloading at the tissue level is increased. Thus, a lower level of hemoglobin in infants and children is just as efficient, in terms of tissue oxygen delivery, as a higher hemoglobin level in adults (Oski, 1973a) (Table 3-3). Table 3-4 compares the hemoglobin concentrations at different ages in terms of equal tissue oxygen unloading (Motoyama et al., 1974). Acceptable Hemoglobin Levels These findings have important clinical implications for anesthesiologists. Until the 1980s it was assumed that children with a hemoglobin level of less than 10 g/dL were not acceptable for general anesthesia and surgery. This level of hemoglobin has been used arbitrarily without the knowledge of different oxygen affinity and tissue oxygen unloading at different ages. It appears from Table 3-4 that if a hemoglobin level of 10 g/dL is acceptable for an adult with a P50 of 27 mm Hg, 8.2 g/dL should theoretically be adequate for an infant older than 8 months of age with an average P50 of 30 mm Hg (without considering the high level of metabolism and oxygen consumption). In contrast, for a 2-month-old premature infant with a P50 of 24 mm Hg, a hemoglobin level of 10 g/dL is equivalent to only 6.8 g/dL in adults, and this may be inadequate to provide sufficient tissue oxygenation in patients with limited cardiac output or oxygen desaturation. With the advent of human immunodeficiency virus (HIV) and acquired immunodeficiency syndrome (AIDS) and the resultant anxiety among the medical community and the lay public about homologous blood transfusion, the criteria for transfusion have changed significantly since the 1980s. At the consensus-developing conference by the National Institutes of Health and the Food and Drug Administration on Perioperative Red Blood Cell Transfusion, it was agreed that the available evidence does not support the “10/30” rule (that is, hemoglobin, 10 g/dL, or hematocrit, 30%), although the literature is remarkable for its lack of carefully controlled, randomized studies that would provide definitive conclusions (Consensus Conference, 1988). Other data suggest that cardiac output does not increase dramatically in healthy adult humans until the hemoglobin value decreases to approximately 7 g/dL. At the Consensus Conference (1988) it was also agreed that the decision to transfuse red blood cells in a specific patient should take into consideration the many factors that comprise clinical judgment. These factors include the duration of anemia, the intravascular volume, the extent of surgery, the probability of massive blood loss, and the presence of coexisting conditions such as impaired cardiopulmonary function and inadequate cardiac output. A general consensus on the acceptable perioperative levels of hemoglobin and hematocrit in infants and young children has not emerged, and the lowest safe limit of hemoglobin for infants less than 2 months of age has not been determined, although in sick infants it is desirable to maintain a hemoglobin level of 12 to 13 g/dL or a hematocrit of 40% (equivalent to 8 g/dL in adults) (see Chapter 14, Blood Conservation). There has been controversy about what constitutes abnormally low oxygen saturation in infants and children postoperatively and what is considered clinically unsafe. Mok and others (1986) reported that during the first week of life, oxygen saturation as monitored with pulse oximetry (SpO2) was noticeably decreased, especially during REM sleep (mean SpO2, 92%) and during feeding (SpO2, 91%). After 4 weeks of age, however, SpO2 was more stable and was maintained at or above 94% during sleep. Thus, an SpO2 of less than 94% can be considered as physiologically abnormal in infants beyond the first week of age. A study in preterm infants (mean gestational age, 33 weeks; postconceptional age, 37 weeks) has shown that median SpO2 at the time of discharge was 99.5% and increased to 100% at follow-up 6 weeks later. The preterm infants had higher baseline saturation and no more incidence of desaturation than full-term infants of equivalent postconceptional ages (Poets et al., 1992). It is generally agreed that SpO2 less than 95% in otherwise healthy infants and children is abnormal and that these patients require oxygen supplementation. The routine use of pulse oximetry has dramatically improved the anesthesiologist's ability to monitor and properly maintain proper oxygenation (Coté et al., 1988, 1991). This is especially true for premature infants, who are susceptible to oxygen toxicity and retinopathy of prematurity, even when breathing room air (Wilson-Mikity syndrome). In premature infants weighing less than 1300 g, the incidence of retinopathy of prematurity increases markedly with exposure to 12 or more hours of Pao2 exceeding 80 mm Hg (Flynn et al., 1992). Arterial oxygen saturation (Sao2) must be adjusted properly so as to maintain Pao2 in the normal neonatal range of 60 to 80 mm Hg (Orzalesi et al., 1967). As mentioned, oxygen affinity to hemoglobin is very high in the neonate and decreases rapidly during the first 3 to 6 months of life (Oski, 1973a, 1973b, 1981). Estimated Pao2 should be adjusted according to age, as shown in Table 3-5. In the newborn, whose P50 is 18 to 20 mm Hg, the range of Sao2 to maintain adequate Pao2 (60 to 80 mm Hg) is 97% to 98% (assuming no transfusion with adult blood has been given), whereas in the adult (P50, 27), it is 91% to 96%. In the neonate, Sao2 of 91% corresponds to Pao2 of 41 mm Hg. Although the values in Table 3-5, based on Severinghaus's nomogram for the Bohr effect, are only estimates, published data comparing arterial Po2 and oxygen saturation seem to agree well with values in the nomogram (Severinghaus, 1966; Ramanathan et al., 1987; Bucher et al., 1989). Unfortunately, another factor compounding the confusion (and clinically too important to ignore) is that the most commonly used pulse oximeters in the United States are artificially set to read 2% to 3% higher at the 90% to 95% range than actual arterial oxygen hemoglobin saturation (as measured by means of cooximetry), and that the pulse oximeters most commonly used in Europe tend to read somewhat lower than actual arterial oxygen saturation (Jennis and Peabody, 1987; Bucher et al., 1989). Unfortunately, a newer and technologically advanced pulse oximeter with less motion artifact, which has increasingly been used in the United States and elsewhere, also h as artificially increased readings; it also reads 2% to 3% higher, matching the reading of the more traditionally used pulse oximeter (M. Patterson, 1995, personal communications). In view of these findings, the range of SpO2 of 93% to 95%, corresponding to an estimated Pao2 of 66 to 74 mm Hg in adults (but only 40 to 50 mm Hg in neonates), often recommended as desirable maintenance levels for neonates and premature infants in intraoperatively or in the intensive care settings, appears much too low for adequate tissue oxygenation. Furthermore, respiratory alkalosis, which may result from assisted or controlled ventilation, would shift the oxygen hemoglobin dissociation curve further to the left (P50, even lower than it already is) and decrease Pao2 and tissue oxygen delivery even further at this range of oxygen saturation (Fig. 3-43). Therefore, in clinical practice, SpO2 levels of 95% to 97% (corresponding Pao2 of 50 to 70 mm Hg in neonates and 60 to 80 mm Hg in infants 1 to 2 months old) but not higher, should be considered. Some anesthetics affect the oxygen affinity of hemoglobin. The presence of cyclopropane (although it has not been used since the 1970s) significantly decreases oxygen affinity and increases P50 by 3 mm Hg without changes in the 2,3-DPG levels, whereas halothane has minimal effects (Orzalesi et al., 1971). Exposure to 50% nitrous oxide, on the other hand, has been reported to produce a marked reversible increase in oxygen affinity; P50 decreased from 26 to 18 mm Hg, a level similar to that of HbF (Fournier and Major, 1984). This finding contrasts with a report based on one patient by Prime (1951), who found no effect with 70% nitrous oxide, and with a study by Smith and others (1970) who reported a 3 mm Hg rightward shift of P50 with an unspecified concentration of nitrous oxide. The finding of Fournier and Major above may be of considerable clinical importance. Although less often used of late, nitrous oxide anesthesia combined with hyperventilation would markedly increase the oxygen affinity of hemoglobin and decrease oxygen unloading at the tissue level. This effect could be hazardous in neonates whose P50 is unusually low even without respiratory alkalosis or nitrous oxide. View chapterExplore book Read full chapter URL: Book2011, Smith's Anesthesia for Infants and Children (Eighth Edition)Etsuro K. Motoyama, Jonathan D. Finder Review article Controversies in Perioperative Care Medicine 2021, Best Practice & Research Clinical AnaesthesiologyKamal Maheshwari MD MPH Factors affecting oxygen uptake At the tissue level, despite adequate FiO2, various physio-chemical factors affect the availability of free oxygen to the tissues. During hypercapnia, acidosis, hyperthermia and increasing 2,3-biphosphoglycerate, the hemoglobin oxygen dissociation curve shifts to right, delivering more oxygen to the tissues. Vascular tone and permeability is affected by drugs or disease state and may affect oxygen diffusion . For example, in sepsis, increased vascular permeability and reduced tone cause tissue edema and limit oxygen diffusion. Tissue may not be able to utilize oxygen, whereas during carbon monoxide poisoning and methemoglobinemia blood oxygen-carrying capacity is reduced. The oxygen requirement may increase under anesthesia, surgery, and during critical illness. While the oxygen consumption is reduced under general anesthesia, recovering to baseline or even increasing in postoperative period . During the use of anesthesia, oxygen extraction in tissues is reduced, necessitating higher delivery . Overall, it is uncertain if some or all surgical patients require supra-normal oxygen delivery in perioperative period. This uncertainty may explain the reported variability in treatment effect with numerous goal-directed fluid therapy protocols, which primarily rely on maximizing stroke volume and thus oxygen delivery. Nonetheless, intraoperative tissue hypoxia is associated with worse outcomes and should be avoided [25,26]. Older patients with many comorbid conditions are at higher risk of imbalance between oxygen delivery and requirement. Many disease states limit oxygen uptake, for example, in chronic obstructive lung disease. Due to physiological shunt and dead space in lungs, 10-fold reduction in oxygen uptake can happen. There are also other diseases, such as congestive heart failure and peripheral vascular disease, limit oxygen delivery. Thus, sicker patient are more sensitive to changes in oxygen delivery and may seem to benefit from higher oxygen delivery and goal-directed therapy. View article Read full article URL: Journal2021, Best Practice & Research Clinical AnaesthesiologyKamal Maheshwari MD MPH Related terms: Carboxyhemoglobin Carbon Monoxide Sodium Bicarbonate Oxygen Saturation Embryonic Hemoglobin Hypothermia Acidosis 2,3 Diphosphoglyceric Acid Hypoxemia Carbon Dioxide View all Topics 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. You can manage your cookie preferences using the “Cookie Settings” link. For more information, see ourCookie Policy 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. 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https://www.quora.com/What-is-the-relationship-between-the-sides-and-angles-of-a-triangle
Something went wrong. Wait a moment and try again. Mathematical Relationship Lines and Angles Geometric Properties Mathematical Structures Geometric Mathematics 5 What is the relationship between the sides and angles of a triangle? · The relationship between the sides and angles of a triangle is governed by several fundamental principles in geometry, primarily encapsulated in the following concepts: Triangle Inequality Theorem The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This can be expressed as: - a+b>c - a+c>b - b+c>a Where a, b, and c are the lengths of the sides of the triangle. Angle-Side Relationships In a triangle, there is a direct relationship between the angles and the lengths of the sides: - Larger Angle = Longer Side: The side opposite the The relationship between the sides and angles of a triangle is governed by several fundamental principles in geometry, primarily encapsulated in the following concepts: Triangle Inequality Theorem The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. This can be expressed as: - a+b>c - a+c>b - b+c>a Where a, b, and c are the lengths of the sides of the triangle. Angle-Side Relationships In a triangle, there is a direct relationship between the angles and the lengths of the sides: - Larger Angle = Longer Side: The side opposite the largest angle is the longest side, and the side opposite the smallest angle is the shortest side. Mathematically, if A, B, and C are the angles opposite to sides a, b, and c respectively, then: - If A>B, then a>b - If A<B, then a<b Law of Sines The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles: asinA=bsinB=csinC This law is particularly useful for solving triangles when you know some combination of sides and angles. Law of Cosines The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles: c2=a2+b2−2ab⋅cosC This is useful for finding a side when you know two sides and the included angle, or for finding an angle when you know all three sides. Sum of Angles The sum of the interior angles of a triangle is always 180∘: A+B+C=180∘ Summary In summary, the sides and angles of a triangle are intricately linked through geometric principles that dictate their relationships. Understanding these relationships is essential for solving various problems in triangle geometry, including finding unknown sides or angles. Alexander Mathey Former Chemical Engineer, retired, lives in Athens, GR · Author has 5.6K answers and 10.8M answer views · 2y Originally Answered: What is the relationship between the sides and angles of a triangle given its side lengths and angle measures? · It is called ‘The law of cosines’. It states that the square of any side equals the sum of the squares of the other two sides minus double their product multiplied by the cosine of the angle opposite to that side. a² = b² + c² - 2bccosA b² = a² + c² - 2accosB c² = a² + b² - 2abcosC If the angle opposite a side is a right angle, then its cosine equals zero, and the ‘Law of cosines’ becomes the familiar form of the Pythagorean theorem. It is called ‘The law of cosines’. It states that the square of any side equals the sum of the squares of the other two sides minus double their product multiplied by the cosine of the angle opposite to that side. a² = b² + c² - 2bccosA b² = a² + c² - 2accosB c² = a² + b² - 2abcosC If the angle opposite a side is a right angle, then its cosine equals zero, and the ‘Law of cosines’ becomes the familiar form of the Pythagorean theorem. Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Reclaim your workday all year with clear, effective communication at your fingertips. Try Grammarly now! Sripad Sambrani Knows Sanskrit · Author has 6.7K answers and 2.9M answer views · 3y The corresponding angle-side pair are related by the Sine rule. Trust this helps. The corresponding angle-side pair are related by the Sine rule. Trust this helps. Related questions What are the relationships between angles, sides, and diagonals of a triangle? What is the relationship between two sides of a triangle, and three angles in it? What is the relationship between the angles of a triangle and the lengths of its sides? What is the relationship between an angle, side, and triangle? In a right angled triangle, what is the relationship between the length of the sides and the angles? Bheema Mudda Former Rtd Director at Government of India (2002–2007) · Author has 4.8K answers and 2.8M answer views · 3y 1] Sine rule 2]Cosine rule give the relations between sides &angles of triangles 1] Sine rule : a/sin A =b/Sin B =c/Sin C a ,b ,c are sides &A,B,&C are angles of triangles 2]Cosine rule a^2 =b^2+c^2 -2 b c Cos A —————-b^2 =c^2+a^2- 2 c a Cos B ——————c^2 =a^2+b^2–2 a b Cos C Ravi Sharma Former Group A Officer From Indian Railways (1973–2009) · Author has 15K answers and 3.7M answer views · 2y Relation between sides and angles of a triangle are defined by Sine and Cosine formulae SINE FORMULA a/SinA= b/SinB= c/SinC COSINE FORMULA CosA= (b²+c²-a²)/2bc CosB= (a²+c²-b²)/2ac CosC= (a²+b²-c²)/2ab Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Jul 31 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. 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What is the relationship between the sides and angles of a triangle given its side lengths and angle measures? What is the relationship between the sides and angles of a right-angled triangle? Why does this relationship hold true for any right-angled triangle? How are angles, sides and area of triangles related? What is the relationship between the interior angles of a triangle and the opposite sides? If the interior angles of a triangle add up to two right angles, are the opposite sides proportional? Why or why not? Girija Warrier Studied at Sufficiently Educated · Author has 5.9K answers and 13.8M answer views · 6y Originally Answered: What is the relation between sides and angle? · Do you mean the relation between sides & number of angles of a polygon? If it is so, the ratio is 1:1 . In any polygon, there are as many sides, as there are number of angles… like in a triangle, 3sides, 3angles. Pentagon has 5sides & 5 angles…. etc etc But if your question is relation between the number of sides & measure of angles. Then these two are not at all related… Sponsored by Amazon Business Solutions and supplies to support learning. Save on essentials and reinvest in students and staff. Easwar Ps teaching 5,6,7 ,8grade students · Author has 162 answers and 92.4K answer views · 3y The beat thing is quantifiable using trigonometry..whereas qualitatively we can say greater side has angle as greater measure and smaller side has as smaller angle opposite to it..best example is Isoceles Triangle..when the Two Adjacent Sides are Equai; angles Opposite to Them are also Equal… Gary Ward MaEd in Education & Mathematics, Austin Peay State University (Graduated 1997) · Author has 4.9K answers and 7.5M answer views · May 15 Related What is the relationship between the circumcenter and the angles of a triangle? What is the relationship between the circumcenter and the angles of a triangle? Radii drawn from corners of the circumscribed triangle to the circumcenter breaks the triangle into three isosceles triangles. Each isosceles triangle has a central angle, c1, c2, and c3. Each of these isosceles triangles has two equal base angles; b1 = (180° - c1)/2, b2 = (180° - c2)/2, and b3 = (180° - c3)/2. The three angles of the circumscribed triangle are b1 + b2, b2 + b3, and b1 + b3. What is the relationship between the circumcenter and the angles of a triangle? Radii drawn from corners of the circumscribed triangle to the circumcenter breaks the triangle into three isosceles triangles. Each isosceles triangle has a central angle, c1, c2, and c3. Each of these isosceles triangles has two equal base angles; b1 = (180° - c1)/2, b2 = (180° - c2)/2, and b3 = (180° - c3)/2. The three angles of the circumscribed triangle are b1 + b2, b2 + b3, and b1 + b3. Promoted by The Hartford The Hartford Updated Aug 22 What is small business insurance? 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It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations. Choosing the right insurance for your small business involves assessing your unique needs and consulting with an advisor to pick from comprehensive policy options. With over 200 years of experience and more than 1 million small business owners served, The Hartford is dedicated to providing personalized solutions that help you focus on growth and success. Get a quote today! Marek Čtrnáct Translator From English · Author has 816 answers and 922.2K answer views · 2y Originally Answered: What is the relationship between the sides and angles of a triangle given its side lengths and angle measures? · The simplest one is probably the law of sines. This law states that a sine of each angle, divided by the length of the opposite side, is the same for all three sides. Mathematically speaking: sin(α)a=sin(β)b=sin(γ)c Frank Jackson BSc(Hons) in Mathematics, University of Bristol (Graduated 1967) · Author has 126 answers and 79.2K answer views · Updated 10mo Related What is the relationship between the angles and sides of a triangle if one angle is equal to one of its adjacent sides? At first glance this seems nonsensical because a side of a triangle has length as a unit of measure whereas an angle when measured in radians is simply a ratio. However it is possible to assign the same real number to a triangle’s side length and to a triangle’s internal angle. However this can be done in two ways. The first way is to have angle x opposite side x with the remaining sides being a and b. The second way is to have angle x opposite side a (or b, it does not matter, since a and b are interchangeable) with the remaining sides being b and x. Case 1. Here we have by cosine rule x2=a2+b At first glance this seems nonsensical because a side of a triangle has length as a unit of measure whereas an angle when measured in radians is simply a ratio. However it is possible to assign the same real number to a triangle’s side length and to a triangle’s internal angle. However this can be done in two ways. The first way is to have angle x opposite side x with the remaining sides being a and b. The second way is to have angle x opposite side a (or b, it does not matter, since a and b are interchangeable) with the remaining sides being b and x. Case 1. Here we have by cosine rule Case 2. Here we have by cosine rule If we keep [math]x[/math] constant and vary [math]a[/math] or [math]b[/math] we get the following locus of points - Green. Case 1. The locus of possible points for vertex X is the circumcircle of triangle ABX. Case 2. The locus of possible points for the vertex A is the infinite line that is the extension of the side XA of the triangle ABX. If we give [math]x[/math] the following values {[math]\pi/6, \pi/3, \pi/2, 2\pi/3, 5\pi/6[/math]} we can plot contours of how [math]b[/math] varies with [math]a[/math] for each of the 5 values given for [math]x[/math]. Contour colours - L.Blue [math]x=\pi/6[/math], Orange [math]x=\pi/3[/math], Green [math]x=\pi/2[/math], Red [math]x=2\pi/3[/math], D.Blue [math]x=5\pi/6[/math]. Case 1. Ellipses Case 2. Hyperbolae Geoff Mangum Founder, PuttingZone in 26 countries worldwide. (2000–present) · Author has 3.8K answers and 3.1M answer views · 2y Any triangle can be scaled up or down. The three sides each change length when scaled, but the three angles remain the same when scaled. So the RATIO of the length of any side to any another also remains the same. Piyush Srivastava B Tech-Marine Engineering from Directorate of Marine Engineering Training, Calcutta (Graduated 2022) · 6y Originally Answered: What is the relation between sides and angle? · Relation between sides and angles of a triangle is given by sine rule ,cosine rule ,tangent rule . To understamd it please go through it…. Related questions What are the relationships between angles, sides, and diagonals of a triangle? What is the relationship between two sides of a triangle, and three angles in it? What is the relationship between the angles of a triangle and the lengths of its sides? What is the relationship between an angle, side, and triangle? In a right angled triangle, what is the relationship between the length of the sides and the angles? What is the difference between an angle and a side of a triangle? What is the relationship between the sides and angles of a triangle given its side lengths and angle measures? What is the relationship between the sides and angles of a right-angled triangle? Why does this relationship hold true for any right-angled triangle? How are angles, sides and area of triangles related? What is the relationship between the interior angles of a triangle and the opposite sides? If the interior angles of a triangle add up to two right angles, are the opposite sides proportional? Why or why not? What is the relationship between the angles and sides in an equilateral triangle? What is the relationship between the sides of a triangle and its interior angles in simplest terms? How would you explain the relationship between angles and sides of a triangle to a 10-year-old? What is the relationship between the angles and sides in an equilateral triangle and a right triangle? What is the relationship between the angles and sides of a triangle if one angle is equal to one of its adjacent sides? 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https://www.hudsonalpha.org/sex-determination-the-x-y-zs-of-sex-chromosomes/
Scroll to top Skip to content Back Everyday DNA 11 min read Sex determination: The X, Y, Z’s of sex chromosomes Author Sarah Sharman Published July 15, 2021 Share Tweet Share Email Share An Everyday DNA blog article Written by: Sarah Sharman, PhD, Science writer Illustrated by: Cathleen Shaw Lately it seems like my social media feed is filled to the brim with gender reveal parties where an expectant couple learns the sex of their baby. Party guests look on as the happy couple cuts into a cake or pops a balloon to reveal pink or blue confetti. Most of us were taught as early as grade school that there is a 50/50 chance that a human baby will be a male or female. Later in our education we learned that an X chromosome from mom and an X chromosome from dad makes a genetic female, while an X chromosome from mom and a Y chromosome from dad makes a genetic male. The human sex determination system is the most familiar to us, but there are several other types of sex determination across the tree of life. The genetic sex of some organisms like turtles and snakes is determined by environmental cues like temperature, pH, or social conditions. However, many more organisms use genetics to determine who becomes a female or a male, just like in humans. Let’s learn about some of the types of genotypic sex determination. Genotypic sex determination Reproduction is the process by which organisms create offspring. There are two main types of reproduction: sexual and asexual. For the purposes of this discussion, we will focus on sexual reproduction. Sexual reproduction requires two parents, one male and one female. Each parent contributes a sex cell, called a gamete, that has half of the DNA as a normal body cell. In most cases, the male cell is called a sperm and the female cell is called an egg. The cells come together to form a zygote in a process called fertilization and the zygote later develops into an embryo. Within the sex cells are one copy of all of the chromosomes from the parent, including one sex chromosome that specifies whether an individual will be genetically male or female. This system of sex determination is called genotypic sex determination and occurs in a majority of species, including humans. There are several types of genotypic sex determination that we will explore below. XY sex determination The XY sex determination system is found in humans, most placental mammals, as well as some insects, snakes, fish, and plants. In this system, the genetic sex of an offspring is determined by a pair of sex chromosomes, always an X from the mother, and either an X or Y from the father. The sex chromosome from the father is considered the sex determining chromosome because it ultimately determines whether the offspring will be genetically male or female. A gene called SRY on the Y chromosome initiates the release of testosterone and the formation of male sex organs. SRY is a transcription factor that can turn on the expression of other genes, serving as a master switch to turn on the suite of “male” genes in a developing organism. ZW sex determination The ZW sex determination system is found in birds, amphibians, some reptiles, and plants like willow, pistachio, and strawberry. In this system the genetic sex of an offspring is determined by a pair of sex chromosomes, always a Z from the father, and either a Z or W from the mother. The ZW system is opposite of the XY system in that the mother’s sex chromosome determines the genetic sex of the offspring. Just as the mammalian Y chromosome carries the male-determining SRY gene, the bird W chromosome carries similar master switches FET1 and ASW, which are necessary for female development of the offspring. XO sex determination The XO sex determination system is found in insects like true bugs, grasshoppers, and locusts. Females have two X chromosomes (XX) and males simply carry a single X, the “O” in “XO” indicates the absence of the second sex chromosome. Males produce two types of sperm, half carrying a single X chromosome and the other half carrying no sex chromosome at all. Eggs fertilized with a sperm carrying the X chromosome will be genetically female, while eggs fertilized with a sperm carrying no sex chromosome will be genetically male. Haplodiploid sex determination The haplodiploid sex determination system is found in several social insects like ants, bees and wasps. Females have a full set of 32 chromosomes and males are haploid, having only 16 chromosomes. Unfertilized eggs develop into males while fertilized eggs develop into females. In this way, it is possible for a male to have a grandfather and grandsons but not a father or sons. UV sex determination The UV sex determination system is found in Bryophytes (mosses, liverwort, hornwort) and some algae. This system is haploid—each individual has only one sex chromosome with females carrying a single U and males carrying a single V. Unlike the XY and ZW systems, both the U and V chromosomes in the UV system are sex determining. In Bryophytes, haploid spores having a U or a V grow into offspring and then the union of eggs and sperm within these offspring allow them to produce more spores. So, it seems like we know a lot about these systems—why are scientists still studying them? Understanding the different types of sex determination systems is critical to the survival and evolution of sexually reproducing organisms. While scientists know a lot about existing systems, sex determination systems can rapidly evolve from one system to another. In organisms other than humans, the genetic sex of offspring might not be revealed with confetti and a big celebration but it is just as important. For example, in crop plants, the sex of individual plants is important for food production because in many crops, like melon, squash, strawberries and grapes, only female flowers make fruits. At the HudsonAlpha Institute for Biotechnology, the lab of Alex Harkess, PhD, Assistant Professor in the Department of Crop, Soil, and Environmental Science at Auburn University and a Faculty Investigator at HudsonAlpha, studies plant reproductive systems and sex determination systems to help breeders and growers produce more robust, durable crops. Sarah Carey, PhD, a postdoctoral researcher in the Harkess lab, studied moss UV sex chromosomes for her graduate thesis. Part of her study was recently published in Science Advances. Carey’s study found that despite the fact that fire moss sex chromosomes are older than the dinosaurs, they have maintained a lot of genes—many of which have been evolving “maleness” or “femaleness” over time. The ability to look at purely male or female genes in these moss may have big implications on the plant world as a whole. Many of the genes on the moss sex chromosomes are known to control fertility in flowering plants and crop plants. Control of reproduction in crop plants is important for food production, and the moss genes may provide new tools to boost female or male linked function. By gaining an even deeper understanding of sex determination systems, scientists hope to help plant breeders more efficiently sex their plants earlier in their lifespan so that those that are commercially valuable, like male asparagus which live longer in fields, can be selected for. Who knows, maybe someday we will see farmers throwing sex reveal parties in the fields for their crops. Related Posts July 21, 2025 7 min read The Many Flavors of Genetic Sequencing: A Sweet Scoop of DNA Discovery An Everyday DNA blog article Written by: Sarah Sharman, PhDIllustrated by: Cathleen... Everyday DNA Read More June 27, 2025 8 min read Everyday DNA Basics: Genetic Risk and Your Health An Everyday DNA blog article Written by: Sarah Sharman, PhDIllustrated by: Cathleen... Everyday DNA Read More
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https://math.stackexchange.com/questions/70938/natural-and-coordinate-free-definition-for-the-riemannian-volume-form
linear algebra - Natural and coordinate free definition for the Riemannian volume form? - 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 Natural and coordinate free definition for the Riemannian volume form? Ask Question Asked 13 years, 11 months ago Modified1 year, 6 months ago Viewed 7k times This question shows research effort; it is useful and clear 52 Save this question. Show activity on this post. In linear algebra and differential geometry, there are various structures which we calculate with in a basis or local coordinates, but which we would like to have a meaning which is basis independent or coordinate independent, or at least, changes in some covariant way under changes of basis or coordinates. One way to ensure that our structures adhere to this principle is to give their definitions without reference to a basis. Often we employ universal properties, functors, and natural transformations to encode these natural, coordinate/basis free structures. But the Riemannian volume form does not appear to admit such a description, nor does its pointwise analogue in linear algebra. Let me list several examples. In linear algebra, an inner product on V V is an element of Sym 2 V∗Sym 2⁡V∗. The symmetric power is a space which may be defined by a universal property, and constructed via a quotient of a tensor product. No choice of basis necessary. Alternatively an inner product can be given by an n×n n×n symmetric matrix. The correspondence between the two alternatives is given by g i j=g(e i,e j)g i j=g(e i,e j). Calculations are easy with this formulation, but one should check (or require) that the matrix transforms appropriately under changes of basis. In linear algebra, a volume form is an element of Λ n(V∗)Λ n(V∗). Alternatively one may define a volume form operator as the determinant of the matrix of the components of n n vectors, relative to some basis. In linear algebra, an orientation is an element of Λ n(V∗)/R>Λ n(V∗)/R>. In linear algebra, a symplectic form is an element of Λ 2(V∗)Λ 2(V∗). Alternatively may be given as some ω i j d x i∧d x j ω i j d x i∧d x j. In linear algebra, given a symplectic form, a canonical volume form may be chosen as vol=ω n vol=ω n. This operation can be described as a natural transformation Λ 2→Λ n Λ 2→Λ n. That is, to each vector space V V, we have a map Λ 2(V)→Λ n(V)Λ 2(V)→Λ n(V) taking ω↦ω n ω↦ω n and this map commutes with linear maps between spaces. In differential geometry, all the above linear algebra concepts may be specified pointwise. Any smooth functor of vector spaces may be applied to the tangent bundle to give a smooth vector bundle. Thus a Riemannian metric is a section of the bundle Sym 2 T∗M Sym 2⁡T∗M, etc. A symplectic form is a section of the bundle Λ 2(M)Λ 2(M), and the wedge product extends to an operation on sections, and gives a symplectic manifold a volume form. This is a global operation; this definition of a Riemannian metric gives a smoothly varying inner product on every tangent space of the manifold, even if the manifold is not covered by a single coordinate patch In differential geometry, sometimes vectors are defined as n n-tuples which transform as v i→v~j∂x i∂x~j v i→v~j∂x i∂x~j under a change of coordinates x→x~x→x~. But a more invariant definition is to say a vector is a derivation of the algebra of smooth functions. Cotangent vectors can be defined with a slightly different transformation rule, or else invariantly as the dual space to the tangent vectors. Similar remarks hold for higher rank tensors. In differential geometry, one defines a connection on a bundle. The local coordinates definition makes it appear to be a tensor, but it does not behave the transformation rules set forth above. It's only clear why when one sees the invariant definition. In differential geometry, there is a derivation on the exterior algebra called the exterior derivative. It may be defined as d σ=∂j σ I d x j∧d x I d σ=∂j σ I d x j∧d x I in local coordinates, or better via an invariant formula d σ(v 1,…,v n)=∑i(−1)i v i(σ(v 1,…,v i^,…,v n))+∑i+j(−1)i+j σ([v i,v j],v 1,…,v i^,…,v j^,…,v n)d σ(v 1,…,v n)=∑i(−1)i v i(σ(v 1,…,v i^,…,v n))+∑i+j(−1)i+j σ([v i,v j],v 1,…,v i^,…,v j^,…,v n) Finally, the volume form on an oriented inner product space (or volume density on an inner product space) in linear algebra, and its counterpart the Riemannian volume form on an oriented Riemannian manifold (or volume density form on a Riemannian manifold) in differential geometry. Unlike the above examples which all admit global basis-free/coordinate-free definitions, we can define it only in a single coordinate patch or basis at a time, and glue together to obtain a globally defined structure. There are two definitions seen in the literature: choose an (oriented) coordinate neighborhood of a point, so we have a basis for each tangent space. Write the metric tensor in terms of that basis. Pretend that the bilinear form is actually a linear transformation (this can always be done because once a basis is chosen, we have an isomorphism to R n R n which is isomorphic to its dual (via a different isomorphism than that provided by the inner product)). Then take the determinant of resulting mutated matrix, take the square root, multiply by the wedge of the basis one-forms (the positive root may be chosen in the oriented case; in the unoriented case, take the absolute value to obtain a density). Choose an oriented orthonormal coframe in a neighborhood. Wedge it together. (Finally take the absolute value in the unoriented case). Does anyone else think that one of these definitions sticks out like a sore thumb? Does it bother anyone else that in linear algebra, the volume form on an oriented inner product space doesn't exist as natural transformation Sym 2→Λ n Sym 2→Λ n? Do the instructions to "take the determinant of a bilinear form" scream out to anyone else that we're doing it wrong? Does it bother anyone else that in Riemannian geometry, in stark contrast to the superficially similar symplectic case, the volume form cannot be defined using invariant terminology for the whole manifold, but rather requires one to break the manifold into patches, and choose a basis for each? Is there any other structure in linear algebra or differential geometry which suffers from this defect? Answer: I've accepted Willie Wong's answer below, but let me also sum it up, since it's spread across several different places. There is a canonical construction of the Riemannian volume form on an oriented vector space, or pseudoform on a vector space. At the level of level of vector spaces, we may define an inner product on the dual space V∗V∗ by g~(σ,τ)=g(u,v)g~(σ,τ)=g(u,v) where u,v u,v are the dual vectors to σ,τ σ,τ under the isomorphism between V,V∗V,V∗ induced by g g (which is nondegenerate). Then extend g~g~ to ⨂k V∗⨂k V∗ by defining g^(a⊗b⊗c,⋯,x⊗y⊗z⋯)=g~(a,x)g~(b,y)g~(c,z)⋯g^(a⊗b⊗c,⋯,x⊗y⊗z⋯)=g~(a,x)g~(b,y)g~(c,z)⋯. Then the space of alternating forms may be viewed as a subspace of ⨂k V∗⨂k V∗, and so inherits an inner product as well (note, however that while the alternating map may be defined canonically, there are varying normalization conventions which do not affect the kernel. I.e. v∧w=k!A l t(v⊗w)v∧w=k!A l t(v⊗w) or v∧w=A l t(v⊗w)v∧w=A l t(v⊗w)). Then g^(a∧b⋯,x∧y⋯)=det[g~(a,x)…]g^(a∧b⋯,x∧y⋯)=det[g~(a,x)…] (with perhaps a normalization factor required here, depending on how Alt was defined). Thus g g extends to an inner product on Λ n(V∗)Λ n(V∗), which is a 1 dimensional space, so there are only two unit vectors, and if V V is oriented, there is a canonical choice of volume form. And in any event, there is a canonical pseudoform. linear-algebra differential-geometry category-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Dec 22, 2011 at 10:27 ziggurismziggurism asked Oct 8, 2011 at 21:09 ziggurismziggurism 17.5k 2 2 gold badges 56 56 silver badges 116 116 bronze badges 8 +1: Good observations. I'd like to know the answer to this too. For the last part: the metric tensor induces an isomorphism between the tangent and cotangent bundles, so with enough abuse of isomorphisms, we can make the coordinate-free definition of determinant work here... but there's no conceptual benefit over thinking of it as a matrix.Zhen Lin –Zhen Lin 2011-10-08 21:40:19 +00:00 Commented Oct 8, 2011 at 21:40 We can use isomorphisms of one non degenerate bilinear form to convert another bilinear form into a linear transformation, but if the two bilinear forms are the same, then the linear transformation is the identity, and so contains no information, and certainly cannot tell you the "determinant" of the bilinear form.ziggurism –ziggurism 2011-10-09 17:05:00 +00:00 Commented Oct 9, 2011 at 17:05 Obviously. But the vector space in question comes with a natural basis, namely the coordinate basis, and so is already equipped with an isomorphism to its dual space.Zhen Lin –Zhen Lin 2011-10-09 20:57:58 +00:00 Commented Oct 9, 2011 at 20:57 2 This is the non-degenerate bilinear form that provides the canonical ("musical") isomorphisms between the spaces of vectors and 1-forms, that you can use now to identify the spaces of bilinear forms and self-adjoint linear transformations. This identification allows you to define the determinant of a bilinear form as the determinant of the corresponding linear transformation.Yuri Vyatkin –Yuri Vyatkin 2011-10-09 21:59:54 +00:00 Commented Oct 9, 2011 at 21:59 1 For the definition to be basis-free you don't need a "natural" basis, but independence of a choice of a basis. This is achieved by using "canonical" (i.e. free of choice) isomorphisms. What you call "natural" basis in R n R n is just a convenient CHOICE of a basis, so called the standard basis in R n R n.Yuri Vyatkin –Yuri Vyatkin 2011-10-09 22:09:02 +00:00 Commented Oct 9, 2011 at 22:09 |Show 3 more comments 4 Answers 4 Sorted by: Reset to default This answer is useful 25 Save this answer. Show activity on this post. A few points: It is necessary to define "Riemannian volume forms" a patch at a time: you can have non-orientable Riemannian manifolds. (Symplectic manifolds are however necessarily orientable.) So you cannot just have a global construction mapping Riemannian metric to Riemannian volume form. (Consider the Möbius strip with the standard metric.) It is however to possible to give a definition of the Riemannian volume form locally in a way that does not depend on choosing a coordinate basis. This also showcases why there cannot be a natural map from S y m 2→Λ n S y m 2→Λ n sending inner-products to volume forms. We start from the case of the vector space. Given a vector space V V, we know that V V and V∗V∗ are isomorphic as vector spaces, but not canonically so. However if we also take a positive definite symmetric bilinear form g∈S y m 2+(V∗)g∈S y m+2(V∗), we can pick out a unique compatible isomorphism ♭:V→V∗♭:V→V∗ and its inverse ♯:V∗→V♯:V∗→V. A corollary is that g g extends to (by abuse of notation) an element of S y m 2+(V)S y m+2(V). Then by taking wedges of g g you get that the metric g g (now defined on V∗V∗) extends to uniquely to a metric 1 on Λ k(V∗)Λ k(V∗). Therefore, up to sign there is a unique (using that Λ n(V∗)Λ n(V∗) is one-dimensional) volume form ω∈Λ n(V∗)ω∈Λ n(V∗) satisfying g(ω,ω)=1 g(ω,ω)=1. But be very careful that this definition is only up to sign. The same construction extends directly to the Riemannian case. Given a differentiable manifold M M. There is a natural map from sections of positive definite symmetric bilinear forms on the tangent space Γ S y m 2+(T∗M)→Γ(Λ n(M)∖{0}/±)Γ S y m+2(T∗M)→Γ(Λ n(M)∖{0}/±) to the non-vanishing top forms defined up to sign. From which the usual topological arguments shows that if you fix an orientation (either directly in the case where M M is orientable or lifting to the orientable double cover if not) you get a map whose image now is a positively oriented volume form. Let me just summarise by giving the punch line again: For every inner product g g on a vector space V V there are two compatible volume forms in Λ n V Λ n V: they differ by sign. Therefore the natural mapping from inner products takes image in Λ n V/±Λ n V/±! Therefore if you want to construct a map based on fibre-wise operations on T M T M sending Riemannian metrics to volume forms, you run the very real risk that, due to the above ambiguity, what you construct is not even continuous anywhere. The "coordinate patch" definition has the advantage that it sweeps this problem under the rug by implicitly choosing one of the two admissible local (in the sense of open charts) orientation. You can do without the coordinate patch if you start, instead, with an orientable Riemannian manifold (M,g,ω)(M,g,ω) and use ω ω to continuously choose one of the two admissible pointwise forms. 1: this used to be linked to a post on MathOverflow, which has since been deleted. So for completeness: the space of k k-tensors is the span of tensors of the form v 1⊗⋯⊗v k v 1⊗⋯⊗v k, and you can extend g g to the space of k k-tensors by setting g(v 1⊗⋯v k,w 1⊗⋯⊗w k):=g(v 1,w 1)g(v 2,w 2)⋯g(v k,w k)g(v 1⊗⋯v k,w 1⊗⋯⊗w k):=g(v 1,w 1)g(v 2,w 2)⋯g(v k,w k) and extending using bilinearity. The space Λ k(V∗)Λ k(V∗) embeds into ⊗k V∗⊗k V∗ in the usual way and hence inherits a inner product. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 29, 2021 at 14:15 answered Nov 30, 2011 at 9:33 Willie WongWillie Wong 75.5k 13 13 gold badges 167 167 silver badges 263 263 bronze badges 6 6 As a corollary: on a orientable Riemannian manifold, there isn't just one, but two Riemannian volume forms. However, on a Kahler manifold, one of the two Riemannian volume forms is not compatible with the symplectic one.Willie Wong –Willie Wong 2011-11-30 09:38:17 +00:00 Commented Nov 30, 2011 at 9:38 Your excellent answer has reminded me of the issue with orientations. So I should never have expected a natural transformation from Sym 2→Λ n Sym 2→Λ n where the codomain is vector spaces. I should either ask for a natural transformation between the corresponding functors defined on oriented vector spaces, or else a natural map Sym 2→|Λ n|Sym 2→|Λ n|, the densities functor. The ambiguity you mention will disappear with either modification, right? So I can still ask for a natural, perhaps global, construction? I will update the question with this change.ziggurism –ziggurism 2011-11-30 13:39:41 +00:00 Commented Nov 30, 2011 at 13:39 So when you write "Then by taking wedges of g g you get that the metric g g [..] extends uniquely to a metric on Λ k(V∗)Λ k(V∗)", that's the construction where you take the factors of a pair of k k-vectors, and fill up a matrix with g g acting on those factors, and then take the determinant, right? I'd like to look closely at that construction, it seems to be the essence of my question. How do we decompose an arbitrary k k-vectors into its factor 1-vectors? Can that be done invariantly?ziggurism –ziggurism 2011-11-30 14:00:06 +00:00 Commented Nov 30, 2011 at 14:00 1 Decomposing an arbitrary k k-vector into its factor 1-vectors is the same as identifying a basis of ⊗k V⊗k V from a basis of V V. You should read MTS's comment (and the link to another of his answers in that comment) on the MO thread I linked to about extending g g to Λ k V Λ k V.Willie Wong –Willie Wong 2011-11-30 16:37:25 +00:00 Commented Nov 30, 2011 at 16:37 Yes, Willie, that is the complete explanation I was looking for. Thank you! And thanks to @MTS by proxy. Exercises remaining include extending the definition of g g to densities, and showing that this construction is indeed a natural transformation.ziggurism –ziggurism 2011-11-30 21:42:03 +00:00 Commented Nov 30, 2011 at 21:42 |Show 1 more comment This answer is useful 12 Save this answer. Show activity on this post. A coordinate-free definition of volume form is in fact well-known and frequently used, e.g. the cited Wikipedia article. I will try to reproduce it the nutshell to the best of my understanding. Let V V be a (real, for certainty) vector space of finite dimension dim V=n dim⁡V=n. The space of n n-forms Λ n(V)Λ n(V) has dimension 1. Thus Λ n(V)Λ n(V) isomorphic to R R, however this isomorphism is not canonical: any choice of non-trivial n n-form ω ω can be mapped to 1∈R 1∈R. A volume form on a finite-dimensional vector space V V is a choice of a top-rank non-trivial exterior form (skew-symmetric n n-linear functional) ω∈Λ n(V)ω∈Λ n(V). I think that this definition is quite coordinate-free. Once such a form has been chosen, it can be used to divide the space of bases in V V into two classes that are called orientations. There are two of them, positive (ω>0 ω>0) and negative (ω<0 ω<0). Having a volume form chosen, one can speak about oriented volumes of parallelotopes, for instance. If for any reason we have an inner product g g in V V we can make this choice canonical. One needs to consider orthonormal frames (with respect to g g). The canonical volume form will take value 1 on positively oriented orthonormal frames. The volume form of an inner-product space (V,g)(V,g) is that canonical choice of a volume form. It can be denoted by V o l g V o l g provided one also keeps in mind that there is a choice of orientation involved. Along these lines one can obtain an understanding of the volume form as the Hodge dual of 1 in a pretty coordinate-free manner. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 9, 2011 at 8:40 Yuri VyatkinYuri Vyatkin 11.7k 2 2 gold badges 47 47 silver badges 75 75 bronze badges 6 The definition of a volume form on an arbitrary vector space is nice: it's an element of Λ n(V∗)Λ n(V∗). That space can be defined via a universal property, and constructed via a quotient of a cartesian product. Never any reference to a basis.ziggurism –ziggurism 2011-11-30 04:02:41 +00:00 Commented Nov 30, 2011 at 4:02 A symplectic vector space is defined by choosing an element ω ω from Λ 2(V∗)Λ 2(V∗), without specifying a basis, or saying how it acts on bases. You can get a volume form from ω ω by just doing a linear operation. That operation, viz. the wedge product, is defined without reference to a basis either via a universal property or else as a projection to a quotient.ziggurism –ziggurism 2011-11-30 04:02:57 +00:00 Commented Nov 30, 2011 at 4:02 The inner product g g is an element of Sym 2(V∗)Sym 2⁡(V∗). To get a volume form from it, I have to specify a basis compatible with g g and then specify how my volume form acts on it. This seems like a much uglier definition to me! I want there to be some fancy multilinear operation from Sym 2(V)→Λ n(V)Sym 2⁡(V)→Λ n(V). I guess if such a thing existed, it would be in the textbooks?ziggurism –ziggurism 2011-11-30 04:03:04 +00:00 Commented Nov 30, 2011 at 4:03 I went to work some of these comments into the original question, but ended up re-writing the question from top to bottom. Hopefully my issue will be clearer?ziggurism –ziggurism 2011-11-30 05:29:28 +00:00 Commented Nov 30, 2011 at 5:29 I think: one can do coordinate free when you just do vector space, but OP (@ziggurism) is asking for manifold?BCLC –BCLC 2020-12-31 12:58:20 +00:00 Commented Dec 31, 2020 at 12:58 |Show 1 more comment This answer is useful 3 Save this answer. Show activity on this post. The metric tensor is symmetric and so can be orthogonally diagonalized, which means that its determinant is just the product, with multiplicity, of its eigenvalues. If we rotate the entire chart such that the coordinate axes at the point of interest follow the the eigenspaces, this means that |g||g| tells us the factor that relates the squared coordinate length of each side of an infinitesimal box with its squared intrinsic length, all multiplied together. Therefore, multiplying with the square root of the determinant will convert a coordinate volumne of an infinitesimal box to the intrinsic volume. The intuition to get here is that a Riemannian manifold, by definition, locally looks like Euclidean space, where the volume of a rectangular box is the product of its side lengths, and we can measure the volume of some particular figure with boxes of any particular orientation and get the same result. We can do this in a manifold too, since the orthogonality of the sides of a box, as well as its side lengths, are all intrinsic. The volume form simply expresses how to do this given any particular chart, by -- for convenience -- using boxes of an orientation that fits with the eigendecomposition of the metric. But the result must be invariant, because the volume of a piece of Euclidean space does not depend on how we turn the coordinate system. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Oct 8, 2011 at 21:49 answered Oct 8, 2011 at 21:42 hmakholm left over Monicahmakholm left over Monica 292k 25 25 gold badges 441 441 silver badges 706 706 bronze badges 3 How do you define an eigenvalue or eigenvector of a bilinear form?ziggurism –ziggurism 2011-10-09 17:13:09 +00:00 Commented Oct 9, 2011 at 17:13 4 I don't define them for the abstract bilinear form. I'm speaking about eigenvalues and -vectors of the matrix that represents the bilinear form in a particular chart. (It stands to reason that this cannot in itself have a coordinate-free interpretation, because the entire point of the g√g factor is to correct for the coordinate-dependent scaling error of d x 1∧⋯∧d x n d x 1∧⋯∧d x n).hmakholm left over Monica –hmakholm left over Monica 2011-10-09 20:44:01 +00:00 Commented Oct 9, 2011 at 20:44 1 This is my point though. Are we really satisfied with a situation where the volume form cannot be defined for the abstract bilinear form? We have to choose a basis, rely on an isomorphism with R n R n and use the canonical metric on R n R n to convert a bilinear form into a linear transformation and then find its eigenvalues. Compare with the symplectic manifold: take the wedge product of the symplectic form with itself n n times. Works on any manifold, without choosing a basis or relying on ad hoc features of R n R n.ziggurism –ziggurism 2011-11-30 04:08:51 +00:00 Commented Nov 30, 2011 at 4:08 Add a comment| This answer is useful 1 Save this answer. Show activity on this post. I guess you were able to piece together from Willie Wong's answer that you can obtain a global construction that yields the natural volume pseudo-form on any (pseudo)Riemannian manifold. The recipe is to use g g to induce a smooth fibrewise inner product on the bundle |Λ|n T M∗|Λ|n T M∗ of top pseudoforms. realise that there are only two n n-pseudoforms ω∈|Ω|n M=Γ(|Λ|n T M∗)ω∈|Ω|n M=Γ(|Λ|n T M∗) that satisfy g(ω,ω)=±1.g(ω,ω)=±1. The sign is an invariant of the metric: either all n n-pseudoforms have g(ω,ω)≥0 g(ω,ω)≥0 or g(ω,ω)≤0 g(ω,ω)≤0. Call this sign s g n(g)s g n(g) and finally let your volume pseudo-form be the one of these two ω ω s whose components are positive in every chart. But there is an alternative global construction that more closely resembles the classical chart-wise construction. It involves taking the "determinant of g g as a bilinear form" but doing so in a coordinate-free way from the beginning. Remember that the bundle of pseudoforms is the tensor product bundle |Λ|n T M∗=Λ n T M∗⊗M P s c M|Λ|n T M∗=Λ n T M∗⊗M P s c M where P s c M P s c M is the pseudoscalar bundle of M M, which is the line bundle associated to the frame bundle of M M via the representation χ(A)r=s g n(det A)r χ(A)r=s g n(det A)r of G L n G L n on R R. Similarly, define the bundle D w D w of (sometimes called "authentic"-) densities of weight w w to be the line bundle associated to the frame bundle of M M by the representation ξ(A)r=(det A)−w r ξ(A)r=(det A)−w r of G L n G L n on R R. Denoting by Λ n T M Λ n T M be the bundle of n n-vectors. Now observe that the two bundles L 1 L 1=Λ n T M∗⊗M D−1 M,=Λ n T M⊗M D 1 M.L 1=Λ n T M∗⊗M D−1 M,L 1=Λ n T M⊗M D 1 M. have the following property. If you take sections s,t s,t of L 1,L 1 L 1,L 1, respectively, and represent them in components first with respect to a local frame e e and then with respect to another local frame e′e′, the components of s s and t t will change by a factor of (det A)−1⋅det A det A⋅(det A)−1=1,=1,(det A)−1⋅det A=1,det A⋅(det A)−1=1, respectively. This means that the components of s s and t t don't change at all when you switch frames. Now, there are unique sections of L 1 L 1 and L 1 L 1 whose first component is 1 1 in every chart: these are the the famous Levi-Civita symbols. Denote them in abstract index notation as ϵ Δ a 1…a n∈Γ(L 1)ϵ a 1…a n Δ∈Γ(L 1)ϵ a 1…a n Δ∈Γ(L 1)ϵ Δ a 1…a n∈Γ(L 1) In other words, they are tensor densities of weight ±1±1. I use the abstract index Δ Δ to indicate the "density"-ness: lower Δ Δ for the bundle D 1 D 1 of weight +1+1 (you know, because it weighs?) and higher Δ Δ for the bundle D−1 D−1 of weight −1−1 (because it floats I guess). Notice that this works for any manifold. I haven't introduced coordinates. Now, the determinant of the metric tensor is the following trace: det(g)Δ Δ=1 n!t r 2 n(ϵ Δ⊗ϵ Δ⊗g⊗⋯⊗g)=1 n!ϵ a 1…a n Δ ϵ b 1…b n Δ g a 1 b 1…g a n b n.det(g)Δ Δ=1 n!t r 2 n(ϵ Δ⊗ϵ Δ⊗g⊗⋯⊗g)=1 n!ϵ Δ a 1…a n ϵ Δ b 1…b n g a 1 b 1…g a n b n. I want to emphasize that there is no choice of frames involved. The rightmost expression is abstract index notation: it just indicates a very specific way to take the traces in the middle. The result is a scalar density of weight 2 2. That is, a section of the bundle D 2 D 2 I defined above: its coordinate expression in a frame is exactly the determinant of the matrix associated to the bilinear form g g in that frame. If you express it in two successive frames, it will pick up a factor of (det A)2(det A)2, where A A is the change of frame matrix. But this coordinate representation may be positive everywhere or negative everywhere, depending on the signature of g g. In fact, its sign is exactly the s g n(g)s g n(g) I defined in point 2 at the beggining of this post. Hence, multiplying by the scalar s g n(g)s g n(g), we obtain a new scalar density of weight 2 2 |det g|Δ Δ=s g n(g)det(g)Δ Δ|det g|Δ Δ=s g n(g)det(g)Δ Δ whose coordinate representation is a positive function at every frame. Taking the positive square roots, we obtain |det g|−−−−−√Ψ Δ|det g|Δ Ψ which now is a section of P s c(M)⊗D 1 P s c(M)⊗D 1, because when you express it in two successive frames, you pick up a factor of |det A|=s g n(det A)(det A)|det A|=s g n(det A)(det A). Notice I used the abstract index Ψ Ψ to indicate the bundle of pseudoscalars. Finally, you can define the volume pseudoform as ω Ψ a 1…a n=|det g|−−−−−√Ψ Δ ϵ Δ a 1…a n.ω a 1…a n Ψ=|det g|Δ Ψ ϵ a 1…a n Δ. Here, the upper and lower Δ Δ s annihilate each other: a density of weight +1+1 times a density of weight −1−1. You can prove that ω ω has the desired property: its inner product will be g(ω,ω)=s g n(g)g(ω,ω)=s g n(g). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Mar 10, 2024 at 7:45 answered Aug 8, 2023 at 8:52 Jackozee HakkiuzJackozee Hakkiuz 6,149 1 1 gold badge 17 17 silver badges 39 39 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 differential-geometry category-theory See similar questions with these tags. 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https://cod.pressbooks.pub/introtologic/chapter/truth-tables/
Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 6 Truth Tables We have seen how challenging it can be to evaluate arguments for their informal qualities. When we turn to evaluating their formal qualities, there are many tools at the logician’s disposal to ease the process. Truth tables are nothing more than a way to visualize the core concepts and formal properties of statements. As such, they also help us quickly evaluate arguments. In this sense, truth tables are nothing more than pictures we draw. We want a way to draw the concepts surrounding truth and possibility. So, if a picture is worth a thousand words, then a truth table is worth all possible worlds… Given our commitment to bivalence, we know that there are only two possible truth values for any given statement, P: it can be either true or false. We have a way of writing this that is called a “truth table.” For an atomic sentence, the truth table is trivial, but when we look at other kinds of sentences, their truth tables will be more complex. The idea of a truth table is to describe the conditions in which a sentence is true or false. We do this by identifying all the atomic sentences that compose that sentence. Remember that larger statements are composed of truth-functional logical operators—so the truth of any compound will depend on the truth value of the smaller atomics out of which the compound is made. Setting Up a Truth Table The set up for a truth table is simple. We will only have four quadrants to our table, so only two lines need to be used to organize the information we need. This will look like this: You might even think of truth tables as being divided in two halves. On the left is the basic information we need, and on the right is the statement(s) we want to evaluate and the information about it. So, on the left side, we stipulate all the possible truth values of these atomic sentences and write these out. On the right side, we then identify under what conditions the sentence (that is composed of the other atomic sentences) is true or false. The idea is that the sentence on the right is dependent on the sentence(s) on the left. So the truth table is filled in like this: | | | --- | | Atomic sentence(s) that compose the dependent sentence on the right | Dependent sentence composed of the atomic sentences on the left | | All possible combinations of truth values of the composing atomic sentences | Resulting truth values for each possible combination of truth values of the composing atomic sentences | One important note before we go any further. When we say we will fill in the upper left quadrant with all the atomic statements that compose our statements on the right, we must follow one simple rule: list the atomics in the SAME order in which they appear in the upper right (when you read from left to right). We do not put them in alphabetic order, and we do not spell out funny words with them. We always put them in exactly the same order in which they appear in the upper right (skipping any repeats). We stipulate all the possible truth values on the bottom left because all of the logical properties that we learned are defined in terms of different possibilities. When you hear most folks talk about this or that being “possible,” they are doing little more than expressing their own uncertainty or ignorance (e.g., “I don’t know! It’s possible.”). This is not how a logician sees it. For us, the claim that “X is possible” is a claim of knowledge. We know. Or rather, we will know for certain that something is possible—because we took the time to check. That’s what the lower left part of a truth table provides: a record of every possible combination of truth values. For an atomic statement, the total number of possible truth values is merely two: true or false. So, this means that if we wish to evaluate a compound statement that is made up entirely of just that one statement, there will also be only two possible combinations of truth values. We can start here, since this will be easy to do while illustrating many key features of how truth tables are set up. So let’s evaluate the following statement: ~ G We start by setting up our truth table, first by laying out our four quadrants and placing the atomic statement(s) in the upper left corner. | | | --- | | G | | Now we will put the statement(s) we wish to evaluate in the upper right corner. | | | --- | | G | ~ G | Now we list all possible truth values for the atomic(s) in our table and number these possibilities: | | | --- | | G | ~ G | | 1. T | | | 2. F | | The statement we want to evaluate is made up of this atomic, so we can transfer over the possible truth values we listed on the left over to the right side, where we see the same atomic. Like this: | | | --- | | G | ~ G | | 1. T | T | | 2. F | F | I know this sounds a bit redundant right now, but that is only because this table is exceedingly simple. When we get to larger statements and larger truth tables, we will be rewarded for taking time on this very simple step. Now we can consider the impact that our logical operator has on the truth of the overall statement. Negations are easy—they simply invert the truth value of whatever statement they deny. When G is true, denying that claim will be false. When G is false, denying that claim will be true. So, we put these inverted values directly under the ~ symbol. Like this: | | | --- | | G | ~ G | | 1. T | F T | | 2. F | F T | This is a bit messy, so we will find it very helpful to clearly indicate which (if any) logical operator is the main connective of the statements we want to evaluate. Remember what we said about the main connective: one of its important qualities is that it is the bearer of truth for the statement. So when we indicate its importance on the truth table, we are graphically showing the truth value for that compound statement. We can either draw a small arrow over the main connective, draw a clear box around it, or both. I prefer to use a box without the arrow, but as we go further you may find that different ways of marking your table work especially well for you. You and/or your professor will come up with an agreeable method. Now our truth table is complete. We have shown every possible relevant scenario in which the truth value of ~ G can be established. We call these scenarios “cases” or “possible worlds,” and numbering them will help us to refer to them later, when we must justify our evaluations. The completed table is nothing more than a picture of possible truth values. This can be used to answer questions about the statement we wanted to evaluate. We’ll discuss this later, but for now just note that we need to first learn how to construct truth tables. Their importance will be discussed once we know how to set them up. So far this has been pretty easy. However, if we have a compound statement to evaluate that has another atomic, then we must add that atomic statement to the upper left side. For example: A v ~ G This doesn’t increase the number of possible truth values for the added statement; however, the number of combinations of truth values has increased. For example, what are the possible combinations of truth values for A and B? After a bit of thought, most folks will be able to figure out that there are four possible combinations: A is true, and G is true A is false, and G is false A is true, but G is false A is false, but G is true If you figured that out, good for you. Now what happens if we add a third atomic statement? For example: M → ( A v ~ G ) Now we have to come up with every possible combination of truth values for those three atomic statements. Q: How many possible combinations are there? Q: What are they? Your job is to answer both questions. Do it correctly, without fail, and do it quickly. Got it already? How about now? No? Well, the answer to our first question is 8 combinations. Now, quickly tell me what those 8 combinations are and don’t mess it up. Also, don’t repeat yourself… Go! No? Okay, you’re going to have to get good at doing this…especially when we add a fourth atomic statement. Like this: ( H v M ) → ( A v ~ G ) Remember what you have to do. Answer the following questions: Q: How many possible combinations are there? Q: What are they? Go! Maybe you figured this out, but the answer to our first question with four atomics is 16 possible combinations. And yes, you have to tell me what they are and put them in your truth table (lower left quadrant). Sounds like fun. If you don’t think this is fun, you just haven’t learned the method yet. We have a simple way to do this—really simple, as in a mechanical way to generate the answers that requires hardly any thought at all. How about that for a combination: guaranteed success + little mental effort Ready? Here’s how we do it. FIRST STEP: Count the number of atomics. You will double the number of possible combinations every time you add an atomic statement. Start with one atomic: you have two possibilities (T/F). Add an atomic (now 2 atomics total) and double the possibilities: from 2 possibilities to 4 possibilities. Add an atomic (now 3 atomics total) and double the possibilities: from 4 possibilities to 8 possibilities. So on and so forth. Remember this phrase: “double it up” That tells you how many cases (or possible worlds) you must account for in your table. Now that we know, we need to detail those cases and list them in the lower left quadrant of our table. This turns out to be surprisingly easy. SECOND STEP: Find the atomic snuggled up against our vertical line. You will simply alternate T and F under it for as many cases as the table requires. Like this: That’s not hard. Now slide over one atomic and “double it up”: You just alternated T and F every other line, so alternate every other two lines. Two true, two false, two true, two false…until all cases are completed. Like this: Now slide over one atomic, and “double it up”: You just alternated T and F every two lines, so alternate ever four lines. Four true, four false, four true, four false…until all cases are completed. Like this: Since we only have 8 cases in this table, this is done in a single pair of four true / four false. However, if we had a table with 16 cases, we would see this pattern repeat more. We would also slide over to our last atomic, and once again “double it up” (alternating T/F every 8 lines). The method chugs along until we are done with all of our atomics. All we have to do is remember three rules: Rule 1: The number of atomics is used to determine the number of cases (possible worlds) Rule 2: Start with the atomic closest to the vertical line Rule 3: “Double it up” The rest of the table requires a bit more knowledge, but the initial set up is pretty mechanical. I should note that Rule 2 is just a convention. There is nothing logically necessary about starting with the atomic next to the vertical line. However, there is great advantage to holding this as a conventional rule for all of us to follow. This makes it easy for us to communicate the information on our table to others who are working on the same problem. Communicating information is critical to justifying your evaluations. So, when you are asked to evaluate a statement or an argument and justify your findings, you will refer to a case (or cases) to do so. If you are communicating in your own “unique” way, the chances are good that you will not get credit for your answer because everyone else is looking at a different line than the one you claim supports your answer. When we all set up our tables with the same conventions, we are all able to communicate clearly and precisely. Put differently, there are no unique snowflakes in logic. We all adhere to the same principles. So, the faster we get on board with the same methods, the easier it is to check that we are following those principles. THIRD STEP: Now that we have identified and exhausted every possible combination of truth values for the atomic statements (that make up the statement[s] we want to evaluate), we can just transfer those values over to the lower right-hand quadrant. Like this: This might sound like a lot of redundant work, but this will make it very easy to complete the table. Before we can do that, we need to know how our logical operators impact the truth values of the component statements. We just saw this with negation, so let’s review that and look at the rest of our logical operators. Negation Negation is simply a denial. So as we saw, you simply invert the value of whatever is negated. A Note on Annotation In the above image, we are using our familiar □ (and later △ for the other operators) to indicate that a negation may apply to any statement at all, be it an atomic or a compound. However, we also used the □ (and later △ for the other operators) in the upper left-hand quadrant of the truth table. While □ and △ are our preferred statement variables, the reader should note that in this quadrant they can only ever be atomic statements. We never put compound statements in the upper left-hand quadrant of a truth table. In this quadrant, □ and △ can only represent atomic statements, whatever they are for the statements we wish to evaluate. Conjunction The conjunction is a union of two assertions. So think of this as a promise: I promise that we will do this and we will do that. To keep my promise (i.e., to ensure I speak the truth), I must fulfill both components. In other words, both sides need to be true in order for such a promise to be kept. Note that conjunctions are typically false. It really is hard to keep such a promise. Memorizing the basic conditions of truth for conjunctions is easy—both sides have to be true for the conjunction to be true (any false conjunct makes the whole thing false). Disjunction The disjunction is a weak expression of possible options. Go back to the notion of keeping a promise. If I promise that at the very least, we will do this or we will do that, well heck, it’s kind of easy to keep that promise. I can claim success in any of three scenarios: (1) we did this, (2) we did that, (3) we did both. Note that disjunctions are typically true. TIP: If you’re going to make a promise, make it an Or-promise. Memorizing the basic conditions of truth for disjunctions is easy—both sides have to be false for the disjunction to be true. Material Conditional The material conditional is a special relationship. We could (and do) casually call it a conditional, but the truth is that this is a logically specific relationship that is often quite different than what most people think of when they think “conditional.” Thus, if you struggle with this, keep calling it a “material conditional” so that you remember it is a technical term and a special kind of conditional. A material conditional is not a claim of causality, nor does it entail any special temporal elements (like “this happens before that”). Many English conditional expressions do include these extra senses of the words, but this is not translated over to the logical operator we know as the material conditional. What our conditional does is nicely capture the relationship of necessary or sufficient conditions. So, let’s consider a story: My dad used to make promises that were dependent on me doing my chores. I was the family dishwasher. I also liked to go rent movies (when there were movie rental stores and we didn’t have streaming services). So, dear old Dad would often promise, “If you do the dishes, we’ll go to rent movies.” Grand idea Dad! Put the ball in my court. On some days, I did the dishes. On other days, I dropped the ball and didn’t do dishes. So what do you think dear old Dad did? Well, on some days, Dad took me to get the movies, and on other days he did not. Q: On which days can we accuse Dad of speaking falsely? Let’s consider only the days I did the dishes. Day 1: He takes me to get movies. Yes! Dad spoke the truth! Day 2: He didn’t take me to get movies. What the!?! Dad lied to me…I can’t believe it, but he spoke falsely. Now let’s consider only the days I did not do the dishes. Day 3: He takes me to get the movies anyway. Can I really be mad at Dad? Nah. (more on this in a bit) Day 4: He didn’t take me to get movies. Well, what was I expecting? I mean, after all, I didn’t do the dishes. Can I accuse Dad of lying? No. So I’ll accept that Dad spoke the truth. If you had to do the dishes in your house, then the material conditional starts to make sense. The trouble for some is the Day 3 scenario. I did NOT do the dishes, so what am I to make of Dad taking me to get the movies anyway? I feel like he misled me. But did he? What was it that Dad said? If you do the dishes, then we’ll go rent movies. From our work on translations, we see that “doing the dishes” is in the antecedent position (the □). This means that doing the dishes is in the place of the sufficient condition—it is supposed to be what logically guarantees that the △ claim is true. Also from our work on translations, we can see that this means that doing the dishes was not put in the place of the necessary condition (that’s the △ part, and in this case it is the claim about going to get the movies). Now we can go back to Dad. Q: Did dear old Dad ever say doing the dishes was necessary? No, he did not. He said it was sufficient (he never said it was necessary). So, when he takes me to get the movies even though I did not do the dishes, he did not propose a false promise. He never said that doing the dishes was necessary for getting the movies; he merely assured me it was sufficient to get him off the couch to drive over there. I may have heard him incorrectly—that’s always a strong possibility—but I cannot accuse him of misleading me simply because I failed to listen carefully. That shortcoming is on me. Thus, the basic conditions of truth for the material conditional are as follows: The easy way to memorize the basic conditions of truth for the material conditional is to remember that you can really only pin a falsehood on it when the antecedent, □, is satisfied (i.e., is true) but the consequent △ was not delivered (i.e., is false). Material Biconditional Remember that biconditionals are what many people hear when someone says a conditional statement. They hear a very tight relationship between both clauses (the □ and the △ ). Our story of doing the dishes illustrates this well. We may have heard Dad make a biconditional promise. We thought he was saying: I’ll take you to get the movies if and only if you do the dishes. That sounds like a Dad-promise. If he had made that promise, then we would expect the truth of his promise to look different than what we saw with the material conditional. And so it is. Now we see in Case 3 (a.k.a. “Day 3” in our story) that there really is something shifty about Dad’s promise if he was planning on going to get the movies all along. In making this biconditional promise, he spoke falsely! Sure, I may not mind (I got the movies), but I sure can’t trust what that guy says. He said doing the dishes was both sufficient and necessary. How can that be true! How can it also be a necessary condition if he gets the movies anyway? Answer: it can’t; Dad speaks with a forked tongue. The easy way to memorize the conditions of truth for a biconditional is to remember: Biconditionals are true when both sides share the same truth value, false if there is a difference on each side. Overview of the Basic Conditions of Truth for the Logical Operators We can capture all of this information in a single truth table for easy reference. You need to have this in your back pocket. We’ll use “A” and “B” as placeholders for □ and △ to keep our table small. This is just a quick reference table that you must memorize: Calculating Truth Values for Statements Have you memorized the truth table for the basic conditions of truth for all of the logical operators? If not, read no further. Stop what you are doing. Go back and reread. You must memorize the truth table for the basic conditions of truth of the logical operators. If you haven’t done so, then from here on out, you are really wasting your time. I know that culturally speaking, many people look down on rote memorization as an approach to learning that is somehow stultifying and soul-crushing. And sure, rote memorization can be soul-crushing if it’s over-emphasized. But sometimes there is just no other way. Sometimes, as with much of the work in doing truth tables, it is actually soul-strengthening. Memorizing information requires you to focus. It can also feel good to flex your memory and see what it can do. If you don’t memorize the truth tables, you will likely find logic to be confusing and depressing. If you do memorize them correctly, you will find much of what follows to be really easy. One hugely important aspect of sentential logic is the calculation of truth values. Once you’ve memorized the truth table definitions for the operators, this turns out to be incredibly easy to do. It can also be somewhat relaxing. Just think of yourself as a mindless computer that takes certain inputs and generates certain outputs. Consider the following statement: ( P → Q ) ↔ ( ~ R → Q ) Earlier we saw that sentential logic is truth functional, which just means that the truth value of any complex statement (such as this one) is determined by the truth values of the simple statements it contains. The basic idea of truth functionality is simple: If you know what truth values to attach to the atomics in the statement above, then you can easily figure out what the truth value of the whole statement must be. In order to see how this works, let’s just stipulate: P: true Q: true R: false What would the truth value of the whole statement be? The statement as a whole is a biconditional, since the main operator is a triple bar. ( P → Q ) ↔ ( ~ R → Q ) If we knew whether the chunk on each side were true or false, then we could consult the truth table for the triple bar to see what the the truth value of the whole statement must be. (And that is why it’s so important to memorize the truth table definitions for the logical operators.) So, let’s take each chunk on its own. The chunk on the left is: P → Q We’ve stipulated that “P” and “Q” are both true, so “P → Q” must be true. So the left-hand chunk is true. ( P → Q ) ↔ ( ~ R → Q ) T The right-hand chunk is just a bit more complicated. We know that “Q” is true. We’ve also stipulated that “R” is false. But that means that “~ R” is true. Since both “~ R” and “Q” are true, it follows that “~ R → Q” is also true. So the right-hand chunk is true as well. ( P → Q ) ↔ ( ~ R → Q ) T T Now recall that a biconditional statement basically asserts that the chunk on the left and the chunk on the right have the same truth value. Here that is indeed the case: the left-hand chunk and the right-hand chunk are both true. So the whole statement is true. We can indicate this by placing a “T” under the main operator. ( P → Q) ↔ ( ~ R → Q) T T T One reason why it’s important to practice these truth value calculations is that this is the key to filling in truth tables for propositions. Suppose we wanted to do a full-blown truth table for the above statement. It would start out looking like this: Here we have already filled out the lower left quadrant with all possible combinations of truth values for the three atomics. We have also transferred those values over to the lower right quadrant for easy reference. Now we need to work through the logical structure of the statement as just discussed. Working through Logical Structure The first thing we need to do is mark the main connective and/or its eventual column. This helps us recognize the main components of the statement. Without knowledge of the main connective, you are lost. You have zero chance of successfully completing the table. So best to review that concept. Practice your statement trees until you are very, very good at recognizing the logical structure of a statement. When you do a truth table, you are effectively working your way up through a statement tree: starting with the truth values for the atomics, then moving up through the statement, until you finally get to the very last operator to create the overall statement. We start on our table by transferring those truth values to the atomics. This has been done above. So next step is to move on to the logical operators who join together or apply to nothing more than atomics. We can start with the left side of the biconditional. The left side is a conditional statement. So the truth of this side will conform to the general conditions of truth for material conditional statements (which, of course, you have memorized by now). We indicate the truth value of this little conditional by putting its truth value directly under the arrow. Like this: The next statement we would like to do is the right side of the biconditional. This too is a little conditional statement. However, this arrow joins an atomic and a negation. So, we cannot fill in that arrow until we know the truth value of the antecedent. Put differently, the right side has the following form: □ → △ We will know the truth of that condition when we know the truth values of □ and △. The △ is an atomic, but the □ is a compound statement. So we cannot fill in the → column until we have a proper column for □. In this case □ is a negation (specifically, inside the box is the statement “~ R”). Fortunately, this tilde applies directly to an atomic statement. So we can fill in the column for the ~. Now we have the right side of the biconditional ready to complete. Our column under the tilde bears the truth of □, so we use that column along with the column under the “Q” to determine the truth values in the column under the right side →. Our table is almost done. We can now see the column of truth values under each side of our biconditional. These two columns will be used to determine the column of truth values under the ↔. Of course, we have memorized the basic conditions of truth for the ↔, so this should be easy. Now we have a complete picture of every possible case in which this statement is true and when it is false. The obvious question is: What are we going to do with this picture? Using Truth Tables to Evaluate Logical Properties Recall in Chapter 1 that we learned about important logical properties of statements. The savvy student remembers that all of these were defined in terms of possibility. So, naturally a truth table is an ideal tool to use for these properties. Evaluating Individual Statements The truth table we just built demonstrates every possible outcome for our previous sample statement: ( P → Q ) ↔ ( ~ R → Q ) Now we can ask questions of possibility. The first, and easiest, might be: Q: Is this statement a logical tautology, contradiction, or contingent statement? If you don’t remember what these properties are, here’s the cheat sheet version: Tautology: a statement which is true under all conditions Contradiction: a statement which is false under all conditions Contingent statement: a statement which is true under some conditions and false under other conditions A truth table provides us with “the conditions” that are important. We call these the “cases” or “possible worlds.” So each numbered case represents a specific condition of a possible world. When we look at a truth table, we are looking at the statement’s main connective column to answer this question. Q: What do you see in the MC column? Is it all Ts, all Fs, or is it a combination? We see that there is a combination of truth values. We don’t see a straight column of nothing but Ts (nor a column of nothing but Fs). If we did see a pure column of Ts, we would say this statement cannot be false; it is always true under all conditions (i.e., it’s a tautology or logically true statement). We would say the opposite if it were a column of pure Fs (i.e., it’s a contradiction or logically false statement). However, in this case we do not see those kinds of columns. We see a column in which sometimes the statement is true and sometimes it is false. So, we say that the statement is a contingent (or logically indeterminate) statement. We might even go deeper into this table. If a friend doubted that this statement is ever true, we can provide them with a very specific set of conditions in which the statement would indeed be true. We can point to specific cases in which it is true. This can give us some sense of what the world must look like in order for the statement to be true or false. Evaluating Sets of Statements Many times, we want to know something about a group of statements. They may not even be an argument, or they may be small parts of an argument. No matter, we just want to know things about a given group. Truth tables can help us explore the logical properties of these groups. Logical Equivalence For example, consider the following: Your friend makes a claim, then later says something different, but insists they said the same thing. In the morning your friend said, “Sure. If I go to the store, I’ll get some milk.” In the afternoon, your friend said, “Okay already! Either I don’t go to the store, or I’ll get some milk.” You think your friend is changing their story to suit themselves. But your friend may defend themselves by insisting that even though they said it differently, their second statement really meant “the same” thing. Well, does it? A truth table can discover if two statements are logically equivalent. Again, if Chapter 1 content does not spring to mind immediately, here’s the cheat sheet version: Logical Equivalence: two statements are logically equivalent if and only if they always have the same truth value as one another under all conditions Since a truth table demonstrates all conditions, let’s go check the table. We need to first translate our friend’s statement. This will do: If I go to the store, I’ll get some milk.S → M Either I don’t go to the store, or I’ll get some milk. ~ S v M We can put multiple statements in a truth table’s upper right-hand quadrant by simply distinguishing them with commas. Like this: With only two atomics, this is a small truth table. The finished version looks like this: With a complete table we can look at the two columns under the main connective of each statement. Q: Are they identical columns, or is there at least one variation in the truth values? Now we can see that our friend really did say the same thing, insofar as their statements reflect the same account of the world. These two statements will always have the same truth value under any and all conditions. CAUTION: Truth tables can be used this way only if we put both statements in the same table. You cannot reliably make one table for the first statement and another table for the second statement. This would run the risk of demonstrating very different possible combinations of truth values (i.e., different possible worlds). If this happens, we cannot make accurate claims about what is and is not possible. We must have all statements under evaluation in a single truth table to make those judgments. Logical Consistency and Inconsistency Let’s look at our other concepts for groups of statements. Remember logical consistency and inconsistency? No worries, here’s the cheat sheet version: Logical Consistency: a set of statements is logically consistent if and only if it is possible for all members of the set to be true under the same conditions Logical Inconsistency: a set of statements is logically inconsistent if and only if it is not possible for all members of the set to be true under the same conditions Truth tables make short work of any question regarding these concepts. Let’s look at an example. Consider the following statements: ~ K → W ~ ( W v K ) W ∙ K Q: Are these statements logically consistent or are they inconsistent? When we put them all together in a truth table, we get the following: Our question is in regard to consistency, so we need to look at the columns under the main connective for all three statements. We look across the table, case by case, at each truth value under the main connectives. We are looking to see if there is any case (at least one) in which all three statements are true. This would demonstrate that it is possible for them to be true under the same conditions. In this case, we do not see a single case in which all statements are true. That is, we never see the possibility of all statements being true, i.e., it is impossible for all members of this set of statements to be true under the same conditions. The set is inconsistent. Let’s try another example. Consider the following statements: ~ B v E ~ ( E v ~ M ) → ~ B M ∙ ~ E Q: Are these statements logically consistent or are they inconsistent? When we put them all together in a truth table, we get the following: Again, we look across the table to read for consistency, focusing only on the main connectives of the three statements in our set. Going case by case, we see that it is rarely the case that all three statements in the set are true under the same conditions. This almost never happens. Rarely, but not impossible! Indeed, in case #7 we see that it is possible for all three statements in our set to be true under the same conditions. What conditions? The condition of the world in which B and E are false while M is true. So this set is logically consistent. Our justification would be “case #7” (or any other case in which this condition was met, if it were met by multiple cases). Note that since consistency merely requires “the possibility” of all true statements, a correct and complete justification does not require that we enumerate every case in which it occurs. Citing just a single case is sufficient. Take note of an important principle of reading truth tables. Completed truth tables demonstrate possibility and impossibility differently: Truth tables demonstrate “POSSIBLE” with at least one case that you can point to with the quality under question. Truth tables demonstrate “IMPOSSIBLE” with the absence of the quality in the entire table. Formal Validity and Invalidity Of course, by far the most important property that interests us occurs when the set of statements is structured as an argument. We can put arguments in a truth table by modifying how we put regular sets in a table. We’ll use commas to distinguish the premises from one another and then a single backslash to separate the conclusion from the other statements. Consider the following argument: ( L v B ) → H S ↔ L ~ B ∙ H ∴ ~ S We can set this up in a single truth table to determine if the argument is valid or invalid. Our cheat sheet definitions will help us focus on what to look for in the completed table. Validity: an argument is valid if and only if it is not possible for all the premises to be true and the conclusion false under the same conditions Invalidity: an argument is invalid if and only if it is possible for all the premises to be true and the conclusion false under the same conditions Here is our completed truth table for the argument: Remember, a truth table shows impossibility when the condition is never met in the entire table. A truth table shows possibility with the condition met in just a single case. So what do you see in this table? Do you see at least one case in which all the premises are true (i.e., under the main connectives, you see a T) and the conclusion is false (there is an F value under its main connective)? If you do see that, then you know the argument is invalid. You should then cite the case number as the justification for your judgment that the argument is invalid. If you never see this, then you make that claim as your justification. Look again at the table to find your answer. This argument is INVALID, case #5. We can see this clearly when we circle the case or put emphasis on it, like this: Note that if I had seen more than one case, I would not need to cite all cases in which the premises are all true and the conclusion false. I just need to cite one case to demonstrate that it is indeed possible for this to happen. Finding many cases would not make the argument “more invalid,” so it would not strengthen my evaluation. A Cautionary Tale Many students new to truth tables struggle with the following observation: I see a case in which all the statements in the argument are true! That’s great. However, what do you think this tells you about the argument? Try this quick quiz: T/F: An argument that has all true statements is valid. Go ahead and take a moment to think about your answer. We’ll wait. Having all true statements in your argument sounds like a good thing. I mean, we are speaking the truth. Isn’t that a good thing? Well, the answer really is: it depends. We have to keep in mind what our intention is with what we say. If all I want to do is accurately report on the conditions of the world, then yes, having a bunch of true statements is a very good thing. However, if I want to do more with what I say, if I want to use some statements as support for the truth of another, then having all true statements is not necessarily such a resounding success. If this is my intention, then I need to judge my efforts by a higher standard than simply having the truth. Take another look at the previous truth table. Do you see any conditions in which we have all true statements in our argument? With the emphasis here we can easily see that, yes, case #14 depicts a situation in which all the statements are true. Cool! Q: What does that tell you about the argument? Q: Does this tell us if the argument is valid? Well, look again at the truth table. Notice that case #5 is still there! The table hasn’t changed simply because we looked at case #14. The table shows that even if you are living in a world described by case #14 and all your statements are in fact true, it is still possible for all the premises to be true and the conclusion false. The possibilities for this argument are the same. Put differently, you might live in the world in which you have a bunch of true statements coming out of your mouth, but the possibility that case #5 describes for those statements isn’t going anywhere. Your argument is still invalid. Shortened Truth Tables We have seen how a complete truth table can help us quickly evaluate statements, sets of statements, and arguments for their formal properties. Often enough, the tables are of manageable size to set up quickly and complete without too much burden. However, once we get past four atomics, truth tables start to get a bit unwieldy in size. Doing a 32- or 64-line or larger truth table is good for the soul—it builds mental toughness. However, as a practical matter, it is very time-consuming. At the end of it all, we really end up only looking at those few lines that demonstrate the possibility of a combination of truth values that interest us. So, this is a clue as to how we can shorten up truth tables. A shortened truth table requires a very high level of mastery over the basic conditions of truth for our logical operators. The “short” in shortened truth table makes it sound easy, and thus appealing; however, you will likely make serious errors if you are not very well-versed in the basic conditions of truth. You are likely to be more successful completing a full truth table than a shortened one until you become really familiar with how the logical operators work. Consider this a warning. We can use shortened truth tables for most any logical property we would like to investigate. However, the most common use is to test for validity. We remember that validity is defined as a certain “impossibility” which requires an entire truth table to demonstrate. However, invalidity is defined as a certain type of “possibility” which requires only one line of a truth table. That line is as follows: | | | | | | --- --- | P1 | P2 | P3 | / | C | | T | T | T | | F | Whatever our premises and conclusion, we know that under the main connective of each must appear this specific combination of truth values to show that the argument is invalid. I don’t really care what the case number is on a truth table; I just want to know if this is a possibility. So…let’s just set them up that way, and then use our knowledge of the basic conditions of truth to see if we can make those values work out. Consider our previous example: ( L v B ) → H S ↔ L ~ B ∙ H ∴ ~ S If we lay this out as it would appear in the upper right-hand quadrant of a truth table, it looks like this: ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S We can then put the possible combination of truth values that we want to test under each statement’s main connective. Since we’re testing for validity, it would look like this: ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S T T T F If this is possible, then we know the argument is invalid. Our next step is to see if any truth values become “forced under this possibility.” Q: Can we look at any of these values and see that some other statement “must be” a specific truth value? We should see one immediately drop into our lap. If our conclusion is ~ S and we are testing the possibility that it is a false statement, then S must be a true statement. So we fill this in everywhere we see an S appear: ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S T T T T F T Notice, because the conclusion forced a value for S, we need to keep that value for all S instances in the argument. So this is why we included this in our table for the second premise. Now we see if this new value also forces any more values. Since our second premise is a biconditional that is true, we know both sides of the biconditional must share the same truth value. We now know that the left side is true, so this forces the right side to be true (in order to keep the T value under the double arrow). Like this: ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S T T T T T T F T As before, we need this same value for all instances of L. The same process repeats. We look for forced truth values with each set of values that we now know “must be the case” under our test for invalidity. Sometimes we see multiple forced values. For example, the savvy student has likely already picked up on two more forced values. The third premise is an And statement that is true. So we know that each conjunct must be true. We also know that the first premise is a conditional statement whose antecedent is itself a disjunction (L v B) which has a true conjunct (the “L”). That forces the “v” to be true as well. Before you get all eager to start dropping these values in, you might want to slow down. Doing a shortened truth table is most successful when we take each small step in its turn. Often in logic, slow is faster. Slow is also more accurate. So, let’s proceed slowly. Remember that we just said our third premise, ( ~ B ∙ H ), is a true And statement, and this means both conjuncts must be true. Let’s do just that one for now: ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S T T T T T T T T T F T Don’t forget that the “H” appears twice in the premises, so once we know that it is true in the third premise we must use that value in all other appearances of the “H” in the table. So far, so good… The savvy student has now started to see where this is going. So far, so good… That’s the worry. If we are able to fill in all the truth values for every atomic statement without any contradiction, then we will be able to go so far as to say it is indeed possible for all the premises to be true and the conclusion false. We are trying to see if these values pan out, or if we run into a situation where we cannot fill in truth values without violating our principle of bivalence (i.e., no statement can be both true and false under the same conditions). Let’s see if we can keep going. The very savvy student has already sniffed out something quirky about our first premise. That premise is a conditional statement whose consequent is true. This means that it doesn’t matter what the truth value of the antecedent is, since the conditional will be true all the same. We also know that we really only need to see if we can give a truth value to the atomic “B” statement that is in the antecedent. Since the truth value of the antecedent no longer matters, we are pretty confident that we can give B any truth value that may be required if one is forced upon it somewhere else in the argument. So we look for other instances of the atomic B. Of course, we find a B in the third premise. Moreover, the value for B there is forced. Because ~ B must be a true statement, the B must be false. ( L v B ) → H , S ↔ L , ~ B ∙ H / ~ S T F T T T T T TF T T F T Now we can see that the truth values of all the atomics have been found without creating any contradictions. We could fill in the truth value for ( L v B ) if we wanted, but as we realized earlier, it really doesn’t matter. The little “v” is true, but the first premise was going to be true anyway. This means we have found a possible scenario in which all premises are true and the conclusion is false. Which scenario? The one in which the truth values of the atomics that define that scenario are as follows: L B H S T F T T Under this case (the case # doesn’t really matter), conditions are such that all premises are true and the conclusion is false. We have demonstrated that this is possible. Thus, the argument is invalid. So what happens when the argument is valid? In this case, we will try as we just did to fill in all combinations of truth values. However, along the way we will realize that we cannot keep going so far. We get forced into claiming that the truth value for a given statement is both true and false. That cannot be possible, thus the argument is valid (because the only way to show all true premises and a false conclusion is to insist that the statements are not “really” those truth values). Here’s a simple example: Y → G Y ∴ G Clearly, this is a valid argument; after all, it is nothing more than an instance of the valid form modus ponens (which we will use extensively in the next chapter). To prove that this is valid, we put it in a shortened truth table: Y → G , Y / G T T F Since one of our premises is already an atomic, this forces all other instances of “Y” to share the same value. So we put that in: Y → G , Y / G T T T F Of course, the same can be said of our conclusion, so that would have been a fine place to start as well. We chose the atomic Y, so let’s stick with that for a bit. When we put the truth value in for the Y that appears in the first premise, we should see that this forces a value for the G in that premise. After all, the first premise is a true conditional statement whose antecedent is true. This means the consequent must be true in order to maintain the T under the arrow. So we put that in: Y → G , Y / G T T T T F So far, so good. Now we have a truth value for Y and we have a truth value for G. But wait! We have two truth values for the atomic G! In the premise, G must be true, but in the conclusion, it is F. We would have to change the truth value in the conclusion to a T, which is conventionally shown like this: Y → G , Y / G T T T T F/T Now our shortened truth table demonstrates that the only way to make these values work out is for G to have two truth values at once. This is not possible. Thus, the argument is valid. Put differently, the only way to make a case in which all premises are true and the conclusion false requires something that is not possible. In practice, when doing shortened truth tables, there may be several ways to show that a combination of truth values is not possible. So your friend’s shortened truth table may appear different than yours, yet both tables are correct. The difference between the two will be the result of your each following up on a different forced value. Last word on shortened truth tables: You do not have to do them. If this all sounds like far too much thought to put into a truth table, fine. You can fill out a complete truth table without much effort; it just takes a bit of time. The full table will demonstrate the same possibility or impossibility. Full truth tables work, and they are not terribly hard to complete. However, if you feel like you have the hang of the basic conditions of truth, shortened truth tables are kind of fun and more engaging than a full table. And yes, if we’re talking about a 128-line truth table, shortening that kind of beast up will save a bit of time too.
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https://www.youtube.com/watch?v=sjN_p_RmOv4
Integral of sin^2(x) from 0 to pi JJHanna 1850 subscribers 174 likes Description 28225 views Posted: 21 Jan 2022 Hey guys, I hope you learned/understood the calculus problem a little better. Feel free to ask me any questions or give me suggestions in the comments below. If you enjoyed the video, please give it a thumbs up. Thanks! Alternate Title: How to Integrate the Definite Integral of sin^2(x) dx between 0 and pi by Trigonometric Integration 27 comments Transcript: what is the integral of sine squared x between zero and pi well we know that we're going to have to use some sort of trig property and upon looking at some we can notice that sine squared of x equals one half times 1 minus cosine 2x this is what we call a half angle identity and using this we can actually now integrate 1 minus cosine of 2x treating one half as the constant see we can't integrate sine squared x automatically since u sub would be useless in this case that would just leave us with another trig variable so essentially what we're going to have to do is now simply plug that in so that will equal the integral between 0 and pi of one-half 1 minus cosine 2x dx and treating one-half as again a coefficient it makes this a little easier to solve we do one-half times the integral between 0 and pi of 1 minus cosine 2x now we're ready to integrate so uh the this will equal one half times the integral of one with respect to x is just x minus one over 2 which is essentially just 1 over whatever coefficient is within this trig function so 1 over 2 times sine of 2x we know that leaving the negative out the integral of cosines just sine and applying the negative it becomes negative one half sine two x and that will be between zero and pi now all we gotta do is plug in some numbers and that'll give us our solution so this equals one-half let's put another parenthesis around all this one-half times pi we're plugging pine for x now minus one-half sine of two times pi so two pi and this will be subtracted by zero plugging now zero in for x the lower bounds minus one half sine of two times zero which you know zero now closing these brackets this will contain our answer so let's do a little bit of more simplification this will equal one half times pi again minus one half sine two pi we know that sine two pi is equivalent to sine of zero which is just zero so pi minus zero and again we're going to subtract zero and what is one half of sine zero again now this be zero but you know you could plus zero it doesn't matter we know it's still going to be 0 and this equates to 1 half times pi and that gives us the answer pi over 2. so the integral of sine squared x between 0 and pi is simply pi over two if you have any questions please leave in the comments below and good luck
11434
https://www.thecurriculumcorner.com/thecurriculumcorner123/making-a-table-to-solve-problems/
Strategy: Make a Table - The Curriculum Corner 123 Skip to Content Home About Us/Ads Reading≺ Readers’ Workshop Management Informational Text Literature Independent Reading & Conferencing Small Group Instruction Fluency Sight Words Word Work Vocabulary Math≺ Math Workshop Management Number Sense Operations Math Facts Fractions Measurement & Data Geometry Problem Solving Writing≺ Writers’ Workshop Management Informational Text Narratives / Storytelling Opinion & Persuasive Poetry Language Conventions Interventions Prompted Writing Management≺ Organization Standards Mornings Distance Learning Seasonal≺ Back to School End of Year Spring Summer Fall Winter Special Celebrations Science/SS≺ Science Social Studies Collections Blog Strategy: Make a Table 6 Comments Practice this math problem solving strategy, Make a Table to Solve a Problem, with the help of these free printable problems. Download this make a table to solve a problem set of word problems for your 1st, 2nd and 3rd grade math students. These worksheets will be a helpful addition to your problem solving collection. This is another free resource for teachers and homeschool families from The Curriculum Corner. Looking to help your students learn to make a table to solve a problem? This math problem solving strategy can be practiced with this set of resources. Math Problem Solving Strategies This is one in a series of resources to help you focus on specific problem solving strategies in the classroom. Within this download, we are offering you a range of word problems for practice. Each page provided contains a single problem solving word problem. Below each story problem you will find a set of four steps for students to follow when finding the answer. This set will focus on the make a table strategy for math problem solving. What are the 4 problem solving steps? After carefully reading the problem, students will: Step 1:Circle the math words. Step 2:Ask yourself: Do I understand the problem? Step 3:Solve the problem using words and pictures below. Step 4:Share the answer along with explaining why the answer makes sense. Draw a Table to Solve a Problem Word Work Questions The problems within this post help children to see how they can make a table when working on problem solving. These problems are for first and second grade students. Within this collection you will find nine different problems. You will easily be able to create additional problems using the wording below as a base. With these word problems students are encouraged to draw pictures, but then to take it a step further by putting the information into a table to help answer the questions. This is a great start to showing students how to organize information as a necessary step in problem solving. The problems include the following selections: Fixing Bikes Flower Petals Lovely Ladybugs Spider Legs Triangles Feet and Inches Counting Nickels Counting Dimes Counting Quarters Quarters in a Dollar You can download this set of Make a Table to Solve a Problem pages here: Problem Solving You might also be interested in the following free resources: Draw a Picture to Solve a Problem Write a Number Sentence to Solve a Problem Addition & Subtraction Word Problem Strategies Fall Problem Solving Winter Problem Solving Spring Problem Solving Summer Problem Solving As with all of our resources, The Curriculum Corner creates these for free classroom use. Our products may not be sold. You may print and copy for your personal classroom use. These are also great for home school families! You may not modify and resell in any form. Please let us know if you have any questions. By Author Jill & Cathy Posted on Last updated: May 10, 2021 Categories Blog, Problem Solving Post navigation Strategy: Write a Number Sentence Problem Solving with Money 6 Comments Δ Comment Name Email Comment Ilona June 29, 2013 at 01:19 PM These are great and will be very useful to me! Thank you. Reply Pam July 10, 2013 at 12:20 PM I love the simplicity of these for my class. I plan to add them to my learning centers. Thank you for sharing them. Reply Diana July 11, 2013 at 05:15 PM Just wanted to let you know that I really appreciate your website and the wealth of activities, checklists, games, center ideas, etc. that are contained in your website. I also appreciate you sharing these things without charging. Thank you for helping educators make a difference in the lives of the students we teach. Reply Tammy Nicholson July 19, 2013 at 11:51 AM Love your worksheets! Thanks so much! Reply chona obregon December 28, 2020 at 01:02 AM Nice worksheets. Thank you for sharing it to us. Reply Jill & Cathy February 1, 2021 at 07:21 PM You’re welcome! Reply Recent Posts 2025 September Digital Calendar 2025 – 2026 First Day of School of Signs August Digital Calendar Search Search for: proud member of: what we do: The Curriculum Corner provides FREE resources for private classroom use only. To find out more about our duplication and sharing & privacy policy click HERE. Copyright ©2025 The Curriculum Corner 123 | Bamboo on Trellis Framework by Mediavine
11435
https://www.chegg.com/homework-help/questions-and-answers/use-integration-parts-find-integral-ln-10x-4-respect-x-q141927968
Solved Use integration by parts to find the integral of | Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Calculus Calculus questions and answers Use integration by parts to find the integral of (ln(10x))^4 with respect to x Your solution’s ready to go! Enhanced with AI, our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: Use integration by parts to find the integral of (ln(10x))^4 with respect to x Use integration by parts to find the integral of (ln(1 0 x))^4 with respect to x There are 4 steps to solve this one.Solution 100%(1 rating) Share Share Share done loading Copy link Step 1 Integrate by parts using the formula ∫u d v=u v−∫v d u , where u=(ln⁡(10 x))4 and d v=1 . (ln⁡(10 x))4 x−∫x 4 ln⁡(10 x)3 x d x Simplify. View the full answer Step 2 UnlockStep 3 UnlockStep 4 UnlockAnswer Unlock Previous questionNext question Not the question you’re looking for? Post any question and get expert help quickly. 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11436
https://pmc.ncbi.nlm.nih.gov/articles/PMC6576128/
Six versus two weeks treatment with doxycycline in Lyme neuroborreliosis: the protocol of a multicentre, non-inferiority, double-blinded and randomised controlled trial - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. PMC Search Update PMC Beta search will replace the current PMC search the week of September 7, 2025. Try out PMC Beta search now and give us your feedback. Learn more Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide New Try this search in PMC Beta Search 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 BMJ Open . 2019 Jun 14;9(6):e027083. doi: 10.1136/bmjopen-2018-027083 Search in PMC Search in PubMed View in NLM Catalog Add to search Six versus two weeks treatment with doxycycline in Lyme neuroborreliosis: the protocol of a multicentre, non-inferiority, double-blinded and randomised controlled trial Anne Marit Solheim Anne Marit Solheim 1 Department of Neurology, Sørlandet Hospital, Kristiansand, Norway 2 Department of Clinical Medicine, University of Bergen, Bergen, Norway Find articles by Anne Marit Solheim 1,2, Unn Ljøstad Unn Ljøstad 1 Department of Neurology, Sørlandet Hospital, Kristiansand, Norway 2 Department of Clinical Medicine, University of Bergen, Bergen, Norway Find articles by Unn Ljøstad 1,2,#, Åse Mygland Åse Mygland 1 Department of Neurology, Sørlandet Hospital, Kristiansand, Norway 2 Department of Clinical Medicine, University of Bergen, Bergen, Norway Find articles by Åse Mygland 1,2,# Author information Article notes Copyright and License information 1 Department of Neurology, Sørlandet Hospital, Kristiansand, Norway 2 Department of Clinical Medicine, University of Bergen, Bergen, Norway ✉ Correspondence to Dr Anne Marit Solheim; anne.marit.solheim@sshf.no Contributed equally. Received 2018 Oct 5; Revised 2019 Apr 16; Accepted 2019 Apr 17; Collection date 2019. © Author(s) (or their employer(s)) 2019. Re-use permitted under CC BY-NC. No commercial re-use. See rights and permissions. Published by BMJ. This is an open access article distributed in accordance with the Creative Commons Attribution Non Commercial (CC BY-NC 4.0) license, which permits others to distribute, remix, adapt, build upon this work non-commercially, and license their derivative works on different terms, provided the original work is properly cited, appropriate credit is given, any changes made indicated, and the use is non-commercial. See: PMC Copyright notice PMCID: PMC6576128 PMID: 31201188 Abstract Introduction Current treatment guidelines for European Lyme neuroborreliosis (LNB) recommend cephalosporins, penicillin or doxycycline for 14–28 days but evidence for optimal treatment length is poor. Treatment lengths in clinical practice tend to exceed the recommendations. Most patients experience a rapid improvement of symptoms and neurological findings within days of treatment, but some report long-term complaints. The underlying mechanisms of remaining complaints are debated, and theories as ongoing chronic infection with Borrelia burgdorferi, dysregulated immune responses, genetic predisposition, coinfection with multiple tick-borne pathogens, structural changes in CNS and personal traits have been suggested. The main purpose of our trial is to address the hypothesis of improved outcome after long-term antibiotic treatment of LNB, by comparing efficacy of treatment with 2 and 6 weeks courses of doxycycline. Methods and analysis The trial has a multicentre, non-inferiority, double-blinded design. One hundred and twenty patients diagnosed with LNB according to European Federation of Neurological Societies (EFNS)guidelines will be randomised to 6 or 2 weeks treatment with oral doxycycline. The patients will be followed for 12 months. The primary endpoint is improvement on a composite clinical score (CCS) from baseline to 6 months after inclusion. Secondary endpoints are improvements in the CCS 12 months after inclusion, fatigue scored on Fatigue Severity Scale, subjective symptoms on the Patient Health Questionnaire-15 scale, health-related quality of life scored on RAND 36-item short form health survey and safety as measured by side effects of the two treatment arms. Blood and cerebrospinal fluid (CSF) are collected from inclusion and throughout the follow-up and a biobank will be established. The study started including patients in November 2015 and will continue throughout December 2019. Ethics and dissemination The study is approved by the Norwegian regional committees for medical and health research ethics and the Norwegian Medicines Agency. Data from the study will be published in peer-reviewed medical journals. Trial registration number 2015-001481-25 Keywords: infectious disease/HIV, adult neurology, infectious diseases Strengths and limitations of this study. The trial has a double-blinded design. The inclusion criteria for Lyme neuroborreliosis are according to the EFNS guidelines. The endpoints of the trial are well-defined. The follow-up period of the included patients is long with registered symptoms, signs and potential side effects. A weakness of the study is that the primary scoring tool, the composite clinical score, is not validated. Introduction European Lyme neuroborreliosis (LNB) is caused by the tick-borne spirochete Borrelia burgdorferi (Bb). LNB can manifest weeks or months after a tick bite that only half of the patients remember. The most common clinical manifestations are subacute painful radiculitis and cranial neuropathy (most often the facial nerve). More rare manifestations are myelitis, encephalitis and peripheral neuropathies. Patients diagnosed with LNB should be treated with antibiotics as early as possible to relieve symptoms and prevent sequelae.1–3 Most patients experience a rapid improvement within days of treatment, but some report long-term complaints.4 The most common long-term complaints are fatigue, pain, concentration and memory problems. Some patients may also have neurological sequela such as sensory disturbances, unsteadiness/vertigo, facial paresis and other paresis.5 The underlying mechanisms of remaining complaints are debated. Theories suggested are ongoing chronic Bb infection, dysregulated immune responses, genetic predisposition, coinfection with multiple tick-borne pathogens, structural changes in CNS and personal traits. Standard treatment for LNB is intravenous ceftriaxone or penicillin, or oral doxycycline for 2–4 weeks.6 Previous studies have shown that 2 weeks of oral doxycycline and intravenous ceftriaxone are equally effective for LNB with painful radiculitis or cranial neuritis and probably also for LNB with symptoms from the central nervous system (myelitis and encephalitis).7–9 Arguments for choosing oral doxycycline are that it is inexpensive and convenient, is found to penetrate the blood–brain barrier and give adequate concentrations in the CSF and is effective against coinfections with other tick-borne agents.10 11 We lack evidence about the optimal duration of antibiotic treatment. Most guidelines recommend treatment for 14–28 days.6 12 13 In Norway, the site for the current study, the guidelines recommend 14 days of treatment. A recent Cochrane review of six randomised treatment studies of adult patients with acute LNB reported improvement in the majority of patients after the initial course of antibiotics and no consistent evidence of treatment failure or need of retreatment.14 In another systematic review, the authors conclude that there is insufficient evidence to determine if extended antibiotic treatment is beneficial to outcome.15 Despite this, and perhaps because of uncertainties surrounding LNB, there are varying treatment regimes in clinical practice, generally with more extensive treatment strategies than recommended in current guidelines. A recent study of the treatment practice of 253 Norwegian LNB patients showed that adherence to guidelines was poor and that two-thirds of the patients received >2 weeks of antibiotic treatment.16 In a time with increasing knowledge and awareness of microbial resistance and other complications of long-term antibiotic treatment, these findings seem like a paradox. The present study therefore seeks to increase the evidence of the current treatment advice by evaluating if treatment with doxycycline for 14 days is inferior or not to treatment for 6 weeks with respect to long-term prognosis of LNB. Method Study design and interventions The study is a randomised, double-blinded, placebo-controlled, multicentre trial with a non-inferiority design. We plan to recruit 120 patients diagnosed with definite or probable LNB according to EFNS guidelines6 at six different hospitals in the southern part of Norway as shown in figure 1. The study is coordinated from Sørlandet Hospital in Kristiansand, Agder County by neurologists connected to the large BorrSci study (Lyme borreliosis; a scientific approach to reduce diagnostic and therapeutic uncertainties). The inclusion and exclusion criteria are shown in box 1. Inclusion started in 2015 and will continue through December 2019 or until the necessary sample size is obtained. Eligibility before inclusion is assessed by, or discussed with, a physician connected to the study and accustomed to evaluating patients with neurological symptoms. The patients are randomised into two treatment arms: (A) doxycycline 200 mg daily for 2 weeks, followed by 4 weeks of placebo; (B) doxycycline 200 mg daily for 6 weeks (figure 2). Figure 1. Open in a new tab Map of Norway with the active centres of recruitment per August 2018 marked in red. Box 1. Inclusion and exclusion criteria. Inclusion criteria Neurological symptoms suggestive of Lyme neuroborreliosis without other obvious reasons, and one or both of CSF pleocytosis (leucocytes≥5/mm 3). Intrathecal Borrelia burgdorferi antibody production. Signed informed consent. Exclusion criteria Age <18 years. Treatment with cephalosporin, penicillin or tetracycline macrolide during the last 14 days before start of doxycycline treatment. Pregnancy, breast feeding and/or women of childbearing potential not using adequate contraception. Adverse reaction to tetracyclines. Serious liver or kidney disease that contraindicates use of doxycyline. Lactose intolerance. Need to use medications contraindicated according to summary of product characteristics (SmPC) of the Investigational Medicinal Product (IMP) (antacid drugs, didanosine, probenecid, phenobarbital, phenytoin, carbamazepine, rifampicin). Figure 2. Open in a new tab Inclusion procedures.LNB,Lyme neuroborreliosis. Allocation and blinding Computerised allocation (stratified according to hospital) is performed at Department of Clinical Research Support, Oslo University Hospital, by an internet-based solution. Maximum objective performance and reporting of the study is achieved by applying a ‘penta-blinded’ approach. The first and second blinding is the traditional double blind design with blinding of participants and investigators. Third, the staff evaluating endpoints and adverse effects is blinded to all other study information. Further, the content of all tables and figures will be fixed before any study data are available. Lastly, the statistical procedures will be performed with the two treatment arms marked as group A and B. Revealing the study arms for the investigators will not take place until all patients have completed the 6-month visit, and for the patients after the 12-month visit. Monitoring and data collection The study is monitored independently according to good clinical practice (GCP) by the Department of Clinical Research. The coordinating investigators at Sørlandet Hospital and investigators at cooperating centres are certified according to GCP. The investigators will enter the data required by protocol into an electronic Case Report Form (Viedoc), also designed by the Department of Clinical Research. The same protocol for data management and monitoring is applied to all collected data. Outcome measures A composite clinical score (CCS) based on subjective symptoms and objective neurological findings from the peripheral and central nervous system (box 2) is registered at baseline, 10 weeks, 6 months and 12 months. Each of the 32 items of the CCS is scored 0=none, 1=mild (without influence on daily life)or 2=severe (with influence on daily life). Maximum total score is 64. The primary endpoint of the study is the difference in CCS sum score at baseline and 6 months after inclusion. Box 2. Composite clinical score. Subjective symptoms related by the patient to the current Lyme neuroborreliosis (LNB): Malaise. Fatigue. Headache. Neck and/or back pain. Abdominal and/or breast pain. Arm pain. Leg pain. Generalised pain located to joints and/or muscles. Memory and/or concentration problems. Other. Peripheral findings related to the current LNB: Facial palsy. Paresis of the eye muscles. Reduced hearing. Other cranial neuropathies. Cervical radicular sensory findings. Cervical radicular paresis.† Thoracic radicular sensory findings. Lumbar radicular sensory findings. Lumbar radicular paresis.† Non-radicular sensory findings.‡ Non-radicular paresis.§ Other. Central findings related to the current LNB: Central findings in one extremity.¶ Central findings in a hemi pattern. Central findings in both legs. Central findings in all extremities. Gait ataxia. Dysphasia/aphasia. Nystagmus. Involuntary movement including tremor. Cognitive impairment. Other. Abnormal sensory pattern in a radicular pattern. †Paresis in a radicular pattern. ‡Sensory findings matching with a peripheral nerve or plexus. §Paresis matching a peripheral nerve or plexus. ¶Central weakness and/or spasticity, impairment in pace or fine motor skills. Secondary endpoints are the difference in CCS at baseline and 12 months after inclusion, fatigue scored according to the questionnaire Fatigue Severity Scale(FSS) at 6 and 12 months, subjective somatic symptoms scores according to the Patient Health Questionnaire(PHQ)-15 at 6 and 12 months and health-related quality of life according to RAND 36-item short form health survey at 6 months, and side effects of the treatment. FSS measures level of agreement from 1 to 7 points with nine statements with the final score representing the mean value of nine items. FSS scores≥5 are regarded as severe fatigue. The FSS has been translated into Norwegian, validated in the general Norwegian population and normative Norwegian data are available.17 PHQ-15 charts prevalence and intensity of 13 somatic symptoms; fatigue/lack of energy and difficulty sleeping during the last 4 weeks. Sum score ranges from 0 to 28 for men and from 0 to 30 for women (only women are asked about menstrual symptoms). The following cut-off values for sum score have been stated for load of somatic symptom, 0–4 points: normal, 5–9 points: mild, 10–14 points: moderate, 15–30 points: severe. The PHQ-15 has been validated in several studies and languages, and normative Swedish data are available.18 RAND 36-item short form health survey consists of 36 questions about different aspects of health-related quality of life. The answer to each question is transformed into a score ranging from 0 to 100, where a higher score indicates better health. The questionnaire is validated in Norwegian, and Norwegian normative data are available.19 Thee patient reported outcome measures were included as secondary endpoints to evaluate the potential impact of residual symptoms on patients daily life. Systemic and CSF inflammation will be assessed with lumbar punctures and blood samples at 6 and 12 months after treatment. There will be established a biobank from this material. Figure 3 depicts a flowchart of the study procedures. Figure 3. Open in a new tab Flowchart of the study procedures.FSS, Fatigue Severity Scale; PHQ, Patient Health Questionnaire. Safety The patients are followed closely during and after treatment to monitor safety. They are contacted by phone 1 week after start of treatment and questioned about symptom severity and possible side effects. Blood sampling with a status of haematology, liver and kidney function to monitor potential side effects takes place at 2 and 4 weeks after start of treatment. The patients are also asked to fill out a patient diary on symptoms and possible side effects once a week for 10 weeks. In cases of disease progression, the patients will be evaluated by a physician and adequate intervention initiated. Disease progression is, in this trial, defined as worsening of the patient’s condition attributed to LNB, despite treatment for 14 days with doxycycline, or serious progression of neurological signs from CNS during treatment. Sample size We used data including the SD from our previous treatment trial on 102 LNB patients treated with either oral doxycyline or intravenous cephtriaxone for 2 weeks and scored with an almost similar clinical scale as the CCS in the power analyses.7 From a clinical point of view, a mean group difference of Δ=0.5 in disfavour of 2 weeks treatment compared with 6 weeks treatment was regarded as an appropriate non-inferiority margin. This non-inferiority margin corresponds to a Cohen’s d effect size of Δ/σ=0.5/1.0=0.5, which is a small and clinical acceptable effect size. With a one-sided test and significance level of 0.05, 50 patients in each treatment group was found to be needed to claim non-inferiority with a non-inferiority margin on mean group difference of 0.5 and a SD of 1.0 with 80% power. To compensate for up to 20% dropouts and non-evaluable patients 120 (ie, 60 in each group) patients will be enrolled. Statistical analysis The main statistical analysis is planned when all patients have completed the 6-month visit. Results will be reported as mean scores with SD or proportions as appropriate. To compare the primary outcome in the two groups we will use a general linear model with treatment group as a factor, and adjustment for duration of symptoms, gender and age. The analysis will be conducted according to the intention-to-treat principle. For other analysis, comparison between groups will be done with, for example, independent samples t-test, non-parametric Mann-Whitney U test or Pearson’s χ 2 test for crosstabs as appropriate. Results from the FSS and PHQ-15 questionnaires will be dichotomised according to predefined cut-offs recommended for case definition and statistically treated as categorical outcomes. P values <0.05 will be considered statistically significant. Ethics and dissemination The trial is registered on Clinicaltrials.org.The study will be conducted in accordance with ethical principles that have their origin in the Declaration of Helsinki and are consistent with Internation Conference of Harmonisation- Good Clinical Pratice (ICH/GCP) (and applicable regulatory requirements. Each patient in the trial is submitted to extensive follow-up as previously described in terms of disease, effect of treatment and side effects to outweigh potential harms. The benefits are considered to outweigh the cons of this trial in the long term, with a potentially more evidence-based treatment of LNB and less extensive use of antibiotics. Data from the study will be published in peer-reviewed medical journals. Patient and public involvement Representatives from the Norwegian patient organisation for Lyme borreliosis (Norsk Lyme Borreliose Forening) were invited and participated in the early stages of planning of the BorrSci project’s design and gave feedback on the drafts of the application for funding. They were also invited to continue work with the project. Inclusion to the study, implications of the intervention and time required to participate is discussed with each individual patient. Local newspapers and other media have been involved in making the project known to the public in different parts of Norway. Supplementary Material Reviewer comments bmjopen-2018-027083.reviewer_comments.pdf (223KB, pdf) Author's manuscript bmjopen-2018-027083.draft_revisions.pdf (866.2KB, pdf) Footnotes UL and ÅM contributed equally. Contributors: AMS contributed in drafting this manuscript, has participated in revisions of the original protocol, includes patients to the study and coordinates the study at Sørlandet Hospital and at the other centres of recruitment. UL and ÅM drafted the original protocol, worked on applications for funding, contributed in drafting this manuscript and include patients to the study. Funding: This work is supported by the Norwegian Multiregional Health Authorities through the BorrSci project (Lyme borreliosis; a scientific approach to reduce diagnostic and therapeutic uncertainties, project 2015113), letter dated 17 April 2015 with case reference 14/01152-4. Competing interests: None declared. Ethics approval: The Norwegian regional committees for medical and health research ethics (REC) and the Norwegian Medicines Agency (SLV) have approved the study. Provenance and peer review: Not commissioned; externally peer reviewed. Patient consent for publication: Not required. Map disclaimer: The depiction of boundaries on the map(s) in this article do not imply the expression of any opinion whatsoever on the part of BMJ (or any member of its group) concerning the legal status of any country, territory, jurisdiction or area or of its authorities. The map(s) are provided without any warranty of any kind, either express or implied. References Ljøstad U, Mygland A. Remaining complaints 1 year after treatment for acute Lyme neuroborreliosis; frequency, pattern and risk factors. Eur J Neurol 2010;17:118–23. 10.1111/j.1468-1331.2009.02756.x [DOI] [PubMed] [Google Scholar] Eikeland R, Mygland Å, Herlofson K, et al. 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https://en.wikipedia.org/wiki/Fourier_transform
Jump to content Search Contents 1 Definition 1.1 Angular frequency (ω) 1.2 Lebesgue integrable functions 2 Background 2.1 History 2.2 Complex sinusoids 2.3 Negative frequency 2.4 Fourier transform for periodic functions 2.5 Sampling the Fourier transform 3 Units 4 Properties 4.1 Basic properties 4.1.1 Linearity 4.1.2 Time shifting 4.1.3 Frequency shifting 4.1.4 Time scaling 4.1.5 Symmetry 4.1.6 Conjugation 4.1.7 Real and imaginary parts 4.1.8 Zero frequency component 4.2 Uniform continuity and the Riemann–Lebesgue lemma 4.3 Plancherel theorem and Parseval's theorem 4.4 Convolution theorem 4.5 Cross-correlation theorem 4.6 Differentiation 4.7 Eigenfunctions 4.8 Inversion and periodicity 4.9 Connection with the Heisenberg group 5 Complex domain 5.1 Laplace transform 5.2 Inversion 6 Fourier transform on Euclidean space 6.1 Uncertainty principle 6.2 Sine and cosine transforms 6.3 Spherical harmonics 6.4 Restriction problems 7 Fourier transform on function spaces 7.1 On other Lp 7.2 Tempered distributions 8 Generalizations 8.1 Fourier–Stieltjes transform on measurable spaces 8.2 Locally compact abelian groups 8.3 Gelfand transform 8.4 Compact non-abelian groups 9 Alternatives 10 Example 11 Applications 11.1 Analysis of differential equations 11.2 Fourier-transform spectroscopy 11.3 Quantum mechanics 11.4 Signal processing 12 Other notations 13 Computation methods 13.1 Discrete Fourier transforms and fast Fourier transforms 13.2 Analytic integration of closed-form functions 13.3 Numerical integration of closed-form continuous functions 13.4 Numerical integration of a series of ordered pairs 14 Tables of important Fourier transforms 14.1 Functional relationships, one-dimensional 14.2 Square-integrable functions, one-dimensional 14.3 Distributions, one-dimensional 14.4 Two-dimensional functions 14.5 Formulas for general n-dimensional functions 15 See also 16 Notes 17 Citations 18 References 19 External links Fourier transform አማርኛ العربية Asturianu Azərbaycanca বাংলা 閩南語 / Bn-lm-gí Беларуская Беларуская (тарашкевіца) Български Boarisch Català Čeština Dansk Deutsch Eesti Ελληνικά Español Esperanto Euskara فارسی Français Galego 한국어 हिन्दी Hrvatski Bahasa Indonesia Íslenska Italiano עברית کٲشُر Қазақша Lietuvių Magyar Македонски Malti Монгол မြန်မာဘာသာ Nederlands 日本語 Norsk bokmål Norsk nynorsk ਪੰਜਾਬੀ Polski Português Romnă Русский Shqip Simple English Slovenčina Slovenščina Српски / srpski Sunda Suomi Svenska தமிழ் Татарча / tatarça ไทย Türkçe Українська Tiếng Việt 吴语 粵語 中文 Kadazandusun ရခိုင် 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 Wikimedia Commons Wikidata item Appearance From Wikipedia, the free encyclopedia Mathematical transform that expresses a function of time as a function of frequency Not to be confused with Fourier method or Fourier's original sine and cosine transforms. | Fourier transforms | | Fourier transform Fourier series Discrete-time Fourier transform Discrete Fourier transform Discrete Fourier transform over a ring Fourier transform on finite groups Fourier analysis Related transforms | In mathematics, the Fourier transform (FT) is an integral transform that takes a function as input, and outputs another function that describes the extent to which various frequencies are present in the original function. The output of the transform is a complex-valued function of frequency. The term Fourier transform refers to both the mathematical operation and to this complex-valued function. When a distinction needs to be made, the output of the operation is sometimes called the frequency domain representation of the original function.[note 1] The Fourier transform is analogous to decomposing the sound of a musical chord into the intensities of its constituent pitches. Functions that are localized in the time domain have Fourier transforms that are spread out across the frequency domain and vice versa, a phenomenon known as the uncertainty principle. The critical case for this principle is the Gaussian function, of substantial importance in probability theory and statistics as well as in the study of physical phenomena exhibiting normal distribution (e.g., diffusion). The Fourier transform of a Gaussian function is another Gaussian function. Joseph Fourier introduced sine and cosine transforms (which correspond to the imaginary and real components of the modern Fourier transform) in his study of heat transfer, where Gaussian functions appear as solutions of the heat equation. The Fourier transform can be formally defined as an improper Riemann integral, making it an integral transform, although this definition is not suitable for many applications requiring a more sophisticated integration theory.[note 2] For example, many relatively simple applications use the Dirac delta function, which can be treated formally as if it were a function, but the justification requires a mathematically more sophisticated viewpoint.[note 3] The Fourier transform can also be generalized to functions of several variables on Euclidean space, sending a function of 3-dimensional "position space" to a function of 3-dimensional momentum (or a function of space and time to a function of 4-momentum). This idea makes the spatial Fourier transform very natural in the study of waves, as well as in quantum mechanics, where it is important to be able to represent wave solutions as functions of either position or momentum and sometimes both. In general, functions to which Fourier methods are applicable are complex-valued, and possibly vector-valued.[note 4] Still further generalization is possible to functions on groups, which, besides the original Fourier transform on R or Rn, notably includes the discrete-time Fourier transform (DTFT, group = Z), the discrete Fourier transform (DFT, group = Z mod N) and the Fourier series or circular Fourier transform (group = S1, the unit circle ≈ closed finite interval with endpoints identified). The latter is routinely employed to handle periodic functions. The fast Fourier transform (FFT) is an algorithm for computing the DFT. Definition [edit] The Fourier transform of a complex-valued on the real line, is the complex valued function , defined by the integral Fourier transform | | | --- | | | Eq.1 | Evaluating the Fourier transform for all values of produces the frequency-domain function. When is (Lebesgue) integrable over the whole real line, the above integral converges at all frequencies to a continuous function (decaying to zero as ). However, interpreting the integral suitably (for example as an improper integral), the Fourier transform also makes sense for some functions that are not necessarily (absolutely) integrable, and even for certain objects that are not functions, like the Dirac delta function (and other tempered distributions). First introduced in Fourier's Analytical Theory of Heat., the corresponding inversion formula for "sufficiently nice" functions is given by the Fourier inversion theorem, i.e., Inverse transform | | | --- | | | Eq.2 | The functions and are referred to as a Fourier transform pair. A common notation for designating transform pairs is: For example, the Fourier transform of the delta function is the constant function : Angular frequency (ω) [edit] When the independent variable () represents time (often denoted by ), the transform variable () represents frequency (often denoted by ). For example, if time has the unit second, then frequency has the unit hertz. The transform variable can also be written in terms of angular frequency, with the unit radian per second. The substitution into Eq.1 produces this convention, where function is relabeled Unlike the Eq.1 definition, the Fourier transform is no longer a unitary transformation, and there is less symmetry between the formulas for the transform and its inverse. Those properties are restored by splitting the factor evenly between the transform and its inverse, which leads to another convention: Variations of all three conventions can be created by conjugating the complex-exponential kernel of both the forward and the reverse transform. The signs must be opposites. Summary of popular forms of the Fourier transform, one-dimensional | ordinary frequency ξ (Hz) | unitary | | | angular frequency ω (rad/s) | unitary | | | non-unitary | | Generalization for n-dimensional functions | ordinary frequency ξ (Hz) | unitary | | | angular frequency ω (rad/s) | unitary | | | non-unitary | | Lebesgue integrable functions [edit] See also: Lp space § Lp spaces and Lebesgue integrals A measurable function is called (Lebesgue) integrable if the Lebesgue integral of its absolute value is finite: If is Lebesgue integrable then the Fourier transform, given by Eq.1, is well-defined for all . Furthermore, is bounded, uniformly continuous and (by the Riemann–Lebesgue lemma) vanishing at infinity. The space is the space of measurable functions for which the norm is finite, modulo the equivalence relation of equality almost everywhere. The Fourier transform on is one-to-one. However, there is no easy characterization of the image, and thus no easy characterization of the inverse transform. In particular, Eq.2 is no longer valid, as it was stated only under the hypothesis that was "sufficiently nice" (e.g., decays with all derivatives). While Eq.1 defines the Fourier transform for (complex-valued) functions in , it is not well-defined for other integrability classes, most importantly the space of square-integrable functions . For example, the function is in but not and therefore the Lebesgue integral Eq.1 does not exist. However, the Fourier transform on the dense subspace admits a unique continuous extension to a unitary operator on . This extension is important in part because, unlike the case of , the Fourier transform is an automorphism of the space . In such cases, the Fourier transform can be obtained explicitly by regularizing the integral, and then passing to a limit. In practice, the integral is often regarded as an improper integral instead of a proper Lebesgue integral, but sometimes for convergence one needs to use weak limit or principal value instead of the (pointwise) limits implicit in an improper integral. Titchmarsh (1986) and Dym & McKean (1985) each gives three rigorous ways of extending the Fourier transform to square integrable functions using this procedure. A general principle in working with the Fourier transform is that Gaussians are dense in , and the various features of the Fourier transform, such as its unitarity, are easily inferred for Gaussians. Many of the properties of the Fourier transform can then be proven from two facts about Gaussians: that is its own Fourier transform; and that the Gaussian integral A feature of the Fourier transform is that it is a homomorphism of Banach algebras from equipped with the convolution operation to the Banach algebra of continuous functions under the (supremum) norm. The conventions chosen in this article are those of harmonic analysis, and are characterized as the unique conventions such that the Fourier transform is both unitary on L2 and an algebra homomorphism from L1 to L∞, without renormalizing the Lebesgue measure. Background [edit] History [edit] Main articles: Fourier analysis § History, and Fourier series § History In 1822, Fourier claimed (see Joseph Fourier § The Analytic Theory of Heat) that any function, whether continuous or discontinuous, can be expanded into a series of sines. That important work was corrected and expanded upon by others to provide the foundation for the various forms of the Fourier transform used since. Complex sinusoids [edit] The red sinusoid can be described by peak amplitude (1), peak-to-peak (2), RMS (3), and wavelength (4). The red and blue sinusoids have a phase difference of θ. In general, the coefficients are complex numbers, which have two equivalent forms (see Euler's formula): The product with (Eq.2) has these forms: which conveys both amplitude and phase of frequency Likewise, the intuitive interpretation of Eq.1 is that multiplying by has the effect of subtracting from every frequency component of function [note 5] Only the component that was at frequency can produce a non-zero value of the infinite integral, because (at least formally) all the other shifted components are oscillatory and integrate to zero (see § Example). It is noteworthy how easily the product was simplified using the polar form, and how easily the rectangular form was deduced by an application of Euler's formula. Negative frequency [edit] See also: Negative frequency § Simplifying the Fourier transform Euler's formula introduces the possibility of negative And Eq.1 is defined Only certain complex-valued have transforms (See Analytic signal. A simple example is ) But negative frequency is necessary to characterize all other complex-valued found in signal processing, partial differential equations, radar, nonlinear optics, quantum mechanics, and others. For a real-valued Eq.1 has the symmetry property (see § Conjugation below). This redundancy enables Eq.2 to distinguish from But it cannot determine the actual sign of because and are indistinguishable on just the real numbers line. Fourier transform for periodic functions [edit] The Fourier transform of a periodic function cannot be defined using the integral formula directly. In order for integral in Eq.1 to be defined the function must be absolutely integrable. Instead it is common to use Fourier series. It is possible to extend the definition to include periodic functions by viewing them as tempered distributions. This makes it possible to see a connection between the Fourier series and the Fourier transform for periodic functions that have a convergent Fourier series. If is a periodic function, with period , that has a convergent Fourier series, then: where are the Fourier series coefficients of , and is the Dirac delta function. In other words, the Fourier transform is a Dirac comb function whose teeth are multiplied by the Fourier series coefficients. Sampling the Fourier transform [edit] For broader coverage of this topic, see Poisson summation formula. The Fourier transform of an integrable function can be sampled at regular intervals of arbitrary length These samples can be deduced from one cycle of a periodic function which has Fourier series coefficients proportional to those samples by the Poisson summation formula: The integrability of ensures the periodic summation converges. Therefore, the samples can be determined by Fourier series analysis: When has compact support, has a finite number of terms within the interval of integration. When does not have compact support, numerical evaluation of requires an approximation, such as tapering or truncating the number of terms. Units [edit] See also: Spectral density § Units The frequency variable must have inverse units to the units of the original function's domain (typically named or ). For example, if is measured in seconds, should be in cycles per second or hertz. If the scale of time is in units of seconds, then another Greek letter is typically used instead to represent angular frequency (where ) in units of radians per second. If using for units of length, then must be in inverse length, e.g., wavenumbers. That is to say, there are two versions of the real line: one which is the range of and measured in units of and the other which is the range of and measured in inverse units to the units of These two distinct versions of the real line cannot be equated with each other. Therefore, the Fourier transform goes from one space of functions to a different space of functions: functions which have a different domain of definition. In general, must always be taken to be a linear form on the space of its domain, which is to say that the second real line is the dual space of the first real line. See the article on linear algebra for a more formal explanation and for more details. This point of view becomes essential in generalizations of the Fourier transform to general symmetry groups, including the case of Fourier series. That there is no one preferred way (often, one says "no canonical way") to compare the two versions of the real line which are involved in the Fourier transform—fixing the units on one line does not force the scale of the units on the other line—is the reason for the plethora of rival conventions on the definition of the Fourier transform. The various definitions resulting from different choices of units differ by various constants. In other conventions, the Fourier transform has i in the exponent instead of −i, and vice versa for the inversion formula. This convention is common in modern physics and is the default for Wolfram Alpha, and does not mean that the frequency has become negative, since there is no canonical definition of positivity for frequency of a complex wave. It simply means that is the amplitude of the wave instead of the wave (the former, with its minus sign, is often seen in the time dependence for sinusoidal plane-wave solutions of the electromagnetic wave equation, or in the time dependence for quantum wave functions). Many of the identities involving the Fourier transform remain valid in those conventions, provided all terms that explicitly involve i have it replaced by −i. In electrical engineering the letter j is typically used for the imaginary unit instead of i because i is used for current. When using dimensionless units, the constant factors might not be written in the transform definition. For instance, in probability theory, the characteristic function Φ of the probability density function f of a random variable X of continuous type is defined without a negative sign in the exponential, and since the units of x are ignored, there is no 2π either: In probability theory and mathematical statistics, the use of the Fourier—Stieltjes transform is preferred, because many random variables are not of continuous type, and do not possess a density function, and one must treat not functions but distributions, i.e., measures which possess "atoms". From the higher point of view of group characters, which is much more abstract, all these arbitrary choices disappear, as will be explained in the later section of this article, which treats the notion of the Fourier transform of a function on a locally compact Abelian group. Properties [edit] Let and represent integrable functions Lebesgue-measurable on the real line satisfying: We denote the Fourier transforms of these functions as and respectively. Basic properties [edit] The Fourier transform has the following basic properties: Linearity [edit] Time shifting [edit] Frequency shifting [edit] Time scaling [edit] The case leads to the time-reversal property: The transform of an even-symmetric real-valued function is also an even-symmetric real-valued function The time-shift, creates an imaginary component, (see § Symmetry. Symmetry [edit] When the real and imaginary parts of a complex function are decomposed into their even and odd parts, there are four components, denoted below by the subscripts RE, RO, IE, and IO. And there is a one-to-one mapping between the four components of a complex time function and the four components of its complex frequency transform: From this, various relationships are apparent, for example: The transform of a real-valued function is the conjugate symmetric function Conversely, a conjugate symmetric transform implies a real-valued time-domain. The transform of an imaginary-valued function is the conjugate antisymmetric function and the converse is true. The transform of a conjugate symmetric function is the real-valued function and the converse is true. The transform of a conjugate antisymmetric function is the imaginary-valued function and the converse is true. Conjugation [edit] (Note: the ∗ denotes complex conjugation.) In particular, if is real, then is even symmetric (aka Hermitian function): If is purely imaginary, then is odd symmetric: Real and imaginary parts [edit] Zero frequency component [edit] Substituting in the definition, we obtain: The integral of over its domain is known as the average value or DC bias of the function. Uniform continuity and the Riemann–Lebesgue lemma [edit] The Fourier transform may be defined in some cases for non-integrable functions, but the Fourier transforms of integrable functions have several strong properties. The Fourier transform of any integrable function is uniformly continuous and By the Riemann–Lebesgue lemma, However, need not be integrable. For example, the Fourier transform of the rectangular function, which is integrable, is the sinc function, which is not Lebesgue integrable, because its improper integrals behave analogously to the alternating harmonic series, in converging to a sum without being absolutely convergent. It is not generally possible to write the inverse transform as a Lebesgue integral. However, when both and are integrable, the inverse equality holds for almost every x. As a result, the Fourier transform is injective on L1(R). Plancherel theorem and Parseval's theorem [edit] Main articles: Plancherel theorem and Parseval's theorem Let f(x) and g(x) be integrable, and let f̂(ξ) and ĝ(ξ) be their Fourier transforms. If f(x) and g(x) are also square-integrable, then the Parseval formula follows: where the bar denotes complex conjugation. The Plancherel theorem, which follows from the above, states that Plancherel's theorem makes it possible to extend the Fourier transform, by a continuity argument, to a unitary operator on L2(R). On L1(R) ∩ L2(R), this extension agrees with original Fourier transform defined on L1(R), thus enlarging the domain of the Fourier transform to L1(R) + L2(R) (and consequently to Lp(R) for 1 ≤ p ≤ 2). Plancherel's theorem has the interpretation in the sciences that the Fourier transform preserves the energy of the original quantity. The terminology of these formulas is not quite standardised. Parseval's theorem was proved only for Fourier series, and was first proved by Lyapunov. But Parseval's formula makes sense for the Fourier transform as well, and so even though in the context of the Fourier transform it was proved by Plancherel, it is still often referred to as Parseval's formula, or Parseval's relation, or even Parseval's theorem. See Pontryagin duality for a general formulation of this concept in the context of locally compact abelian groups. Convolution theorem [edit] Main article: Convolution theorem The Fourier transform translates between convolution and multiplication of functions. If f(x) and g(x) are integrable functions with Fourier transforms f̂(ξ) and ĝ(ξ) respectively, then the Fourier transform of the convolution is given by the product of the Fourier transforms f̂(ξ) and ĝ(ξ) (under other conventions for the definition of the Fourier transform a constant factor may appear). This means that if: where ∗ denotes the convolution operation, then: In linear time invariant (LTI) system theory, it is common to interpret g(x) as the impulse response of an LTI system with input f(x) and output h(x), since substituting the unit impulse for f(x) yields h(x) = g(x). In this case, ĝ(ξ) represents the frequency response of the system. Conversely, if f(x) can be decomposed as the product of two square integrable functions p(x) and q(x), then the Fourier transform of f(x) is given by the convolution of the respective Fourier transforms p̂(ξ) and q̂(ξ). Cross-correlation theorem [edit] Main articles: Cross-correlation and Wiener–Khinchin_theorem In an analogous manner, it can be shown that if h(x) is the cross-correlation of f(x) and g(x): then the Fourier transform of h(x) is: As a special case, the autocorrelation of function f(x) is: for which Differentiation [edit] Suppose f(x) is differentiable almost everywhere, and both f and its derivative f′ are integrable (in ). Then the Fourier transform of the derivative is given by More generally, the Fourier transformation of the nth derivative f(n) is given by Analogously, , so By applying the Fourier transform and using these formulas, some ordinary differential equations can be transformed into algebraic equations, which are much easier to solve. These formulas also give rise to the rule of thumb "f(x) is smooth if and only if f̂(ξ) quickly falls to 0 for |ξ| → ∞." By using the analogous rules for the inverse Fourier transform, one can also say "f(x) quickly falls to 0 for |x| → ∞ if and only if f̂(ξ) is smooth." Eigenfunctions [edit] See also: Mehler kernel and Hermite polynomials § Hermite functions as eigenfunctions of the Fourier transform The Fourier transform is a linear transform which has eigenfunctions obeying with A set of eigenfunctions is found by noting that the homogeneous differential equation leads to eigenfunctions of the Fourier transform as long as the form of the equation remains invariant under Fourier transform.[note 6] In other words, every solution and its Fourier transform obey the same equation. Assuming uniqueness of the solutions, every solution must therefore be an eigenfunction of the Fourier transform. The form of the equation remains unchanged under Fourier transform if can be expanded in a power series in which for all terms the same factor of either one of arises from the factors introduced by the differentiation rules upon Fourier transforming the homogeneous differential equation because this factor may then be cancelled. The simplest allowable leads to the standard normal distribution. More generally, a set of eigenfunctions is also found by noting that the differentiation rules imply that the ordinary differential equation with constant and being a non-constant even function remains invariant in form when applying the Fourier transform to both sides of the equation. The simplest example is provided by which is equivalent to considering the Schrödinger equation for the quantum harmonic oscillator. The corresponding solutions provide an important choice of an orthonormal basis for L2(R) and are given by the "physicist's" Hermite functions. Equivalently one may use where Hen(x) are the "probabilist's" Hermite polynomials, defined as Under this convention for the Fourier transform, we have that In other words, the Hermite functions form a complete orthonormal system of eigenfunctions for the Fourier transform on L2(R). However, this choice of eigenfunctions is not unique. Because of there are only four different eigenvalues of the Fourier transform (the fourth roots of unity ±1 and ±i) and any linear combination of eigenfunctions with the same eigenvalue gives another eigenfunction. As a consequence of this, it is possible to decompose L2(R) as a direct sum of four spaces H0, H1, H2, and H3 where the Fourier transform acts on Hek simply by multiplication by ik. Since the complete set of Hermite functions ψn provides a resolution of the identity they diagonalize the Fourier operator, i.e. the Fourier transform can be represented by such a sum of terms weighted by the above eigenvalues, and these sums can be explicitly summed: This approach to define the Fourier transform was first proposed by Norbert Wiener. Among other properties, Hermite functions decrease exponentially fast in both frequency and time domains, and they are thus used to define a generalization of the Fourier transform, namely the fractional Fourier transform used in time–frequency analysis. In physics, this transform was introduced by Edward Condon. This change of basis functions becomes possible because the Fourier transform is a unitary transform when using the right conventions. Consequently, under the proper conditions it may be expected to result from a self-adjoint generator via The operator is the number operator of the quantum harmonic oscillator written as It can be interpreted as the generator of fractional Fourier transforms for arbitrary values of t, and of the conventional continuous Fourier transform for the particular value with the Mehler kernel implementing the corresponding active transform. The eigenfunctions of are the Hermite functions which are therefore also eigenfunctions of Upon extending the Fourier transform to distributions the Dirac comb is also an eigenfunction of the Fourier transform. Inversion and periodicity [edit] Further information: Fourier inversion theorem and Fractional Fourier transform Under suitable conditions on the function , it can be recovered from its Fourier transform . Indeed, denoting the Fourier transform operator by , so , then for suitable functions, applying the Fourier transform twice simply flips the function: , which can be interpreted as "reversing time". Since reversing time is two-periodic, applying this twice yields , so the Fourier transform operator is four-periodic, and similarly the inverse Fourier transform can be obtained by applying the Fourier transform three times: . In particular the Fourier transform is invertible (under suitable conditions). More precisely, defining the parity operator such that , we have: These equalities of operators require careful definition of the space of functions in question, defining equality of functions (equality at every point? equality almost everywhere?) and defining equality of operators – that is, defining the topology on the function space and operator space in question. These are not true for all functions, but are true under various conditions, which are the content of the various forms of the Fourier inversion theorem. This fourfold periodicity of the Fourier transform is similar to a rotation of the plane by 90°, particularly as the two-fold iteration yields a reversal, and in fact this analogy can be made precise. While the Fourier transform can simply be interpreted as switching the time domain and the frequency domain, with the inverse Fourier transform switching them back, more geometrically it can be interpreted as a rotation by 90° in the time–frequency domain (considering time as the x-axis and frequency as the y-axis), and the Fourier transform can be generalized to the fractional Fourier transform, which involves rotations by other angles. This can be further generalized to linear canonical transformations, which can be visualized as the action of the special linear group SL2(R) on the time–frequency plane, with the preserved symplectic form corresponding to the uncertainty principle, below. This approach is particularly studied in signal processing, under time–frequency analysis. Connection with the Heisenberg group [edit] The Heisenberg group is a certain group of unitary operators on the Hilbert space L2(R) of square integrable complex valued functions f on the real line, generated by the translations (Ty f)(x) = f (x + y) and multiplication by ei2πξx, (Mξ f)(x) = ei2πξx f (x). These operators do not commute, as their (group) commutator is which is multiplication by the constant (independent of x) ei2πξy ∈ U(1) (the circle group of unit modulus complex numbers). As an abstract group, the Heisenberg group is the three-dimensional Lie group of triples (x, ξ, z) ∈ R2 × U(1), with the group law Denote the Heisenberg group by H1. The above procedure describes not only the group structure, but also a standard unitary representation of H1 on a Hilbert space, which we denote by ρ : H1 → B(L2(R)). Define the linear automorphism of R2 by so that J2 = −I. This J can be extended to a unique automorphism of H1: According to the Stone–von Neumann theorem, the unitary representations ρ and ρ ∘ j are unitarily equivalent, so there is a unique intertwiner W ∈ U(L2(R)) such that This operator W is the Fourier transform. Many of the standard properties of the Fourier transform are immediate consequences of this more general framework. For example, the square of the Fourier transform, W2, is an intertwiner associated with J2 = −I, and so we have (W2f)(x) = f (−x) is the reflection of the original function f. Complex domain [edit] The integral for the Fourier transform can be studied for complex values of its argument ξ. Depending on the properties of f, this might not converge off the real axis at all, or it might converge to a complex analytic function for all values of ξ = σ + iτ, or something in between. The Paley–Wiener theorem says that f is smooth (i.e., n-times differentiable for all positive integers n) and compactly supported if and only if f̂ (σ + iτ) is a holomorphic function for which there exists a constant a > 0 such that for any integer n ≥ 0, for some constant C. (In this case, f is supported on [−a, a].) This can be expressed by saying that f̂ is an entire function which is rapidly decreasing in σ (for fixed τ) and of exponential growth in τ (uniformly in σ). (If f is not smooth, but only L2, the statement still holds provided n = 0.) The space of such functions of a complex variable is called the Paley—Wiener space. This theorem has been generalised to semisimple Lie groups. If f is supported on the half-line t ≥ 0, then f is said to be "causal" because the impulse response function of a physically realisable filter must have this property, as no effect can precede its cause. Paley and Wiener showed that then f̂ extends to a holomorphic function on the complex lower half-plane τ < 0 which tends to zero as τ goes to infinity. The converse is false and it is not known how to characterise the Fourier transform of a causal function. Laplace transform [edit] See also: Laplace transform § Fourier transform The Fourier transform f̂(ξ) is related to the Laplace transform F(s), which is also used for the solution of differential equations and the analysis of filters. It may happen that a function f for which the Fourier integral does not converge on the real axis at all, nevertheless has a complex Fourier transform defined in some region of the complex plane. For example, if f(t) is of exponential growth, i.e., for some constants C, a ≥ 0, then convergent for all 2πτ < −a, is the two-sided Laplace transform of f. The more usual version ("one-sided") of the Laplace transform is If f is also causal, and analytical, then: Thus, extending the Fourier transform to the complex domain means it includes the Laplace transform as a special case in the case of causal functions—but with the change of variable s = i2πξ. From another, perhaps more classical viewpoint, the Laplace transform by its form involves an additional exponential regulating term which lets it converge outside of the imaginary line where the Fourier transform is defined. As such it can converge for at most exponentially divergent series and integrals, whereas the original Fourier decomposition cannot, enabling analysis of systems with divergent or critical elements. Two particular examples from linear signal processing are the construction of allpass filter networks from critical comb and mitigating filters via exact pole-zero cancellation on the unit circle. Such designs are common in audio processing, where highly nonlinear phase response is sought for, as in reverb. Furthermore, when extended pulselike impulse responses are sought for signal processing work, the easiest way to produce them is to have one circuit which produces a divergent time response, and then to cancel its divergence through a delayed opposite and compensatory response. There, only the delay circuit in-between admits a classical Fourier description, which is critical. Both the circuits to the side are unstable, and do not admit a convergent Fourier decomposition. However, they do admit a Laplace domain description, with identical half-planes of convergence in the complex plane (or in the discrete case, the Z-plane), wherein their effects cancel. In modern mathematics the Laplace transform is conventionally subsumed under the aegis Fourier methods. Both of them are subsumed by the far more general, and more abstract, idea of harmonic analysis. Inversion [edit] Still with , if is complex analytic for a ≤ τ ≤ b, then by Cauchy's integral theorem. Therefore, the Fourier inversion formula can use integration along different lines, parallel to the real axis. Theorem: If f(t) = 0 for t < 0, and |f(t)| < Cea|t| for some constants C, a > 0, then for any τ < −⁠a/2π⁠. This theorem implies the Mellin inversion formula for the Laplace transformation, for any b > a, where F(s) is the Laplace transform of f(t). The hypotheses can be weakened, as in the results of Carleson and Hunt, to f(t) e−at being L1, provided that f be of bounded variation in a closed neighborhood of t (cf. Dini test), the value of f at t be taken to be the arithmetic mean of the left and right limits, and that the integrals be taken in the sense of Cauchy principal values. L2 versions of these inversion formulas are also available. Fourier transform on Euclidean space [edit] The Fourier transform can be defined in any arbitrary number of dimensions n. As with the one-dimensional case, there are many conventions. For an integrable function f(x), this article takes the definition: where x and ξ are n-dimensional vectors, and x · ξ is the dot product of the vectors. Alternatively, ξ can be viewed as belonging to the dual vector space , in which case the dot product becomes the contraction of x and ξ, usually written as ⟨x, ξ⟩. All of the basic properties listed above hold for the n-dimensional Fourier transform, as do Plancherel's and Parseval's theorem. When the function is integrable, the Fourier transform is still uniformly continuous and the Riemann–Lebesgue lemma holds. Uncertainty principle [edit] Further information: Uncertainty principle Generally speaking, the more concentrated f(x) is, the more spread out its Fourier transform f̂(ξ) must be. In particular, the scaling property of the Fourier transform may be seen as saying: if we squeeze a function in x, its Fourier transform stretches out in ξ. It is not possible to arbitrarily concentrate both a function and its Fourier transform. The trade-off between the compaction of a function and its Fourier transform can be formalized in the form of an uncertainty principle by viewing a function and its Fourier transform as conjugate variables with respect to the symplectic form on the time–frequency domain: from the point of view of the linear canonical transformation, the Fourier transform is rotation by 90° in the time–frequency domain, and preserves the symplectic form. Suppose f(x) is an integrable and square-integrable function. Without loss of generality, assume that f(x) is normalized: It follows from the Plancherel theorem that f̂(ξ) is also normalized. The spread around x = 0 may be measured by the dispersion about zero defined by In probability terms, this is the second moment of |f(x)|2 about zero. The uncertainty principle states that, if f(x) is absolutely continuous and the functions x·f(x) and f′(x) are square integrable, then The equality is attained only in the case where σ > 0 is arbitrary and C1 = ⁠4√2/√σ⁠ so that f is L2-normalized. In other words, where f is a (normalized) Gaussian function with variance σ2/2π, centered at zero, and its Fourier transform is a Gaussian function with variance σ−2/2π. Gaussian functions are examples of Schwartz functions (see the discussion on tempered distributions below). In fact, this inequality implies that: In quantum mechanics, the momentum and position wave functions are Fourier transform pairs, up to a factor of the Planck constant. With this constant properly taken into account, the inequality above becomes the statement of the Heisenberg uncertainty principle. A stronger uncertainty principle is the Hirschman uncertainty principle, which is expressed as: where H(p) is the differential entropy of the probability density function p(x): where the logarithms may be in any base that is consistent. The equality is attained for a Gaussian, as in the previous case. Sine and cosine transforms [edit] Main article: Sine and cosine transforms Fourier's original formulation of the transform did not use complex numbers, but rather sines and cosines. Statisticians and others still use this form. An absolutely integrable function f for which Fourier inversion holds can be expanded in terms of genuine frequencies (avoiding negative frequencies, which are sometimes considered hard to interpret physically) λ by This is called an expansion as a trigonometric integral, or a Fourier integral expansion. The coefficient functions a and b can be found by using variants of the Fourier cosine transform and the Fourier sine transform (the normalisations are, again, not standardised): and Older literature refers to the two transform functions, the Fourier cosine transform, a, and the Fourier sine transform, b. The function f can be recovered from the sine and cosine transform using together with trigonometric identities. This is referred to as Fourier's integral formula. Spherical harmonics [edit] Let the set of homogeneous harmonic polynomials of degree k on Rn be denoted by Ak. The set Ak consists of the solid spherical harmonics of degree k. The solid spherical harmonics play a similar role in higher dimensions to the Hermite polynomials in dimension one. Specifically, if f(x) = e−π|x|2P(x) for some P(x) in Ak, then f̂(ξ) = i−k f(ξ). Let the set Hk be the closure in L2(Rn) of linear combinations of functions of the form f(|x|)P(x) where P(x) is in Ak. The space L2(Rn) is then a direct sum of the spaces Hk and the Fourier transform maps each space Hk to itself and is possible to characterize the action of the Fourier transform on each space Hk. Let f(x) = f0(|x|)P(x) (with P(x) in Ak), then where Here J(n + 2k − 2)/2 denotes the Bessel function of the first kind with order ⁠n + 2k − 2/2⁠. When k = 0 this gives a useful formula for the Fourier transform of a radial function. This is essentially the Hankel transform. Moreover, there is a simple recursion relating the cases n + 2 and n allowing to compute, e.g., the three-dimensional Fourier transform of a radial function from the one-dimensional one. Restriction problems [edit] See also: Fourier extension operator In higher dimensions it becomes interesting to study restriction problems for the Fourier transform. The Fourier transform of an integrable function is continuous and the restriction of this function to any set is defined. But for a square-integrable function the Fourier transform could be a general class of square integrable functions. As such, the restriction of the Fourier transform of an L2(Rn) function cannot be defined on sets of measure 0. It is still an active area of study to understand restriction problems in Lp for 1 < p < 2. It is possible in some cases to define the restriction of a Fourier transform to a set S, provided S has non-zero curvature. The case when S is the unit sphere in Rn is of particular interest. In this case the Tomas–Stein restriction theorem states that the restriction of the Fourier transform to the unit sphere in Rn is a bounded operator on Lp provided 1 ≤ p ≤ ⁠2n + 2/n + 3⁠. One notable difference between the Fourier transform in 1 dimension versus higher dimensions concerns the partial sum operator. Consider an increasing collection of measurable sets ER indexed by R ∈ (0,∞): such as balls of radius R centered at the origin, or cubes of side 2R. For a given integrable function f, consider the function fR defined by: Suppose in addition that f ∈ Lp(Rn). For n = 1 and 1 < p < ∞, if one takes ER = (−R, R), then fR converges to f in Lp as R tends to infinity, by the boundedness of the Hilbert transform. Naively one may hope the same holds true for n > 1. In the case that ER is taken to be a cube with side length R, then convergence still holds. Another natural candidate is the Euclidean ball ER = {ξ : |ξ| < R}. In order for this partial sum operator to converge, it is necessary that the multiplier for the unit ball be bounded in Lp(Rn). For n ≥ 2 it is a celebrated theorem of Charles Fefferman that the multiplier for the unit ball is never bounded unless p = 2. In fact, when p ≠ 2, this shows that not only may fR fail to converge to f in Lp, but for some functions f ∈ Lp(Rn), fR is not even an element of Lp. Fourier transform on function spaces [edit] See also: Riesz–Thorin theorem The definition of the Fourier transform naturally extends from to . That is, if then the Fourier transform is given by This operator is bounded as which shows that its operator norm is bounded by 1. The Riemann–Lebesgue lemma shows that if then its Fourier transform actually belongs to the space of continuous functions which vanish at infinity, i.e., . Furthermore, the image of under is a strict subset of . Similarly to the case of one variable, the Fourier transform can be defined on . The Fourier transform in is no longer given by an ordinary Lebesgue integral, although it can be computed by an improper integral, i.e., where the limit is taken in the L2 sense. Furthermore, is a unitary operator. For an operator to be unitary it is sufficient to show that it is bijective and preserves the inner product, so in this case these follow from the Fourier inversion theorem combined with the fact that for any f, g ∈ L2(Rn) we have In particular, the image of L2(Rn) is itself under the Fourier transform. On other Lp [edit] For , the Fourier transform can be defined on by Marcinkiewicz interpolation, which amounts to decomposing such functions into a fat tail part in L2 plus a fat body part in L1. In each of these spaces, the Fourier transform of a function in Lp(Rn) is in Lq(Rn), where q = ⁠p/p − 1⁠ is the Hölder conjugate of p (by the Hausdorff–Young inequality). However, except for p = 2, the image is not easily characterized. Further extensions become more technical. The Fourier transform of functions in Lp for the range 2 < p < ∞ requires the study of distributions. In fact, it can be shown that there are functions in Lp with p > 2 so that the Fourier transform is not defined as a function. Tempered distributions [edit] Main article: Distribution (mathematics) § Tempered distributions and Fourier transform See also: Rigged Hilbert space One might consider enlarging the domain of the Fourier transform from by considering generalized functions, or distributions. A distribution on is a continuous linear functional on the space of compactly supported smooth functions (i.e. bump functions), equipped with a suitable topology. Since is dense in , the Plancherel theorem allows one to extend the definition of the Fourier transform to general functions in by continuity arguments. The strategy is then to consider the action of the Fourier transform on and pass to distributions by duality. The obstruction to doing this is that the Fourier transform does not map to . In fact the Fourier transform of an element in can not vanish on an open set; see the above discussion on the uncertainty principle. The Fourier transform can also be defined for tempered distributions , dual to the space of Schwartz functions . A Schwartz function is a smooth function that decays at infinity, along with all of its derivatives, hence and: The Fourier transform is an automorphism of the Schwartz space and, by duality, also an automorphism of the space of tempered distributions. The tempered distributions include well-behaved functions of polynomial growth, distributions of compact support as well as all the integrable functions mentioned above. For the definition of the Fourier transform of a tempered distribution, let and be integrable functions, and let and be their Fourier transforms respectively. Then the Fourier transform obeys the following multiplication formula, Every integrable function defines (induces) a distribution by the relation So it makes sense to define the Fourier transform of a tempered distribution by the duality: Extending this to all tempered distributions gives the general definition of the Fourier transform. Distributions can be differentiated and the above-mentioned compatibility of the Fourier transform with differentiation and convolution remains true for tempered distributions. Generalizations [edit] Fourier–Stieltjes transform on measurable spaces [edit] See also: Bochner–Minlos theorem, Riesz–Markov–Kakutani representation theorem, and Fourier series § Fourier-Stieltjes series The Fourier transform of a finite Borel measure μ on Rn, given by the bounded, uniformly continuous function: is called the Fourier–Stieltjes transform due to its connection with the Riemann-Stieltjes integral representation of (Radon) measures. If is the probability distribution of a random variable then its Fourier–Stieltjes transform is, by definition, a characteristic function. If, in addition, the probability distribution has a probability density function, this definition is subject to the usual Fourier transform. Stated more generally, when is absolutely continuous with respect to the Lebesgue measure, i.e., then and the Fourier-Stieltjes transform reduces to the usual definition of the Fourier transform. That is, the notable difference with the Fourier transform of integrable functions is that the Fourier-Stieltjes transform need not vanish at infinity, i.e., the Riemann–Lebesgue lemma fails for measures. Bochner's theorem characterizes which functions may arise as the Fourier–Stieltjes transform of a positive measure on the circle. One example of a finite Borel measure that is not a function is the Dirac measure. Its Fourier transform is a constant function (whose value depends on the form of the Fourier transform used). Locally compact abelian groups [edit] Main article: Pontryagin duality The Fourier transform may be generalized to any locally compact abelian group, i.e., an abelian group that is also a locally compact Hausdorff space such that the group operation is continuous. If G is a locally compact abelian group, it has a translation invariant measure μ, called Haar measure. For a locally compact abelian group G, the set of irreducible, i.e. one-dimensional, unitary representations are called its characters. With its natural group structure and the topology of uniform convergence on compact sets (that is, the topology induced by the compact-open topology on the space of all continuous functions from to the circle group), the set of characters Ĝ is itself a locally compact abelian group, called the Pontryagin dual of G. For a function f in L1(G), its Fourier transform is defined by The Riemann–Lebesgue lemma holds in this case; f̂(ξ) is a function vanishing at infinity on Ĝ. The Fourier transform on T = R/Z is an example; here T is a locally compact abelian group, and the Haar measure μ on T can be thought of as the Lebesgue measure on [0,1). Consider the representation of T on the complex plane C that is a 1-dimensional complex vector space. There are a group of representations (which are irreducible since C is 1-dim) where for . The character of such representation, that is the trace of for each and , is itself. In the case of representation of finite group, the character table of the group G are rows of vectors such that each row is the character of one irreducible representation of G, and these vectors form an orthonormal basis of the space of class functions that map from G to C by Schur's lemma. Now the group T is no longer finite but still compact, and it preserves the orthonormality of character table. Each row of the table is the function of and the inner product between two class functions (all functions being class functions since T is abelian) is defined as with the normalizing factor . The sequence is an orthonormal basis of the space of class functions . For any representation V of a finite group G, can be expressed as the span ( are the irreps of G), such that . Similarly for and , . The Pontriagin dual is and for , is its Fourier transform for . Gelfand transform [edit] Main article: Gelfand representation The Fourier transform is also a special case of Gelfand transform. In this particular context, it is closely related to the Pontryagin duality map defined above. Given an abelian locally compact Hausdorff topological group G, as before we consider space L1(G), defined using a Haar measure. With convolution as multiplication, L1(G) is an abelian Banach algebra. It also has an involution given by Taking the completion with respect to the largest possibly C-norm gives its enveloping C-algebra, called the group C-algebra C(G) of G. (Any C-norm on L1(G) is bounded by the L1 norm, therefore their supremum exists.) Given any abelian C-algebra A, the Gelfand transform gives an isomorphism between A and C0(A^), where A^ is the multiplicative linear functionals, i.e. one-dimensional representations, on A with the weak- topology. The map is simply given by It turns out that the multiplicative linear functionals of C(G), after suitable identification, are exactly the characters of G, and the Gelfand transform, when restricted to the dense subset L1(G) is the Fourier–Pontryagin transform. Compact non-abelian groups [edit] The Fourier transform can also be defined for functions on a non-abelian group, provided that the group is compact. Removing the assumption that the underlying group is abelian, irreducible unitary representations need not always be one-dimensional. This means the Fourier transform on a non-abelian group takes values as Hilbert space operators. The Fourier transform on compact groups is a major tool in representation theory and non-commutative harmonic analysis. Let G be a compact Hausdorff topological group. Let Σ denote the collection of all isomorphism classes of finite-dimensional irreducible unitary representations, along with a definite choice of representation U(σ) on the Hilbert space Hσ of finite dimension dσ for each σ ∈ Σ. If μ is a finite Borel measure on G, then the Fourier–Stieltjes transform of μ is the operator on Hσ defined by where U(σ) is the complex-conjugate representation of U(σ) acting on Hσ. If μ is absolutely continuous with respect to the left-invariant probability measure λ on G, represented as for some f ∈ L1(λ), one identifies the Fourier transform of f with the Fourier–Stieltjes transform of μ. The mapping defines an isomorphism between the Banach space M(G) of finite Borel measures (see rca space) and a closed subspace of the Banach space C∞(Σ) consisting of all sequences E = (Eσ) indexed by Σ of (bounded) linear operators Eσ : Hσ → Hσ for which the norm is finite. The "convolution theorem" asserts that, furthermore, this isomorphism of Banach spaces is in fact an isometric isomorphism of C-algebras into a subspace of C∞(Σ). Multiplication on M(G) is given by convolution of measures and the involution defined by and C∞(Σ) has a natural C-algebra structure as Hilbert space operators. The Peter–Weyl theorem holds, and a version of the Fourier inversion formula (Plancherel's theorem) follows: if f ∈ L2(G), then where the summation is understood as convergent in the L2 sense. The generalization of the Fourier transform to the noncommutative situation has also in part contributed to the development of noncommutative geometry.[citation needed] In this context, a categorical generalization of the Fourier transform to noncommutative groups is Tannaka–Krein duality, which replaces the group of characters with the category of representations. However, this loses the connection with harmonic functions. Alternatives [edit] In signal processing terms, a function (of time) is a representation of a signal with perfect time resolution, but no frequency information, while the Fourier transform has perfect frequency resolution, but no time information: the magnitude of the Fourier transform at a point is how much frequency content there is, but location is only given by phase (argument of the Fourier transform at a point), and standing waves are not localized in time – a sine wave continues out to infinity, without decaying. This limits the usefulness of the Fourier transform for analyzing signals that are localized in time, notably transients, or any signal of finite extent. As alternatives to the Fourier transform, in time–frequency analysis, one uses time–frequency transforms or time–frequency distributions to represent signals in a form that has some time information and some frequency information – by the uncertainty principle, there is a trade-off between these. These can be generalizations of the Fourier transform, such as the short-time Fourier transform, fractional Fourier transform, Synchrosqueezing Fourier transform, or other functions to represent signals, as in wavelet transforms and chirplet transforms, with the wavelet analog of the (continuous) Fourier transform being the continuous wavelet transform. Example [edit] The following figures provide a visual illustration of how the Fourier transform's integral measures whether a frequency is present in a particular function. The first image depicts the function which is a 3 Hz cosine wave (the first term) shaped by a Gaussian envelope function (the second term) that smoothly turns the wave on and off. The next 2 images show the product which must be integrated to calculate the Fourier transform at +3 Hz. The real part of the integrand has a non-negative average value, because the alternating signs of and oscillate at the same rate and in phase, whereas and oscillate at the same rate but with orthogonal phase. The absolute value of the Fourier transform at +3 Hz is 0.5, which is relatively large. When added to the Fourier transform at -3 Hz (which is identical because we started with a real signal), we find that the amplitude of the 3 Hz frequency component is 1. However, when you try to measure a frequency that is not present, both the real and imaginary component of the integral vary rapidly between positive and negative values. For instance, the red curve is looking for 5 Hz. The absolute value of its integral is nearly zero, indicating that almost no 5 Hz component was in the signal. The general situation is usually more complicated than this, but heuristically this is how the Fourier transform measures how much of an individual frequency is present in a function Real and imaginary parts of the integrand for its Fourier transform at +5 Hz. Magnitude of its Fourier transform, with +3 and +5 Hz labeled. To re-enforce an earlier point, the reason for the response at Hz is because and are indistinguishable. The transform of would have just one response, whose amplitude is the integral of the smooth envelope: whereas is Applications [edit] See also: Spectral density § Applications Linear operations performed in one domain (time or frequency) have corresponding operations in the other domain, which are sometimes easier to perform. The operation of differentiation in the time domain corresponds to multiplication by the frequency,[note 7] so some differential equations are easier to analyze in the frequency domain. Also, convolution in the time domain corresponds to ordinary multiplication in the frequency domain (see Convolution theorem). After performing the desired operations, transformation of the result can be made back to the time domain. Harmonic analysis is the systematic study of the relationship between the frequency and time domains, including the kinds of functions or operations that are "simpler" in one or the other, and has deep connections to many areas of modern mathematics. Analysis of differential equations [edit] Perhaps the most important use of the Fourier transformation is to solve partial differential equations. Many of the equations of the mathematical physics of the nineteenth century can be treated this way. Fourier studied the heat equation, which in one dimension and in dimensionless units is The example we will give, a slightly more difficult one, is the wave equation in one dimension, As usual, the problem is not to find a solution: there are infinitely many. The problem is that of the so-called "boundary problem": find a solution which satisfies the "boundary conditions" Here, f and g are given functions. For the heat equation, only one boundary condition can be required (usually the first one). But for the wave equation, there are still infinitely many solutions y which satisfy the first boundary condition. But when one imposes both conditions, there is only one possible solution. It is easier to find the Fourier transform ŷ of the solution than to find the solution directly. This is because the Fourier transformation takes differentiation into multiplication by the Fourier-dual variable, and so a partial differential equation applied to the original function is transformed into multiplication by polynomial functions of the dual variables applied to the transformed function. After ŷ is determined, we can apply the inverse Fourier transformation to find y. Fourier's method is as follows. First, note that any function of the forms satisfies the wave equation. These are called the elementary solutions. Second, note that therefore any integral satisfies the wave equation for arbitrary a+, a−, b+, b−. This integral may be interpreted as a continuous linear combination of solutions for the linear equation. Now this resembles the formula for the Fourier synthesis of a function. In fact, this is the real inverse Fourier transform of a± and b± in the variable x. The third step is to examine how to find the specific unknown coefficient functions a± and b± that will lead to y satisfying the boundary conditions. We are interested in the values of these solutions at t = 0. So we will set t = 0. Assuming that the conditions needed for Fourier inversion are satisfied, we can then find the Fourier sine and cosine transforms (in the variable x) of both sides and obtain and Similarly, taking the derivative of y with respect to t and then applying the Fourier sine and cosine transformations yields and These are four linear equations for the four unknowns a± and b±, in terms of the Fourier sine and cosine transforms of the boundary conditions, which are easily solved by elementary algebra, provided that these transforms can be found. In summary, we chose a set of elementary solutions, parametrized by ξ, of which the general solution would be a (continuous) linear combination in the form of an integral over the parameter ξ. But this integral was in the form of a Fourier integral. The next step was to express the boundary conditions in terms of these integrals, and set them equal to the given functions f and g. But these expressions also took the form of a Fourier integral because of the properties of the Fourier transform of a derivative. The last step was to exploit Fourier inversion by applying the Fourier transformation to both sides, thus obtaining expressions for the coefficient functions a± and b± in terms of the given boundary conditions f and g. From a higher point of view, Fourier's procedure can be reformulated more conceptually. Since there are two variables, we will use the Fourier transformation in both x and t rather than operate as Fourier did, who only transformed in the spatial variables. Note that ŷ must be considered in the sense of a distribution since y(x, t) is not going to be L1: as a wave, it will persist through time and thus is not a transient phenomenon. But it will be bounded and so its Fourier transform can be defined as a distribution. The operational properties of the Fourier transformation that are relevant to this equation are that it takes differentiation in x to multiplication by i2πξ and differentiation with respect to t to multiplication by i2πf where f is the frequency. Then the wave equation becomes an algebraic equation in ŷ: This is equivalent to requiring ŷ(ξ, f) = 0 unless ξ = ±f. Right away, this explains why the choice of elementary solutions we made earlier worked so well: obviously f̂ = δ(ξ ± f) will be solutions. Applying Fourier inversion to these delta functions, we obtain the elementary solutions we picked earlier. But from the higher point of view, one does not pick elementary solutions, but rather considers the space of all distributions which are supported on the (degenerate) conic ξ2 − f2 = 0. We may as well consider the distributions supported on the conic that are given by distributions of one variable on the line ξ = f plus distributions on the line ξ = −f as follows: if Φ is any test function, where s+, and s−, are distributions of one variable. Then Fourier inversion gives, for the boundary conditions, something very similar to what we had more concretely above (put Φ(ξ, f) = ei2π(xξ+tf), which is clearly of polynomial growth): and Now, as before, applying the one-variable Fourier transformation in the variable x to these functions of x yields two equations in the two unknown distributions s± (which can be taken to be ordinary functions if the boundary conditions are L1 or L2). From a calculational point of view, the drawback of course is that one must first calculate the Fourier transforms of the boundary conditions, then assemble the solution from these, and then calculate an inverse Fourier transform. Closed form formulas are rare, except when there is some geometric symmetry that can be exploited, and the numerical calculations are difficult because of the oscillatory nature of the integrals, which makes convergence slow and hard to estimate. For practical calculations, other methods are often used. The twentieth century has seen the extension of these methods to all linear partial differential equations with polynomial coefficients, and by extending the notion of Fourier transformation to include Fourier integral operators, some non-linear equations as well. Fourier-transform spectroscopy [edit] Main article: Fourier-transform spectroscopy The Fourier transform is also used in nuclear magnetic resonance (NMR) and in other kinds of spectroscopy, e.g. infrared (FTIR). In NMR an exponentially shaped free induction decay (FID) signal is acquired in the time domain and Fourier-transformed to a Lorentzian line-shape in the frequency domain. The Fourier transform is also used in magnetic resonance imaging (MRI) and mass spectrometry. Quantum mechanics [edit] The Fourier transform is useful in quantum mechanics in at least two different ways. To begin with, the basic conceptual structure of quantum mechanics postulates the existence of pairs of complementary variables, connected by the Heisenberg uncertainty principle. For example, in one dimension, the spatial variable q of, say, a particle, can only be measured by the quantum mechanical "position operator" at the cost of losing information about the momentum p of the particle. Therefore, the physical state of the particle can either be described by a function, called "the wave function", of q or by a function of p but not by a function of both variables. The variable p is called the conjugate variable to q. In classical mechanics, the physical state of a particle (existing in one dimension, for simplicity of exposition) would be given by assigning definite values to both p and q simultaneously. Thus, the set of all possible physical states is the two-dimensional real vector space with a p-axis and a q-axis called the phase space. In contrast, quantum mechanics chooses a polarisation of this space in the sense that it picks a subspace of one-half the dimension, for example, the q-axis alone, but instead of considering only points, takes the set of all complex-valued "wave functions" on this axis. Nevertheless, choosing the p-axis is an equally valid polarisation, yielding a different representation of the set of possible physical states of the particle. Both representations of the wavefunction are related by a Fourier transform, such that or, equivalently, Physically realisable states are L2, and so by the Plancherel theorem, their Fourier transforms are also L2. (Note that since q is in units of distance and p is in units of momentum, the presence of the Planck constant in the exponent makes the exponent dimensionless, as it should be.) Therefore, the Fourier transform can be used to pass from one way of representing the state of the particle, by a wave function of position, to another way of representing the state of the particle: by a wave function of momentum. Infinitely many different polarisations are possible, and all are equally valid. Being able to transform states from one representation to another by the Fourier transform is not only convenient but also the underlying reason of the Heisenberg uncertainty principle. The other use of the Fourier transform in both quantum mechanics and quantum field theory is to solve the applicable wave equation. In non-relativistic quantum mechanics, the Schrödinger equation for a time-varying wave function in one-dimension, not subject to external forces, is This is the same as the heat equation except for the presence of the imaginary unit i. Fourier methods can be used to solve this equation. In the presence of a potential, given by the potential energy function V(x), the equation becomes The "elementary solutions", as we referred to them above, are the so-called "stationary states" of the particle, and Fourier's algorithm, as described above, can still be used to solve the boundary value problem of the future evolution of ψ given its values for t = 0. Neither of these approaches is of much practical use in quantum mechanics. Boundary value problems and the time-evolution of the wave function is not of much practical interest: it is the stationary states that are most important. In relativistic quantum mechanics, the Schrödinger equation becomes a wave equation as was usual in classical physics, except that complex-valued waves are considered. A simple example, in the absence of interactions with other particles or fields, is the free one-dimensional Klein–Gordon–Schrödinger–Fock equation, this time in dimensionless units, This is, from the mathematical point of view, the same as the wave equation of classical physics solved above (but with a complex-valued wave, which makes no difference in the methods). This is of great use in quantum field theory: each separate Fourier component of a wave can be treated as a separate harmonic oscillator and then quantized, a procedure known as "second quantization". Fourier methods have been adapted to also deal with non-trivial interactions. Finally, the number operator of the quantum harmonic oscillator can be interpreted, for example via the Mehler kernel, as the generator of the Fourier transform . Signal processing [edit] The Fourier transform is used for the spectral analysis of time-series. The subject of statistical signal processing does not, however, usually apply the Fourier transformation to the signal itself. Even if a real signal is indeed transient, it has been found in practice advisable to model a signal by a function (or, alternatively, a stochastic process) which is stationary in the sense that its characteristic properties are constant over all time. The Fourier transform of such a function does not exist in the usual sense, and it has been found more useful for the analysis of signals to instead take the Fourier transform of its autocorrelation function. The autocorrelation function R of a function f is defined by This function is a function of the time-lag τ elapsing between the values of f to be correlated. For most functions f that occur in practice, R is a bounded even function of the time-lag τ and for typical noisy signals it turns out to be uniformly continuous with a maximum at τ = 0. The autocorrelation function, more properly called the autocovariance function unless it is normalized in some appropriate fashion, measures the strength of the correlation between the values of f separated by a time lag. This is a way of searching for the correlation of f with its own past. It is useful even for other statistical tasks besides the analysis of signals. For example, if f(t) represents the temperature at time t, one expects a strong correlation with the temperature at a time lag of 24 hours. It possesses a Fourier transform, This Fourier transform is called the power spectral density function of f. (Unless all periodic components are first filtered out from f, this integral will diverge, but it is easy to filter out such periodicities.) The power spectrum, as indicated by this density function P, measures the amount of variance contributed to the data by the frequency ξ. In electrical signals, the variance is proportional to the average power (energy per unit time), and so the power spectrum describes how much the different frequencies contribute to the average power of the signal. This process is called the spectral analysis of time-series and is analogous to the usual analysis of variance of data that is not a time-series (ANOVA). Knowledge of which frequencies are "important" in this sense is crucial for the proper design of filters and for the proper evaluation of measuring apparatuses. It can also be useful for the scientific analysis of the phenomena responsible for producing the data. The power spectrum of a signal can also be approximately measured directly by measuring the average power that remains in a signal after all the frequencies outside a narrow band have been filtered out. Spectral analysis is carried out for visual signals as well. The power spectrum ignores all phase relations, which is good enough for many purposes, but for video signals other types of spectral analysis must also be employed, still using the Fourier transform as a tool. Other notations [edit] Other common notations for include: In the sciences and engineering it is also common to make substitutions like these: So the transform pair can become A disadvantage of the capital letter notation is when expressing a transform such as or which become the more awkward and In some contexts such as particle physics, the same symbol may be used for both for a function as well as it Fourier transform, with the two only distinguished by their argument I.e. would refer to the Fourier transform because of the momentum argument, while would refer to the original function because of the positional argument. Although tildes may be used as in to indicate Fourier transforms, tildes may also be used to indicate a modification of a quantity with a more Lorentz invariant form, such as , so care must be taken. Similarly, often denotes the Hilbert transform of . The interpretation of the complex function f̂(ξ) may be aided by expressing it in polar coordinate form in terms of the two real functions A(ξ) and φ(ξ) where: is the amplitude and is the phase (see arg function). Then the inverse transform can be written: which is a recombination of all the frequency components of f(x). Each component is a complex sinusoid of the form e2πixξ whose amplitude is A(ξ) and whose initial phase angle (at x = 0) is φ(ξ). The Fourier transform may be thought of as a mapping on function spaces. This mapping is here denoted F and F(f) is used to denote the Fourier transform of the function f. This mapping is linear, which means that F can also be seen as a linear transformation on the function space and implies that the standard notation in linear algebra of applying a linear transformation to a vector (here the function f) can be used to write F f instead of F(f). Since the result of applying the Fourier transform is again a function, we can be interested in the value of this function evaluated at the value ξ for its variable, and this is denoted either as F f(ξ) or as (F f)(ξ). Notice that in the former case, it is implicitly understood that F is applied first to f and then the resulting function is evaluated at ξ, not the other way around. In mathematics and various applied sciences, it is often necessary to distinguish between a function f and the value of f when its variable equals x, denoted f(x). This means that a notation like F(f(x)) formally can be interpreted as the Fourier transform of the values of f at x. Despite this flaw, the previous notation appears frequently, often when a particular function or a function of a particular variable is to be transformed. For example, is sometimes used to express that the Fourier transform of a rectangular function is a sinc function, or is used to express the shift property of the Fourier transform. Notice, that the last example is only correct under the assumption that the transformed function is a function of x, not of x0. As discussed above, the characteristic function of a random variable is the same as the Fourier–Stieltjes transform of its distribution measure, but in this context it is typical to take a different convention for the constants. Typically characteristic function is defined As in the case of the "non-unitary angular frequency" convention above, the factor of 2π appears in neither the normalizing constant nor the exponent. Unlike any of the conventions appearing above, this convention takes the opposite sign in the exponent. Computation methods [edit] The appropriate computation method largely depends how the original mathematical function is represented and the desired form of the output function. In this section we consider both functions of a continuous variable, and functions of a discrete variable (i.e. ordered pairs of and values). For discrete-valued the transform integral becomes a summation of sinusoids, which is still a continuous function of frequency ( or ). When the sinusoids are harmonically related (i.e. when the -values are spaced at integer multiples of an interval), the transform is called discrete-time Fourier transform (DTFT). Discrete Fourier transforms and fast Fourier transforms [edit] Sampling the DTFT at equally-spaced values of frequency is the most common modern method of computation. Efficient procedures, depending on the frequency resolution needed, are described at Discrete-time Fourier transform § Sampling the DTFT. The discrete Fourier transform (DFT), used there, is usually computed by a fast Fourier transform (FFT) algorithm. Analytic integration of closed-form functions [edit] Tables of closed-form Fourier transforms, such as § Square-integrable functions, one-dimensional and § Table of discrete-time Fourier transforms, are created by mathematically evaluating the Fourier analysis integral (or summation) into another closed-form function of frequency ( or ). When mathematically possible, this provides a transform for a continuum of frequency values. Many computer algebra systems such as Matlab and Mathematica that are capable of symbolic integration are capable of computing Fourier transforms analytically. For example, to compute the Fourier transform of cos(6πt) e−πt2 one might enter the command integrate cos(6pit) exp(−pit^2) exp(-i2pift) from -inf to inf into Wolfram Alpha.[note 8] Numerical integration of closed-form continuous functions [edit] Discrete sampling of the Fourier transform can also be done by numerical integration of the definition at each value of frequency for which transform is desired. The numerical integration approach works on a much broader class of functions than the analytic approach. Numerical integration of a series of ordered pairs [edit] If the input function is a series of ordered pairs, numerical integration reduces to just a summation over the set of data pairs. The DTFT is a common subcase of this more general situation. Tables of important Fourier transforms [edit] The following tables record some closed-form Fourier transforms. For functions f(x) and g(x) denote their Fourier transforms by f̂ and ĝ. Only the three most common conventions are included. It may be useful to notice that entry 105 gives a relationship between the Fourier transform of a function and the original function, which can be seen as relating the Fourier transform and its inverse. Functional relationships, one-dimensional [edit] The Fourier transforms in this table may be found in Erdélyi (1954) or Kammler (2000, appendix). | Function | Fourier transform unitary, ordinary frequency | Fourier transform unitary, angular frequency | Fourier transform non-unitary, angular frequency | Remarks | --- --- | | | | | | Definitions | | 101 | | | | | Linearity | | 102 | | | | | Shift in time domain | | 103 | | | | | Shift in frequency domain, dual of 102 | | 104 | | | | | Scaling in the time domain. If |a| is large, then f(ax) is concentrated around 0 andspreads out and flattens. | | 105 | | | | | The same transform is applied twice, but x replaces the frequency variable (ξ or ω) after the first transform. | | 106 | | | | | nth-order derivative. As f is a Schwartz function | | 106.5 | | | | | Integration. Note: is the Dirac delta function and is the average (DC) value of such that | | 107 | | | | | This is the dual of 106 | | 108 | | | | | The notation f ∗ g denotes the convolution of f and g — this rule is the convolution theorem | | 109 | | | | | This is the dual of 108 | | 110 | For f(x) purely real | | | | Hermitian symmetry. z indicates the complex conjugate. | | 113 | For f(x) purely imaginary | | | | z indicates the complex conjugate. | | 114 | | | | | Complex conjugation, generalization of 110 and 113 | | 115 | | | | | This follows from rules 101 and 103 using Euler's formula: | | 116 | | | | | This follows from 101 and 103 using Euler's formula: | Square-integrable functions, one-dimensional [edit] The Fourier transforms in this table may be found in Campbell & Foster (1948), Erdélyi (1954), or Kammler (2000, appendix). | Function | Fourier transform unitary, ordinary frequency | Fourier transform unitary, angular frequency | Fourier transform non-unitary, angular frequency | Remarks | --- --- | | | | | | Definitions | | 201 | | | | | The rectangular pulse and the normalized sinc function, here defined as sinc(x) = ⁠sin(πx)/πx⁠ | | 202 | | | | | Dual of rule 201. The rectangular function is an ideal low-pass filter, and the sinc function is the non-causal impulse response of such a filter. The sinc function is defined here as sinc(x) = ⁠sin(πx)/πx⁠ | | 203 | | | | | The function tri(x) is the triangular function | | 204 | | | | | Dual of rule 203. | | 205 | | | | | The function u(x) is the Heaviside unit step function and a > 0. | | 206 | | | | | This shows that, for the unitary Fourier transforms, the Gaussian function e−αx2 is its own Fourier transform for some choice of α. For this to be integrable we must have Re(α) > 0. | | 208 | | | | | For Re(a) > 0. That is, the Fourier transform of a two-sided decaying exponential function is a Lorentzian function. | | 209 | | | | | Hyperbolic secant is its own Fourier transform | | 210 | | | | | Hn is the nth-order Hermite polynomial. If a = 1 then the Gauss–Hermite functions are eigenfunctions of the Fourier transform operator. For a derivation, see Hermite polynomial. The formula reduces to 206 for n = 0. | Distributions, one-dimensional [edit] The Fourier transforms in this table may be found in Erdélyi (1954) or Kammler (2000, appendix). | Function | Fourier transform unitary, ordinary frequency | Fourier transform unitary, angular frequency | Fourier transform non-unitary, angular frequency | Remarks | --- --- | | | | | | Definitions | | 301 | | | | | The distribution δ(ξ) denotes the Dirac delta function. | | 302 | | | | | Dual of rule 301. | | 303 | | | | | This follows from 103 and 301. | | 304 | | | | | This follows from rules 101 and 303 using Euler's formula: | | 305 | | | | | This follows from 101 and 303 using | | 306 | | | | | This follows from 101 and 207 using | | 307 | | | | | This follows from 101 and 207 using | | 308 | | | | | Here it is assumed is real. For the case that alpha is complex see table entry 206 above. | | 309 | | | | | Here, n is a natural number and δ(n)(ξ) is the nth distribution derivative of the Dirac delta function. This rule follows from rules 107 and 301. Combining this rule with 101, we can transform all polynomials. | | 310 | | | | | Dual of rule 309. δ(n)(ξ) is the nth distribution derivative of the Dirac delta function. This rule follows from 106 and 302. | | 311 | | | | | Here sgn(ξ) is the sign function. Note that ⁠1/x⁠ is not a distribution. It is necessary to use the Cauchy principal value when testing against Schwartz functions. This rule is useful in studying the Hilbert transform. | | 312 | | | | | ⁠1/xn⁠ is the homogeneous distribution defined by the distributional derivative | | 313 | | | | | This formula is valid for 0 > α > −1. For α > 0 some singular terms arise at the origin that can be found by differentiating 320. If Re α > −1, then |x|α is a locally integrable function, and so a tempered distribution. The function α ↦ |x|α is a holomorphic function from the right half-plane to the space of tempered distributions. It admits a unique meromorphic extension to a tempered distribution, also denoted |x|α for α ≠ −1, −3, ... (See homogeneous distribution.) | | | | | | | Special case of 313. | | 314 | | | | | The dual of rule 311. This time the Fourier transforms need to be considered as a Cauchy principal value. | | 315 | | | | | The function u(x) is the Heaviside unit step function; this follows from rules 101, 301, and 314. | | 316 | | | | | This function is known as the Dirac comb function. This result can be derived from 302 and 102, together with the fact thatas distributions. | | 317 | | | | | The function J0(x) is the zeroth order Bessel function of first kind. | | 318 | | | | | This is a generalization of 317. The function Jn(x) is the nth order Bessel function of first kind. The function Tn(x) is the Chebyshev polynomial of the first kind. | | 319 | | | | | γ is the Euler–Mascheroni constant. It is necessary to use a finite part integral when testing ⁠1/|ξ|⁠ or ⁠1/|ω|⁠against Schwartz functions. The details of this might change the coefficient of the delta function. | | 320 | | | | | This formula is valid for 1 > α > 0. Use differentiation to derive formula for higher exponents. u is the Heaviside function. | Two-dimensional functions [edit] | Function | Fourier transform unitary, ordinary frequency | Fourier transform unitary, angular frequency | Fourier transform non-unitary, angular frequency | Remarks | --- --- | 400 | | | | | The variables ξx, ξy, ωx, ωy are real numbers. The integrals are taken over the entire plane. | | 401 | | | | | Both functions are Gaussians, which may not have unit volume. | | 402 | | | | | The function is defined by circ(r) = 1 for 0 ≤ r ≤ 1, and is 0 otherwise. The result is the amplitude distribution of the Airy disk, and is expressed using J1 (the order-1 Bessel function of the first kind). | | 403 | | | | | This is the Hankel transform of r−1, a 2-D Fourier "self-transform". | | 404 | | | | | | Formulas for general n-dimensional functions [edit] | Function | Fourier transform unitary, ordinary frequency | Fourier transform unitary, angular frequency | Fourier transform non-unitary, angular frequency | Remarks | --- --- | 500 | | | | | | | 501 | | | | | The function χ[0, 1] is the indicator function of the interval [0, 1]. The function Γ(x) is the gamma function. The function J⁠n/2⁠ + δ is a Bessel function of the first kind, with order ⁠n/2⁠ + δ. Taking n = 2 and δ = 0 produces 402. | | 502 | | | | | See Riesz potential where the constant is given byThe formula also holds for all α ≠ n, n + 2, ... by analytic continuation, but then the function and its Fourier transforms need to be understood as suitably regularized tempered distributions. See homogeneous distribution.[note 9] | | 503 | | | | | This is the formula for a multivariate normal distribution normalized to 1 with a mean of 0. Bold variables are vectors or matrices. Following the notation of the aforementioned page, Σ = σ σT and Σ−1 = σ−T σ−1 | | 504 | | | | | Here Re(α) > 0 | See also [edit] Analog signal processing Beevers–Lipson strip Constant-Q transform Discrete Fourier transform DFT matrix Fast Fourier transform Fourier integral operator Fourier inversion theorem Fourier multiplier Fourier series Fourier sine transform Fourier–Deligne transform Fourier–Mukai transform Fractional Fourier transform Indirect Fourier transform Integral transform Hankel transform Hartley transform Laplace transform Least-squares spectral analysis Linear canonical transform List of Fourier-related transforms Mellin transform Multidimensional transform NGC 4622, especially the image NGC 4622 Fourier transform m = 2. Nonlocal operator Quantum Fourier transform Quadratic Fourier transform Short-time Fourier transform Spectral density Spectral density estimation Symbolic integration Time stretch dispersive Fourier transform Transform (mathematics) Notes [edit] ^ Sentence structure is often sufficient to distinguish the intended meaning: for example, "Apply the Fourier transform to [an input]" refers to the operation, whereas "The Fourier transform of [an input]" refers to its output. ^ Depending on the application a Lebesgue integral, distributional, or other approach may be most appropriate. ^ Vretblad (2000) provides solid justification for these formal procedures without going too deeply into functional analysis or the theory of distributions. ^ In relativistic quantum mechanics one encounters vector-valued Fourier transforms of multi-component wave functions. In quantum field theory, operator-valued Fourier transforms of operator-valued functions of spacetime are in frequent use, see for example Greiner & Reinhardt (1996). ^ A possible source of confusion is the frequency-shifting property; i.e. the transform of function is The value of this function at is meaning that a frequency has been shifted to zero (also see Negative frequency). ^ The operator is defined by replacing by in the Taylor expansion of ^ Up to an imaginary constant factor whose magnitude depends on what Fourier transform convention is used. ^ The direct command fourier transform of cos(6pit) exp(−pit^2) would also work for Wolfram Alpha, although the options for the convention (see Fourier transform § Other conventions) must be changed away from the default option, which is actually equivalent to integrate cos(6pit) exp(−pit^2) exp(iomegat) /sqrt(2pi) from -inf to inf. ^ In Gelfand & Shilov 1964, p. 363, with the non-unitary conventions of this table, the transform of is given to be from which this follows, with . Citations [edit] ^ Pinsky 2002, p. 91. ^ Fourier 1822, p. 525 ^ Fourier 1878, p. 408 ^ Jordan (1883) proves on pp. 216–226 the Fourier integral theorem before studying Fourier series. ^ Titchmarsh 1986, p. 1 ^ Rahman 2011, p. 10. ^ Oppenheim, Schafer & Buck 1999, p. 58 ^ Stade 2005, pp. 298–299. ^ Howe 1980. ^ Folland 1989 ^ Fourier 1822 ^ Arfken 1985 ^ a b Pinsky 2002 ^ Proakis, John G.; Manolakis, Dimitris G. (1996). Digital Signal Processing: Principles, Algorithms, and Applications (3rd ed.). Prentice Hall. p. 291. ISBN 978-0-13-373762-2. ^ Katznelson 2004, p. 153. ^ Stein & Weiss 1971, p. 2. ^ a b c d e f Stein & Weiss 1971 ^ Rudin 1987, p. 187 ^ Rudin 1987, p. 186 ^ Folland 1992, p. 216 ^ Wolf 1979, p. 307ff ^ Folland 1989, p. 53 ^ Celeghini, Gadella & del Olmo 2021 ^ a b Duoandikoetxea 2001 ^ a b Boashash 2003 ^ Condon 1937 ^ Wolf 1979, p. 320 ^ a b Wolf 1979, p. 312 ^ Folland 1989, p. 52 ^ Howe 1980 ^ Paley & Wiener 1934 ^ Gelfand & Vilenkin 1964 ^ Kirillov & Gvishiani 1982 ^ Clozel & Delorme 1985, pp. 331–333 ^ de Groot & Mazur 1984, p. 146 ^ Champeney 1987, p. 80 ^ a b c Kolmogorov & Fomin 1999 ^ Wiener 1949 ^ Champeney 1987, p. 63 ^ Widder & Wiener 1938, p. 537 ^ Pinsky 2002, chpt. 2.4.3 The Uncertainty Principle ^ Stein & Shakarchi 2003, chpt. 5.4 The Heisenberg uncertainty principle ^ Chatfield 2004, p. 113 ^ Fourier 1822, p. 441 ^ Poincaré 1895, p. 102 ^ Whittaker & Watson 1927, p. 188 ^ Grafakos 2004 ^ Grafakos & Teschl 2013 ^ Stein & Weiss 1971, pp. 1–2. ^ Rudin 1987, pp. 182–183. ^ Chandrasekharan 1989, pp. 7–8, 84. ^ More generally, one can take a sequence of functions that are in the intersection of L1 and L2 and that converges to f in the L2-norm, and define the Fourier transform of f as the L2 -limit of the Fourier transforms of these functions. ^ "Applied Fourier Analysis and Elements of Modern Signal Processing Lecture 3" (PDF). January 12, 2016. Archived from the original (PDF) on 2020-10-03. Retrieved 2019-10-11. ^ Stein & Weiss 1971, Thm. 2.3. ^ a b Katznelson 2004. ^ Mallat 2009, p. 45. ^ Strichartz 1994, p. 150. ^ Hunter 2014. ^ Pinsky 2002, p. 256. ^ Rudin 1991, p. 15. ^ Edwards 1982, pp. 53, 67, 72–73. ^ Katznelson 2004, p. 173 The typical conventions in probability theory take eiξx instead of e−i2πξx. ^ Billingsley 1995, p. 345. ^ Katznelson 2004, pp. 40, 155, 164. ^ Edwards 1982, p. 53. ^ Hewitt & Ross 1970, Chapter 8 ^ Knapp 2001 ^ Correia, L. B.; Justo, J. F.; Angélico, B. A. (2024). "Polynomial Adaptive Synchrosqueezing Fourier Transform: A method to optimize multiresolution". Digital Signal Processing. 150: 104526. Bibcode:2024DSPRJ.15004526C. doi:10.1016/j.dsp.2024.104526.{{cite journal}}: CS1 maint: article number as page number (link) ^ Gradshteyn et al. 2015 ^ Press et al. 1992 ^ Bailey & Swarztrauber 1994 ^ Lado 1971 ^ Simonen & Olkkonen 1985 ^ "The Integration Property of the Fourier Transform". The Fourier Transform .com. 2015 . Archived from the original on 2022-01-26. Retrieved 2023-08-20. ^ Stein & Weiss 1971, Thm. IV.3.3 ^ Easton 2010 ^ Stein & Weiss 1971, Thm. 4.15 ^ Stein & Weiss 1971, p. 6 References [edit] Arfken, George (1985), Mathematical Methods for Physicists (3rd ed.), Academic Press, ISBN 9780120598205 Bailey, David H.; Swarztrauber, Paul N. (1994), "A fast method for the numerical evaluation of continuous Fourier and Laplace transforms" (PDF), SIAM Journal on Scientific Computing, 15 (5): 1105–1110, Bibcode:1994SJSC...15.1105B, CiteSeerX 10.1.1.127.1534, doi:10.1137/0915067, archived from the original (PDF) on 2008-07-20, retrieved 2017-11-01 Billingsley, Patrick (1995), Probability and measure, New York, NY: Wiley, ISBN 978-0-471-00710-4 Boashash, B., ed. (2003), Time–Frequency Signal Analysis and Processing: A Comprehensive Reference, Oxford: Elsevier Science, ISBN 978-0-08-044335-5 Bochner, S.; Chandrasekharan, K. (1949), Fourier Transforms, Princeton University Press Bracewell, R. N. (2000), The Fourier Transform and Its Applications (3rd ed.), Boston: McGraw-Hill, ISBN 978-0-07-116043-8 Campbell, George; Foster, Ronald (1948), Fourier Integrals for Practical Applications, New York: D. Van Nostrand Company, Inc. Celeghini, Enrico; Gadella, Manuel; del Olmo, Mariano A. (2021), "Hermite Functions and Fourier Series", Symmetry, 13 (5): 853, arXiv:2007.10406, Bibcode:2021Symm...13..853C, doi:10.3390/sym13050853 Champeney, D.C. (1987), A Handbook of Fourier Theorems, Cambridge University Press, Bibcode:1987hft..book.....C Chandrasekharan, Komaravolu (1989), Classical Fourier Transforms, Berlin, Heidelberg: Springer Berlin Heidelberg, doi:10.1007/978-3-642-74029-9, ISBN 978-3-540-50248-7 Chatfield, Chris (2004), The Analysis of Time Series: An Introduction, Texts in Statistical Science (6th ed.), London: Chapman & Hall/CRC, ISBN 9780203491683 Clozel, Laurent; Delorme, Patrice (1985), "Sur le théorème de Paley-Wiener invariant pour les groupes de Lie réductifs réels", Comptes Rendus de l'Académie des Sciences, Série I, 300: 331–333 Condon, E. U. (1937), "Immersion of the Fourier transform in a continuous group of functional transformations", Proc. Natl. Acad. Sci., 23 (3): 158–164, Bibcode:1937PNAS...23..158C, doi:10.1073/pnas.23.3.158, PMC 1076889, PMID 16588141 de Groot, Sybren R.; Mazur, Peter (1984), Non-Equilibrium Thermodynamics (2nd ed.), New York: Dover Duoandikoetxea, Javier (2001), Fourier Analysis, American Mathematical Society, ISBN 978-0-8218-2172-5 Dym, H.; McKean, H. (1985), Fourier Series and Integrals, Academic Press, ISBN 978-0-12-226451-1 Easton, Roger L. Jr. (2010), Fourier Methods in Imaging, John Wiley & Sons, ISBN 978-0-470-68983-7, retrieved 26 May 2020 Edwards, R. E. (1979). Fourier Series. Graduate Texts in Mathematics. Vol. 64. New York, NY: Springer New York. doi:10.1007/978-1-4612-6208-4. ISBN 978-1-4612-6210-7. Edwards, R. E. (1982). Fourier Series. Graduate Texts in Mathematics. Vol. 85. New York, NY: Springer New York. doi:10.1007/978-1-4613-8156-3. ISBN 978-1-4613-8158-7. Erdélyi, Arthur, ed. (1954), Tables of Integral Transforms, vol. 1, McGraw-Hill Feller, William (1971), An Introduction to Probability Theory and Its Applications, vol. II (2nd ed.), New York: Wiley, MR 0270403 Folland, Gerald (1989), Harmonic analysis in phase space, Princeton University Press Folland, Gerald (1992), Fourier analysis and its applications, Wadsworth & Brooks/Cole Fourier, J.B. Joseph (1822), Théorie analytique de la chaleur (in French), Paris: Firmin Didot, père et fils, OCLC 2688081 Fourier, J.B. Joseph (1878) , "The Analytical Theory of Heat", Nature, 18 (451), translated by Alexander Freeman, The University Press: 192, Bibcode:1878Natur..18Q.192., doi:10.1038/018192a0 (translated from French) Gradshteyn, Izrail Solomonovich; Ryzhik, Iosif Moiseevich; Geronimus, Yuri Veniaminovich; Tseytlin, Michail Yulyevich; Jeffrey, Alan (2015), Zwillinger, Daniel; Moll, Victor Hugo (eds.), Table of Integrals, Series, and Products, translated by Scripta Technica, Inc. (8th ed.), Academic Press, ISBN 978-0-12-384933-5 Grafakos, Loukas (2004), Classical and Modern Fourier Analysis, Prentice-Hall, ISBN 978-0-13-035399-3 Grafakos, Loukas; Teschl, Gerald (2013), "On Fourier transforms of radial functions and distributions", J. Fourier Anal. Appl., 19 (1): 167–179, arXiv:1112.5469, Bibcode:2013JFAA...19..167G, doi:10.1007/s00041-012-9242-5, S2CID 1280745 Greiner, W.; Reinhardt, J. (1996), Field Quantization, Springer, ISBN 978-3-540-59179-5 Gelfand, I.M.; Shilov, G.E. (1964), Generalized Functions, vol. 1, New York: Academic Press (translated from Russian) Gelfand, I.M.; Vilenkin, N.Y. (1964), Generalized Functions, vol. 4, New York: Academic Press (translated from Russian) Hewitt, Edwin; Ross, Kenneth A. (1970), Abstract harmonic analysis, Die Grundlehren der mathematischen Wissenschaften, Band 152, vol. II: Structure and analysis for compact groups. Analysis on locally compact Abelian groups, Springer, MR 0262773 Hörmander, L. (1976), Linear Partial Differential Operators, vol. 1, Springer, ISBN 978-3-540-00662-6 Howe, Roger (1980), "On the role of the Heisenberg group in harmonic analysis", Bulletin of the American Mathematical Society, 3 (2): 821–844, doi:10.1090/S0273-0979-1980-14825-9, MR 0578375 Hunter, J.K. (2014), "Appendix: The Fourier transform", Lecture Notes on PDEs, retrieved January 12, 2025 James, J.F. (2011), A Student's Guide to Fourier Transforms (3rd ed.), Cambridge University Press, Bibcode:2011sgft.book.....J, ISBN 978-0-521-17683-5 Jordan, Camille (1883), Cours d'Analyse de l'École Polytechnique, vol. II, Calcul Intégral: Intégrales définies et indéfinies (2nd ed.), Paris{{citation}}: CS1 maint: location missing publisher (link) Kaiser, Gerald (1994), "A Friendly Guide to Wavelets", Physics Today, 48 (7): 57–58, Bibcode:1995PhT....48g..57K, doi:10.1063/1.2808105, ISBN 978-0-8176-3711-8 Kammler, David (2000), A First Course in Fourier Analysis, Prentice Hall, ISBN 978-0-13-578782-3 Katznelson, Yitzhak (2004), An Introduction to Harmonic Analysis, Cambridge University Press, doi:10.1017/cbo9781139165372, ISBN 978-0-521-83829-0 Khare, Kedar; Butola, Mansi; Rajora, Sunaina (2023), "Chapter 2.3 Fourier Transform as a Limiting Case of Fourier Series", Fourier Optics and Computational Imaging (2nd ed.), Springer, doi:10.1007/978-3-031-18353-9, ISBN 978-3-031-18353-9, S2CID 255676773 Kirillov, Alexandre; Gvishiani, Alexei D. (1982) , Theorems and Problems in Functional Analysis, Springer (translated from Russian) Knapp, Anthony W. (2001), Representation Theory of Semisimple Groups: An Overview Based on Examples, Princeton University Press, ISBN 978-0-691-09089-4 Kolmogorov, Andrey Nikolaevich; Fomin, Sergei Vasilyevich (1999) , Elements of the Theory of Functions and Functional Analysis, Dover (translated from Russian) Lado, F. (1971), "Numerical Fourier transforms in one, two, and three dimensions for liquid state calculations", Journal of Computational Physics, 8 (3): 417–433, Bibcode:1971JCoPh...8..417L, doi:10.1016/0021-9991(71)90021-0 Mallat, Stéphane (2009), A wavelet tour of signal processing: the sparse way, Amsterdam Boston: Elsevier/Academic Press, doi:10.1016/B978-0-12-374370-1.X0001-8, ISBN 978-0-12-374370-1 Müller, Meinard (2015), The Fourier Transform in a Nutshell. (PDF), Springer, doi:10.1007/978-3-319-21945-5, ISBN 978-3-319-21944-8, S2CID 8691186, archived from the original (PDF) on 2016-04-08, retrieved 2016-03-28; also available at Fundamentals of Music Processing, Section 2.1, pages 40–56 Oppenheim, Alan V.; Schafer, Ronald W.; Buck, John R. (1999), Discrete-time signal processing (2nd ed.), Upper Saddle River, N.J.: Prentice Hall, ISBN 0-13-754920-2 Paley, R.E.A.C.; Wiener, Norbert (1934), Fourier Transforms in the Complex Domain, American Mathematical Society Colloquium Publications, Providence, Rhode Island: American Mathematical Society Pinsky, Mark (2002), Introduction to Fourier Analysis and Wavelets, Brooks/Cole, ISBN 978-0-534-37660-4 Poincaré, Henri (1895), Théorie analytique de la propagation de la chaleur, Paris: Carré Polyanin, A. D.; Manzhirov, A. V. (1998), Handbook of Integral Equations, Boca Raton: CRC Press, ISBN 978-0-8493-2876-3 Press, William H.; Flannery, Brian P.; Teukolsky, Saul A.; Vetterling, William T. (1992), Numerical Recipes in C: The Art of Scientific Computing, Second Edition (2nd ed.), Cambridge University Press Proakis, John G.; Manolakis, Dimitri G. (1996). Digital Signal Processing: Principles, Algorithms and Applications (3 ed.). New Jersey: Prentice-Hall International. Bibcode:1996dspp.book.....P. ISBN 9780133942897. sAcfAQAAIAAJ. Rahman, Matiur (2011), Applications of Fourier Transforms to Generalized Functions, WIT Press, ISBN 978-1-84564-564-9 Rudin, Walter (1991), Fourier Analysis on Groups, New York, NY: John Wiley & Sons, ISBN 978-0-471-52364-2 Rudin, Walter (1987), Real and Complex Analysis (3rd ed.), Singapore: McGraw Hill, ISBN 978-0-07-100276-9 Simonen, P.; Olkkonen, H. (1985), "Fast method for computing the Fourier integral transform via Simpson's numerical integration", Journal of Biomedical Engineering, 7 (4): 337–340, doi:10.1016/0141-5425(85)90067-6, PMID 4057997 Smith, Julius O. "Mathematics of the Discrete Fourier Transform (DFT), with Audio Applications --- Second Edition". ccrma.stanford.edu. Retrieved 2022-12-29. We may think of a real sinusoid as being the sum of a positive-frequency and a negative-frequency complex sinusoid. Stade, Eric (2005). Fourier Analysis. Wiley. doi:10.1002/9781118165508. ISBN 978-0-471-66984-5. Stein, Elias; Shakarchi, Rami (2003), Fourier Analysis: An introduction, Princeton University Press, ISBN 978-0-691-11384-5 Stein, Elias; Weiss, Guido (1971), Introduction to Fourier Analysis on Euclidean Spaces, Princeton, N.J.: Princeton University Press, ISBN 978-0-691-08078-9 Strichartz, Robert S. (1994), A guide to distribution theory and Fourier transforms, Boca Raton: CRC Press, ISBN 0-8493-8273-4 Taneja, H.C. (2008), "Chapter 18: Fourier integrals and Fourier transforms", Advanced Engineering Mathematics, vol. 2, New Delhi, India: I. K. International Pvt Ltd, ISBN 978-8189866563 Titchmarsh, E. (1986) , Introduction to the theory of Fourier integrals (2nd ed.), Oxford University: Clarendon Press, ISBN 978-0-8284-0324-5 Vretblad, Anders (2000), Fourier Analysis and its Applications, Graduate Texts in Mathematics, vol. 223, New York: Springer, ISBN 978-0-387-00836-3 Whittaker, E. T.; Watson, G. N. (1927), A Course of Modern Analysis (4th ed.), Cambridge University Press Widder, David Vernon; Wiener, Norbert (August 1938), "Remarks on the Classical Inversion Formula for the Laplace Integral", Bulletin of the American Mathematical Society, 44 (8): 573–575, doi:10.1090/s0002-9904-1938-06812-7 Wiener, Norbert (1949). Extrapolation, Interpolation, and Smoothing of Stationary Time Series: With Engineering Applications. MIT Press. ISBN 9780262257190. {{cite book}}: ISBN / Date incompatibility (help) Wilson, R. G. (1995), Fourier Series and Optical Transform Techniques in Contemporary Optics, New York: Wiley, ISBN 978-0-471-30357-2 Wolf, Kurt B. (1979), Integral Transforms in Science and Engineering, Springer, doi:10.1007/978-1-4757-0872-1, ISBN 978-1-4757-0874-5 Yosida, K. (1968), Functional Analysis, Springer, ISBN 978-3-540-58654-8 External links [edit] Media related to Fourier transformation at Wikimedia Commons Encyclopedia of Mathematics Weisstein, Eric W. "Fourier Transform". MathWorld. Fourier Transform in Crystallography | Authority control databases | | International | | | National | United States France BnF data Japan Czech Republic Israel | | Other | Yale LUX | Retrieved from " Categories: Fourier analysis Integral transforms Unitary operators Joseph Fourier Mathematical physics Hidden categories: CS1 maint: article number as page number Articles with short description Short description matches Wikidata Pages using multiple image with auto scaled images All articles with unsourced statements Articles with unsourced statements from May 2009 CS1 French-language sources (fr) CS1: long volume value CS1 maint: location missing publisher CS1 errors: ISBN date Commons category link from Wikidata Fourier transform Add topic
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https://brainly.com/question/45634659
[FREE] The surface area of a right-circular cone with radius r and height h is given by: S = \pi r \sqrt{r^2 - brainly.com 3 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 +53,6k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +45,1k Ace exams faster, with practice that adapts to you Practice Worksheets +6,7k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified The surface area of a right-circular cone with radius r and height h is given by: S=π r r 2+h 2​ and its volume is: V=3 1​π r 2 h (a) Determine h and r for the cone with a given surface area S=3 and maximal volume V. h=, r= (b) What is the ratio h/r for a cone with a given volume V=5 and minimal surface area S? h/r= 1 See answer Explain with Learning Companion NEW Asked by jokereq5455 • 01/07/2024 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 53970222 people 53M 0.0 0 Upload your school material for a more relevant answer The surface area of a right-circular cone of radius r and height h is S=πr(sqrt(r2+H_2) and its volume is V=1/3(πr2h). (a) h = 3/sqrt(2), r = 3/(sqrt(2pi)) (b) h/r = sqrt(5/(3pi)) Explanation For part (a), to maximize the volume, we differentiate the volume formula V = (1/3)πr²h with respect to h and r and set the resulting equations to zero. Solving, we find h = 3/sqrt(2) and r = 3/(sqrt(2pi)). This maximizes the volume under the constraint of surface area S = 3. For part (b), to minimize surface area with fixed volume, we again use calculus. The volume V = (1/3)πr²h and surface area S = πr(sqrt(r² + h²)) are related. By setting the derivative of S with respect to r equal to zero, we get h² = r². Substituting this into the volume formula and given V = 5, we find r = sqrt(5/(3pi)) and h = sqrt(5/3). The ratio h/r is then sqrt(5/(3pi)). In summary, for the cone with maximal volume under the given surface area constraint, h = 3/sqrt(2) and r = 3/(sqrt(2pi)). For the cone with minimal surface area under the given volume constraint, h/r = sqrt(5/(3pi)). These results are obtained through calculus optimization techniques, ensuring accurate solutions. Answered by Neetoo •33.7K answers•54M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 53970222 people 53M 0.0 0 Graduate Classical Mechanics - Michael Fowler Fundamentals of Calculus - Joel Robbin, Sigurd Angenent Classical Mechanics - Peter Dourmashkin Upload your school material for a more relevant answer For part (a), the height and radius of the cone are h=2​3​ and r=2 π​3​. For part (b), the ratio r h​=3 π 5​​. Explanation (a) To determine the height h and radius r of the cone for the given surface area S=3 that maximizes the volume V, we start with the equations for surface area and volume: S=π r r 2+h 2​ and V=3 1​π r 2 h. Given the surface area, we can express h in terms of r: h=(π r S​)2−r 2​. Plugging in S=3: h=(π r 3​)2−r 2​. Substituting this into the volume equation and then finding the critical points using calculus will yield the values for h and r. By carrying out this process, we find: h=3/2​,r=3/(2 π​). (b) For part (b), to find the ratio r h​ for a cone with given volume V=5 that minimizes surface area, we follow a similar method. We express h in terms of r using the volume equation: h=(1/3)π r 2 5​=π r 2 15​. Substitute this expression for h into the surface area equation and find the derivative to minimize it. This leads to finding the relationship between h and r and, ultimately, we derive that: r h​=3 π 5​​. Examples & Evidence For example, if we have a cone with a known surface area, we can calculate the necessary dimensions that maximize the volume by using the relationships established through the equations for surface area and volume. Similarly, with a fixed volume, we can manipulate the surface area equation to derive the height-to-radius ratio for the most efficient cone shape. The calculations involved are based on fundamental principles of calculus, applying derivatives to optimize functions under constraints related to surface area and volume for geometric shapes. Thanks 0 0.0 (0 votes) Advertisement jokereq5455 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer the surface area of a right-circular cone of radius and height h is S = πr√(r²+h²), and its volume is v = 1/3πr²h. (a) determine h and r for the cone with given surface area s = 8 and maximal volume v. H = _, r = __ Community Answer a right circular cone has volume v = p/3 r^2 h and surface area s = pi r sqrt r^2 h^2. find the dimensions of the cone with surface area 1 and maximum volume Community Answer 10 The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What is the largest volume of a right circular cone if its surface area is S? Use Lagrange multipliers to solve the problem. Community Answer 1 Find the dimensions of a right cone with surface area S=4π (not including the base) that has the largest volume. (The volume of a circular cone is V ˉ = 3 1 ​ πr 2 h and the surface areas is S=πrl where l= r 2 +h 2 ​ is the slant height of the cone.) Community Answer ​​​​​​​ The surface area of a right-circular cone of radius r and height h is S=π r \sqrt{r^{2}+h^{2}}, and its volume is V=\frac{1}{3} π r^{2} h (a) Determine h and r 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? New questions in Mathematics Explain the association and provide an explanation for the relationship between: • The number of letters in a person's name and the last digit of their phone number. • The number of ice cubes in a drink and the temperature of the drink. The system has no solution. The system has a unique solution: System A $\begin{array}{r} -x-4 y=4 \ x+4 y=4 \end{array}$ (x,y)=(1,□) The system has infinitely many solutions. They must satisfy the following equation: y=□ Consider the following system of equations: System B $\begin{array}{c} -x-3 y=-6 \ x+3 y=6 \end{array}$ Which of the following statements describes the solution to this system? A. The system has no solution. B. The system has a unique solution: (x,y)=C□ C. The system has infinitely many solutions. They must satisfy the following equation: y=□ Identify the expression that uses adding the additive inverse to rewrite the expression 16−7+22−12. A. −16+(−7)+(−22)+(−12) B. −16+7+(−22)+12 C. 16+(−7)+22+(−12) D. 16+(−7)+22+12 (5 3⋅9542​)3 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
11439
https://mathworld.wolfram.com/PermutationCycle.html
Permutation Cycle -- from Wolfram MathWorld TOPICS AlgebraApplied MathematicsCalculus and AnalysisDiscrete MathematicsFoundations of MathematicsGeometryHistory and TerminologyNumber TheoryProbability and StatisticsRecreational MathematicsTopologyAlphabetical IndexNew in MathWorld Discrete Mathematics Combinatorics Permutations Permutation Cycle Download Wolfram Notebook A permutation cycle is a subset of a permutation whose elements trade places with one another. Permutations cycles are called "orbits" by Comtet (1974, p.256). For example, in the permutation group, (143) is a 3-cycle and (2) is a 1-cycle. Here, the notation (143) means that starting from the original ordering , the first element is replaced by the fourth, the fourth by the third, and the third by the first, i.e., . There is a great deal of freedom in picking the representation of a cyclic decomposition since (1) the cycles are disjoint and can therefore be specified in any order, and (2) any rotation of a given cycle specifies the same cycle (Skiena 1990, p.20). Therefore, (431)(2), (314)(2), (143)(2), (2)(431), (2)(314), and (2)(143) all describe the same permutation. The following table gives the set of representations for each element of the symmetric group on three elements, , sorted in lowest canonical order (first by cycle length, and then by lowest initial order of elements). permutation of notation (1)(2)(3) (1)(23) (3)(12) (123) (132) (2)(13) The cyclic decomposition of a permutation can be computed in the Wolfram Language with the function PermutationCycles[p] and the permutation corresponding to a cyclic decomposition can be computed with PermutationList[c]. Here, the individual cycles are represented using the function Cycles. In previous versions, the cyclic decomposition could be computed less efficiently using ToCycles[p] in the Wolfram Language package Permutationsand the permutation corresponding to a cyclic decomposition could be computed using FromCycles[Image 18: {_c1_, ..., _cn_Image 19: {] in the Wolfram Language package Permutations . According to Vardi (1991), the Wolfram Language code for ToCycles is one of the most obscure ever written. Every permutation group on symbols can be uniquely expressed as a product of disjoint cycles (Skiena 1990, p.20). A cycle decomposition of a permutation can be viewed as a class of a permutation group. The number of -cycles in a permutation group of order is given by (1) where are the Stirling numbers of the first kind. More generally, let be the number of permutations of having exactly cycles all of which are of length . are sometimes called the associated Stirling numbers of the first kind (Comtet 1974, p.256). The quantities appear in a closed-form expression for the coefficients of in Stirling's series (Comtet 1974, p.257 and 267). The following table gives the triangles for . Sloane 1A0082751; 1, 1; 2, 3, 1; 6, 11, 6, 1; 24, 50, 35, 10, 1; ... 2A0083061; 2; 6, 3; 24, 20; 120, 130, 15; 720, 924, 210; ... 3A0502112; 6; 24; 120, 40; 720, 420; 5040, 3948; 40320, ... 4A0502126; 24; 120; 720; 5040, 1260; 40320, 18144; ... 5A05021324; 120; 720; 5040; 40320; 362880, 72576; ... The functions are given by the recurrence relation (2) where is the falling factorial, combined with the initial conditions (3) (4) (Riordan 1958, p.85; Comtet 1974, p.257). See also Golomb-Dickman Constant, Permutation, Permutation Group, Stirling Number of the First Kind, Stirling's Series, Subset Explore with Wolfram|Alpha More things to try: 1.05 12,000 directrix of parabola x^2+3y=16 logistic map r=4 References Biggs, N. Discrete Mathematics, rev. ed. Oxford, England: Clarendon Press, 1993.Comtet, L. Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, p.257, 1974.Graham, R.L.; Knuth, D.E.; and Patashnik, O. Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley, 1994.Knuth, D.E. The Art of Computer Programming, Vol.1: Fundamental Algorithms, 3rd ed. Reading, MA: Addison-Wesley, 1997.Riordan, J. Combinatorial Identities. New York: Wiley, 1958.Skiena, S. "The Cycle Structure of Permutations." §1.2.4 in Implementing Discrete Mathematics: Combinatorics and Graph Theory with Mathematica. Reading, MA: Addison-Wesley, pp.20-24, 1990.Sloane, N.J.A. Sequences A008275, A008306, A050211, A050212, A050213 in "The On-Line Encyclopedia of Integer Sequences."Stanton, D. and White, D. Constructive Combinatorics. New York: Springer-Verlag, 1986.Vardi, I. Computational Recreations in Mathematica. Redwood City, CA: Addison-Wesley, p.223, 1991. Referenced on Wolfram|Alpha Permutation Cycle Cite this as: Weisstein, Eric W. "Permutation Cycle." From MathWorld--A Wolfram Resource. Subject classifications Discrete Mathematics Combinatorics Permutations About MathWorld MathWorld Classroom Contribute MathWorld Book wolfram.com 13,278 Entries Last Updated: Sun Sep 28 2025 ©1999–2025 Wolfram Research, Inc. Terms of Use wolfram.com Wolfram for Education Created, developed and nurtured by Eric Weisstein at Wolfram Research Created, developed and nurtured by Eric Weisstein at Wolfram Research
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https://www.cityofrosemead.org/cms/one.aspx?portalid=10035075&pageid=11981660
Unit:1 Unit Objective(s)/Big Ideas: Analyze proportional relationships and use them to solve real world and mathematical problems. Apply and extend previous understandings of operations with fractions to add, subtract, multiply, and divide rational numbers. Solve real -life and mathematical problems using numerical and algebraic expressions and equations. Draw, construct and describe geometrical figures and describe the relationships between them. Lesson: 1 of 1 Continued Date: 10 -13 -2016 Lesson Objective(s): 7.RP.2 Recognize and represent proportional relationships between quantities. 7.RP.2a Decide whether two quantities are in a proportional rel ationship, e.g., by testing for equivalent ratios in a table or graphing on a coordinate plane and observing whether the graph is a straight line through the origin. 7.RP.2b Identify the constant of proportionality (unit rate) in tables, graphs, equations, diagrams, and verbal descriptions of proportional relationships. Students will Use direct variation to solve problems Essential Question(s): How can I show that 2 objects are proportional? I = Introduction T = Teacher Directed S = Student Centered Learning Activities C = Closure H = Homework Mark on the chart below the instructional methods and strategies used in each section. Instructional Delivery Method Instructional Strategy Method I T S C Strategy I T S C H Class discussion Similarities & Differences Guided practice/modeling Summarizing & Note Taking x Hands on/experiment/lab x Reinforcing Effort & Recognition x Learning centers Homework and Practice x Lecture Nonlinguistic Representations Peer evaluation Cooperative Learning Seatwork x Setting Objectives & Feedback x x x Student presentations Generating and Testing Hypotheses Technology x Questions, Cues, Advance Organizers Cooperative Learning Vocabulary x x Culturally Responsive Teaching Technology x Other: ___ Culturally Responsive Teaching Section Detailed Explanation I As students enter the classroom, have them begin data binder for math. Introductio n Prior Knowledge They will also spend time completing their quiz over slope and proportionality. Students will also be working on ALEKS when finished with today's lesson for approx 30 minutes. T Teacher Directed Instruction There are two ways to tell if two ratios form a proportion. You must either show that the ratios are equivalent fractions or show that the cross products are equal. It is usually easier to simplify ratios before finding equivalent fractions or multiplying to find cross products. A proportion can be written in a variety of ways as long as there is a relationship among the ratios involved. A relation that does not have these characteristics is not a proportion. The rate ratio is written as a fraction, which implies the operation of division. Remind students that this is a clue for how to find the unit rate. A rate of change describes how one quantity changes in relatio n to another. S Student Learning Activities Ask the scaffolded questions for each example to differentiate instruction. page 81 -88 Find a constant ratio. ● How does the graph show that there is a constant rate of change? The graph is a straight line. ● How do you determine if there is a constant ratio between the two quantities? Divide the height by the time. ● Are the quantities proportional? Explain . Yes, they are proportional because they have a constant ratio. ● How does the graph show that the relationship is proportional? It is a straight line that passes through the origin. ● What is the difference between a constant rate of change and a constant of proportionality? The constant rate of change is the change in y compared to the chang e in x. The constant of proportionality is the ratio of y to x. If there is a constant rate of change, there may or may not be a constant of proportionality. Need Another Example? The amount of money Serena earns at her job is shown in the graph. Determine the rate in dollars per hour. $10 per hour Identify and explain what the constant of proportionality represents. ● Does the amount of money Julio earns each hour change? No , it is always $10 per hour. ● What does “k” represent? the constant rate of change ● In your own words, what does the “constant of proportionality” mean? Sample answer: The two quantities are always in the same (constant) proportion (ratio) to one another. ● Describe what this relationship would look like, if graphed. a straight line that passes through the origin ● What would be some of the ordered pairs on the line? Sample answer: (0,0), (1, 10), (2, 20), (3, 30), (4, 40) Need Another Example? Neil is practi cing for his typing test. The equation y = 45x represents the total number of words y he can type in x minutes. Identify the constant of proportionality. Then explain what it represents in this situation. 45; Neil can type 45 words per minute. Determine direct variation. ● How do you find each cost in the table? Multiply the number of pizzas by $8. Then add $3 for the delivery charge. ● How do you find each ratio? Divide each cost by the number of pizzas. ● How do you know that this relationship is not a direct variation? The ratios are not the same and the line does not pass through the origin. ● How could you alter the scenario so that the relationship would be a direct variation? Sample answer: Eliminate the $3 delivery charge. Need Another Exa mple? A photographer charges a $30 sitting fee and then $6 for each photograph ordered. Make a table and a graph to show the cost of 1, 2, 3, and 4 photographs. Is there a direct variation? Explain. no; Sample answer: ; Because there is no constant ratio and the line does not go through the origin, there is no direct variation. Determine direct variation ● Are all of the ratios the same? If so, what are they? yes; or 12 ● Does the amount of money earned each hour remain constant? Explain. Yes, the amount of money earned is a constant $12 per hour. ● Does this table represent a direct variation? Explain. Yes, the ratios are the same. ● Why is the constant of proportionality a “constant”? The ratio always remains the same. Need Another Example? Determine whether the relationship is a direct variation. If so, state the constant of proportionality. no continue from here Closure page 85 and 87 and 88 Materials Needed: glencoe text pages 81 -88 Assessment: HSD 7th grade Activity #5 Special Notes: ALEKS is assigned as homework for 30min a night 5 times a week vocabulary: coordinate plane quadrants ordered pair x-coordinate y-coordinate y-axis origin x-axis direct variation constant of variation constant of proportionalilty Quality Questions: imbedded in lesson plan HSD Lesson Plan Protocol Teacher: Pearce Subject: Math Gr:7
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https://math.stackexchange.com/questions/1666880/how-to-prove-b-c-if-ab-ac-cancellation-law-in-groups
How to prove $b=c$ if $ab=ac$ (cancellation law in groups)? - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How to prove b=c b=c if a b=a c a b=a c (cancellation law in groups)? Ask Question Asked 9 years, 7 months ago Modified9 years, 7 months ago Viewed 13k times This question shows research effort; it is useful and clear 25 Save this question. Show activity on this post. I want to prove for a group G G, that if a∘b=a∘c a∘b=a∘c then this is true b=c b=c I started with b=b∘e b=b∘e, but this didn't help me at all. Next I tried with this: (a∘b)∘c=a∘(b∘c)(a∘b)∘c=a∘(b∘c) but I don't know/understand how to go further. How can I prove this equation? group-theory Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Feb 24, 2016 at 6:38 user99914 asked Feb 22, 2016 at 10:24 greedsingreedsin 591 1 1 gold badge 5 5 silver badges 11 11 bronze badges 10 3 See also: Cancellation property in groups and Proof of the right and left cancellation laws for GroupsMartin Sleziak –Martin Sleziak 2016-02-22 12:27:53 +00:00 Commented Feb 22, 2016 at 12:27 1 Divide both sides by a.user191074 –user191074 2016-02-22 15:51:34 +00:00 Commented Feb 22, 2016 at 15:51 1 I think you have to make sure a a is not 0 0.kleineg –kleineg 2016-02-22 16:33:18 +00:00 Commented Feb 22, 2016 at 16:33 1 When I was teaching, I wouldn't let my students use the word "cancel". If they did, I would have them identify if they were using the additive inverse or the multiplicative inverse (I wanted them to actually think about what they were doing and why). If this case, you're using the Distributive Property.CharlieHorse –CharlieHorse 2016-02-22 20:41:42 +00:00 Commented Feb 22, 2016 at 20:41 3 @kleineg, what is this 0 0 that you talk about?Carsten S –Carsten S 2016-02-22 23:40:39 +00:00 Commented Feb 22, 2016 at 23:40 |Show 5 more comments 6 Answers 6 Sorted by: Reset to default This answer is useful 39 Save this answer. Show activity on this post. Suppose a⋅b=a⋅c a⋅b=a⋅c Let a−1 a−1 be the inverse element of a a in G G (s.t. a−1⋅a=a⋅a−1=e a−1⋅a=a⋅a−1=e where e e is the identity element), which must exist by the axioms of groups. Now consider a−1⋅(a⋅b)=a−1⋅(a⋅c)a−1⋅(a⋅b)=a−1⋅(a⋅c) By associativity, we have (a−1⋅a)⋅b=(a−1⋅a)⋅c(a−1⋅a)⋅b=(a−1⋅a)⋅c By the definition of inverse, we have e⋅b=e⋅c e⋅b=e⋅c where e e is the identity element (s.t. e⋅x=x⋅e=x e⋅x=x⋅e=x for all x∈G x∈G). By the definition of the identity element, b=c b=c Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 22, 2016 at 10:46 Yiyuan LeeYiyuan Lee 14.6k 5 5 gold badges 41 41 silver badges 72 72 bronze badges 13 23 Because that's what you have to prove.yellon –yellon 2016-02-22 16:01:28 +00:00 Commented Feb 22, 2016 at 16:01 8 Here's a counter-example of why you can't just "cancel on both sides": Let A A be a matrix. If A v⃗=A w⃗A v→=A w→, does it follow that v⃗=w⃗v→=w→? Answer: not unless A A has an inverse, and not all matrices have inverses.Michael Seifert –Michael Seifert 2016-02-22 16:35:20 +00:00 Commented Feb 22, 2016 at 16:35 22 @user1717828 There is actually substance to this proof - i.e. it says that if we have inverses then we have cancellation. There are common systems without cancellation - like modular arithmetic - so we definitely shouldn't treat it as an axiom. If we start with 3⋅3≡3⋅1(mod 6)3⋅3≡3⋅1(mod 6) Then we just cancel on both sides, then we get 3≡1(mod 6).3≡1(mod 6). Whoops.Milo Brandt –Milo Brandt 2016-02-22 16:36:31 +00:00 Commented Feb 22, 2016 at 16:36 8 @user1717828 I don't think this has anything to do with being a physicist - I've always thought it was a symptom of bad (or insufficiently deep) math education. It's very common among students of all quantitative fields that they memorize the rule of cancellation without understanding why things can be cancelled, and that gets them into trouble when dealing with more sophisticated situations like matrices, modular arithmetic, differential operators, etc.David Z –David Z 2016-02-22 21:12:40 +00:00 Commented Feb 22, 2016 at 21:12 7 @DarrelHoffman: but the question concerns a general group with operator ∘∘. We don't usually introduce the // notation for anything other than multiplication, and it'd be wrong to say a−1 a−1 is equal to 1/a 1/a in the case where ∘∘ is addition.Steve Jessop –Steve Jessop 2016-02-22 22:15:19 +00:00 Commented Feb 22, 2016 at 22:15 |Show 8 more comments This answer is useful 8 Save this answer. Show activity on this post. Hint: If you know that 4⋅x=4⋅y 4⋅x=4⋅y, how do you prove that x=y x=y? Hint 2: Think about inverses Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 22, 2016 at 10:25 5xum5xum 127k 6 6 gold badges 136 136 silver badges 213 213 bronze badges 4 Because N is closed under multiplication.djechlin –djechlin 2016-02-22 21:38:37 +00:00 Commented Feb 22, 2016 at 21:38 I'm sorry when I said multiplication I meant multiplication, not the operation you just made up.djechlin –djechlin 2016-02-22 21:52:31 +00:00 Commented Feb 22, 2016 at 21:52 Because Z is an ACRU with a nonempty subset N with trichotomy that is closed under addition and multiplication.djechlin –djechlin 2016-02-22 21:57:37 +00:00 Commented Feb 22, 2016 at 21:57 Associative commutative ring with unit. My point was that you first hint rather begs the question.djechlin –djechlin 2016-02-22 21:59:59 +00:00 Commented Feb 22, 2016 at 21:59 Add a comment| This answer is useful 6 Save this answer. Show activity on this post. G G is a group. One of the axioms of a group is that every element has an inverse. This means that a∈G a∈G has an inverse a−1∈G a−1∈G. This will help a lot. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 22, 2016 at 10:35 AliAli 427 2 2 silver badges 8 8 bronze badges Add a comment| This answer is useful 6 Save this answer. Show activity on this post. Ok, we know a,b,c∈G a,b,c∈G b=e∘b=(a−1∘a)∘b=a−1∘(a∘b)=a−1∘(a∘c)=(a−1∘a)∘c=c b=e∘b=(a−1∘a)∘b=a−1∘(a∘b)=a−1∘(a∘c)=(a−1∘a)∘c=c Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Feb 22, 2016 at 10:39 answered Feb 22, 2016 at 10:29 openspaceopenspace 6,801 3 3 gold badges 20 20 silver badges 38 38 bronze badges 5 Where is e e is neutral element in G G openspace –openspace 2016-02-22 10:30:28 +00:00 Commented Feb 22, 2016 at 10:30 And where you use ′′ to denote the inverse (this is probably a bad idea in general).Tobias Kildetoft –Tobias Kildetoft 2016-02-22 10:31:58 +00:00 Commented Feb 22, 2016 at 10:31 @TobiasKildetoft why? We know that in G G∃a,a′:a∘a′=a′∘a=e∃a,a′:a∘a′=a′∘a=e openspace –openspace 2016-02-22 10:34:15 +00:00 Commented Feb 22, 2016 at 10:34 1 It is a bad idea because the inverse is denoted by a−1 a−1. This is universal when working with groups written multiplicatively. Using ' is something done temporarily until it has been established that the inverse is unique and two-sided. This also frees up ' to denote alternative elements.Tobias Kildetoft –Tobias Kildetoft 2016-02-22 10:36:29 +00:00 Commented Feb 22, 2016 at 10:36 @TobiasKildetoft ok, I understand you openspace –openspace 2016-02-22 10:39:16 +00:00 Commented Feb 22, 2016 at 10:39 Add a comment| This answer is useful 4 Save this answer. Show activity on this post. By the group properties each element has an inverse. So you can just multiply your equation on the left by a−1 a−1. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 22, 2016 at 11:23 CAT0CAT0 63 4 4 bronze badges Add a comment| This answer is useful 3 Save this answer. Show activity on this post. Multiply both sides of the given equation a∘b=a∘c a∘b=a∘c on the left by the inverse of a a to get the desired result. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 22, 2016 at 10:48 JoeJoe 12.1k 2 2 gold badges 20 20 silver badges 53 53 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 group-theory 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 Linked 5Proof of the right and left cancellation laws for Groups 4Cancellation property in groups 0Prove that it is impossible that there exists a,b,c∈G a,b,c∈G such that a≠b a≠b but a∘c=b∘c a∘c=b∘c. 0Why is the 'Law of Cancellation' for groups only an implication? Related 6How to find finite groups 2meanable groups 1C⊆D C⊆D. Prove P(C)P(C) is a subgroup of P(D)P(D). 0Algebraic groups? 2Direct sum of Abelian Finitely generated groups 1Prove the following function set is a group under function composition. 1A question about bracketing in generalized associative law for groups Hot Network Questions Should I let a player go because of their inability to handle setbacks? My dissertation is wrong, but I already defended. How to remedy? For every second-order formula, is there a first-order formula equivalent to it by reification? Countable and uncountable "flavour": chocolate-flavoured protein is protein with chocolate flavour or protein has chocolate flavour Why do universities push for high impact journal publications? Lingering odor presumably from bad chicken How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? 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https://learnius.com/slp/5+Model+of+Speech+Production/3+Transfer+Function/time+shifting+property+of+the+z-transform
time shifting property of the z-transform - learnius learnius ec ss slp Spoken Language Processing 1 Course Logistics 2 Introduction 3 Speech Production, Perception and Phonetics 4 Speech Signal Representations 5 Model of Speech Production Model of Speech Production 1 System Modelling 2 LTI Systems 3 Transfer Function frequency response z-transform time shifting property of the z-transform unit delay transfer function rational transfer function zeros of the transfer function poles of the transfer function relation between the z-transform and the DTFT inverse z-transform z-transform pairs convolution property of the z-transform 4 Filtering 5 Acoustical Model 6 Linear Prediction 7 Problems 9 Speech Synthesis eitt llms learnius.com learnius time shifting property of the z-transform If the discrete-time signalx(n) is shifted by n 0 samples in the time domain, its z-transform becomes: x(n−n 0)→Z z−n 0 X(z) This is a direct result of the application of the z-transform equation to the signal y(n)=x(n−n 0): Y(z)=∑n=−∞+∞y(n)z−n=∑n=−∞+∞x(n−n 0)z−n=∑m=−∞+∞x(m)z−(m+n 0)=z−n 0∑m=−∞+∞x(m)z−m=z−n 0 X(z) Links to this page Model of Speech Production rational transfer function second-order all-pole IIR system ztran-o01 two real poles time shifting property of the z-transform Not found This page does not exist Interactive graph On this page Powered by Obsidian Publish
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https://pages.uoregon.edu/jschombe/glossary/logarithm.html
Logarithm Logarithm: Logarithm is the exponent or power to which a base must be raised to yield a given number. An example of a logarithm is as follows. In the expression b x = N, if b is the base and equal to 10 and N a number, equal to 100, then x is equal to 2 and is said to be the logarithm of 100 to the base 10. This is written: log 100 = 2, in which it is understood that log means logarithm to the base 10. The latter is also called a common logarithm. Logarithms that employ the base e, in which e = 2.71828 ... are called natural, or Napierian, logarithms; the notation used is ln, to distinguish natural logarithms from common logarithms (log). When a common logarithm of a number is written as the sum of an integer and a positive decimal (e.g., 2.3147), the integer--called the characteristic--serves to locate the decimal point in the number, and the decimal--called the mantissa--indicates the digits in the number. The latter are determined from tables of logarithms, which relate mantissas to numbers. When the number is greater than or equal to 1, the characteristic is 1 less than the number of digits to the left of the decimal point; when the number is less than 1, the characteristic is negative and is 1 more than the number of zeros following the decimal point. For example, the number 365.0 has the characteristic 2; the number 0.005 has the characteristic -3. Excerpt from the Encyclopedia Britannica without permission.
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https://www.youtube.com/watch?v=xGsHqSUbnLY
College Math Lesson 1.5 - Venn Diagrams with Three Sets Farrington Math 1450 subscribers 2 likes Description 381 views Posted: 21 Jul 2020 Objective: At the end of this lesson, students will be able to draw a Venn diagram with three overlapping circles to represent three sets and describe the members of each region. Notes: Venn diagrams are particularly useful for dealing with three sets that may overlap one another. Because the circles overlap one another, they form a total of eight regions (including the region outside all three circles). Example: Use a Venn diagram to describe these three sets: teachers, redheads, and dancers. Describe the members of each region. 1 comments Transcript: Intro lesson 1.5 venn diagrams with three sets so our objective for this lesson is to draw a venn diagram with three overlapping circles representing three sets and describe the members of each region Lesson all right so let's start by drawing a venn diagram with three sets so to do that we'll draw a circle here and then an overlapping circle here and then a circle that overlaps both of them down here and so we'll call this this circle the set a and this is the set b and this is the set c and now notice that we've created eight different regions by drawing these three circles overlapping like this so the first region is going to be right here second region right here third is down here and then we've got a region here called four region here is five and six right in the center is seven and then the eighth region is outside of all three of our sets so that's that's the region where none of these three sets uh applies all right so so let's let's label what's in each of these sets so in or in each of these regions so in region one we have just a just set a not b and c in set two we're going to have just b in set three we'll have just c and then in four in this region 4 we have both a and b but not c so 4 is a and b and five is b and c six is a and c and seven is all three a b and c and then eight is none eight is uh neither a b nor c so none of these applies out in this region eight as you can see there's eight regions and that's um these are the sets that are represented in each of those regions Example Problem all right for our example problem we're asked to use a venn diagram to describe these three sets teachers redheads and dancers and we're asked to label each of the regions or to describe the members in each of these regions okay so um the reason i chose these three sets is because each member in my family belongs in a different region in this so we'll start up here with this region which up um these are folks who are teachers but not redheads or dancers so we'll call these uh non-redhead uh non-dancer teachers that's that red region there and that would be me in this region i'm not a redhead i'm not a dancer i am a teacher all right um over here we have people who are redheads but not teachers or dancers these would be uh non-redhead no sorry non-dancer non-teacher redheads so my oldest son would belong in that region up there he has red hair but he's not a teacher and he's not a dancer all right down in this region these folks are just dancers they're not teachers or redheads so this would be um non-redhead non-teacher dancers that would be my youngest son down in this region he is a dancing fool but he uh is not red-headed and he's not a teacher um okay next would be um let's go with this region right here in the center so maybe let's draw an arrow up to this region to show the center and we'll put it down here so these uh in the center these are teachers who are also red-headed and dancers so these are going to be um redhead uh redhead dancing teachers so that would be my wife in the center there she is a redhead she is a dancer and she is a teacher um okay now let's do uh outside here this would be people who are none of the above so these would be um let's see non-redhead uh non-dancer non-teachers out in that region that would be my middle son he is not a redhead he's not a dancer and he's not a teacher um whoever got one two three four five there's six people in my family my daughter would fall into this region so let's draw an arrow out here to this region um these are people who are redheads and dancers but not teachers so let's go with redhead non-teacher answers so that would be my daughter there she is a redhead she loves to dance but she's not a teacher all right let's do this region here so this region would be teachers who are redheads but not dancers so what are we going to call those people uh redhead non-dancer teachers redhead non-dancer teachers uh so who belongs in that region maybe mrs fraley i don't know i've never seen mrs fraley dance maybe she's an excellent dancer i don't know um but then we have one last region that's right here these are people who are teachers and dancers but not redheads so let's call these non-redhead uh dancing teachers uh so this region might be uh mr bazios who has legendary dance moves is clearly a teacher but is not a redhead so that's how we're going to label uh the eight regions in this spin diagram all right that's it for this week for extra practice think about how you could create a venn diagram with three sets so that every member of your family would be in a different region
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https://courses.grainger.illinois.edu/ece205/sp2021/notes/Lecture7-8.pdf
Lecture 7 - 8: Circuit Analysis - KCL, Node Voltage Analysis Learning Objectives: 1. Define Kirchhoff's current law (KCL) 2. Compute voltages in simple circuits using KCL 3. Use node voltage analysis method to compute node voltages 4. Derive current division formula and analyze the limitations of of current divider Last few lectures we used loop analysis technique to determine currents in a circuit. Loop analysis technique is based on the Kirchhoff's voltage law. Node voltage analysis, based on Kirchhoff's Current Law (KCL), is another procedural method that can be used to determine the voltage at a various nodes in a circuit. Kirchhoff's Current Law (KCL ) 1. KCL states that the algebraic sum of currents entering or exiting a node is zero. As we discussed earlier a node is a point in a circuit where two or more elements are connected together. Consider some circuits shown in Fig. 1.1 below, the algebraic sum of currents in each case is also shown. As shown in this examples we take currents coming out of a node as positive. Note that the actual current might be coming into the node but that will not affect our computations. Fig. 1.1(a) Fig. 1.1(b) െ𝐢𝟏൅𝐢𝟐െ𝐢𝟑ൌ𝟎 െ𝐢𝟏െ𝐢𝟐൅𝐢𝟑ൌ𝟎 Fig 1.1. Illustration of KCL Notation: The currents leaving a node are taken as positive. Node voltage analysis 2. We will now apply KCL to larger circuits with many nodes and compute voltages as different nodes in the circuit. As mentioned earlier this procedure is called node voltage analysis. Thus, loop analysis method discussed earlier is based on KVL and node voltage analysis is based on KCL. It should be noted that since in Example 1.1: Apply KCL to node 1 shown in circuit below and compute current 𝐢𝟏. Solution: െ𝐢𝟏൅𝟏െ𝟐ൌ𝟎 ⇒𝐢𝟏ൌെ𝟏 𝐦𝐀 Note that in example 1.1, current 𝐢𝟏came out negative. This just means that the actual current leaving node 1. We will see in further examples that assuming a certain direction of current does not constrain us in any way. Lecture 7-8 Page 1 node voltage analysis we are solving for the voltage at a node, we must identify a ground node as a reference point. For most of the problems we discuss in this class, the ground node will be clearly specified. Node voltage analysis involves the following steps: Identify the nodes in a circuit. • For each node assume that the currents are leaving the node. An incorrect assumption will still yield a correct voltage as long as KCL is correctly applied. • Set up node equations. • Solve node equations to obtain node voltages. • Example 2.1: Compute voltages 𝐕𝟏, 𝐕𝟐, 𝐕𝟑in the circuit shown below. Solution: By inspection: 𝐕𝟑ൌ𝟑 𝐕 Node 1 (KCL): 𝐢𝟏𝟐൅𝐢𝟏𝐠ൌ𝟎 By Ohm's Law: 𝐢𝟏𝟐ൌ 𝐕𝟏𝟐 𝟐𝐤𝛀 ⎯⎯⎯ൌ 𝐕𝟏ି𝐕𝟐 𝟐 ⎯⎯⎯⎯⎯ By Inspection: 𝐢𝟏𝐠ൌെ𝟐𝐦𝑨 Hence Node 1 equation can be written as, Node 1: 𝐕𝟏ି𝐕𝟐 𝟐 ⎯⎯⎯⎯⎯ െ𝟐ൌ𝟎⇒𝐕𝟏െ𝐕𝟐ൌ𝟒 (1) Node 2: 𝐢𝟐𝟏൅𝐢𝟐𝐠൅𝐢𝟐𝟑ൌ𝟎 Ohms' Law: െ𝟐൅ 𝐕𝟐ି𝟎 𝟒 ⎯⎯⎯⎯൅ 𝐕𝟐ି𝐕𝟑 𝟐 ⎯⎯⎯⎯⎯ ൌ𝟎 𝟓𝐕𝟐െ𝟏𝟐ൌ𝟖⇒𝐕𝟐ൌ𝟒 𝐕 From (1), 𝐕𝟏ൌ𝟖𝑽 Example 2.2: Find voltages 𝐕𝟏, 𝐕𝟐, 𝐕𝟑in the circuit shown below. Solution: Node 1 (KCL): 𝐢𝟏𝟐൅𝐢𝟏𝐠ൌ𝟎 By Ohm's Law: 𝐢𝟏𝟐ൌ 𝐕𝟏𝟐 𝟐𝛀 ⎯⎯⎯ൌ 𝐕𝟏ି𝐕𝟐 𝟐 ⎯⎯⎯⎯⎯ By Inspection: 𝐢𝟏𝐠ൌ𝟑𝑨 Node 1: 𝐕𝟏ି𝐕𝟐 𝟐 ⎯⎯⎯⎯⎯ ൅𝟑ൌ𝟎⇒െ𝐕𝟏൅𝐕𝟐ൌ𝟔 (1) Lecture 7-8 Page 2 Node 2 (KCL): 𝐢𝟐𝟏൅𝐢𝟐𝐠൅𝐢𝟐𝟑ൌ𝟎 By Ohm's Law: 𝐢𝟐𝟏ൌ 𝐕𝟐𝟏 𝟐𝛀 ⎯⎯⎯ൌ 𝐕𝟐ି𝐕𝟏 𝟐 ⎯⎯⎯⎯⎯ 𝐢𝟐𝟑ൌ 𝐕𝟐𝟑 𝟑𝛀 ⎯⎯⎯ൌ 𝐕𝟐ି𝐕𝟑 𝟑 ⎯⎯⎯⎯⎯ By Inspection: 𝐢𝟐𝐠ൌെ𝟏𝑨 Node 2: 𝐕𝟐ି𝐕𝟏 𝟐 ⎯⎯⎯⎯⎯ ൅ 𝐕𝟐ି𝐕𝟑 𝟑 ⎯⎯⎯⎯⎯ െ𝟏ൌ𝟎⇒െ𝟑𝐕𝟏൅𝟒𝐕𝟐ൌ𝟔 (2) Solving (1) and (2), ൤െ𝟏 𝟏 െ𝟑 𝟒൨൤𝐕𝟏 𝐕𝟐൨ൌ൤𝟔 𝟔൨⇒൤𝐕𝟏 𝐕𝟐൨ൌ൤െ𝟏 𝟏 െ𝟑 𝟒൨ ି𝟏 ቂ𝟏𝟎 𝟐ቃ ൤𝐕𝟏 𝐕𝟐൨ൌቂെ𝟏𝟖 െ𝟏𝟐ቃ 𝐕 Example 2.3: Compute voltages 𝐕𝟏, 𝐕𝟐, 𝐕𝟑in the circuit shown below. Solution: Solved in class as part of worksheet. Lecture 7-8 Page 3 Supernodes 3. Earlier we had discussed the concept of superloops or "bigger loops". This concept came up when we had a current source in a branch that was common to two loops. The concept of supernodes is similar to superloops. We will use the idea of supernodes when there are multiple voltage sources in a circuit and some of them are not referenced to the ground in the circuit. We will see that similar to superloops, we will form a "bigger node" and then apply KCL to all currents coming out of the "bigger node". Let us look at the following example to get familiar with supernodes. Example 3.1: Compute voltages 𝐕𝟏, 𝐕𝟐, 𝐕𝟑in the circuit shown below. Solution: By inspection: 𝐕𝟏ൌ𝟏𝟐 𝐕 Supernode Equation (KCL): Look at all the currents coming out of the circled supernode. 𝐢𝟐𝟏൅𝐢𝟐𝐠൅𝐢𝟑𝐠𝟏 ൅𝐢𝟑𝐠𝟐ൌ𝟎 𝐕𝟐െ𝟏𝟐 𝟐 ⎯⎯⎯⎯⎯⎯⎯൅𝐕𝟐 𝟐 ⎯⎯൅𝐕𝟑 𝟏 ⎯⎯െ𝟒ൌ𝟎 𝐕𝟐൅𝐕𝟑ൌ𝟏𝟎 Voltage relationships at supernode: 𝐕𝟐െ𝐕𝟑ൌ𝟐𝟒 Solving ሺ1ሻ and ሺ2ሻ, ቂ𝟏 𝟏 𝟏 െ𝟏ቃ൤𝐕𝟐 𝐕𝟑൨ൌቂ𝟏𝟎 𝟐𝟒ቃ⇒൤𝐕𝟐 𝐕𝟑൨ൌቂ𝟏 𝟏 𝟏 െ𝟏ቃ ି𝟏 ቂ𝟏𝟎 𝟐𝟒ቃ ൤𝐕𝟏 𝐕𝟐൨ൌቂ𝟏𝟕 െ𝟕ቃ 𝐕 Supernode Example 3.2: Compute voltages 𝐕𝟏, 𝐕𝟐, 𝐕𝟑, 𝐕𝟒 in the circuit shown below. Solution: By inspection: 𝐕𝟑ൌ𝐕𝟒ൌ𝟖𝐕 Supernode Equation (KCL): 𝑖ଵ௚൅𝑖ଶଷൌ0 𝑉 ଵെ0 1 ⎯⎯⎯⎯⎯⎯ ൅𝑉 ଶെ𝑉 ଷ 3 ⎯⎯⎯⎯⎯⎯ൌ0 ⇒3𝑉 ଵ൅𝑉 ଶൌ8 (1) Lecture 7-8 Page 4 Voltage relationships at supernode: െ𝐕𝟏൅𝐕𝟐ൌ𝟒 (2) Solving (1) and (2), ቂ𝟑 𝟏 െ𝟏 𝟏ቃ൤𝐕𝟏 𝐕𝟐൨ൌቂ𝟖 𝟒ቃ⇒൤𝐕𝟏 𝐕𝟐൨ൌቂ𝟑 𝟏 െ𝟏 𝟏ቃ ି𝟏 ቂ𝟖 𝟒ቃ ⇒ ൤𝐕𝟏 𝐕𝟐൨ൌ൤𝟏 𝟓൨ 𝐕 Alternate method: When we first look at example 3.2, it seems to be a supernode problem. As we just saw applying the supernode node method to the problem does work! At the same time the problem can be simplified. This is shown in Fig. 3.1 below. As shown in Fig. 3.1 (a), the highlighted elements are series connected and hence their order can be interchanged. Fig. 3.1 (b) shows the circuit after rearranging the elements. The problem can be solved using node voltage method by setting up one equation for node 2. Lecture 7-8 Page 5
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https://www.historyofcreativity.com/mid80/renaissance-literature
Renaissance Literature Close × Close ###### The History of Creativity Explore History Creators Movements Skills Works About Us Contact Us Clear History Creators Movements Skills Works Renaissance Literature ; "Renaissance Literature") Movement:Renaissance Literature Dates:c. 1400 - c. 1500 Historical Context Renaissance literature refers to European literature which was influenced by the intellectual and cultural tendencies associated with the Renaissance. The literature of the Renaissance was written within the general movement of the Renaissance, which arose in 14th-century Italy and continued until the 16th century while being diffused into the rest of the western world. It is characterized by the adoption of a humanist philosophy and the recovery of the classical Antiquity. It benefited from the spread of printing in the latter part of the 15th century. Overview For the writers of the Renaissance, Greco-Roman inspiration was shown both in the themes of their writing and in the literary forms they used. The world was considered from an anthropocentric perspective. Platonic ideas were revived and put to the service of Christianity. The search for pleasures of the senses and a critical and rational spirit completed the ideological panorama of the period. New literary genres such as the essay (Montaigne) and new metrical forms such as the Spenserian stanza made their appearance. The impact of the Renaissance varied across the continent; countries that were predominantly Catholic or Protestant experienced the Renaissance differently. Areas where the Eastern Orthodox Church was culturally dominant, as well as those areas of Europe under Islamic rule, were more or less outside its influence. The period focused on self-actualization and one's ability to accept what is going on in one's life. The earliest Renaissance literature appeared in Italy in the 14th century; Petrarch, Machiavelli, and Ariosto are notable examples of Italian Renaissance writers. From Italy the influence of the Renaissance spread at different times to other countries and continued to spread around Europe through the 17th century. The English Renaissance and the Renaissance in Scotland date from the late 15th century to the early 17th century. In northern Europe, the scholarly writings of Erasmus, the plays of William Shakespeare, the poems of Edmund Spenser and the writings of Sir Philip Sidney may be considered Renaissance in character. The development of the printing press (using movable type) by Johannes Gutenberg in the 1440 encouraged authors to write in their local vernacular instead of Greek or Latin classical languages, thus widening the reading audience and promoting the spread of Renaissance ideas. See also Allegory in Renaissance literature British literature#The Renaissance Elizabethan literature English Renaissance theatre Renaissance in Croatia Dutch Renaissance and Golden Age literature French Renaissance literature German literature#German Renaissance and Reformation Italian Renaissance literature Polish Renaissance literature Portuguese Renaissance and Portuguese literature#First classical phase: The Renaissance Scottish Renaissance literature Spanish Renaissance literature Swedish literature#Renaissance literature Literature by century 15th century in literature 16th century in literature Poetry by century 15th century in poetry 16th century in poetry External links General Resources in the Renaissance & 17th Century Content provided by Wikipedia Creators Giovanni Boccaccio 1313 - 1375 1353 - The Decameron Renaissance Literature Italian Geoffrey Chaucer 1343 - 1400 c. 1400 - The Canterbury Tales Medieval Literature English Niccolò Machiavelli 1469 - 1527 1532 - The Prince Renaissance Literature Italian Thomas More 1478 - 1535 1516 - Utopia Renaissance Literature English Francesco Petrarca 1304 - 1374 1374 - Triumphs Renaissance Literature Italian François Rabelais c. 1494 - 1553 c. 1564 - Gargantua and Pantagruel Renaissance Literature French Giorgio Vasari 1511 - 1574 1568 - Lives of the Most Excellent Painters, Sculptors, and Architects Renaissance Literature Italian Our Mission The History of Creativity is a visual encyclopaedia that allows you to time travel to any time and place in the past or present. About UsContact UsPrivacy © 2025 The History of Creativity
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https://www.quora.com/What-are-the-typical-values-of-strain-rates-of-steel-tensile-testing-the-auto-makers-are-interested-in-for-crash-worthiness
What are the typical values of strain rates of steel tensile testing the auto makers are interested in for crash worthiness? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Engineering Crashworthiness Steel Strain Rate Automakers Tensile Testing Material Science and Tech... Automotive Engineering Mechanical Engineering 5 What are the typical values of strain rates of steel tensile testing the auto makers are interested in for crash worthiness? All related (32) Sort Recommended Assistant Bot · 1y In the context of automotive crashworthiness, the strain rates of steel during tensile testing are typically in the range of 0.001 to 1 s⁻¹. However, for crash testing specifically, higher strain rates are often of interest, as they better simulate the conditions experienced during a collision. Typical strain rates for crash testing can range from 1 to 1000 s⁻¹. These higher rates account for the dynamic loading conditions that materials experience in real-world crash scenarios. The specific strain rate used can depend on the particular design and safety requirements of the vehicle, as well as Continue Reading In the context of automotive crashworthiness, the strain rates of steel during tensile testing are typically in the range of 0.001 to 1 s⁻¹. However, for crash testing specifically, higher strain rates are often of interest, as they better simulate the conditions experienced during a collision. Typical strain rates for crash testing can range from 1 to 1000 s⁻¹. These higher rates account for the dynamic loading conditions that materials experience in real-world crash scenarios. The specific strain rate used can depend on the particular design and safety requirements of the vehicle, as well as the type of steel being tested (e.g., high-strength steels, advanced high-strength steels). Understanding how steel behaves under these varying strain rates is crucial for automakers to optimize vehicle structures for safety and performance during crashes. Upvote · Related questions More answers below Why does the tensile test steel specimen breaks at 45 degrees? What is the reason for a higher yield point in a high-strain rate tensile test as compared to low-strain rate tensile tensile testing? What is the importance of gauge length in the tensile test of steel? At what point during a tensile test does the calculation of true stress and strain become invalid? What do you mean by tensile strain at break? Anindya Das materials engineer · Author has 142 answers and 742.1K answer views ·6y In general, as per the ASTM E8(M) standard for tensile testing of metallic materials, the strain rate relevant for knowing the material’s tensile properties should be within 0.0001 to 0.01 /s (engineering strain rate). For a comparison of the tensile properties of different materials or for different conditions, the user must perform the tensile tests at a fixed strain rate, which can be between 0.0001 to 0.01 /s. Generally, it is convenient and widely followed to select a strain rate of 0.001 /s for the tensile tests. Now ASTM E8(M) prescribes the strain rates which are quasi-static in nature. Continue Reading In general, as per the ASTM E8(M) standard for tensile testing of metallic materials, the strain rate relevant for knowing the material’s tensile properties should be within 0.0001 to 0.01 /s (engineering strain rate). For a comparison of the tensile properties of different materials or for different conditions, the user must perform the tensile tests at a fixed strain rate, which can be between 0.0001 to 0.01 /s. Generally, it is convenient and widely followed to select a strain rate of 0.001 /s for the tensile tests. Now ASTM E8(M) prescribes the strain rates which are quasi-static in nature. You can find the categories of ranges of strain rates in ASM handbook. In general, the strain rates that we apply to evaluate tensile properties are quasi-static in nature (higher than creep strain rates but lower than intermediate or high strain rates). As you said ‘crashworthiness’ which involves deformation similar to crash conditions. A crash is an impact kind of situation, where the local strain rates are very high. It has been found that during an automobile crash, the local strain rates can be as high as 100 to 1000 /s. This range of strain rates (100 to 1000 /s) comes under intermediate/high strain rate range. In this range, the standard ASTM E8(M) is no longer valid. Recently ISO has come up with a standard for tensile testing at this regime of strain rates. To sum up, automobile makers and other material industries try to evaluate the properties of the steels at this range (100 to 1000 /s) in order to assess the crashworthiness of the automobile components. P.S. I did my PhD exactly in this field, ‘High strain rate Deformation’. Upvote · 9 4 Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) ·Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. Continue Reading This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Upvote · 999 485 999 103 99 17 Annamalai Former Chief Mechanical Engineer , Engg Consultant · Author has 5.4K answers and 3.4M answer views ·4y Related What is the suggested stress strain curve for high tensile steel? Mild steel is normally have 45000 to 55000 psi Tensile Stress. High tensile steel will have 60000 psi and above. Some are low alloy steels. There are thousands of high tensile strength steels. Each material , having different treatments(cast, rolling/pressing & heat treatment), will vary in the shape of Stress strain curve. So, it is better to test a sample in the same batch of product. Sample Stress-strain curve for some of the High tensile stress is given here. Continue Reading Mild steel is normally have 45000 to 55000 psi Tensile Stress. High tensile steel will have 60000 psi and above. Some are low alloy steels. There are thousands of high tensile strength steels. Each material , having different treatments(cast, rolling/pressing & heat treatment), will vary in the shape of Stress strain curve. So, it is better to test a sample in the same batch of product. Sample Stress-strain curve for some of the High tensile stress is given here. Upvote · 9 1 David Barry Design Engineer (2012–present) · Author has 130 answers and 310.9K answer views ·8y Related What is a 'coupon' in steel coupon testing for a tensile strength test? The coupon is literally whats in the video. It’s a specially milled steel sample that necks down to an exact diameter prior to tensile testing. Continue Reading The coupon is literally whats in the video. It’s a specially milled steel sample that necks down to an exact diameter prior to tensile testing. Upvote · 9 8 9 3 Related questions More answers below How do you increase the tensile strength of steel? How do you calculate tensile stress and strain for steel bars? What is the tensile strength of steel? What are the charcteristics of high tensile steel? What is the maximum allowable tensile strength for steel bars? Nand Kishore Gajarajan Influential Automotive Engineer,Impact BioMechanics Advocate · Author has 424 answers and 282.8K answer views ·2y Related Why are true stress and true strain, rate dependent curves used in any vehicle crash analysis? True stress and true strain curves are used in forming and vehicle crash simulations (LS-Dyna, etc.) because they give a better interpretation of how a material performs when it is being deformed. The stress-strain curve provides engineers with a list of important considerations needed for structural stiffness, durability and bending, including mechanical properties such as strength, toughness, el Continue Reading True stress and true strain curves are used in forming and vehicle crash simulations (LS-Dyna, etc.) because they give a better interpretation of how a material performs when it is being deformed. The stress-strain curve provides engineers with a list of important considerations needed for structural stiffness, durability and bending, including mechanical properties such as strength, toughness, elasticity, yield point, strain energy, resilience, and elongation during load. To obtain trustworthy vehicle crash CAE simulation results, precise material behaviors at i... Upvote · Promoted by The Hartford The Hartford We help protect over 1 million small businesses ·Updated Sep 19 What is small business insurance? Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickl Continue Reading Small business insurance is a comprehensive type of coverage designed to help protect small businesses from various risks and liabilities. It encompasses a range of policies based on the different aspects of a business’s operations, allowing owners to focus on growth and success. The primary purpose of small business insurance is to help safeguard a business’s financial health. It acts as a safety net, helping to mitigate financial losses that could arise from the unexpected, such as property damage, lawsuits, or employee injuries. For small business owners, it’s important for recovering quickly and maintaining operations. Choosing the right insurance for your small business involves assessing your unique needs and consulting with an advisor to pick from comprehensive policy options. With over 200 years of experience and more than 1 million small business owners served, The Hartford is dedicated to providing personalized solutions that help you focus on growth and success. Learn about our coverage options! Upvote · 999 555 9 1 9 3 Ankit Khurana Lives in Canada (2019–present) · Author has 107 answers and 116.9K answer views ·8y Related What is the characteristic of tensile stress-strain curve for high-strength metal with minimal ductility? Tensile stress strain curves shows the variation of stress and strain when a body is subjected​to tensile loading. Minimal ductility of the material means it doesn't have the tendency to elongate much and the material fails by sudden fracture. Such materials are brittle in nature and their characteristic curve has its highest point as ultimate tensile stress in stress-strain curve as they don't show much of the plastic deformation. Low ductility doesn't mean low strength. For example : ceramic a brittle material with high strength. Image source : Google Continue Reading Tensile stress strain curves shows the variation of stress and strain when a body is subjected​to tensile loading. Minimal ductility of the material means it doesn't have the tendency to elongate much and the material fails by sudden fracture. Such materials are brittle in nature and their characteristic curve has its highest point as ultimate tensile stress in stress-strain curve as they don't show much of the plastic deformation. Low ductility doesn't mean low strength. For example : ceramic a brittle material with high strength. Image source : Google Upvote · 9 1 Gene Couch Former Union Glazier for 35 Years · Author has 4.6K answers and 1.5M answer views ·1y Related Why is the steel mechanical test performed always a tensile test and not a compression test, bend test, or flexural test? Is it because cracks propagate easily in tensile? FYI, you’re question is based of a false premise. The FACT is, while tensile strength may be the most COMMON test asked for, it’s certainly not the ONLY test for steel. Also, understanding how steel is made gives clues to what tests are important. FYI, steel is a mixture and is made into ingots and then rolled to desired thickness and width. ANy flaws in the steel- gases, foreign objects, folds (laminations), etc. are simply rolled out, in the steel, so there’re always imperfections and it is the luck of the draw if the steel you buy has any inclusions or not,. IN FACT, 8 different tests are ava Continue Reading FYI, you’re question is based of a false premise. The FACT is, while tensile strength may be the most COMMON test asked for, it’s certainly not the ONLY test for steel. Also, understanding how steel is made gives clues to what tests are important. FYI, steel is a mixture and is made into ingots and then rolled to desired thickness and width. ANy flaws in the steel- gases, foreign objects, folds (laminations), etc. are simply rolled out, in the steel, so there’re always imperfections and it is the luck of the draw if the steel you buy has any inclusions or not,. IN FACT, 8 different tests are available on steel, for different applications , one or all may be asked for. I know that in the Aerospace industry, ALL of them are asked for. Basically, While strength is the amount of force material can withstand before deforming or failing, and toughness is the ability to resist fracturing under force, hardness is the ability to withstand friction and abrasion. FYI, the 8 are: 1) of course, Tensile- the “pull test”; 2) Yield- imagine a bookshelf. When the shelf bows due to too many heavy books, it has reached or passed it’s “yield point”- point to where the steel doesn’t return to it’s original shape; 3) Elongation or “Ductile”- the difference between tensile and yield. The greater the elongation, the more ductility a material has; 4) Sheer- the force where steel shears- typically used on fasteners, but in some applications, sheets, rods, plates, too; 5) Ultrasonic testing; 6) Xray. 7) Charpy test- for toughness- a heavy headed pendulum is swung at a sheet of steel, until it bends or snaps- that point is measured. 8) Brinell hardness test- For example, a diamond is very difficult to scratch (high hardness) but is relatively easy to shatter (low toughness). Mild A36 steel has a BHN of 133, whereas an abrasion resistant grade of steel has a BHN greater than 330. IN my trade- Glazing, the sheer strength was most important, as our frames and glass had to withstand the sheer force of tornadoes and hurricanes. 2nd important is toughness - as the most and worst damage in a tornado or hurricane (BTW , a hurricane can spawn 100s of tornadoes) is flying debris. Upvote · 9 2 9 1 Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder ·Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. 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WithKashKick, you can actually get paid to play. No weird surveys, no endless ads, just real money for playing games you’d probably be playing anyway. Some people are even making over $1,000 a month just doing this! Oh, and here’s a little pro tip: If you wanna cash out even faster, spending $2 on an in-app purchase to skip levels can help you hit your first $50+ payout way quicker. Once you’ve got $10, you can cash out instantly through PayPal—no waiting around, just straight-up money in your account. Seriously, you’re already playing—might as well make some money while you’re at it.Sign up for KashKick and start earning now! Upvote · 20K 20K 1.6K 1.6K 999 446 Mark Barton Gravitational Wave Researcher at University of Glasgow (2019–present) · Author has 18.3K answers and 24.6M answer views ·Aug 9 Related Do the shear, torsional, and flexural strengths of steels have values comparable to the ultimate tensile strength? How close is comparable? There’s a bunch of different elastic moduli that are relevant depending on how the material is constrained when you’re stressing it, and from any two you can calculate all the others. The most common basic two are the Young’s modulus (for a wire under tension but free to contract in the other two directions as you stretch it), and the shear modulus (for pure shear or torsion). Then for metals, the tensile strength is often around one hundredth of the Young’s modulus, and the shear modulus is a tenth or so. But there are exceptions: cast iron is really good for Young’s m Continue Reading How close is comparable? There’s a bunch of different elastic moduli that are relevant depending on how the material is constrained when you’re stressing it, and from any two you can calculate all the others. The most common basic two are the Young’s modulus (for a wire under tension but free to contract in the other two directions as you stretch it), and the shear modulus (for pure shear or torsion). Then for metals, the tensile strength is often around one hundredth of the Young’s modulus, and the shear modulus is a tenth or so. But there are exceptions: cast iron is really good for Young’s modulus but completely crappy for tensile strength. (So it’s a cost effective choice for the bases of heavy machines, but not much else.) And for rubbery materials the shear modulus is much lower, like two orders of magnitude or more below the Young’s modulus. Upvote · 9 3 Philip Howie postdoctoral researcher in materials science; academic · Author has 3K answers and 13M answer views ·6y Related What do you mean by tensile strain at break? Tensile means in tension. While being pulled. Image source. Strain is change in length divided by original length: ε=Δ l l 0 ε=Δ l l 0 At break means at the point where the sample fractures into two. Image source. So tensile strain at break is the fractional change in length at the point where a sample being pulled apart finally fractures. On this stress-strain curve, it’s the strain at that point right at the far end, marked with a big X. Image source. Continue Reading Tensile means in tension. While being pulled. Image source. Strain is change in length divided by original length: ε=Δ l l 0 ε=Δ l l 0 At break means at the point where the sample fractures into two. Image source. So tensile strain at break is the fractional change in length at the point where a sample being pulled apart finally fractures. On this stress-strain curve, it’s the strain at that point right at the far end, marked with a big X. Image source. Upvote · 9 5 9 1 Promoted by US Auto Insurance Now US Auto Insurance Now Helping Drivers Find Great Car Insurance Deals ·Sep 23 How do I find a cheap car insurance if I am a new young driver? Do I need to call each insurance and ask? When you're a new driver, that first insurance quote can feel like a punch to the gut. The rates often seem astronomical, and it's easy to feel like you're being priced out of something that's a necessity. This happens because insurance companies view new drivers as high-risk, given their lack of an established driving history. It's frustrating, but with the right strategy, you can find a policy that's both fair and affordable. The biggest mistake new drivers make is accepting the first quote they get. Not all insurance companies calculate risk the same way. What one company considers a high-ri Continue Reading When you're a new driver, that first insurance quote can feel like a punch to the gut. The rates often seem astronomical, and it's easy to feel like you're being priced out of something that's a necessity. This happens because insurance companies view new drivers as high-risk, given their lack of an established driving history. It's frustrating, but with the right strategy, you can find a policy that's both fair and affordable. The biggest mistake new drivers make is accepting the first quote they get. Not all insurance companies calculate risk the same way. What one company considers a high-risk factor, another might offer a discount for. Some insurers have specific programs tailored to young drivers, rewarding them for things like good grades, taking a defensive driving course, or even using a telematics device to track safe driving habits. The challenge is finding which companies offer these savings. That's where an online tool like this one becomes your best friend. This platform is specifically designed to help you navigate this complex landscape. Instead of spending hours on the phone or a dozen different websites, you simply enter your information once. From there, it shows you competitive quotes from a wide network of insurers, including those that might be more favorable to new or young drivers. This allows you to easily compare options and find hidden discounts you might have otherwise missed. Using a platform that compares insurance quotes for you is an essential first step in your search for a budget-friendly policy. It empowers you to see all your options in one place, ensuring you don't pay more than you have to for the coverage you need. Upvote · 99 26 9 3 Glen Schult 55+ years of experiance riding, driving and fixing machines. · Author has 579 answers and 1.2M answer views ·2y Related What is the normal value of stress at which failure occurs in steel bolts and nuts used in automobiles? There isn’t a direct answer for this question. There are many different grades of bolts that have different yield to stress values, different shear stress values, and all that also depends on the size of the hardware. Age also is a consideration when it comes to corrosion that effects the strength. You’d need to get deep into the “Machinery’s Handbook” to get the real answer you might be looking for. Materials such as steel, aluminum, brass are all going to have different strengths and different reasons for their uses. Select a size, it’s usage, the material, then you can figure out what grade Continue Reading There isn’t a direct answer for this question. There are many different grades of bolts that have different yield to stress values, different shear stress values, and all that also depends on the size of the hardware. Age also is a consideration when it comes to corrosion that effects the strength. You’d need to get deep into the “Machinery’s Handbook” to get the real answer you might be looking for. Materials such as steel, aluminum, brass are all going to have different strengths and different reasons for their uses. Select a size, it’s usage, the material, then you can figure out what grade of bolt you’ll want to use. Upvote · Srivenkata Subramanian knows a bit of Metallurgy and Materials!! · Author has 130 answers and 316.4K answer views ·9y Related Is there any difference between the toughness value got from tension test and impact tests for mild steel? Yes, there is a difference. The area under the curve of the stress stain plot will indicate the toughness of the material, i.e, higher the area under the curve, larger is the toughness of the part. However, in a tensile testing the forces are biaxial and there is no intial crack or sharp corners introduced. Secondly, the loading is standard progressive load and there is no impact loads. The primary objective of a tensile testing is to measure the load vs deflection and not the toughness. Testing temperature also plays a role in deciding the mechanism of failure. However, in impact resistance t Continue Reading Yes, there is a difference. The area under the curve of the stress stain plot will indicate the toughness of the material, i.e, higher the area under the curve, larger is the toughness of the part. However, in a tensile testing the forces are biaxial and there is no intial crack or sharp corners introduced. Secondly, the loading is standard progressive load and there is no impact loads. The primary objective of a tensile testing is to measure the load vs deflection and not the toughness. Testing temperature also plays a role in deciding the mechanism of failure. However, in impact resistance testing, the loading is short term and we measure the energy for propogation of the crack by finding the energy lost by the impacter on impacting the test material. Lower the value, lower the toughness of the part. The loads are not bixial and the triaxiality at the tip of the crack takes precedence. Here also, testing temperature plays an important role. Impact testing is especially useful in determining the ductile to brittle transition temperatures and type of fracture at different conditions. It is ok to compare the toughness of the material with the same testing method, but it will not be correct to compare the energy for failure in tensile testing and impact testing. Upvote · 9 9 9 2 Joshua Williams B.S. in Manufacturing Engineering&Welding Technology, Weber State University (Graduated 2019) · Author has 1.2K answers and 1.4M answer views ·1y Related Does strain hardening of material increase the tensile strength or does it change the yield strength only? This depends on the metal. But, for most practical examples, cold work will indeed increase the ultimate tensile strength, in addition to the yield strength. However the increase in UTS is less than the increase in YS, and cold working will reduce the total elongation before breakage. Moreover the Strength Ratio of UTS/YS will go down which is sometimes a specified property in ASTM standards for example. It's also worth noting that for heavily cold worked materials, I.E. cold rolled stainless or carbon steel sheets, properties may be significantly anisotropic. Meaning that the yield strength par Continue Reading This depends on the metal. But, for most practical examples, cold work will indeed increase the ultimate tensile strength, in addition to the yield strength. However the increase in UTS is less than the increase in YS, and cold working will reduce the total elongation before breakage. Moreover the Strength Ratio of UTS/YS will go down which is sometimes a specified property in ASTM standards for example. It's also worth noting that for heavily cold worked materials, I.E. cold rolled stainless or carbon steel sheets, properties may be significantly anisotropic. Meaning that the yield strength parallel to the direction of rolling is greater than perpendicular to the rolling direction. This fact can sometimes cause issues in bent or stamped parts. Upvote · 9 2 9 2 David Goodman Author has 1.9K answers and 14.9M answer views ·8y Related What is tensile strength testing? What is tensile strength testing? Tensile strength testing is pulling on 2 ends of a material to measure how much force is required to stretch the material to failure. Tensile strength is how well a material resists stretching and breaking when enlongation force is applied, as opposed to compressive strength where measures a material’s resistance to compression. An example of a material with high tensile strength and low compressive strength is nylon rope. An example of a material with high compressive strength and low tensile strength is a stack of bricks. A human limb bone averages a tensile st Continue Reading What is tensile strength testing? Tensile strength testing is pulling on 2 ends of a material to measure how much force is required to stretch the material to failure. Tensile strength is how well a material resists stretching and breaking when enlongation force is applied, as opposed to compressive strength where measures a material’s resistance to compression. An example of a material with high tensile strength and low compressive strength is nylon rope. An example of a material with high compressive strength and low tensile strength is a stack of bricks. A human limb bone averages a tensile strength of about 130 MPa, while graphene has been tested to 130,000 MPa. Here an photo of a set of testing vices used to apply and measure tensile strength of a material. Upvote · 9 8 9 1 Related questions Why does the tensile test steel specimen breaks at 45 degrees? How do you calculate the tensile strength of steel beams? What is density for high tensile steel bars? What is the allowable tensile stress in steel as per the AISC code? What is the difference between tensile and compressive stress on a bar of steel? How do you increase the tensile strength of steel? How do you calculate tensile stress and strain for steel bars? What is the tensile strength of steel? What are the charcteristics of high tensile steel? What is the maximum allowable tensile strength for steel bars? Can you explain the process of plastic deformation in steel under tensile loading? Why is the tensile strength of steel low? What is the reason for a higher yield point in a high-strain rate tensile test as compared to low-strain rate tensile tensile testing? What is the type of strain developed in the pre-tensioned steel in pre-stressed concrete after cutting of tendons? Is it compressive or tensile strain? Why can an usual construction steel withstand more compression than tensile force (compression vs tensile test)? Related questions Why does the tensile test steel specimen breaks at 45 degrees? What is the reason for a higher yield point in a high-strain rate tensile test as compared to low-strain rate tensile tensile testing? What is the importance of gauge length in the tensile test of steel? At what point during a tensile test does the calculation of true stress and strain become invalid? What do you mean by tensile strain at break? Why can an usual construction steel withstand more compression than tensile force (compression vs tensile test)? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.oxfordinternationalenglish.com/common-english-grammar-mistakes/
Year-Round Groups Online Courses Accommodation Social Programme Prices & Payment Handbooks & Forms English Level Test Virtual Learning Campus Academic Support Student Support FAQs Blog Accreditations & Inspection Reports School Policies Our Team Book for a Student Become an Agent Marketing Portal Morning 18/20 Afternoon 18/20 Semi Intensive 23/26 Intensive 27/30 Super Intensive 36/40 Specialised Electives Business English Electives Evening Business English UK Pathway Programmes Canada University Pathways USA University Pathways Australia Pathways Halifax University College Pathways IELTS Trinity CertTESOL Foreign Youth Canadian Youth London Greenwich Oxford Brighton Edinburgh Vancouver Toronto Halifax New York City Melbourne Book Now Enquire Now Common English Grammar Mistakes and How to Fix Them One of the most common challenges experienced by people who are learning English as a second language is choosing the right word for the right context. The English language is well known for being the language of international communication in the modern world – and wherever you originate from, and whatever native tongue you speak, it’s likely that learning English will be invaluable in both your personal and professional lives. Of course, the English language frequently frustrates new learners with various grammatical hurdles and stumbling blocks. Signing up for our courses at Oxford International English School is a great way to improve your English language skills. Common English Grammar Mistakes 1) Present and Past Tense Present tenses in English are used to talk about the present, the future and to summarise a book, film or play when telling a story in the present tense. There are four present tense forms in the English language. Present Simple:I Work Present Continuous: I am Working Present Perfect: I have worked Present perfect continuous: I have been working Rules: You can use the past tense to talk about events or situations that have finished. You can also use past tense in English to talk about long-standing events and situations that have already happened in the past. For example: When I was a young child, I lived in the countryside. Here are some frequently used examples of verbs in past simple: are, was, were and went. 2) How To Avoid the Overuse of Adverbs Adverbs are a varied class of words that work in many different ways to express many different kinds of meaning.  This can make adverbs a useful word group. You should, however, avoid overusing these words to describe actions and events.  The most commonly overused adverbs are manner adverbs, this particular type of adverb modifies the verb. For example: Emily Scott shook her head vigorously. He was in a good mood now, smiling broadly as he grabbed his mug of tea. A common issue in story writing occurs when you rely too heavily on manner adverbs in your stories. For example: The curtain opened quickly, and Ben came slowly into the room. He saw Emma looking flirtatiously with Jack and walked over to her aggressively. ‘Why are you here?’ he screamed angrily. Here is the same extract with the manner verbs highlighted: The curtain opened quickly, and Ben came slowly into the room. He saw Emma looking flirtatiously with Jack and walked over to her aggressively. ‘Why are you here?’ he screamed angrily. The correct use of adverbs is to show not tell the reader what is happening in the story. 3) Your/You’re These words are also troublesome homophones that cause many problems. Rules: “Your” indicates a possession – and defines that something belongs to you. “You’re” is short for “You are”. Here is how not to use these words: Yourbeautiful. Do you know when your going? Can I have you’re coat? How to get it right: You’re beautiful. Do you know when you’re going? Can I have your coat? 4) Misplacing Apostrophes You find apostrophes a little tricky, but once you follow the rules, it will become easy. Putting an apostrophe in the wrong place is a common mistake. Rules: Apostrophes indicate something belongs to something or is owned by someone else. To show that something belongs to one person, place the apostrophe before the letter ‘S.’ For example – “The girl’s sheep”. To show that something belongs to more than one person, you need to place the apostrophe after the letter ‘S’. For example – “The girls’ sheep”. Apostrophes are also used in contracted words such as “Can’t”to indicate that the ‘O’ is missing from “Cannot.” Apostrophes should never be used to make a word plural. 5) There / Their /They’re You may find that these pesky homophones, a little bit of a headache. Rules: Use “There” to refer to a place that isn’t here, for example, “Over there.” Use “Their” to refer to how owns something – showing that something belongs to that person. Use “They’re” is a shortened version of “They are”. Here is how not to use these words: Their going to be here soon. We should contact they’refriend. Can we use there house? They’re is is an argument that says. Here is how you use these words correctly: They’re going to be here soon. We should contacttheir friend. Can we use their house? There is an argument that says. 6) Confusing similar spellings and words The English language is quite rich in words which sound similar, or are spelled similarly, but which have different meanings and need to be used in different contexts. Perhaps the most common stumbling block experienced by people who are learning English as a second language is making sure to use the right word in the right context, rather than a similar but improper one. Our English language courses are designed to address this issue, helping learners to confidently choose the correct words in various contexts. The only way to avoid this issue is to learn which words fit in which context, on a case-by-case basis. Here are some words people often mix up: “Two,” “too,” and “to” “Here” and “hear” “Your” and “you’re” “Weather” and “whether” 7) Using incomplete comparisons Many words in the English language imply a comparison – and using them without “completing the comparison” is a common grammatical mistake. Here’s an example of an incomplete comparison: “It was much hotter today.” To make this example grammatically correct, you would need to complete this comparison. Here’s one way you could do that: “It was much hotter today than yesterday.” 8) Getting adjectives and adverbs confused Confusing your adjectives and adverbs often results in speech or writing that comes off as very informal, and even uneducated – and it’s a great way of infuriating many English teachers. Often, you’ll notice this issue happening with words that end in “-ly.” Here are a couple of grammatically incorrect examples: “It was a real nice day today.” “I ran quick to the bus stop.” And here’s how these two examples would look if they were made grammatically correct: “It was a really nice day today.” “I ran quickly to the bus stop.” 9) Misplacing your modifiers Language would be pretty dull without words to add a bit of extra flavour to sentences and descriptive speech. This is exactly where modifiers come in. With modifiers, “the tiger” can become “the fearsome tiger,” “the sunrise,” can become “the beautiful sunrise,” and so on. The issue is that these modifiers need to be placed very close to the word they’re modifying, or else the meaning falls apart. “Misplacing your modifiers” means that you are putting these modifiers too far away from the terms they are meant to be modifying, in your sentence. The result is confusion. In fact, misplaced modifiers can even completely change the meaning of your sentence in unintended ways. Here’s an example of a misplaced modifier: “He almost walked for the entire day.” And here’s how this example would read with the modifier in the right place: “He walked for almost the entire day.” In the example with the misplaced modifier, it is not clear if he “crawled”, “ran slowly”, or simply “thought about walking” for the entire day. In the correct example, the meaning is clear. 10) Falling into pronoun disagreement A common grammatical mistake for English learners is for their pronouns and nouns to disagree, when dealing with singular and plural examples. The straightforward rule is that singular pronouns must go with singular nouns, and plural pronouns must go with plural nouns. So, for example: “Every boy must sign in when they arrive” is incorrect. “Boy” is singular, and “they” is plural. The correct phrasing here would be: “Every boy must sign in when he arrives.” Grammar and punctuation are essential in the English language and gaining confidence in how to avoid any grammatical errors is a valuable part of your learning journey.  You should practice developing your grammar daily; it will help you to become a confident writer with a firm grasp on the English language. One of the most common challenges experienced by people who are learning English as a second language is choosing the right word for the right context.
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https://www.youtube.com/watch?v=wsT2f3WlOog
Class 10 | Coordinate Geometry 03 | Slope | Equation of line | Pearson IIT Foundation AKBAR CLASSES 35200 subscribers 52 likes Description 1720 views Posted: 12 Jun 2021 📗 Class 10 Pearson IIT Foundation Series Mathematics ☑ Coordinate Geometry ➡ Slope ➡ Equations of line in different forms Contact us: ✆ 7366863696 ✉ akbarclasses@gmail.com Support us : Account Number - 33307235385 IFSC- SBIN0000117 MD ZAREEF AKBAR FOLLOW US ON: Instagram: Twitter: Facebook: YouTube: AkbarClasses #PearsonIITFoundation 11 comments Transcript: हुआ है हेलो हेलो बच्चों कैसे हैं आप लोग आशा करता हूं कि आप लोग अच्छे होंगे हमने शुरू किया था पोज मिडल क्लास 10th पीएसएनडी फाऊंडेशन सीरीज और हम यह भी कहा था कि जो सीबीएसई के बच्चे हैं जो सीबीएसई एनसीआरटी को फॉलो करते हैं उनके लिए भी यह फायदेमंद हैं आज के इस लेक्चर में अगर आप देखें एनसीईआरटी के सिलेबस को उठाकर तो उसमें यह सारी बातें नहीं है लेकिन आप क्लास टेंथ में तो इतनी सारी बातें आपको जाननी चाहिए आपको फायदा करेगा तो शुरू करते हैं आज क्या-क्या पढ़ने वाले हैं ठीक है आज हम लोग जो पढ़ने जा रहे हैं सबसे पहले जो हम लोग डिस्कस करेंगे वह क्या डिस्कस करेंगे स्ट्रेट लाइंस के बारे में क्या डिस्कस करेंगे इस स्ट्रेट लाइंस हमारा टॉपिक के पहले स्ट्रेट लाइंस अच्छा ठीक है स्प्रेड नाइंथ में सबसे पहले बात करेंगे इंक्लिनेशन ऑफ लाइन ऐसे मैं चाहूं तो इसे अपने हिसाब से पढ़ाओ लेकिन बुक में जैसे-जैसे है वह मैं थ्योरी को कंप्लीट करवाना चाहता हूं तो सबसे पहला ऐप इंक्लिनेशन आफ फ्लाइंग किस को कहते इंक्लिनेशन आफ फ्लाइंग जैसे एक ही आप जानते हैं यह आपका और टिफिन प्लेन है एक्स एक्स इस वायरस इन थ्री बनता है कोई भी यह लाइन अगर आप कोई भी लाइनें स्वागत है ठीक है वह लाइन पॉजिटिव एक्स के साथ जो एंटी क्लॉकवाइज एंग्री बनाता है इसको कहते हैं हम लोग एंगल आफ इंक्लिनेशन यह कहते हैं एक लाइन जो आप ड्रायर करते हैं जो आप बनाते हैं कहीं पर काटे ध्यान प्लेट में पॉजिटिव डायरेक्शन आफ एक्स एक्स इससे एंटी क्लॉकवाइज बगल बनता है ऐसे भी हो सकता है क्लॉकवाइज एंटी क्लॉक वाइज वह उसका नाम है एंगल आफ पेंशन सिंपल के बाद समझ गए होंगे वॉटर्स एनवेलपिंग एलाइनेशन कोई भी लाइन हो भिलाईनगर देखें यह लाइन एक्स बढ़ रहा है और वह भी वर्ड एक्स के साथ हवाई अड्डे एक बढ़ेगा वहीं बढ़ेगा या घटेगा घटेगा यह पॉजिटिव मतलब यह नौकरी से कम है यह क्या नाइस लुक आएगा लोगों के बारे में प्रेम करते थे होती है अब एक लाइन ऐसा है यह लाइन पॉजिटिव डायरेक्शन आफ एक्सिस से कैसे आगे बढ़ाया मॉडयूलूस यह भी आपका क्या एंजिला पेंशन पेंशन दो तरह का हो सकता है एक होगा एक्यूट और एक क्या होगा अब ट्यूशन तीसरे तरह का क्या हो सकता हूं बताइए एक हो सकता है कि एक्सेस फॉर में हो या एक्सेस के पैनल हो जाए कि एक हो सकता है मारा ऐसा हो जाए ठीक है जो एक्स के साथ कितना कि बनाया घुमा ही जीरो डिग्री और एक ऐसा हो सकता है जो कितना ही बड़ा डिग्री तो कितने तरह की हमारे निलंबन की अनुशंसा सो सकते हैं एक क्यूट मॉड्यूल्स डिग्री और नौ डिग्री क्या होता है न इंक्लिनेशन पॉजिटिव डायरेक्शन आफ एक्सिस एंटी क्लॉकवाइज नाइन का जंगल होगा उसको करेंगे हम लोग एंड विल ऑफ इंक्लिनेशन उम्मीद करता हूं कि आप समझ गए होंगे समझे बाबू सब यह हुई बात एंगल ऑफ़ इंक्लिनेशन की नेक्स्ट क्या बात होगी बताइए नेक्स्ट की स्लोप और ग्रेडियंट ऑफ लाइन कि क्या जानेंगे इस लोग और ग्रेडिएंट ऑफलाइन सिंपल सिंपल सी बात है ठीक है थे स्नो ख्या कि ग्रेडियंट ए फ्लाइंग ठीक है बहुत इंपोर्टेंट है कैसे सिंपल सा ठीक है यह जो हमने अभी जाना क्या जाना हमने एक लाइन है यह लाइन एक्सेस के साथ जितना भी बनाएगा इसको ढकेंगे थी कि आपका एंगल आफ इंक्लिनेशन और स्लॉट का सिंबल है एम होता है अटेंडेंट आफ थिस टो अटैंड क्या होता है परपेंडिकुलर बटे भेज दिए जाएंगे यदि इसी को हम क्या करेंगे इस लोग क्या कहेंगे लोग कुछ लोग क्या होता है एंजॉयमेंट आफ एंगल आफ इंक्लिनेशन क्या होता है बताइए अटेंडेंट आफ एंड लाफ इन भिलाई नेशन अगर यह का काव्य कितना हो जाए पांच डिग्री कितना हो जाए पांच डिग्री तो क्या हो जाएगा स्लो हो जाएगा 1025 बैंजो किसके बराबर होता वन के बराबर यहीं पर बैठा है तो क्या कर दूं अभी जो मैंने बोला कि आपका स्लो पॉजिटिव अगर इस तरीके का हो तो ऐसे एंगल कितना बनाएगा कर इधर सेंडल मान लेते हैं से पांच डिग्री हो 1st - क्या होगा बताइए फुटपाथ बिक्री यह का 135वीं इधर से अगर प्लस मन होता है दर्द क्या कहा - 153 के लिए तो बेसिकली आपको याद रखना टैन 135 यहां जसलोक है यह क्या होगा - मन ठीक है इधर से पांच विकेट 138 के चैन 130 का वैल्यू - में याद रखना है या आप यह याद रख लें आप मेरे ट्रिगोनोमेट्रिक लैटर देखेंगे तो समझ आ जाएगा टैन 180 डिग्री - थीटा यह होता है माइनर टैंक का मतलब यह 1035 ले सकते हैं 183 - 5310 होने के बाद तेरा क्या होगा - 10.51 ठीक है स्लो पॉजिटिव हो सकता है डिपो सकता है ₹100 हो सकता पॉजिटिव और निगेटिव और तीसरा क्या हो सकता हूं बताइए मैं कुछ लोगों की बात करूं यह क्या-क्या हो सकता है श्लोक 12 बात समझ आ गई पॉजिटिव प्लस बनाया - बनाया और क्या-क्या हो सकता है अगर मैं यह लूं यह आपका ऐसे ही है तो अबे क्या होगा लाइफ जी 0 क्या होता है 0 और अगर ऐसी हो जाए इसका शुरू हो गया ऐसा और अगर यह हो जाए तो अनिल जैन दिल्ली बनारस के साथ तो स्लो क्या हो जाएगा 1093 जो किसके बराबर नॉट डिफरेंट बनाकर पूछ लो क्या हो सकता पॉजिटिव एंड नेगेटिव वि0 का होगा जब एक्सप्रेस के पैनल होगा नॉट डिफाइंड का होगा जब वह एक्सेस के पैनल होगा ठीक है इतनी बात आपको समझ है आ गई होगी ही चलें इस लोकप्रिय होता है 10थ क्या होता है एंगल आफ इंक्लिनेशन को टीका कुछ रिजल्ट्स निकले में जो मैंने पहली डिस्कस कर दिया है पीछे के मेकर्स आप देखेंगे तो आपको समझ में आ जाएगा 126 उसके हिसाब से मैं बोल देता हूं स्लो को हर रोज वाटर लाइन है में निकली सबसे पहला क्या लिखा है दुश्मनो को ऑफ होरिज़ॉन्टल रिज़ोल्यूशन ने बताया कि अच्छा ठीक है तुम क्या करेंगे बताओ कि विष्णु अभिषेक को अगर पहला इसमें लिखूं तो अगर मैं एक्ट्रेस की बात करो किसकी बात करो एक्स वाई इसी एक्सेस में हम है टैक्स के साथ कितना गेट बंद है इस लाइन का दिए रोडवेज स्टाफ एक्स एक्स 800 लोफ एक्सप्रेस 30 इसके पहले भी और सभी एक्स एक्स रे पियाला निर्भय नाम रा जी रोड स्लोप आफ वायरेसेस जीरो क्यों क्योंकि स्लॉट मतलब टाटा कितना जीरो डिग्री कितना होगा सिर्फ सिंपल सी बात दूसरी बात किया करिए इन्होंने इन्होंने यह बात करी है स्लॉट वर्टिकल लाइन मतलब वाला लाइन यह प्रैक्टिस यह वही है वर्टिकल लाइन एक्सप्रेस में नंबर एट यानि बना रहे हैं तो क्या हो जाएगा बताइए लो पॉलीटिकल लाइंस ब्रो ए स्लो फॉर योर मेडिकल शॉप से दवाई एक्सेस और उनकी कमबैक है और हम लोग क्या माने एक्सेस के साथ एंटी क्लॉकवाइज बनेगा उस माइक उसी को क्या बोलेंगे डेफिनेशन और मेंटेन लगा दें तो क्या हो जाएगा वह स्लो स्लो ऑफ ए वर्टिकल लाइन इज नॉट डिफाइंड डे टेस्ट स्लोप आफ वायरेसेस इज इंफिनिटी स्लो फॉर्म मेरे वॉइस जरूर है लेकिन स्लोप आफ एनी साइनिफिकेंट वॉइस स्लोप आफ एनी लाइन यह पेंडेंट 2xx इस को देखने x-axis 50 होगा और ऐक्सिस के पैनल कोई भी लाइन हो उसका भी सुधार होगा उसी तरीके से वर्टिकल लाइन का स्लोप इंफिनिटी होगा और वायरस का आरोप इंफिनिटी और गया होगा बाइक्स इसके पहले जो भी लाइन होगा उसका लोग क्योंकि डेबिट चाहिए हो यह ठीक है आपको मैं इस सारी बातें ऐसी बता देता हूं एक्स एक्स इसका 01 सेक्शन के पैनल कोई लाइन उसका तिलक जीरो वायरस का स्लॉट इंफिनिटी या वायरस के फैलने कोई भी लाइन हो उसका स्लॉट इंफिनिटी होगा ठीक है इतनी सारी बातें हो गई अब थ्योरम वन इस बुक के हिसाब से लिखे थ्योरम वन फटाफट कर आ रहा हूं प्यूरी क्रॉप्ड अप नहीं है क्या है अमूमन बताइए 154 कह रहा है ट्यून वर्टिकल लाइन हाउ टू ने वर्टिकल लाइन है और पैरालिसिस एंड ओनली आफ लाइफ एंड ओनली आफ मधेश 9999 ए बी एम थे स्लोप्स आफ इक्वल टू अच्छा ठीक है देखिए अब बात किया है मानते हैं कि हमारा यह 1 से और इसी वजह से इंग्लिश उत्तम और 2 लाइंस है ठीक है अब दोनों अब यहां पर बना रहा है और यह के साथ यह और अगर यह है अगर यह राघवन का फ्लेम लो कर दो अगर बराबर होगा अच्छा ठीक है या आप कह सकते हैं दोनों तरफ से एक्टिव गतिविधियों के बराबर की राशि के बराबर का मतलब एलाइनमेंट पहल है यह लाइक कर रहा है कि इन थे शैडो का ट्वेंटी-20 क्रिकेट के बराबर होगा तो 1031 बराबर भी क्या होगा 10:30 2001 बराबर क्या होगा सिलेक्ट तू तो स्लो अगर बराबर हो तो लाइंस क्या होगा पहला यह गिर पड़ा होगा तो लोग बराबर होगा यह बात क्लियर है और अगर यह तो एक बात हो गई दूसरी बात यह कट क्लॉक को मैं बराबर लिखूं तो लिख सकता है 1031 बराबर 1032 तो क्या होगा विटामिन बराबर टीटू और यह दोनों अगर बराबर होगा तो कॉरस्पॉडिंग बराबर होने का मतलब यह है कि एल्बम किसके अप्रैल लोग अल्टो का मतलब आप लोग पैदल होगा तो यह जरूरी है कि लाइंस पैरेलल होगा और अगर लाइंस बारे में तो यह जरूरी है कि इस लोक में होगा क्लियर है है तो यह दोनों तरफ से थ्रू है स्लॉट्स अगर बराबर होगा तो समझ जाइए कि नाइंथ रिजल्ट होंगे और अगर लाइंस पैरेलल होंगे तो उस लोग बराबर होगा दोनों तरफ से वेलन है ठीक है इतना उतार लीजिए चलिए दिवस मजा गई स्लो बराबर तो गाइस ब्लाइंड प्रदूषणों बराबर चुके होंगे इसलिए मैं अनिल सिंह ढूंढ ठीक है चली गई नेक्स्ट जो रंग तो क्या जाता है थ्योरम तू कह रहा है इन वर्टिकल लाइंस साइड पर पर्टिकुलर अ 129 ए वर्टिकल लाइंस बट क्लाइंट को छोड़कर क्यों बाघ का 9 को परपेंडिकुलर टो रिटर्न गिफ्ट हाउ टू इस बीच अधर टिप्स कि एंड ओनली फॉर प्रोडक्ट इन एंड कि ऑनली इफ कि दब्बू प्रोडक्ट्स अब आधे श्लोक सिर्फ - मैन अ कर दो है - वर्ल्ड टॉप क्वेश्चन में क्या कह रहा है मध्य में थ्योरम यह कह रहा है कि दो वर्टिकल लाइंस हैं कितने लाइन से दो वर्टिकल लाइन से ना समझ आया तो आप कमेंट सेक्शन में जरूर बताएं ठीक है दो वर्टिकल लाइंस मतलब एक लाइन है यह फाइल दो भिन्न वर्टिकल लाइन से दोनों एक दूसरे का पर निकले एग मतलब यह लाइन है एलिमेंट और एक लाइन है अल्टो दोनों के बीच कितना इन थे डिक्री यह कह रहा है ठीक है यह नौकरी दोनों के बीच ट्रेंगल है और यह एक के साथ मानते हैं गलती टावर बना रहा है और यह कितना बना रहा है कि टैटू तो श्लोक-1 कितना होगा यह होगा 10.51 श्लोक-2 क्या होगा 10322 चीख इतना समझ आ गई यह ट्यूब तो यह कितना होगा बताइए यह होगा 180 डिग्री - ठाट टू क्या होगा यह होगा 180 डिग्री - कि यह तो क्या हो जाएगा 580 ए प्लस कितना डिग्री नौ डिग्री प्लस कितना 132 ओ हु इज इक्वल टू 180 डिग्री ठीक है तो क्या हो जाएगा 1818 कैंसिल तो हम क्या लिख सकते हैं थी था वन क्लास नाइंथ डिग्री बराबर थी टैटू और आपको अगर पता नहीं हो तो यह जान लीजिए अब मेरे ट्रिगोनोमेट्रिक अलेक्सिस देख सकते हैं नौ डिग्री प्रति सैकंड ठंड के साथ चारों तरफ और में बदल जाता है यह उत्तम माइनस कौस थीटा एक्स रखेंगे तो यहां पर भी डाल देंगे ठीक है 1020 प्लस एक्स क्या होता है - कॉन्टैक्ट्स तो क्या हो जाएगा इधर भी 3 लगा देते हैं हैं और इधर भी टेंट लगा देते हैं तो क्या मिलेगा में रख - कोट 58 बराबर 10322 है कि अब इसको अगर हम प्रदर लिखें आगे तो क्या होगा अब देखिए को ठीक है उनको ले सकता - वन नंबर 1031 क्योंकि कॉटन का रहती प्रोफाइल है इज इक्वल टू क्या होगा बताइए 1032 मतलब यह इसको अगर मैं यहां पर लिखूं आप ध्यान से देखें योगा माइनस वन फटे वववववव M2 डैम वन और शैंपू का जो प्रोडक्ट है वह कितना माइनस वन मतलब यह कि दो लोग होंगे लाइंस के अगर हम दोनों का क्लॉक का जो प्रोडक्ट है करो - बना रहा है दैट मींस लाइंस वर्टिकल है ठीक है तो लाइंस वर्टिकल होंगे हम सॉरी लाइंस पर्टिकुलर होंगे 2 लाइंस एक दूसरे के बीच जो मेंटल है वन इंटीरियर एंगल बनाएगा कब जब दोनों के स्लॉट्स का जो प्रोडक्ट है वह - मन होगा या दोनों के स्नैक्स कागज प्रोडक्ट - मन होगा तो लाइंस परपेंडिकुलर होगा लाइन पर वेंट्रिकुलर होगा तो यह जरूरी है कि दोनों के स्लो कार्य प्रोडक्ट्स उसका जो प्रोडक्ट - है या दोनों के लोगों का प्रयोग न होगा तो लाइंस पर्टिकुलर होगा दोनों तरफ से बॉईल है उतार लीजिए बहुत ही आसान ठीक है हमने क्या जाना है कि 2 लाइंस अगर आपस में फेडरल हो तो स्लोप्स आफ इसमें बराबर होंगे और 2 लाइंस अगर आपस में पॉइंट कूलर हो तो दोनों के स्लॉट का जो प्रोडक्ट है वह - बंट होगा ठीक है इसका इस्तेमाल हम लोग कभी न कभी करेंगे तो चलिए नेक्स्ट है कुछ लोग लाइन पासिंग थ्रू द पॉइंट्स तो बस लोक में आ 90 25 सिंह को चू 22 प्वाइंट्स कि एक्टिव वन एंड X2 कॉमेडी 2 लाइंस है वह पॉइंट्स आप 1512 उनसे पास करने वाले वाला जो लाइन होगा कौन सा लाइन ऐसा यह हमारा एक लाइन है यह पास कर रहा है दो पॉइंट से X1 प्रो m1 से और दूसरा पॉइंट X2 को Y2 कोई बात नहीं यह जो पॉइंट है यह पॉइंट से पास करने वाला जो लाइन है इस लाइन का स्लोप चाहिए मतलब क्या होता है के साथ जंगल बनाएगा थीटा उसका टाइम जेंडर टेस्ट लो क्या होगा 10:30 था इसको निकालने के लिए हम क्या करेंगे हम कुछ भी करेंगे तो करेंगे देखते रहिए आप यह आपका x12 यह होगा यहां से यहां तक यह पॉइंट X1 है यह पॉइंट से एक्टिव तो यहां से और इसे यहां तक डिस्ट्रेक्शन है और यहां तक एडिसन सेक्टर-2 है उसी तरह से वुमन है और यह वाइट इस बात पर जानते होंगे यह वाइट वाइट वाइट बन है तो यह बताइए कितना होगा वॉइस टू माइनस वन X2 है और यह यौवन है तो यह कितना होगा बताइए योगा X2 - ऐक्सिडेंट यह आपका इंस्ट्रूमेंटल सैक्सफोन तो यह कितना होगा next 9 - शिक्षण यह कितना होगा वह टू माइनस वन यह हो जाएगा आपका X2 - 10 नंबर डिग्री और योगा 12951 यह अगर ठीक है तो यह भी कितना होगा बताइए थी तब तो बताइए 10:30 था क्या हो जाएगा 10:30 था मतलब स्लो यह जाएगा वाइट टू माइनस वन फटे एक्स्ट्रा - इलेक्ट्रॉनिक यह आपका हो गया स्लो स्लो कैसे निकलेगा हवाई के वाइको डिजिट का डिफरेंस बाइक कोऑर्डिनेटर डिफ्रेंस है न और कोऑर्डिनेटर डिफरेंस का जो रेशू होगा अगर वाइट वाइट वाइट बन कर लेंगे तो नीचे कमेंट तक बंध अगर ऊपर वाइबर नोटिफिकेशन मैनें चीज का ध्यान रखें कि आपको जो आपका जिसमे से आप करेंगे जिसको पहले मानेंगे ऊपर नीचे भी पहले मानेंगे समझ रहे हैं यह सिर्फ सेकंड वाला घर यह पहले ले रहे हैं तो सेकंड वाला यहां पर भी डालना पड़ेगा फर्स्ट ऑयलाइनर पहले थोड़ा पहले आपको लेना पड़ेगा नहीं तो फिर हो जाएगा कलेक्ट ठीक है तो क्या हो गया हमारा फॉर्मूला वन टीम मैनेजमेंट बटे एक्सटेंशन वर्टिकल चाहिए बट यह राजकुमार जैन का श्रेष्ठ वस्तुओं स्नो किया है 1033 गया ऐंगल आप इन क्लाउन ठीक है अब इस लुक की बात हुई तो आप रोक की बात कही तरह से होगी इधर से भी हो सकता है कि लीडरशिप सकता भीम ने जाना मान लेते हम थी टाल दिए कि ट्रायंगल और यह 10:30 चाल है ठीक है तो यह आपको कुछ याद रख लेना है व्यूज अच्छा ठीक है जीरो डिग्री में जो सिरा होता है है और पार्किग मैं आपको पता है कितना होता है वन तो मुजफ्फरनगर कांड होगा तो 135 डिटेल्स 180 डिग्री - आपको याद होना चाहिए था वह पांच यह दिखेगा - न भी याद रखना है ठीक है उसके बाद आपको पता थर्ड डिग्री में कितना होता है यह भी रूटीन ठीक है और 180 डिग्री - अतिरिक्त ना वह शिफ्ट डिग्री यह हो जाएगा कितना में में 150 डिग्री में - एक बटे रूटीन उसी तरीके से अगर आप जान हैं कितना डिग्री का साथ डिग्री का तो ट्रेंड साथ डिग्री होता है रूट 3 तो अगर मैं क्या लिखूं कि कितना होगा बताइए 120 डिग्री तो 120 में हो जाएगा - रूटीन 60 मुझे अपना रूटीन कितना आप याद कर लें इतनी आपको जरूरत पड़ेगी और 1000 होता है और 1000 के नॉट डिफरेंट होता है उतार लीजिए कि अब NET अ कि यह बात हो गई किसकी 22 प्वाइंट्स गिवन है उससे पास करने वाला जो लाइन होगा स्पष्ट रुख क्या होगा 12951 बटे एक्स्ट्रा - एक्सप्लेन अब यह पॉइंट से भी बीएमसी सिर्फ पॉइंट ठेकमा बिल्कुल मार्कशीट और पोली नियर मे कि को इंडियन के दिन क्या होगा खोलें मतलब देखिए एक पॉइंट के हैं एक पॉइंट बीए और बीएससी दोनों को एक ही लाइन में आएगा तो बताइए यह लाइन में होगा तो यह भी स्वागत तोड़कर आगे बढ़ाएंगे अभिषेक को आगे बढ़ाएंगे लाइन बनेगा दोनों लाइन सेम है जब लाइन Jio SIM होगा तो उसका स्लॉट भी सेम होगा डेज क्या होगा स्लो वर्क आफ ए बी इज इक्वल टू स्लो फॉर्म ऑफ बी सी स्लो क्योंकि आपस में बराबर में तो इससे स्लो कर इसे स्लो वर्क बराबर-बराबर क्या होगा सिंपल सी बात है ठीक है ठीक है नेक्स्ट क्वेश्चन करके फिर और करने की कोशिश करेंगे तो जल्दी जल्दी आ आप मुझे बहुत अच्छे क्वेश्चन है - लोग लाइन है और क्या चाहिए से लोग लाइन जॉइनिंग टू प्वाइंट्स ओं में जॉइनिंग द पॉइंट को पॉज 5 - 30 कर सकते हैं या आप अभी जो आपने बोला था ना कि लोग कैसे निकलेगा का मतलब यह डिफरेंस - 4 8 - इसका कितना - तीन बटे एक्स का सेवन - 151 - को प्लस 3 - 2 - इग्नोर किया गया निगेटिव हो गया स्लो किया गया निकट मतलब इसका मंगल होगा वह ट्यूसडे होगा क्या होगा नौकरी से ज्यादा होगा चली गई निकालना बहुत सिंपल है वहीं के कोऑर्डिनेटर डिफरेंस बटे के कोर्डिनेटर का डिफरेंस दूसरा एक्सांपल हैं मैं सिर्फ 10 लोगों का इन इंग्लिश लोग ऑफ लाइन है जुआ नहीं द प्वाइंट के कमा पूर ा कि यह काम आफ आर्ट एंड - 9 - टू इन्हीं अ क्वेश्चन है बोलो हैं कि हाफ क्या होगा बताइए इसका स्लोगन एवं कितना ज्ञान है वन और स्लो क्या होता है वहीं के को-आर्डिनेटर डिफरेंस - टू - फॉर बर्थडे - 3 - क्वाइट डिफरेंट बराबर कर दिया हुआ है स्लो कितना है अब आप यहां से निकाल सिंपल - - - - - - - - - Plus के हो जाएगा बारे में गया ना हो तो आप कर सकते हैं तो चलिए यहां पर हम रहो गया नेक्स्ट एग्जांपल है फ्रंट लो फ्लैम या निकालने का बिल्कुल सिंपल सिंपल है आज शाम का वैल्यू निकालने और क्वेश्चन में कहा है इस लाइन जॉइनिंग टू प्वाइंट्स 126 ए एंड बी [संगीत] ये हमको मार फ्रॉम क्लास 5th कि यह एक लाइन है लाइन वंश इस परपेंडिकुलर लाइन जॉइनिंग टू द प्वाइंट पी - 2 का MP3 कि थैंक्यू - 49 - 5 मतलब यह तो बोला है कि क्वेश्चन क्वेश्चन एंड पॉइंट्स पेट में पॉइंट पर जाना होगा उसके जो प्रोडक्ट होगा तो लोक के रेस्क्यू डिफरेंट डिफरेंट करें - 05 - - - - - 3 एक बटे - 4 - - टू इज इक्वल टू माइनस वन तो क्या हो जाएगा बताइए टेंपलर्स पेस्ट पिंपल S8 बट A1 - अल्बर्ट माइनस प्लस 2 - 281 आयोग - मंदिर चौक 84 हो गया फेल्ट टिप और एंब्रेस ठाट 1432 इजी वे टू माइनस वन डेट मींस के एम का वैल्यू हमारा हो जाएगा फोरम बराबर - - 3210 हो जाएगा - पति व पुत्र मतलब अगर आप फिट कर दें आप एम्पलाई निकालेंगे इसे भी और बेटियों का जो आप ड्रॉप्स निकालेंगे इसका जो प्रोडक्ट हो - मन होगा या आप अलग-अलग लोग निकाल लेंगे फिर अलग-अलग अगर उसका लाइन बनाएं तो दुल्हन हेग के बैंगलोर बनाएगा वह नौकरी बनाएगा उम्मीद करता हूं कि आप समझ पाए होंगे अच्छा ठीक है धार्मिक कार्यों गेम तो चलिए कोई बात नहीं बहुत अच्छे अच्छे बच्चे लोग अब आते हैं एक्सांपल में पॉइंट ऑफ कि अ पॉइंट्स - 31 मार्च 6th है - 9 मेज़ कि मैं के वीरों का मर्फी नकुल नियर मी 98 कि स्कूल जूनियर ने पांडे ओं कि मैं मंडे को शेयर का क्या मतलब जैसे पॉइंट मानोगे और दूसरा पॉइंट कमांड बी तीसरा पॉइंट या मारुति यह - 36 - नौ माह के और यह जो लोग डिस्टेंस फॉर थे लेकिन यह बहुत ही इजी हो इसके बराबर हो ना तो ठीक - - - - ए अच्छा ठीक है वह के को डांस का डिफरेंस फॉरेक्स के को-आर्डिनेटर डिफ्रेंस खजूर एक हुए कुछ लोगों ने अपनी - है यहां पर वह बटे जीरो - - नहीं है हैं या आप चाहे तो इसको ले सकते हैं ऐसी का भी ले सकते हैं ठीक हो जाएगा तीन बटे - 8 प्लस माइनस एक्स इज इक्वल टू पुष्टि - हे बट एस वे ऑल नो कितना हो जाएगा 10000 ठीक तो क्या हो जाएगा हो जाएगा 326 - 29 2009 मैं क्यों जाएगा बताइए नहीं सीक्वल टू - थर्टी प्लस 2 ए 9 प्लस 30 इंच यौन मैंने कुछ गलत करें चीजों को कि यह - टीन कि यह - 15th जी हां में कहीं ना कहीं गलती हुई है थोड़ा सा चेक कर लेते हैं 8 - तीन को मार 6 - 9th मे 0f 10 थंब - इनको जॉइंट अमेज्ड है जीरो माफी टीम - 6 मोंठ बराबर अच्छा ठीक है यह - Twitter - - - 3 थोड़ा आराम से करते हैं उस पार्ट - हे बट 10 - - 9 कितना हो जाएगा यह - सिक्स तो यह था जी हां यह है कि मानसा प्लस तरीके होगा - 6 बराबर 15 - के फटे नोट फिर को सॉल्व करेंगे तो नॉएडा - नो चक्र 54 - 15 चैनल - 90 प्लस 6 हो जाएगा नोएडा - 6 बराबर - 90 प्लस0 पति ने बराबर इसमें 40 - 6 - कि हमारा - सिंपल सी बात है उतार लीजिए तो चलिए हो गया अब हम लोग बात करेंगे इंटरसेप्ट कि किसकी बात करेंगे सिंपल सी बातें इंटरनेट वाले से लिखने की जरूरत नहीं है तभी मैं बता दूं जैसे को मतलब एक्सेस को जहां लाइन फट करेगा वह एक सीनियर सकता है वहीं को जहां कट करेगा क्या वह इंट्रस्ट है तो इंटरसेप्ट और इंटर सेट ऑफ स्ट्रैट लाइन टो इंटरसेप्ट स्ट्रैट लाइन का क्या होगा मानते हैं कि स्ट्रैट लाइन है ऐसा यह कैसा है यहां पर जो कट करेगा योगा एक्स इंटरसेप्ट पर यह जो कट करेगा रिमाइंडर सका जो पॉइंट ठीक है तो एक्ट में कट करने वाला पॉइंट हो गए थे और एक्सपीर वह का मतलब क्या होता है जीरो एक्सप्रेस वाया हेलो क्या होता है जीरो फ्री हो जाएगा टैक्स इंटरसेप्ट में एकमा जीरो और वाइन टेस्टिंग में क्या हो जाएगा वहीं को कट करेगा देखा है लोगों का दूसरों का मा 10 और वह कुछ होगा इसके बाद आपको समझ है आ गई होगी आ गई होगी ठीक है कि अब NET यहां पर बहुत सारी बातें में खरीद ली है लेकिन तब भी यहां चुकी थी जो गिवर है वह बता दूं क्वेश्चन अप्लाई नहीं जंगल फॉर्म अ क्वेश्चन ऑयलाइनर जनरल सॉन्ग अच्छा ठीक है सबसे पहला क्वेश्चन ऑफ लाइन थे जनरल फॉर्म फॉर्म बताइए कि जनरल फॉर्म यूथ एक्स प्लस बी स्क्वायर प्लस सीट 120 ठीक है इसको अगर आप इसको बनाएंगे स्ट्रैट लाइन है स्ट्रैट लाइन तो इक्वेशन आफ स्पिरिट लाइन क्या यह जनरल फॉर्म में और सफेद में अब यहां पर कंडीशन है कंडीशन या के मोड ऐप प्लस मूड भी क्या नहीं होना चाहिए इसका सं0 नहीं होना चाहिए कंडीशन है क्या कंडीशन है कि मूडे और मूड भी का जिस समय वह 0 क्यों क्योंकि अगर एक और 2010 हो गया तो जीरो पंजा जीरो किया जाएगा जीरो हो जाएगा क्या हो जाएगा जीरो हो जाएगा अगर यह काव्य हुआ यह भी पॉसिबल है जब यह वि0 वि0 भी डुबो हो जाए तो अक्सर वाला चैप्टर में हो तो वेनिश हो जाएगा ना तो सिर्फ फर्स्ट नंबर देगा तो हमारा एक स्ट्रांग रहेगा इन क्वेश्चन बनने का कोई सवाल बनता नहीं है याद रहे कि एक्सप्रेस वाराणसी बराबर जीरो में रखिए और बीच का जो वैल्यू है एक साथ दोनों सिरों नहीं हो सकता अगर यह जीरो होगा तो सिर्फ वही बच्चे का यह वाटर आधे रोगों से देखे लेफ्टओवर जाएगा तो यह कुछ जनरल सैम क्वेश्चन है उसमें और बीच जो यहां पर यह कौन है एक्स का क्वेश्चन भी कौन है भाई को क्वेश्चन मतलब यह और b2x रॉय की कॉपी सेंड हैं उस फाइनल टेनिस लि0 नहीं सकता गरीबी की हो गए 2001 क्वेश्चन हो गया तो सिर्फ बदलेगा तो हमारा मामला हो जाएगा खिलाफ ठीक है कि बाद आप समझ गए एक इक्वेशन ऑफ थे इक्वेशन ऑफ नंबर हिम्मत सीरवी इक्वेशन ऑफ एक्स एक्स एक्स एक्स इसका मतलब यह मैंने बता दिया है एडवाइजर मेघवाल एक्सेस पॉइंट होता है उसी तरह से इक्वेशन आफ वायरेसेस क्या होगा और और अगर बात करें तो यह क्या होगा उसी तरीके से अगर वायरस के पैनल की बात करें तो क्या होगा होगा अगर हम एक सिंपल सी बात है आधुनिक ऑफलाइन में इसमें आते हैं और बैक साइड में जहां पर आपका कोई भी रूप नहीं किया है उसको मैं समझाने की कोशिश करता हूं ठीक है सबसे पहला के है ऑब्लिक लाइन में और दलित कल्याण है नेक्स्ट पैडल हो और निवेश के पहलू ठीक है ग्रेडियंट फॉर्म और स्लो फॉर्म को सबसे पहले से कि अमलीडी बात हुई किसकी बात हो रही है और यह लिख लाइन ठीक इसमें कौन सा फॉर्म कि डिफरेंट फॉर्म्स रेडियंट फॉर्म और स्लो फॉर्म ग्रेडियंट और प्रूफ फॉर ठीक है इसमें क्या बताइए अगर यह कितने पास करें और जिन से पास करें पासिंग थ्रू द ऑरिजिन पासिंग थ्रू थे और एजेंट उसके बाद स्लोप इंटरेस्ट फ्रॉम यहां पर फॉर्म यह पाल सिंह को चोदा कि आर्य जी आप चिंता करने की जरूरत नहीं है बिल्कुल सिंपल सी बात है अगर मैं बात करूं यह एक्सेस है वायरस है और एक लाइन है ठीक है वह किससे पास कर रहा है और इस दिन से और इसका स्लोप गिवेन है एम क्या स्लो है एम का मतलब क्या है अगर इससे पास करें तो कंफर्म होना के जीरो का यह जीरो और अगर एक पॉइंट हम जंगल पॉइंट देने एक्सेस ऑयल यह भी हो सकता है और वह स्ट्राइक भी जीरो भी होगा और वह 200 से कम 10 होगा मैं यह कह रहा हूं कि जो एक्स्ट्रा ऑइल है यह जनरल पॉइंट है इस लाइन का ठीक है यह लाइन का नाम है जिसका स्लोप हम है और स्लो क्या होगा वहीं के कोऑर्डिनेटर डिफरेंट अगर यहां पर अक्षत तो इस इंस्टैंट में यहां पर मैं एक हैं और यह कितना डिस्टेंस टेक्स्ट और यह कितने वाइट तो यह एक्स और यह क्या है बताइए मोबाइल पर कि उस लोप हो जाएगा है यह बट यह नाइस पेनल्टी बन रहा है एम बराबर 10:30 टकराव जाएगा परपेंडिकुलर बड़े व्यस्त हवाई पट्टियां सिंपल सी बात तो वह बराबर हो जाएगा नोट इससे आसान होता है क्या वह बराबर मिनट क्या है इक्वेशन ऑफ लाइन है स्लो फॉर्म और जो लाइन है वह किसके पास कर रहा है और इस दिन से पास करना है इससे सिंपल और कुछ बात नहीं होती चले उतार लीजिए जैसी किताब में जैसे देशों के अनुसार यह बता देता हूं बस और कुछ नहीं चाहिए तो अपने हिसाब से लेकर आ जाते हैं थे सेकंड में क्या बोल है बताइए कौन सा फॉर्म पॉइंट स्लोकाम यह पॉइंट स्लो फॉर्म कितने सारे फॉर्मेट के बने उसी को में एक्सप्लेन कर रहा हूं ठीक है अब मान लेते हैं यह एक हमारा हुआ है ठीक है कि अ मानने की यह हमारा ठीक और कि यह लाइन है जिसमें पॉइंट गिवर है पॉइंट है और गैंस आफ थे उसे जरूरत नहीं एक्सेस ऑयल और लोग यह जानते हैं और कोई भी पॉइंट को इस पॉइंट को देख लो कि यह कितना होगा बताइए कि समझ रहे हैं आप यह आपका एक्सेस ऑयल तो यह हो गया एक्स और यह क्या हो गया नेक्स्ट वन यह क्या हो जाएगा ना अच्छा ठीक है योगा वहीं और एक हो जाएगा वाइट विनेगर वाइट वाइट बन तो यह कितना हो जाएगा वन - मई यह कितना हो जाएगा बताइए यह जाएगा एक्स - एक्स है तू यह जाएगा योग्य वाह वाह वाह वाह वाह वाह वाह वाह तो यह कितना होगा अगर आपको समझ में परेशानी है तो मैं अच्छे से कर दे रहा हूं ठीक है आपको लगता होगा कि आप करना तो मैं थोड़ा बड़ा बना देता हूं ठीक है एग से और यह वही है अगर हम ऐसे बना एक लाइन है वह और यह से पास नहीं करना है ठीक है देख लीजिएगा दूसरा पॉइंट पूरा कर दीजिएगा तो यह पॉइंट आपका एक्सेस ऑयल जनरल पॉइंट और एक ही पॉइंट एक्टिव व तो क्या होगा यह आपका नेक्स्ट और यह क्या है X1 यह ऑरेंज शैड और इजेक्शन तो यह कितना हो जाएगा एक्स वन माइनस एक्स यह भी कितना होगा X1 - है कि उसके बाद यह कितना होगा वहीं और यह कितना होगा वन तो यह हो जाएगा आपका वन - वाइफ यह भी कितना होगा वॉल - वाइट और यह अंदर कितना बना रहा थीटा और स्लॉट ही किसके बराबर 10:30 के बराबर स्लॉट मतलब हम यह क्या होता है वर्टिकल बट ए होरिजेंटल मतलब यह इसको आप उसे चंद्र युद्ध पर मेडिकल वेदव्यास ने हवन - माई बटे X1 - डेटिव व्यायाम X1 - एक्ट यह क्या करें इसको इधर कर दें - 151 - को लेकर यहां पर - एक्स - एक्स - एक्स - एक्स = क्या होता है बराबर-बराबर किया है बार ऐसा करने से आपको थोड़ा सा लगा कि ऐसा कुछ भी नहीं होता है घृणा है अच्छा ठीक है तीसरा फ्लैशलाइट इंटरेस्ट फ्रॉम फ्रॉम snap.com मैं जो बोलता है वह गिवर न समझे थर्ड नंबर स्लो इंटरसेप्ट ठीक है मारा लाइन है ठीक है जो मानते हैं कि किसको की वाइफ को यहां पर इंटरसेप्ट करने और यह जो गिवर है यह सीट डिस्टेंस कि मैं ठीक है मान लेते हैं यह पॉइंट यह होगा एक्स 1200 सीसी और कहीं भी लाइन में किसी भी पॉइंट को इस पॉइंट को इस पॉइंट बताइए बताइए और यह है है और यह कितना हुआ है तो यह हो जाएगा आपका वॉल - सी तो यह कितना हो जाएगा वॉल - सी और यह अंकल खींचता है तो श्लोक जो होगा 10:30 था खुसरो कि बहन ठीक है व हो जाएगा वहीं - सी बटे एक्टिव व्हील - कि बराबर हम एक्स्ट्रा यह बराबर क्या होगा हम एक्सप्रेस की उम्मीद करता हूं कि आप समझ पाए होंगे उतार लीजिए नहीं समझ आए तो आप कमेंट सेक्शन में मुझे जरूर बता सकते हैं अच्छा ठीक है अब उधर टू प्वाइंट फॉर में कौन सा फॉर्म 2.1 अच्छा ठीक है कौन सा फॉर्म हाउ टू प्वाइंट्स एक फोन डिपॉजिट स्कीम बने अब नहीं बोला कि लोग बस यह जिसका नाम है और इस तरीके से दो पॉइंट है यह पॉइंट यहां ठीक है यह का यह माप लेते हैं और यह है यह 1615 व और यह क्या होगा वह क्या होगा - समझ गए लेकिन टू पॉइंट टू पॉइंट लोग नहीं लेते हैं इसको पॉइंट से अच्छा ठीक है तू क्या हो जाएगा स्लोप और भी कैसे लिख सकते हैं यह इस लोगों का सारा वाली - अिववािहत - व्वे एक्स - एक्स इन दोनों बराबर हो जाएगा वहीं माइनस वन बट ए टेक्स्ट मैसेज सेक्शन बराबर वन टू - यौनिक बट एस टू माइनस वन या हवाई माइनस वन बराबर होगा वह टू - मा यन बट एस टू माइनस वन इन टू एक्स माइनस सेक्शन यह एक आपका फॉर्म हो गया किसका ऐसा लिखिए ऐसा लिखिए यह टू पॉइंट स्पर्म हो गया फ्रॉम बेसिकली स्लॉट आपको याद रखना है वर्टिकल चेंज बट यह रवा का डिफरेंस हवाई कोऑर्डिनेटर डिफरेंस और कॉर्ड का डिफरेंस का जो रेशू होगा वही होगा आपका स्लो इसके बाद नहीं सारी नौटंकी इस गहन नौटंकी जो पिता बे गिवन ए ए हवा से intersect फॉर्म-कौनसा फॉर्म इंट्रस्टेड फैट कि फाइंडर फॉर क्या कहता है यह के यह एक्स्ट्रा है और यह दो जगह के एंटर करें मानते हैं कि इंटरसेप्ट किया यह डिस्टेंस है वह यह है कि होगा इसका वैल्यू एक और वह का 102 और यह जो डिस्टेंस है यह भी है ट्रैक्स काव्य काव्य कि अब क्या होगा स्लो होगा स्लो इंटरवल तक प्योर घी मैंने बस कुछ लोग क्या होगा एक्स्ट्रा डिफरेंट डिफरेंट सुबह - 20 एक बटे जीरो - ही तो बात पूछूं कि टैम बराबर क्या हो जाएगा - भी बैठे कुछ लोग तो यह निकला एक पॉइंट हम लेते हैं एक्सेस ऑयल तो क्या हो जाएगा बिल्कुल भी निकलेगा स्लो क्या निकलेगा बताइए वाली - जीरो फटे एक्स माइनस है तो क्या हो जाएगा बराबर वाइबर टेस्ट - के तो दो जगह मारा लाख इस तरह से निकला तो क्या लिख सकते हैं - बी बर्थडे बराबर वाइबर टेक्स्ट - से इसको फोल्ड करेंगे कि हो जाएगा - बी एक्स मैं प्लस जॉब भी मल्टिप्लाई कर दिया हमने बराबर का यह वाइट इसको अगर सॉल्व करें तो क्या हो जाएगा यह - भी एक्सपायर लेकर चले जाते हैं ना कर सके वहीं तो यह हो जाएगा यह भी बराबर वैवाहिक प्लस बी डोमिनेंट इधर भी हमने एबी से divide किया और इधर भी किसी डिवाइड कि यह भी से तो क्या फायदा होगा यह तो कैंसल भी यहां पर देख सकते हैं बट ए प्लस बराबर यह बहुत ही सुंदर में आ गया उम्मीद करता हूं कि सारी बातें आपको समझ गया ठीक है जल्दी आ तो चलिए इसमें एक और इंपोर्टेंट बात इसका एरिया जो सिंपल अधिक रहा है यह भी है प्रिया कितना होगा इसका देखिए फिर धागा हाफ यह भी हाफ बस ड्राइड लेकिन अगर डिवाइस जहाज मोड लगा देंगे सिंपल सी बात है अच्छा ठीक है चलिए क्वेश्चन करते हम लोग ओं के लिए विंटर सिर्फ अमेरिका लिया या बनेगा एक्स वाई और यह जो पॉइंट है है उससे मिलकर अ क्वेश्चन अ पॉइंट क्वेश्चन लाइन पर लाइक शेयर और कि फाइंड थे इक्वेशन ऑफ लाइन ब्राइडल 2xx इस एंड पासिंग थ्रू थे गवर्नमेंट फॉर पोस्टिंग मैं सॉफ्ट कॉपी - को किस से बात कर रहा है तीन को - 4 से और किसके पैरेलल है वह एक्सेस के पैनल है ट्रैक्टर के पैनल का मतलब क्या क्वेश्चन है अक्षय यह वाक्य एक्स के पर लाइक और स्टैंड तो अगर कॉस्ट होगा और वॉइस मेल कितना गिव बेनिफिट से कर रहा है - 4 से तो बेसिकली वह यह तो नहीं होगा इससे पहले बैक टू ऊपर से होगा नीचे से लेकिन वह का हेलीकॉप्टर कितने - 4th नीचे हो गई वाला हो गया - 6 होगा ठीक है तो वही होगा - 4 या वाइफ प्लस पॉइंट 120 यह क्वेश्चन ऑफ लाइन है तू किससे बात कर रहा है टीम को - 4 से एक्सेस कपड़ा टैक्स के पैरों को हाई कॉस्ट है और वह है कि - अर्थवर्क हमारे शरीर पोषण तत्व प्रचुर वृद्धि प्रोफेसर भी लिख सकते हैं हम लोग उम्मीद करता हूं कि आप समझ पाए होंगे नेक्स्ट है बहुत ही सिंपल है चिंता करने की जरूरत नहीं है ठीक है इंडिकेशन आफ नॉट हेविंग कि फाइंड थे इक्वेशन ऑफ लाइन हेविंग अलार्म क्लॉक ऑफ - 384 स्नो प्वाइंट - 3rd फ्लोर एंड पासिंग थ्रू एंड ए पासिंग थ्रू 13 कहा - 49 कि मैं ठीक है स्लो गिवर और पॉइंट गिव ए पुश योर प्वाइंट फॉर मजा आएगा तो पॉइंट्स लो फ्लैम वाला क्वेश्चन हो क्या था वहीं माइनस वन बराबर एम एक्स - एक्स स्टूडेंट तो क्या होगा बताइए वाई वाई तो इश्वर तो हमारा जंगल पॉइंट से तो रहेगा यह क्वेश्चन में माइनस वन कितना - 4 जगह बने स्लो कितने है - 3 4 एक्स - एक्स वन कितना गवर्नर तीन मतलब यह तो वाइफ प्लस फोर इक्वल टू माइनस 3 व 4 एक्स माइनस 3 नरेश पोरवाल प्लस 6 बराबर - 3 एक्स प्लस नीतियां नो तो हो जाएगा पूर्व आई प्लस रिएक्शन में सोलह-सोलह के साथ उपलक्ष्य में रिकॉर्ड यह तो सीरियस प्रॉब्लम है वह रिकॉर्ड जी बिल्कुल सही आंसर आया उतार लीजिए उम्मीद करता हूं कि आप समझ पाए होंगे आप डायग्राम के साथ ही कर सकते हैं ठीक है कि चेन्नई एक्सप्रेस अ क्वेश्चन क्या कह रहा है कनेक्शन प्राइम मिनिस्टर्स हैव अ यह पॉइंट इक्वेशन ऑफ लाइन थे मेकिंग इंटरसेप्ट्स मेकिंग को इंटरसेप्ट 139 - 43 एंड माइनस 40 को डायलिसिस ऑन द कि कॉर्डिनेट एक्सप्रेस कि रिस्पेक्टिवली ओं अच्छा ठीक है अब क्वेश्चन जैसा है उस हिसाब से आप फिगर भी बना ही ले सकते हैं मन करे तो किस-किस में देकर तीन और - स्टार ऐसा कुछ यह वाला होगा जतिन पर कट कर रहा है हमारे सर यहां से भी बना सकते हैं यह बेसिकली क्या है यह एक्स्ट्रा बेलो जीरो वाट का बल्ब - है एक्स्ट्रा व्यक्ति इन 100 बीसी का आपको क्या चाहिए क्वेश्चन ऑफलाइन तो कोई भी पॉइंट आसमान के एक मोबाइल नहीं सकते हैं या क्या करें जो आप को मिला क्या मिला बताइए एक्स बर्थडे प्लेस व्हाईल बट एबी बराबर वन फ्रेंड्स हमारे चैनल पॉइंट और यह कितना ट्री प्लस वाई वी कितना - पूर्व इज इक्वल टू वन इसको आप अच्छे से लिख सकते हैं - 12 12312 4 एक्स माइनस 3 वायु बराबर वन रिस्पांस - त्यौहार व बिल्कुल सही आंसर आया है या पैसे कैसे कर सकते अल्टरनेटिव मैथर्ड सैफुल इस्लाम निकालने स्लो कितना होगा 0 - - 4 एक बटे खेल 3 - जीरो कुछ लोग कितना निकल जाएगा हमारे पास प्लस 4 बटे 3.43 हो गया अब लिख सकते हैं वहीं - जीरो थे मैटेरियल विद स्लो है ना एक बटे एक्स माइनस 3 टी हो गया वहीं बटे एक्ट - 30 - 3 बराबर क्या होगा 43 इसको सॉल्व करेंगे तो बराबर 41 - बार है तो वह माइनस प्लस 1200 अब तू ही करेगा सेम आएगा 41 - इन्हें बराबर बार एसएम आएगा तो आप ऐसे भी फिगर के साथ कर सकते हैं या फॉर्मूला को ट्राय कर सकते हैं जो आपका रेडिमेड आया नेक्स्ट हमारा होगा एरिया ऑफ ट्रायंगल और सेक्शन फॉर्मूला फॉर डिस्कशन व हम लोग नेक्स्ट क्लास में करेंगे और आप यह जान लें कि अगर आप सीवीसी और एनसीआरटी को फॉलो करते हैं तो भी आपके लिए बहुत इंपोर्टेंट आगे जाकर बहुत फायदा करेगा आपका वक्त भी रिप्लाई नहीं दिया ठीक है मिलते हैं नेक्स्ट वीडियो में थैंक यू वेरी मच ऊ
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Home > Stratified Sampling CRO Glossary Stratified Sampling Definition last updated: Definition first published: Discover that stratified sampling is, how to calculate it and how it stacks up to other types of sampling. What is Stratified Random Sampling? Stratified sampling is a structured sampling technique that enhances representation and accuracy by dividing a population into distinct subgroups, or strata, based on shared characteristics. Each stratum is then randomly sampled in proportion to its presence in the overall population, ensuring that no group is underrepresented or overrepresented. This method is particularly effective when a population has heterogeneous traits, as it improves statistical precision and reduces sampling bias compared to simple random sampling. For instance, if a university wants to assess student satisfaction, categorizing students by academic year—such as freshmen, sophomores, juniors, and seniors—allows for a more balanced and reliable analysis. By ensuring that input from each group is proportionally represented, stratified sampling provides a comprehensive and equitable perspective on the overall student experience. When to Use Stratified Random Sampling Stratified random sampling is ideal when the population has distinct subgroups that may influence the results of the study. Here are specific scenarios where it is most applicable: Ensuring Representation: When certain subgroups in a population are underrepresented or overrepresented, stratified sampling can balance their presence in the sample.Example: In a political survey, ensuring representation from different ethnic groups or income brackets. Example: In a political survey, ensuring representation from different ethnic groups or income brackets. Analyzing Subgroup Variability: When researchers are interested in analyzing variations among subgroups.Example: Studying customer satisfaction across different age groups to understand how preferences differ by age. Example: Studying customer satisfaction across different age groups to understand how preferences differ by age. Increasing Precision: Stratified sampling reduces variability and improves the accuracy of estimates compared to simple random sampling, especially when subgroups differ significantly.Example: In healthcare research, stratifying by gender to study treatment efficacy across male and female patients. Example: In healthcare research, stratifying by gender to study treatment efficacy across male and female patients. Example: In a political survey, ensuring representation from different ethnic groups or income brackets. Example: Studying customer satisfaction across different age groups to understand how preferences differ by age. Example: In healthcare research, stratifying by gender to study treatment efficacy across male and female patients. By carefully selecting subgroups, researchers can gain deeper insights and avoid biased or skewed results. How to Conduct Stratified Random Sampling: Step-by-Step Guide Conducting stratified random sampling involves several precise steps to ensure an accurate representation of the population. This method is especially useful when you need to analyze specific subgroups within a population while still obtaining reliable overall results. Below is a step-by-step guide explaining the process in detail. 1. Define the Population The first step in stratified random sampling is to clearly define the population you want to study. This involves setting boundaries such as geographical location, demographics, or behaviors to ensure the group is well-defined and relevant to your research objectives. For example, if a retail company wants to survey customer satisfaction, the population might include all customers who made purchases within the last year. Establishing this definition ensures your sample accurately represents the intended audience and avoids ambiguity or irrelevant data. 2. Identify the Strata Next, divide the population into distinct subgroups or strata based on shared characteristics. Strata should reflect variables relevant to your study, such as age, gender, income, education, or location. For instance, a healthcare study might group participants by age ranges, such as 18–29, 30–49, and 50+. The strata must be mutually exclusive, meaning no individual should belong to more than one group, and collectively exhaustive so that every individual is part of a stratum. Proper stratification is critical to ensuring that no subgroup is overrepresented or underrepresented in the final sample. 3. Determine the Sample Size Once the population and strata are defined, calculate the total sample size needed for your study. This can be done using statistical formulas or tools to ensure the sample is large enough for reliable results. For example, if your population consists of 10,000 individuals, you might decide that a sample size of 1,000 is required to achieve a 95% confidence level with a 5% margin of error. Choosing an appropriate sample size is essential to maintain the balance between accuracy and resource constraints. 4. Allocate Samples Proportionally Distribute the sample size across the strata in proportion to their representation within the population. For example, if 60% of your population resides in urban areas and 40% in rural areas, you would allocate 600 samples to urban residents and 400 to rural residents for a total sample size of 1,000. This proportional approach ensures that each subgroup is fairly represented in the study, which is particularly important when analyzing the characteristics or behaviors of diverse populations. 5. Randomly Select Participants Within Each Stratum Within each stratum, use a random sampling method to select participants. This could involve random number generators, lottery systems, or specialized statistical software to ensure the selection process is unbiased and every individual has an equal chance of being included. For example, if you have a list of 5,000 urban residents, you could randomly select 600 individuals from that list to form the urban stratum sample. Random selection ensures that the sample accurately represents the diversity within each stratum. 6. Collect Data from Participants After selecting participants, gather data through surveys, interviews, experiments, or other appropriate methods. Ensure that the data collection process is consistent across all strata to avoid introducing bias. For instance, if you’re measuring customer satisfaction, you might send identical surveys to participants in each subgroup and collect responses via email or phone. Consistency in data collection methods helps maintain the validity of your study. 7. Analyze Results Compile and analyze the data while maintaining the stratification structure. This ensures that each subgroup’s results are considered in proportion to its representation in the population. For example, you might calculate the average customer satisfaction score for each stratum (e.g., age groups) and then combine these results to determine the overall satisfaction level. Weighted averages are often used in this step to account for the size differences between strata, ensuring the analysis accurately reflects the entire population. 8. Validate the Sample Finally, validate your sample by cross-checking it against the population characteristics to ensure it meets the initial criteria and proportions. If discrepancies are found, adjustments can be made during the analysis stage to correct for overrepresentation or underrepresentation. Documenting each step of the process is crucial for transparency and allows for the methodology to be replicated in future studies. Example of Stratified Random Sampling Imagine that HubSpot wants to survey to assess user satisfaction among its customer base. Their customer population includes three key segments: small businesses, mid-sized businesses, and enterprise-level companies. Since these segments vary significantly in size and usage patterns, stratified random sampling ensures that insights are representative of each group’s experience. Population and Strata Breakdown The customer base is divided as follows: The survey aims to sample 1,000 customers while maintaining proportional representation across these segments. Sample Allocation Using proportional allocation, the number of customers to survey from each segment is calculated: Random Selection Within Strata Next, HubSpot uses random sampling techniques within each stratum to select participants: Small Businesses (Stratum 1): A random number generator selects 500 businesses from the list of 50,000 small-business customers. Mid-Sized Businesses (Stratum 2): From the mid-sized customer list of 30,000, 300 are chosen at random. Enterprise Businesses (Stratum 3): Lastly, 200 enterprise customers are randomly sampled from their segments. Visualization of Results Once data is collected, HubSpot analyzes the satisfaction scores for each segment to gain tailored insights and aggregate overall metrics. Here’s an example of how the results might be structured: The overall satisfaction score of 8.26 reflects a weighted average, ensuring that each segment’s voice is proportionally represented. Advantages of Stratified Random Sampling Stratified random sampling is a highly effective method for capturing the diversity within a population. By dividing the population into meaningful strata, researchers can ensure representation across various subgroups, which leads to more precise and reliable results. This approach also allows for deeper insights into specific subsets, reducing variability and potential bias. Here is a summary of its advantages: Improved Accuracy: Stratification reduces sampling error by ensuring subgroups are proportionately represented. Better Insights: It enables detailed analysis of subgroup characteristics, which might be overlooked in other sampling methods. Resource Efficiency: By focusing on specific subgroups, resources are used effectively to gather meaningful data. Fair Representation: Avoids the risk of bias that might result from over- or under-sampling certain population segments. Disadvantages of Stratified Random Sampling While it offers numerous benefits, stratified random sampling can be resource-intensive. It requires detailed knowledge of the population, additional time to define and organize strata, and a careful approach to ensure proportional sampling. This complexity can make it costly and challenging to execute, especially with large or poorly defined populations. Here is a summary of its disadvantages: Complexity: Stratifying the population and conducting random sampling for each subgroup can be time-consuming and labor-intensive. Requires Detailed Data: Accurate stratification depends on having prior knowledge of subgroup characteristics, which may not always be available. Limited to Measurable Variables: This method works best for variables like age or income but may not capture abstract traits like preferences or attitudes. Types of Stratified Random Sampling Stratified random sampling can be categorized based on how sample sizes are allocated to the strata. These types are particularly useful in tailoring sampling methods to meet the objectives of a study. 1. Proportional Stratified Sampling In proportional stratified sampling, the sample size from each stratum is proportional to its size in the overall population. This ensures that larger subgroups contribute more to the sample, accurately representing the entire population. Example: Suppose a SaaS company like HubSpot is conducting customer feedback research across four segments: small businesses, mid-sized businesses, enterprise-level clients, and freelancers. The total population consists of 6,000 customers distributed as follows: Small businesses: 500 customers Mid-sized businesses: 1,000 customers Enterprise clients: 1,500 customers Freelancers: 3,000 customers If the total sample size is 600, with a uniform sampling fraction of 1/10, the sample size for each stratum will be calculated as: This approach ensures representation proportional to each segment’s size in the population. 2. Disproportional Stratified Sampling Disproportional sampling involves allocating sample sizes to strata based on research priorities, rather than their proportion of the population. This approach is useful when specific strata need more focus or when smaller groups require adequate representation. Example: Continuing with the SaaS company scenario, imagine the company wants to emphasize feedback from small businesses and freelancers due to unique challenges in these groups. Sampling fractions for each stratum are adjusted accordingly: Here, the final sample size for each stratum is equal (250 respondents), even though the population sizes differ significantly. This approach allows for greater representation of strata deemed more critical to the research objectives. Simple vs. Stratified Random Sampling Simple random sampling and stratified random sampling are both effective tools in statistical analysis, but they serve different purposes and work best in distinct scenarios. Simple Random Sampling A simple random sample involves selecting individuals purely at random from the entire population, ensuring that each member has an equal chance of being included. This method is often chosen when: There is limited knowledge about the population’s characteristics. The population is homogeneous or lacks clear subgroups. The goal is to represent the entire population without focusing on specific subgroups. Example: Imagine a candy company with a customer base of 10,000 people. If it wants to study customer buying habits, it might randomly select 100 individuals and use the findings to make generalizations about the entire customer base. Stratified Random Sampling In stratified random sampling, the population is divided into subgroups, or strata, based on shared characteristics such as age, income, or geographic location. A sample is then taken from each stratum, ensuring proportional representation. This method is ideal when: The population is diverse, with distinct subgroups that might behave differently. Researchers aim to study differences or trends across subgroups. Precision and reduced variability are crucial. Example: Using the same candy company, suppose the company divides its customers into strata by age groups: children, young adults, and seniors. Each group might have different purchasing behaviors, so stratified sampling ensures all these differences are captured by sampling proportionally from each group. Key Differences To Wrap Things Up Stratified random sampling is a powerful method for improving the accuracy and relevance of research. By dividing the population into meaningful subgroups and sampling proportionately, researchers can reduce bias, gain deeper insights, and make better-informed decisions. This method is particularly useful in fields such as market research, healthcare, and social sciences, where subgroup differences can significantly impact outcomes. FAQs What is the difference between stratified random sampling and cluster sampling? Stratified sampling divides the population into subgroups and samples proportionately from each. Cluster sampling, on the other hand, divides the population into clusters and randomly selects entire clusters for study. What Are the Two Types of Stratified Random Sampling? The two main types are proportional stratified sampling, where sampling matches population proportions, and disproportional stratified sampling, where samples are drawn equally or based on specific needs. What is the difference between simple random sampling and stratified random sampling? Simple random sampling selects participants randomly from the entire population. Stratified random sampling divides the population into strata and samples within each subgroup. Sign up to our bi-monthly newsletter! Actionable eCommerce insights only. Master what matters most in eCommerce ✅ Get more loyal customers ✅ Improve Customer Lifetime Value ✅ Maximize profits Discover all features 30-day free trial, no credit card necessary. 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https://math.libretexts.org/Bookshelves/Algebra/Advanced_Algebra/09%3A_Sequences_Series_and_the_Binomial_Theorem/9.03%3A_Geometric_Sequences_and_Series
Skip to main content 9.3: Geometric Sequences and Series Last updated : Mar 22, 2024 Save as PDF 9.2: Arithmetic Sequences and Series 9.4: Binomial Theorem Page ID : 6249 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Identify the common ratio of a geometric sequence. Find a formula for the general term of a geometric sequence. Calculate the nth partial sum of a geometric sequence. Calculate the sum of an infinite geometric series when it exists. Geometric Sequences A geometric sequence18, or geometric progression19, is a sequence of numbers where each successive number is the product of the previous number and some constant r. an=ran−1GeometricSequence And because anan−1=r, the constant factor r is called the common ratio20. For example, the following is a geometric sequence, 9,27,81,243,729… Here a1=9 and the ratio between any two successive terms is 3. We can construct the general term an=3an−1 where, a1=9a2=3a1=3(9)=27a3=3a2=3(27)=81a4=3a3=3(81)=243a5=3a4=3(243)=729⋮ In general, given the first term a1 and the common ratio r of a geometric sequence we can write the following: a2=ra1a3=ra2=r(a1r)=a1r2a4=ra3=r(a1r2)=a1r3a5=ra3=r(a1r3)=a1r4⋮ From this we see that any geometric sequence can be written in terms of its first element , its common ratio , and the index as follows: an=a1rn−1GeometricSequence In fact, any general term that is exponential in n is a geometric sequence. Example 9.3.1: Find an equation for the general term of the given geometric sequence and use it to calculate its 10th term: 3,6,12,24,48… Solution Begin by finding the common ratio , r=63=2 Note that the ratio between any two successive terms is 2. The sequence is indeed a geometric progression where a1=3 and r=2. an=a1rn−1=3(2)n−1 Therefore, we can write the general term an=3(2)n−1 and the 10th term can be calculated as follows: a10=3(2)10−1=3(2)9=1,536 Answer: an=3(2)n−1;a10=1,536 The terms between given terms of a geometric sequence are called geometric means21. Example 9.3.2: Find all terms between a1=−5 and a4=−135 of a geometric sequence. In other words, find all geometric means between the 1st and 4th terms. Solution Begin by finding the common ratio r. In this case, we are given the first and fourth terms: an=a1rn−1Usen=4a4=a1r4−1a4=a1r3 Substitute a1=−5 and a4=−135 into the above equation and then solve for r. −135=−5r327=r33=r Next use the first term a1=−5 and the common ratio r=3 to find an equation for the nth term of the sequence. an=a1rn−1an=−5(3)n−1 Now we can use an=−5(3)n−1 where n is a positive integer to determine the missing terms. a1=−5(3)1−1=−5⋅30=−5a2=−5(3)2−1=−5⋅31=−15a3=−5(3)3−1=−5⋅32=−45a4=−5(3)4−1=−5⋅33=−135}geometricmeans Answer: −15,−45 The first term of a geometric sequence may not be given. Example 9.3.3: Find the general term of a geometric sequence where a2=−2 and a5=2125. Solution To determine a formula for the general term we need a1 and r. A nonlinear system with these as variables can be formed using the given information and an=a1rn−1:: {a2=a1r2−1a5=a1r5−1⟹{−2=a1rUsea2=−2.2125=a1r4Usea5=2125. Solve for a1 in the first equation, −2=a1r⇒−2r=a1 2125=a1r4 Substitute a1=−2r into the second equation and solve for r. 2125=a1r4 2125=(−2r)r4 2125=−2r3 −1125=r3 −15=r Back substitute to find a1: a1=−2r=−2(−15)=10 Therefore, a1=10 and r=−15. Answer: an=10(−15)n−1 Exercise 9.3.1 Find an equation for the general term of the given geometric sequence and use it to calculate its 6th term: 2,43,89,… Answer : an=2(23)n−1;a6=64243 www.youtube.com/v/IGPEl9vloLY Geometric Series A geometric series22 is the sum of the terms of a geometric sequence. For example, the sum of the first 5 terms of the geometric sequence defined by an=3n+1 follows: S5=5∑n=13n+1=31+1+32+1+33+1+34+1+35+1=32+33+34+35+36=9+27+81+35+36=1,089 Adding 5 positive integers is manageable. However, the task of adding a large number of terms is not. Therefore, we next develop a formula that can be used to calculate the sum of the first n terms of any geometric sequence. In general, Sn=a1+a1r+a1r2+…+a1rn−1 Multiplying both sides by r we can write, rSn=a1r+a1r2+a1r3+…+a1rn Subtracting these two equations we then obtain, Sn−rSn=a1−a1rn Sn(1−r)=a1(1−rn) Assuming r≠1 dividing both sides by (1−r) leads us to the formula for the nth partial sum of a geometric sequence23: Sn=a1(1−rn)1−r(r≠1) In other words, the nth partial sum of any geometric sequence can be calculated using the first term and the common ratio . For example, to calculate the sum of the first 15 terms of the geometric sequence defined by an=3n+1, use the formula with a1=9 and r=3. S15=a1(1−r15)1−r=9⋅(1−315)1−3=9(−14,348,906)−2=64,570,077 Example 9.3.4: Find the sum of the first 10 terms of the given sequence: 4,−8,16,−32,64,… Solution Determine whether or not there is a common ratio between the given terms. r=−84=−2 Note that the ratio between any two successive terms is −2; hence, the given sequence is a geometric sequence. Use r=−2 and the fact that a1=4 to calculate the sum of the first 10 terms, Sn=a1(1−rn)1−rS10=4[1−(−2)10]1−(−2)]=4(1−1,024)1+2=4(−1,023)3=−1,364 Answer: S10=−1,364 Example 9.3.5: Evaluate: ∑6n=12(−5)n. Solution In this case, we are asked to find the sum of the first 6 terms of a geometric sequence with general term an=2(−5)n. Use this to determine the 1st term and the common ratio r: a1=2(−5)1=−10 To show that there is a common ratio we can use successive terms in general as follows: r=anan−1=2(−5)n2(−5)n−1=(−5)n−(n−1)=(−5)1=−5 Use a1=−10 and r=−5 to calculate the 6th partial sum . Sn=a1(1−rn)1−rS6=−10[1−(−5)6]1−(−5)=−10(1−15,625)1+5=−10(−15,624)6=26,040 Answer: 26,040 Exercise 9.3.2 Find the sum of the first 9 terms of the given sequence: −2,1,−1/2,… Answer : S9=−171128 www.youtube.com/v/v-t3P95rWe8 If the common ratio r of an infinite geometric sequence is a fraction where |r|<1 (that is −1<r<1), then the factor (1−rn) found in the formula for the nth partial sum tends toward 1 as n increases. For example, if r=110 and n=2,4,6 we have, 1−(110)2=1−0.01=0.99 1−(110)4=1−0.0001=0.9999 1−(110)6=1−0.00001=0.999999 Here we can see that this factor gets closer and closer to 1 for increasingly larger values of n. This illustrates the idea of a limit, an important concept used extensively in higher-level mathematics, which is expressed using the following notation: limn→∞(1−rn)=1 where |r|<1 This is read, “the limit of (1−rn) as n approaches infinity equals 1.” While this gives a preview of what is to come in your continuing study of mathematics, at this point we are concerned with developing a formula for special infinite geometric series . Consider the nth partial sum of any geometric sequence, Sn=a1(1−rn)1−r=a11−r(1−rn) If |r|<1 then the limit of the partial sums as n approaches infinity exists and we can write, Sn=a11−r(1−rn)⟹n→∞S∞=a11−r⋅1 Therefore, a convergent geometric series24 is an infinite geometric series where |r|<1; its sum can be calculated using the formula: S∞=a11−r Example 9.3.6: Find the sum of the infinite geometric series : 32+12+16+118+154+… Solution Determine the common ratio , Since the common ratio r=13 is a fraction between −1 and 1, this is a convergent geometric series . Use the first term a1=32 and the common ratio to calculate its sum S∞=a11−r=321−(13)=3323=32⋅32=94 Answer: S∞=94 Note In the case of an infinite geometric series where |r|≥1, the series diverges and we say that there is no sum. For example, if an=(5)n−1 then r=5 and we have S∞=∑∞n=1(5)n−1=1+5+25+⋯ We can see that this sum grows without bound and has no sum. Exercise 9.3.3 Find the sum of the infinite geometric series : ∑∞n=1−2(59)n−1 Answer : −92 www.youtube.com/v/KxsPVUyle\_A A repeating decimal can be written as an infinite geometric series whose common ratio is a power of 1/10. Therefore, the formula for a convergent geometric series can be used to convert a repeating decimal into a fraction . Example 9.3.7: Write as a fraction : 1.181818… Solution Begin by identifying the repeating digits to the right of the decimal and rewrite it as a geometric progression . 0.181818…=0.18+0.0018+0.000018+…=18100+1810,000+181,000,000+… In this form we can determine the common ratio , r=1810,00018100=1810,000×10018=1100 Note that the ratio between any two successive terms is 1100. Use this and the fact that a1=18100 to calculate the infinite sum: S∞=a11−r=181001−(1100)=1810090100=18100⋅10099=211 Therefore, 0.181818…=211 and we have, 1.181818…=1+211=1211 Answer: 1211 Example 9.3.8: A certain ball bounces back to two-thirds of the height it fell from. If this ball is initially dropped from 27 feet, approximate the total distance the ball travels. Solution We can calculate the height of each successive bounce: 27⋅23=18 feet Heightofthefirstbounce18⋅23=12 feetHeightofthesecondbounce12⋅23=8 feet Heightofthethirdbounce The total distance that the ball travels is the sum of the distances the ball is falling and the distances the ball is rising. The distances the ball falls forms a geometric series , 27+18+12+…Distancetheballisfalling where a1=27 and r=23. Because r is a fraction between −1 and 1, this sum can be calculated as follows: S∞=a11−r=271−23=2713=81 Therefore, the ball is falling a total distance of 81 feet. The distances the ball rises forms a geometric series , 18+12+8+⋯Distancetheballisrising where a1=18 and r=23. Calculate this sum in a similar manner: S∞=a11−r=181−23=1813=54 Therefore, the ball is rising a total distance of 54 feet. Approximate the total distance traveled by adding the total rising and falling distances: 81+54=135 feet Answer: 135 feet Key Takeaways A geometric sequence is a sequence where the ratio r between successive terms is constant. The general term of a geometric sequence can be written in terms of its first term a1, common ratio r, and index n as follows: an=a1rn−1. A geometric series is the sum of the terms of a geometric sequence. The nth partial sum of a geometric sequence can be calculated using the first term a1 and common ratio r as follows: Sn=a1(1−rn)1−r. The infinite sum of a geometric sequence can be calculated if the common ratio is a fraction between −1 and 1 (that is |r|<1) as follows: S∞=a11−r. If |r|≥1, then no sum exists. Exercise 9.3.4 Write the first 5 terms of the geometric sequence given its first term and common ratio . Find a formula for its general term. a1=1;r=5 a1=1;r=3 a1=2;r=3 a1=5;r=4 a1=2;r=−3 a1=6;r=−2 a1=3;r=23 a1=6;r=12 a1=1.2;r=0.6 a1=−0.6;r=−3 Answer : 1. 1,5,25,125,625;an=5n−1 3. 2,6,18,54,162;an=2(3)n−1 5. 2,−6,18,−54,162;an=2(−3)n−1 7. 3,2,43,89,1627;an=3(23)n−1 9. 1.2,0.72,0.432,0.2592,0.15552;an=1.2(0.6)n−1 Exercise 9.3.5 Given the geometric sequence, find a formula for the general term and use it to determine the 5th term in the sequence. 7,28,112,… −2,−10,−50,… 2,12,18,… 1,25,425,… 8,4,2,… 6,2,23,… −1,23,−49,… 2,−32,98,… 13,−2,12,… 25,−2,10,… −3.6,−4.32,−5.184,… 0.8,−2.08,5.408,… Find the general term and use it to determine the 20th term in the sequence: 1,x2,x24,… Find the general term and use it to determine the 20th term in the sequence: 2,−6x,18x2… The number of cells in a culture of a certain bacteria doubles every 4 hours. If 200 cells are initially present, write a sequence that shows the population of cells after every nth 4-hour period for one day. Write a formula that gives the number of cells after any 4-hour period. A certain ball bounces back at one-half of the height it fell from. If this ball is initially dropped from 12 feet, find a formula that gives the height of the ball on the nth bounce and use it to find the height of the ball on the 6th bounce. Given a geometric sequence defined by the recurrence relation an=4an−1 where a1=2 and n>1, find an equation that gives the general term in terms of a1 and the common ratio r. 18. Given the geometric sequence defined by the recurrence relation an=6an−1 where a1=12 and n>1, find an equation that gives the general term in terms of a1 and the common ratio r. Answer : 1. an=7(4)n−1,a5=1,792 3. an=2(14)n−1,a5=1128 5. an=8(12)n−1,a5=12 7. an=−(−23)n−1,a5=−1681 9. an=13(−6)n−1,a5=432 11. an=−3.6(1.2)n−1,a5=−7.46496 13. an=(x2)n−1;a20=x19219 15. 400 cells; 800 cells; 1,600 cells; 3,200 cells; 6,400 cells; 12,800 cells; pn=400(2)n−1 cells 17. an=2(4)n−1 Exercise 9.3.6 Given the terms of a geometric sequence, find a formula for the general term. a1=−3 and a6=−96 a1=5 and a4=−40 a1=−2 and a8=−164 a1=34 and a4=−136 a2=18 and a5=486 a2=10 and a7=320 a4=−2 and a9=64 a3=−43 and a6=3281 a5=153.6 and a8=9,830.4 a4=−2.4×10−3 and a9=−7.68×10−7 Answer : 1. an=−3(2)n−1 3. an=−2(12)n−1 5. an=6(3)n−1 7. an=14(−2)n−1 9. an=0.6(4)n−1 Exercise 9.3.7 Find all geometric means between the given terms. a1=2 and a4=250 a1=13 and a6=−196 a2=−20 and a5=−20,000 a3=49 and a6=−16,807 Answer : 1. 10,50 3. −200;−2,000 Exercise 9.3.8 Calculate the indicated sum. an=2n+1;S12 an=(−2)n+1;S12 an=(12)n;S7 an=(23)n−1;S6 an=5(−3)n−1;S5 an=−7(−4)n;S5 an=2(−14)n;S5 an=13(2)n+1;S10 ∑5n=15n ∑6n=1(−4)n ∑10k=12k+1 ∑14k=12k−1 ∑10k=1−2(3)k ∑8k=15(−2)k ∑5n=12(12)n+2 ∑4n=1−3(23)n an=(15)n;S∞ an=(23)n−1;S∞ an=2(−34)n−1;S∞ an=3(−16)n;S∞ an=−2(12)n+1;S∞ an=−13(−12)n;S∞ ∑∞n=12(13)n−1 ∑∞n=1(15)n ∑∞n=13(2)n−2 ∑∞n=1−14(3)n−2 ∑∞n=112(−16)n ∑∞n=113(−25)n Answer : 1. 16,380 3. 127128 5. 305 7. −205512 9. 3,905 11. 4,092 13. −177,144 15. 3164 17. 14 19. 87 21. −1 23. 3 25. No sum 27. −114 Exercise 9.3.9 Write as a mixed number. 1.222… 5.777… 2.252525… 3.272727… 1.999… 1.090909… Suppose you agreed to work for pennies a day for 30 days. You will earn 1 penny on the first day, 2 pennies the second day, 4 pennies the third day, and so on. How many total pennies will you have earned at the end of the 30 day period? What is the dollar amount? An initial roulette wager of $100 is placed (on red) and lost. To make up the difference , the player doubles the bet and places a $200 wager and loses. Again, to make up the difference , the player doubles the wager to $400 and loses. If the player continues doubling his bet in this manner and loses 7 times in a row, how much will he have lost in total? 9. A certain ball bounces back to one-half of the height it fell from. If this ball is initially dropped from 12 feet, approximate the total distance the ball travels. 10. A golf ball bounces back off of a cement sidewalk three-quarters of the height it fell from. If the ball is initially dropped from 8 meters, approximate the total distance the ball travels. 11. A structured settlement yields an amount in dollars each year, represented by n, according to the formula pn=6,000(0.80)n−1. What is the total amount gained from the settlement after 10 years? 12. Beginning with a square, where each side measures 1 unit, inscribe another square by connecting the midpoints of each side. Continue inscribing squares in this manner indefinitely, as pictured: Find the sum of the area of all squares in the figure. (Hint: Begin by finding the sequence formed using the areas of each square.) Answer : 1. 129 3. 22599 5. 2 7. 1,073,741,823 pennies; $10,737,418.23 9. 36 feet 11. $26,778.77 Exercise 9.3.10 Categorize the sequence as arithmetic, geometric, or neither. Give the common difference or ratio, if it exists. −12,24,−48,… −7,−5,−3,… −3,−11,−19,… 4,9,16,… 2,32,43,… 43,89,1627,… 16,−16,−12,… 13,14,316,… 12,14,16… −110,−15,−310,… 1.26,0.252,0.0504,… 0.02,0.08,0.18,… 1,−1,1,−1,… 0,0,0,… Answer : 1. Geometric; r=−2 3. Arithmetic; d=−8 5. Neither 7. Arithmetic; d=−13 9. Neither 11. Geometric; r=0.2 13. Geometric; r=−1 Exercise 9.3.11 Categorize the sequence as arithmetic or geometric, and then calculate the indicated sum. an=3(5)n−1;S8 an=5−6n;S22 an=2n;S14 an=2n;S10 an=−2(17)n−1;S∞ an=−2+17n;S8 Answer : 1. Geometric; 292,968 3. Arithmetic; 210 5. Geometric; −73 Exercise 9.3.12 Calculate the indicated sum. ∑50n=1(3n−5) ∑25n=1(4−8n) ∑12n=1(−2)n−1 ∑∞n=15(−12)n−1 ∑40n=15 ∑∞n=10.6n Answer : 1. 3,575 3. −1,365 5. 200 Exercise 9.3.13 Use the techniques found in this section to explain why 0.999…=1. Construct a geometric sequence where r=1. Explore the nth partial sum of such a sequence. What conclusions can we make? Answer : 1. Answer may vary Footnotes 18A sequence of numbers where each successive number is the product of the previous number and some constant r. 19Used when referring to a geometric sequence. 20The constant r that is obtained from dividing any two successive terms of a geometric sequence; anan−1=r. 21The terms between given terms of a geometric sequence. 22The sum of the terms of a geometric sequence. 23The sum of the first n terms of a geometric sequence, given by the formula: Sn=a1(1−rn)1−r,r≠1. 24An infinite geometric series where |r|<1 whose sum is given by the formula:S∞=a11−r. 9.2: Arithmetic Sequences and Series 9.4: Binomial Theorem
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https://www.learner.org/series/learning-math-geometry/what-is-geometry/part-c-folding-paper-40-minutes/
Skip to main content Skip to main content Private: Learning Math: Geometry Private: What Is Geometry? Part C: Folding Paper (40 minutes) Session 1, Part C Geometers distinguish between a drawing and a construction. Drawings are intended to aid memory, thinking, or communication, and they needn’t be much more than rough sketches to serve this purpose quite well. The essential element of a construction is that it is a kind of guaranteed recipe. It shows how a figure can be accurately drawn with a specified set of tools. A construction is a method, while a picture merely illustrates the method. The most common tools for constructions in geometry are a straightedge (a ruler without any markings on it) and a compass (used for drawing circles). In the problems below, your tools will be a straightedge and patty paper. You can fold the patty paper to create creases. Since you can see through the paper, you can use the folds to create geometric objects. Though your “straightedge” might actually be a ruler, don’t measure! Use it only to draw straight segments. See Note 4 below. Throughout this part of the session, use just a pen or pencil, your straightedge, and patty paper to complete the constructions described in the problems. Here is a sample construction with patty paper to get you started: To construct the midpoint of a line segment, start by drawing a line segment on the patty paper. Next, fold the paper so that the endpoints of the line segment overlap. This creates a crease in the paper. The intersection of the crease and the original line segment is the midpoint of the line segment. Constructions To construct a perpendicular line, consider that a straight line is a 180° angle. Can you cut that angle in half (since perpendicular lines form right angles, or 90° angles)? To construct a parallel line, you may need to construct another line before the parallel to help you. Problem C1 Draw a line segment. Then construct a line that is perpendicular to it parallel to it the perpendicular bisector of the segment (A perpendicular bisector is perpendicular to the segment and bisects it; that is, it goes through the midpoint of the segment, creating two equal segments.) Problem C2 Draw an angle on your paper. Construct its bisector. (An angle bisector is a ray that cuts the angle exactly in half, making two equal angles.) Constructing Triangles Problem C3 Illustrate each of these definitions with a sketch using four different triangles. Try to draw four triangles that are different in significant ways — different side lengths, angles, and types of triangles. The first one in definition (a) is done as an example. A triangle has three altitudes, one from each vertex. (An altitude of a triangle is a line segment connecting a vertex to the line containing the opposite side and perpendicular to that side.) A triangle has three medians. (A median is a segment connecting any vertex to the midpoint of the opposite side.) A triangle has three midlines. (A midline connects two consecutive midpoints.) When you construct medians, you need to do two things: First find the midpoint; then fold or draw a segment connecting that point to the opposite vertex. Except in the case of special triangles (such as an equilateral triangle, and one median in an isosceles triangle), you can’t construct a median with just one fold. When you construct altitudes, you need to construct a perpendicular to a segment, but not necessarily at the midpoint of that segment. And remember that the altitude may fall outside the triangle, so you might want to draw or fold an extension of the sides of the triangle to help you. Problem C4 Draw five triangles, each on its own piece of patty paper. Use one triangle for each construction below. Carefully construct the three altitudes of the first triangle. Carefully construct the three medians of the second triangle. Carefully construct the three midlines of the third triangle. Carefully construct the three perpendicular bisectors of the fourth triangle. Carefully construct the three angle bisectors of the fifth triangle. | | | --- | | | VIDEO SEGMENT In this video segment, participants construct the altitudes, medians, and midlines of their triangles. Compare your solutions to Problem C4 with those in this video segment. What are the similarities and differences in your results? What conjectures can you make about the constructions you’ve just completed? You can find this segment on the session video approximately 15 minutes and 40 seconds after the Annenberg Media logo. | Problems C3 and C4 and the Video Segment problems taken from Connected Geometry, developed by Educational Development Center, Inc. p. 32. © 2000 Glencoe/McGraw-Hill. Used with permission. www.glencoe.com/sec/math Concurrencies in Triangles When three or more lines meet at a single point, they are said to be concurrent. The following surprising facts are true for every triangle: The medians are concurrent; they meet at a point called the centroid of the triangle. (This point is the center of mass for the triangle. If you cut a triangle out of a piece of paper and put your pencil point at the centroid, you would be able to balance the triangle there.) The perpendicular bisectors are concurrent; they meet at the circumcenter of the triangle. (This point is the same distance from each of the three vertices of the triangles.) The angle bisectors are concurrent; they meet at the incenter of the triangle. (This point is the same distance from each of the three sides of the triangles.) The altitudes are concurrent; they meet at the orthocenter of the triangle. Triangles are the only figures where these concurrencies always hold. (They may hold for special polygons, but not for just any polygon of more than three sides.) We’ll revisit these points in a later session and look at some explanations for why some of these lines are concurrent. You’ll explore the derivation of such terms as incenter and circumcenter later in Session 5 of this course. TAKE IT FURTHER Problem C5 For each construction in parts a-d, start with a freshly drawn segment on a clean piece of patty paper. Then construct the following shapes: an isosceles triangle with your segment as one of the two equal sides an isosceles triangle whose base is your segment a square based on your segment an equilateral triangle based on your segment More Constructions Think about how you might construct the exact side lengths needed for these squares. For example, the first square will need a side length exactly one half the original. The third square is very difficult! Problem C6 Start with a square sheet of paper. Construct a square with exactly one-fourth the area of your original square. How do you know that the new square has one-fourth the area of the original square? Construct a square with exactly one half the area of your original square. How do you know that the new square has one half the area of the original square? Construct a square with exactly three-fourths the area of your original square. Problem C7 Recall that the centroid is the center of mass of a geometric figure. How could you construct the centroid of a square? Take It Further Problem C8 When you noticed concurrencies in the folds, were you sure that the segments were concurrent? What would convince you that, for example, the medians of every triangle really are concurrent? Notes Note 4 If you are working in a group, you may choose to do all of the construction problems as a group activity. Watch for someone with appropriate solutions (for example, folding the two endpoints to each other, rather than measuring). Ask that person to share the solution and explain why it will always work on any segment. That’s the goal for these problems: to come up with general methods that will always work and that don’t rely on measurement. At the end, leave at least 10 minutes to share methods, even if not everyone is done. Then make a list of conjectures that come from that problem. Solutions Problem C1 Start by drawing a line segment. Then do the following: a. Fold the paper so that one of the endpoints of the line segment lies somewhere on the line segment. The crease created defines a line perpendicular to the original line segment. b.Use the process above to construct a perpendicular line. Then use the same process to construct a line perpendicular to the new line, making sure that this second perpendicular is a different line from the original. Since this third line and the original are each perpendicular to the second line, they are parallel. c. Fold the paper so that the endpoints of the line segment overlap. Draw a line segment along the crease, intersecting the original line segment. This new line segment is perpendicular to the original one and bisects it, because we used the same process that we used to construct the midpoint in the sample construction. Problem C2 Draw an angle on a piece of paper. Next, fold the paper so that the two sides of the angle overlap. The crease created defines a bisector of the angle. Problem C3 For parts (a)-(c), draw several triangles, at least one of which has an obtuse angle (to see that the definitions make sense in general). Then draw in the altitudes. Repeat with medians. Repeat with midlines. a. altitudes: b. medians: c.midlines: Problem C4 Draw five triangles on separate pieces of patty paper, and then do the following: a.Pick a side. Fold the paper so that the crease is perpendicular to the side [see Problem C1(a)] and so that it goes through the vertex opposite the side. You may have to extend the line segments of your triangle if the triangle has an angle larger than 90°. (See illustration for an example of what this looks like.) Connect the side with the vertex along the crease. The line segment drawn is the altitude corresponding to the side chosen. Now repeat with the other two sides. b. Pick a side. Fold the paper so that the endpoints of the chosen side overlap. The midpoint of the side is the point where the side intersects the crease. Using a straightedge, connect the midpoint of the side with the vertex opposite it. Repeat with the other two sides. c. Pick a side. Find the midpoint of the side by following the construction of question (b). Repeat this construction with the other two sides. Using a straightedge, connect the consecutive midpoints. d. Pick a side. Construct a perpendicular bisector of the chosen side using the construction from Problem C1(c). Repeat with the other two sides. e. Pick an angle. Fold the paper so that the two sides of this angle overlap. The crease defines a ray that bisects the chosen angle. Repeat with the other two angles. Problem C5 Draw a line segment, and then do the following: a.Make a crease that goes through one of the endpoints of the original line segment. The crease will extend to the edges of the paper. Fold the paper along the crease and mark where the second endpoint of the line segment overlaps with the folded piece. Then unfold the paper and connect the marked point with the point where the original line segment and the crease intersect. Next, connect the two “free” ends of the two line segments with a straight line. The result is an isosceles triangle, where the original line segment is one of its two equal sides. b.Construct a perpendicular bisector of the line segment. Choose any point on the perpendicular bisector and connect it with the endpoints of the original line segment. The resulting triangle is isosceles and has the original line segment as its base. c. Extend the line segment to form a line, being sure to mark the original endpoints of the line segment. Use this line to construct a perpendicular line through the endpoints of the original line segment [see Problem C1(a)]. You should now have two parallel lines, each perpendicular to the original line segment at the endpoints. Then fold the paper so that the original line segment overlaps the first perpendicular line you drew. Mark the point on the perpendicular line where the second endpoint falls on this line. (This defines one of the equal, perpendicular sides.) Perform the same process on the second perpendicular line to define the third side of the square. Finally, use the straightedge to connect the two points you marked on the perpendicular lines to define the fourth side of the square. d.Construct the perpendicular bisector of the segment. Pick one of the two endpoints of the original line segment. Fold the paper along a line that contains the endpoint such that the other endpoint falls on the perpendicular bisector. Flip your patty paper, and then mark that spot on the bisector. Connect the marked spot with the two endpoints of the original line segment. Problem C6 a. Fold the paper in half to make a rectangle. Fold it in half again by bisecting the longer sides of the rectangle. The resulting square has one-fourth the area of the original one. There are exactly four squares that fit exactly on top of each other, so they must have the same area. Since together they completely make up the original square, each must be one-fourth of the original square. b. Find the midpoints of all four sides. Connect the consecutive midpoints. The resulting square has one-half the area of the original square. To see this, connect the diagonals of the new square. You will see four triangles inside the square and four triangles outside, all of which have the same area. Half the area of the original square is inside the new square. c. In order to obtain a square with exactly three-fourths the area of the original square (sides = 1) we need to calculate the sides of the new square: a • a = 3/4 a2 = 3/4 a = √3/2 So we are looking to construct a square whose sides are equal to √3/2. Start with a piece of patty paper. Think of the bottom edge as the base of an equilateral triangle. Fold the vertical midline. The third vertex of the equilateral triangle will be on this midline. Bring the lower right vertex up to the midline, so that the entire length of the bottom edge is copied from the lower left vertex to the midline. The distance from the bottom of the midline to this mark is √3/2. It is the height of an equilateral triangle with the side length equal to 1. This can also be easily calculated using the Pythagorean theorem for the right triangle with the sides 1 and 1/2. Next, fold down from this mark. This is the height of the square. Fold down the corner so that you can fold the right side in the same amount as the top. Fold over the right side. This square has area 3/4 of the original square. Problem C7 One way to do this is to use a straightedge to draw the two diagonals of the square. The centroid is the point of their intersection. Another is to draw the perpendicular bisectors of two consecutive sides of the square [the same construction as Problem C6(a)]. The intersection of these bisectors is the same centroid. Problem C8 Noticing what appear as concurrencies in the folds may lead one to conjecture that concurrencies occur in general. Keep in mind any one construction that suggests this is a special case. Therefore, in order to convince ourselves that they do occur in general, we need to construct a formal proof. Series Directory Private: Learning Math: Geometry 2 Triangles and Quadrilaterals 1 Part A: Different Triangles (20 minutes) 2 Part B: Linkage-Strip Constructions (40 minutes) 3 Part C: Building Towers (45 minutes) 4 Homework 3 Polygons 1 Part A: Hidden Polygons (20 minutes) 2 Part B: Classifying Polygons (40 minutes) 3 Part C: Definitions and Proof (45 minutes) 4 Homework 4 Parallel Lines and Circles 1 Part A: Introduction to Geometer’s Sketchpad (25 minutes) 2 Part B: Parallel Lines (45 minutes) 3 Part C: Circles 4 Part D: Thinking with Technology (15 minutes) 5 Homework 5 Dissections and Proof 1 Part A: Tangrams (15 minutes) 2 Part B: Cutting Up (50 minutes) 3 Part C: The Midline Theorem (55 minutes) 4 Homework 6 The Pythagorean Theorem 1 Part A: The Pythagorean Theorem (20 minutes) 2 Part B: Proving the Pythagorean Theorem (65 minutes) 3 Part C: Applications of the Pythagorean theorem (35 minutes) 4 Homework 7 Symmetry 1 Part A: Line Symmetry (30 minutes) 2 Part B: Rotation Symmetry (30 minutes) 3 Part C: Translation Symmetry and Frieze Patterns (60 minutes) 4 Homework 8 Similarity 1 Part A: Scale Drawings (50 minutes) 2 Part B: Similar Triangles (35 minutes) 3 Part C: Trigonometry (35 minutes) 4 Homework 9 Solids 1 Part A: Platonic Solids (45 minutes) 2 Part B: Nets (30 minutes) 3 Part C: Cross Sections (45 minutes) 4 Homework 10 Classroom Case Studies, K-2 1 Part A: Geometry as a Problem Solving Process (25 minutes) 2 Part B: Developing Geometric Reasoning (40 minutes) 3 Part C: Activities That Illustrate Geometric Reasoning (55 minutes) 4 Homework 11 Classroom Case Studies, 3-5 1 Part A: Geometry as a Problem-Solving Process (25 minutes) 2 Part B: Developing Geometric Reasoning (40 minutes) 3 Part C: Problems That Illustrate Geometric Reasoning (55 minutes) 4 Homework 12 Classroom Case Studies, 6-8 1 Part A: Geometry and Reasoning (25 minutes) 2 Part B: Developing Geometric Reasoning (40 minutes) 3 Part C: Problems That Illustrate Geometric Reasoning (55 minutes) 4 Homework Credits Produced by WGBH Educational Foundation. 2003. Closed Captioning ISBN: 1-57680-597-2 Sections ### 1.1 Part A: Quick Images (10 minutes) Session 1, Part A ### 1.2 Part B: Building from Directions (35 minutes) ### 1.3 Part C: Folding Paper (40 minutes) Session 1, Part C ### 1.4 Part D: Basic Objects (20 minutes) Session 1, Part D In Part C, you used dots on paper to represent points, segments drawn by a pen, and folds to represent ideal line segments and even lines. ### 1.5 Homework Session 1, Homework ### Supplementary: Learning Math: Geometry — Perspective: What is Geometry? Suggested Reading Sessions ### Session 2 Triangles and Quadrilaterals Learn about the classifications of triangles, their different properties, and relationships between them. Examine concepts such as triangle inequality, triangle rigidity, and side–side–side congruence, and look at the conditions that cause them. Compare how these concepts apply to quadrilaterals. Explore properties of triangles and quadrilaterals through practical applications such as building structures. ### Session 3 Polygons Explore the properties of polygons through puzzles and games, then proceed into a more formal classification of polygons. Look at mathematical definitions more formally, and explore how terms can have different but equivalent definitions. ### Session 4 Parallel Lines and Circles Use dynamic geometry software to construct figures with given characteristics, such as segments that are perpendicular, parallel, or of equal length, and to examine the properties of parallel lines and circles. Look past formal definitions and discover the properties and relationships among geometric figures for yourself. ### Session 5 Dissections and Proof Review and explore transformations such as translation, reflection, and rotation. Apply these ideas to solve more complex geometric problems. Use your knowledge of properties of figures to reason through, solve, and justify your solutions to problems. Analyze and prove the midline theorem. ### Session 6 The Pythagorean Theorem Continue to examine the idea of mathematical proof. Look at several geometric or algebraic proofs of one of the most famous theorems in mathematics: the Pythagorean theorem. Explore different applications of the Pythagorean theorem, such as the distance formula. ### Session 7 Symmetry Investigate symmetry, one of the most important ideas in mathematics. Explore geometric notions of symmetry by creating designs and examining their properties. Investigate line symmetry and rotation symmetry; then learn about frieze patterns. ### Session 8 Similarity Examine your intuitive notions of what makes a "good copy" and then progress toward a more formal definition of similarity. Explore similar triangles and look into some applications of similar triangles, including trigonometry. ### Session 9 Solids Explore various aspects of solid geometry. Examine platonic solids and why there are a finite number of them. Investigate nets and cross-sections for solids as a way of establishing the relationships between two–dimensional and three–dimensional geometry. ### Session 10 Classroom Case Studies, K-2 This is the final session of the Geometry course! In this session, we will examine how geometry as a problem-solving process might look when applied to situations in your own classroom. This session is customized for three grade levels. ### Session 11 Classroom Case Studies, 3-5 This is the final session of the Geometry course! In this session, we will examine how geometry as a problem-solving process might look when applied to situations in your own classroom. This session is customized for three grade levels. ### Session 12 Classroom Case Studies, 6-8 This is the final session of the Geometry course! In this session, we will examine how geometry as a problem-solving process might look when applied to situations in your own classroom. This session is customized for three grade levels. Watch Videos 11 and 12 in the 10th session for grade 6–8 teachers. Explore how the concepts developed in this course can be applied through case studies of grade 6–8 teachers (former course participants) who have adapted their new knowledge to their classrooms.
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https://math.stackexchange.com/questions/2116529/finding-the-last-two-digits-of-17198
Skip to main content Finding the last two digits of 17198 Ask Question Asked Modified 8 years, 7 months ago Viewed 4k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Question 1: What are the last two digits of 17198? My Attempt: The pdf hinted to reduce it via mod 100. So my work is as follows: Since 172≡289≡−11mod100, we have 17198≡(−11)99≡−1199mod100 However, I'm not sure how to compute −1199 without too much of a hassle. Questions: How do you complete this problem? Why use mod 100? Why not mod 50 or some other number? modular-arithmetic Share CC BY-SA 3.0 Follow this question to receive notifications edited Jan 27, 2017 at 15:05 Crescendo asked Jan 27, 2017 at 14:56 CrescendoCrescendo 4,44711 gold badge1818 silver badges4141 bronze badges 4 You want to use −1 rather than −11. But working in mod 10 will only give you the last digit. A number mod 100 gives the last 2 digits. Michael – Michael 2017-01-27 15:02:44 +00:00 Commented Jan 27, 2017 at 15:02 2 The hint was to work mod 100 but you used mod 10 RGS – RGS 2017-01-27 15:04:37 +00:00 Commented Jan 27, 2017 at 15:04 @RSerrao That mod 10 thing was just a stupid mistake made by me... It should be fixed now... Crescendo – Crescendo 2017-01-27 15:06:15 +00:00 Commented Jan 27, 2017 at 15:06 Try doing mod 100 of a few arbitrary numbers with different number of digits. kingW3 – kingW3 2017-01-27 15:10:44 +00:00 Commented Jan 27, 2017 at 15:10 Add a comment | 6 Answers 6 Reset to default This answer is useful 3 Save this answer. Show activity on this post. An alternative approach, exploiting the Chinese remainder theorem and the fact that 2 is a generator in Z/(25Z)∗: 17198≡(−8)198≡23⋅198≡214≡16⋅1024≡−16≡9(mod25) 17198≡1(mod4) together imply 17198≡9(mod100). Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 27, 2017 at 15:23 Jack D'AurizioJack D'Aurizio 371k4141 gold badges418418 silver badges883883 bronze badges Add a comment | This answer is useful 2 Save this answer. Show activity on this post. Reducing an integer mod. 100 consists exactly in taking its last two digits ! Indeed, if we consider n∈N and divide it by 100, we get something like n=100q+r with 0≤r<100. Now, for your example, you mean mod. 100 and not mod. 10 ... 17198≡−1199(mod.100) But 1199=(113)33=(1331)33≡3133=(313)11=2979111≡9111 and 912=8281, so 912≡81. Then 9110=(912)5≡815=3486784401≡1 Finally 1199≡91(mod.100) and so 17198≡9(mod.100) Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 27, 2017 at 15:14 AdrenAdren 8,0961212 silver badges2929 bronze badges Add a comment | This answer is useful 2 Save this answer. Show activity on this post. Note 172=−1+10k⇒1720≡(−1+10k)10≡(−1)10Binomial Theorem≡1(mod100) thus we conclude 17198≡(1720)10172≡1−11≡9−99≡9(mod100) Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jan 27, 2017 at 15:20 answered Jan 27, 2017 at 15:14 Bill DubuqueBill Dubuque 283k4242 gold badges338338 silver badges1k1k bronze badges 1 The first step is special case of LTE = Lifting The Exponent [172≡−1(mod10)]10⇒1720≡(−1)10(mod102) Bill Dubuque – Bill Dubuque 2024-10-23 17:31:03 +00:00 Commented Oct 23, 2024 at 17:31 Add a comment | This answer is useful 1 Save this answer. Show activity on this post. Hint: First note that −11≡−1(mod10). There's only two possible integer powers of −1. For your second question, we use (mod100) because the last two digits of any integer is exactly the remainder when divided by 100. Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 27, 2017 at 15:04 Yiyuan LeeYiyuan Lee 14.6k55 gold badges4141 silver badges7272 bronze badges 3 So does that mean that mod 1000 gives you the last three digits, mod 10000 gives you the last four digits, and so on? Crescendo – Crescendo 2017-01-27 15:06:53 +00:00 Commented Jan 27, 2017 at 15:06 @Crescendo yepper The Count – The Count 2017-01-27 15:09:01 +00:00 Commented Jan 27, 2017 at 15:09 Okay, cool. And also, what about a number of the form abcmodn? How would you reduce? Is it the same as ab where you reduce a and then simplify? Or is there a quicker way? Crescendo – Crescendo 2017-01-27 15:18:56 +00:00 Commented Jan 27, 2017 at 15:18 Add a comment | This answer is useful 1 Save this answer. Show activity on this post. For the first question, you may want to refer to this Wikipedia page where you are presented with some different methods. My favorite one and the most pratical is under the title "Right-to-left binary method". As for the second question, what is the remainder of 56 when divided by 100? And what about 543? And 642058311? Note that 56=0⋅100+56,543=5⋅100+43,642058311=6420583⋅100+11. If you take those equalities mod100, everything but the last two digits of each number is reduced to 0. Share CC BY-SA 3.0 Follow this answer to receive notifications edited Jan 27, 2017 at 15:20 answered Jan 27, 2017 at 15:11 RGSRGS 9,88322 gold badges2121 silver badges3535 bronze badges Add a comment | This answer is useful -1 Save this answer. Show activity on this post. Let G=(Z/100Z)∗. Since |G|=40 and 17∈G, we have 17198=1740⋅5⋅17−2=1⋅17−2 mod 100 To find 17−2 mod 100 we note first that 17⋅3=1 mod 10. Then, we only have to solve 17⋅(10x+3)=1 mod 100 for x∈[0,9]∩Z, in order to find the inverse. Doing some computations, we find 70x=50 mod 100 x=5 solves the equation and one can verify directly that 17⋅53=1 mod 100. We deduce 17198=532=9 mod 100 Share CC BY-SA 3.0 Follow this answer to receive notifications answered Jan 27, 2017 at 15:35 Veridian DynamicsVeridian Dynamics 2,5211111 silver badges1515 bronze badges 1 Generally Hensel lifting can be done very simply for inverses, e.g. here mod10: i.e. ⇒ mod102: 17−1≡3 17⋅3≡1+5⋅10117≡× 3|31+5⋅10≡3(1−5⋅10)≡53 Bill Dubuque – Bill Dubuque 2024-10-23 17:58:51 +00:00 Commented Oct 23, 2024 at 17:58 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 modular-arithmetic See similar questions with these tags. Featured on Meta stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 2 Computing inverses modnk by Hensel lifting or Newton's method 0 Stuck on middle step of proof that ap(p−1)=1(modp2) -2 Is it true that if gcd(a,c)=1 then ab(modc) reduces to ab(modφ(c))(modc)? 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elementary number theory - Find all $a,b,c \in \Bbb N, a,b,c \gt 1$, such that $a|(bc - 1) , b|(ac - 1)$ and $c|(ab - 1)$. - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Find all a,b,c \in \Bbb N, a,b,c \gt 1, such that a|(bc - 1) , b|(ac - 1) and c|(ab - 1). Ask Question Asked 3 months ago Modified3 months ago Viewed 144 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. \begingroup Find all a,b,c \in \Bbb N, a,b,c \gt 1, such that a|(bc - 1) , b|(ac - 1) and c|(ab - 1). This is my first post here. I enjoy grappling with number theory type questions. I saw this problem in an old edition of the Mathematical Gazette (since lost). By trial and error, I found a solution, {2,3,5}. I think it's true to say that, for any such a,b \text{ and } c then abc\ |\ (ab + bc + ac - 1), but I can't use this to yield any further information about a,b \text{ or } c. My feeling is that there are no other solutions, but I can't prove it. I tried using AI, but it just quoted the trivial solutions {1,1,1}. Any hints/help would be great, thank you. elementary-number-theory Share Cite Follow Follow this question to receive notifications edited Jun 29 at 12:50 Bowei Tang 4,235 5 5 gold badges 15 15 silver badges 37 37 bronze badges asked Jun 29 at 12:44 The GooseThe Goose 31 3 3 bronze badges \endgroup 3 2 \begingroup I would expect this to be a duplicate, though I can't find a match. In any case, divide by abc to get an inequality for \frac 1a+\frac 1b+\frac 1c-\frac 1{abc}.\endgroup lulu –lulu 2025-06-29 13:37:57 +00:00 Commented Jun 29 at 13:37 \begingroup Welcome to Math SE. Like lulu, I also expect this to be a duplicate. Regarding your result of abc\mid (ab+bc+ac-1), related to lulu's suggestion, consider if it's possible for the minimum of a, b and c to be \ge 3.\endgroup John Omielan –John Omielan 2025-06-29 14:03:05 +00:00 Commented Jun 29 at 14:03 \begingroup Thank you for the welcome and reply John, and to everyone else who responded. It did not occur to me to consider magnitudes, being too focused on the properties of division.\endgroup The Goose –The Goose 2025-07-04 07:52:34 +00:00 Commented Jul 4 at 7:52 Add a comment| 5 Answers 5 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. \begingroup Okay, you got abc|ab+bc+ac -1. (I presume by abc|(bc-1)(ac-1)(ab-1)=a^2b^2c^2 -a^2bc - ab^2c -abc^2+ ab + bc + ac -1) That means that \frac {ab+bc+ac - 1}{abc}=\frac 1a + \frac 1b + \frac 1c -\frac 1{abc}=k is an integer. Now a,b,c > 1 so a,b,c \ge 2 so \frac 1a, \frac 1b, \frac 1c \le \frac 12 and so k < \frac 32. And as abc > a,b,c we have 0 < \frac 1a + \frac 1b +\frac 1c -\frac 1{abc} =k < \frac 32. So \frac 1a+\frac 1b + \frac 1c -\frac 1{abc} =1 Note if a,b,c \ge 3 then \frac 1a + \frac 1b + \frac 1c -\frac 1{abc} < \frac 13 + \frac 13 + \frac 13 =1 so we must have at least one of a,b,c equal to 2. Wolog, let a=2. Now we have \frac 12 + \frac 1b + \frac 1c -\frac 1{2bc}=1 or \frac 1b + \frac 1c -\frac 1{2bc}=\frac 12 Now 2bc > c so \frac 12 = \frac 1b + (\frac 1c -\frac 2bc) > \frac 1b. So b> 2 and by the same argument c > 2 so b,c \ge 3. If bc \ge 4 we get \frac 12 =\frac 1b + \frac 1c -\frac 1{2bc}> \frac 14 + \frac 14 = \frac 12 so at least one term must be equal to 3. Wolog b=3. So now we have \frac 13 +\frac 1c - \frac 1{6c} =\frac 12. Well, just solve for c. \frac 5{6c} = \frac 16 and c = 5. So {2,3,5} are the only solutions where all are greater than 1. But note. We haven't tried to solve for any non-trivial answer where they aren't all greater than one. To do that our assuming that a,b,c \ge 2 and \frac 1a+\frac 1b + \frac 1c -\frac 1bc < \frac 32 is not valid. But we do have \frac 1a+\frac 1b + \frac 1c -\frac 1bc < 3. If we set a=1 and k=1 we get \frac 1b + \frac 1c = \frac 1{bc} which is easily seen to be impossible. So k \ne 1. So setting a=1 and k=2 we get \frac 1b + \frac 1c = 1+\frac 1{bc}. We can't have both b,c \ge 2 so wolog b = 1 and then we get \frac 1c = \frac 1c and {1,1,c} will yield 1|c-1 and c|0. Will hold. So all solutions are: {1,1,c} for any c, and {2,3,5}. Share Cite Follow Follow this answer to receive notifications edited Jun 29 at 15:39 answered Jun 29 at 15:32 fleabloodfleablood 132k 5 5 gold badges 52 52 silver badges 142 142 bronze badges \endgroup Add a comment| This answer is useful 0 Save this answer. Show activity on this post. \begingroup I'll leave a possible answer to the problem here in case you want to read it. If I were to give you just a hint on how to continue, I'd say you should turn the relation abc \mid (ab+bc+ac-1) into an inequality and proceed from there. If a,b,c > 1, then the numbers must be pairwise different, so let's assume w.l.o.g. that 1 < a < b < c. From abc \mid (ab + bc + ac - 1), we deduce that bc \mid b + c + \frac{bc-1}{a}. In particular, turning this relation into an inequality, we get (after rearranging a bit) (bc-1) < \frac{a}{a-1}(b+c), so bc < \frac{2a}{a-1}c. In particular, b < \frac{2a}{a-1} \leq 4, so b \leq 3 and a = 2. Therefore, it suffices to find all solutons s.t. a = 2 and b=3. If this is the case, then a \mid (bc-1) implies that c is odd, so 5 \leq c, whereas c\mid(ab-1) implies c\leq 5. So indeed, {2,3,5} is the only solution. Share Cite Follow Follow this answer to receive notifications answered Jun 29 at 14:10 AndrésAndrés 342 10 10 bronze badges \endgroup Add a comment| This answer is useful 0 Save this answer. Show activity on this post. \begingroup You can say 2\le a\leq b\le c and we have abc\le ab+bc+ca-1 so if a+b <ab then c\le {ab-1\over ab-a-b} If f(x) = {ax-1\over ax-x-a} we see this rational function is decresing on [2,\infty) so c\leq f(b)\leq f(2)= {2a-1\over a-2} = g(a) Since g is decresing (2,\infty) we see that if a>3 we have a\leq c \leq g(4) = {7\over 3} So \boxed{a\le 3}. If a=3 we get c\mid 3b-1 so 3b-1=kc and k\leq 2. If k=2 then 2c=3b-1\implies b\mid 2(3c-1)=6c-2= 9b-5\implies b\mid 5\implies b=5 Then c=7 but 3\nmid 5\cdot 7-1 = 34, so no solution. If k=1 then c=3b-1\implies b\mid 9b-4\implies b\mid 4\implies b=2\vee b=4 Then c=5 if b=2 and this WORKS but b=4 does not. If a=2 we get c\mid 2b-1 so 2b-1=c. Since b\mid 2c-1 = 4b-3 we get b=3 and c=5 hich e already know WORKS. Share Cite Follow Follow this answer to receive notifications answered Jun 29 at 14:12 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges \endgroup Add a comment| This answer is useful 0 Save this answer. Show activity on this post. \begingroup Wlog a\leq b\leq c. Rewrite divisibility as ab+bc+ca-1 = kabc for some positive integer k. Clearly k\leq 2. If k=2 then c(2ab-a-b)=ab-1 Since 2ab-a-b>ab-1\implies c<1 If k=1 then c(ab-a-b)=ab-1 \implies ab-a-b\mid ab-1\implies ab-a-b \mid a+b-1\;\;()\implies ab-a-b \leq a+b-1 \implies b\le {2a-1\over a-2}\;\;{\rm if}\;\; a\ne 2 Say a\geq 4. Since f(a)= {2a-1\over a-2} is decreasing we have a\leq b\leq f(a) \leq f(4)= 7/3 a contradicition. So a=3 then b\leq 5 and we see: c= {3b-1\over 2b-3} that nothing works. We are left to see what happend if a=2 in (). In this case b-2\mid 2b-1\implies b-2\mid 3\implies b\in{-1,1,3,5} e see that only b=3 with c=5 works. Share Cite Follow Follow this answer to receive notifications answered Jun 29 at 15:26 nonusernonuser 92.1k 20 20 gold badges 110 110 silver badges 218 218 bronze badges \endgroup Add a comment| This answer is useful 0 Save this answer. Show activity on this post. \begingroup It is better to solve the problem in a general way. \left{!\begin{aligned} & ab+T=cq \ & ac+T=bt \ & bc+T=af \end{aligned}\right. We factor out the number T as follows. T=Ap((B-A)t-p(A+1)^2) Then the solutions can be written as follows. \left{!\begin{aligned} & a=(A+1)p \ & b=A(B+1)p \ & c=A(A+1)p+At \ & q=(B-A)p \ & t=t \ & f=A(AB-1)p+ABt \end{aligned}\right. Share Cite Follow Follow this answer to receive notifications answered Jun 29 at 17:21 individindivid 4,435 1 1 gold badge 16 16 silver badges 26 26 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 elementary-number-theory See similar questions with these tags. 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Published Time: 2023-05-24 名为 "取样偏差 "的问题 - Mind the Graph 博客 GalleryPoster MakerTemplatesPricingCustom DesignAbout UsBlog Login现在开始 Our Anniversary Party ends in 0-492 days 20 hours 51 minutes 50 seconds Learn More 溯源 一个叫做抽样偏差的问题 在统计学、社会科学和流行病学等学科内进行研究时，抽样偏差是一个重要的考虑因素。 Gilberto de Abreu 7分钟阅读 05/24/2023 无论使用什么方法或研究什么学科，研究人员都需要确保他们使用的是反映他们所研究人群特征的代表性样本。本文将探讨抽样偏见的概念，它的不同类型和应用方式，以及减轻其影响的最佳做法。 什么是取样偏差？ 抽样偏见是指人口中的某些个人或群体比其他个人或群体更有可能被纳入样本，从而导致有偏见或不具代表性的样本。这种情况的发生有多种原因，如非随机抽样方法、自我选择偏差或研究人员偏差。 换句话说，抽样偏见会破坏研究结果的有效性和普遍性，因为它使样本偏向于某些可能不代表更大人口的特征或观点。 理想情况下，你必须以随机的方式选择所有的调查参与者。然而，在实践中，由于成本和受访者的可用性等限制，可能很难做到随机选择参与者。即使你不做随机的数据收集，意识到你的数据中可能存在的偏差也是至关重要的。 采样偏差的一些例子包括： 志愿者偏向:自愿参与研究的参与者可能与不自愿参与的参与者有不同的特征，导致样本的非代表性。 非随机抽样:如果研究者只从某些地方选择参与者，或只选择具有某些特征的参与者，就会导致样本的偏颇。 生存能力的偏差:当一个样本只包括在某一特定情况下生存或成功的个人，而不包括那些没有生存或失败的人，就会出现这种情况。 方便取样:这种类型的抽样涉及到选择那些容易接近的参与者，比如那些碰巧在附近的人，或者那些回复在线调查的人，这可能不代表更大的人群。 确认性偏见:研究人员可能会无意识地或有意地选择支持其假设或研究问题的参与者，从而导致有偏见的结果。 霍桑效应:参与者在知道自己被研究或观察时，可能会改变自己的行为或反应，导致结果不具有代表性。 如果你意识到这些偏差，你可以在分析中考虑它们，做偏差校正，更好地理解你的数据所代表的人群。 采样偏差的类型 选择偏差当样本不能代表人口时就会发生。 测量偏差当收集到的数据不准确或不完整时，就会发生。 报告偏差当受访者提供不准确或不完整的信息时，就会出现这种情况。 非应答偏差不具代表性的样本：当人口中的一些成员没有对调查作出回应，导致不具代表性的样本。 导致抽样偏差的原因 方便取样挑选样本：基于方便而不是使用科学方法。 自我选择的偏差：只包括那些自愿参加调查的人，这可能不代表人口。 采样框架的偏差当用于选择样本的抽样框架不能代表人口时。 存活率偏差当只有人口中的某些成员参与时，会导致样本的不代表性。例如，如果研究人员只对活着的人进行调查，他们可能不会收到在研究进行之前就已经死亡的人的意见。 由于缺乏知识而产生的抽样偏差：没有认识到可能导致有偏见的估计的变异性来源。 由于样本管理的错误导致的抽样偏差未能使用适当的或运作良好的抽样框架，或拒绝参与研究，导致对样本的选择有偏见。 临床试验中的取样偏差 临床试验负责测试一种新的治疗方法或药物对特定人群的有效性。它们是药物开发过程中的一个重要部分，在向公众发布之前确定一种治疗方法是否安全和有效。然而，临床试验也很容易出现选择偏差。 当用于研究的样本不能代表所要代表的人群时，就会出现选择偏差。就临床试验而言，当参与者被有选择地选择参与或自我选择时，就会出现选择偏差。 假设一家制药公司正在进行一项临床试验，测试一种新的癌症药物的疗效。他们决定通过在医院、诊所和癌症支持团体的广告，以及通过在线申请来招募研究参与者。然而，他们收集的样本可能偏向于那些更愿意参加试验的人或患有某种类型癌症的人。这可能会使研究结果难以推广到更大的人群。 为了尽量减少临床试验中的选择偏差，研究人员必须执行严格的纳入和排除标准以及随机选择程序。这将确保为研究选择的参与者样本能够代表更大的人群，最大限度地减少所收集的数据中的任何偏差。 由于抽样偏差而产生的问题 采样偏差是有问题的，因为对样本计算的统计数字有可能是系统性错误的。它可能导致对人口中相应参数的系统性高估或低估。这在实践中会发生，因为实际上不可能确保抽样的完全随机性。 如果失实程度较小，那么可以将样本作为一个合理的近似于随机样本。此外，如果样本在被测量的数量上没有明显的差异，那么有偏见的样本仍然可以是一个合理的估计。 虽然有些人可能故意使用有偏见的样本来产生误导性的结果，但更多时候，有偏见的样本只是反映了难以获得真正有代表性的样本，或者是对其测量或分析过程中的偏见一无所知。 外推法：超出范围 在统计学中，对超出数据范围的事物得出结论被称为外推法。从有偏见的样本中得出结论是外推的一种形式：因为抽样方法系统地排除了所考虑的人口的某些部分，所以推论只适用于抽样的子人口。 例如，如果将基于大学本科生样本的推断应用于老年人或仅受过八年级教育的成年人，也会出现外推现象。推断是应用或解释统计学的一个常见错误。有时，由于很难或不可能获得好的数据，推断是我们能做的最好的事情，但它总是需要至少带着一粒盐--而且常常带着大剂量的不确定性。 从科学到伪科学 正如维基百科上提到的，一个关于无知的偏见如何存在的例子是在广泛使用比值（又称折叠变化）作为生物学差异的衡量标准中。因为用两个有一定差异的小数字实现大比率比较容易，而用两个有较大差异的大数字实现大比率相对来说比较困难，所以在比较相对较大的数字测量时，可能会遗漏较大的显著差异。 有些人把这称为 "分界偏见"，因为使用比值（除法）而不是差值（减法）使分析结果从科学变为伪科学。 一些样本使用了有偏见的统计设计，但还是可以对参数进行估计。例如，美国国家卫生统计中心在其许多全国性调查中故意对少数民族人口进行超额采样，以便在这些群体中获得足够的估计精度。 这些调查需要使用样本加权，以产生所有民族群体的适当估计。如果满足某些条件（主要是加权的计算和使用正确），这些样本可以准确地估计人口参数。 减轻抽样偏见的最佳做法 选择适当的抽样方法以确保所得到的数据准确反映所研究的人群是至关重要的。 随机抽样技术:使用随机抽样技术可以提高样本代表人口的概率。这种技术有助于确保样本尽可能地代表相关人群，因此，不太可能包含偏见。 样本量的计算:应进行样本量计算，以便有足够的力量来检验有统计意义的假设。样本量越大，人口的代表性就越强。 趋势分析:寻求替代的数据来源，分析任何观察到的可能未被选择的数据趋势。 检查偏见:应监测偏见的发生，以确定系统地排除或过度纳入特定数据点。 注意样品 在进行研究时，抽样偏差是一个重要的考虑因素。无论使用什么方法或研究什么学科，研究人员都需要确保他们使用的是反映他们所研究人群特征的代表性样本。 在创建研究报告时，密切关注样本选择过程，以及用于收集样本数据的方法是至关重要的。应采用随机抽样技术、样本量计算、趋势分析和检查偏见等最佳做法，以确保研究结果的有效性和可靠性，从而使其更有可能影响政策和实践。 在几分钟内制作出引人注目的科学信息图表 Mind the Graph 是一个强大的在线工具，供需要创建高质量科学图形和插图的科学家使用。该平台对用户友好，具有不同技术水平的科学家都可以使用，使其成为需要为其出版物、报告和其他科学交流材料创建图形的研究人员的理想解决方案。 无论您是生命科学、物理科学还是工程领域的研究人员，Mind the Graph都能提供广泛的资源，帮助您以清晰和有视觉冲击力的方式交流您的研究成果。 开始免费创建信息图表 相关文章 如何为研究选择正确的取样类型 出版偏见：你需要知道的一切 研究中的性别偏见。它是什么，有什么影响？ 订阅我们的通讯 关于有效视觉的独家高质量内容 科学中的交流。 独家指南 设计提示 科学新闻和趋势 教程和模板 订阅 Mind The Graph About Us朋友介绍褒奖成功案例 插图 大脑细胞RNA鼠标查看全部 Templates 幻灯片演示图形化摘要信息图表海报查看全部 订阅 首发学生研究员团队和实验室 资源 Poster Maker定制设计帮助中心 内容 科学插图：通向视觉科学世界的钥匙如何在研究论文中包含数字如何创建有效的图表式摘要查看全部 连接社交媒体 2025 仙人掌通信。版权所有 隐私政策|饼干政策|使用条款|职业生涯 Chinese EnglishCzechDutchFrenchGermanItalianIndonesianJapaneseKoreanPolishPortugueseRussianSpanishTurkishUkrainianSwedishRomanianBulgarianFinnishGreekHungarianNorwegianDanishEstonianSlovenianLithuanianLatvianSlovakChinese
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https://medium.com/@AlexanderObregon/checking-if-a-number-is-within-a-range-in-java-da2729ea3e05
Sitemap Open in app Sign in Sign in Checking if a Number Is Within a Range in Java Alexander Obregon 8 min read11 hours ago Working with numbers often means checking if a value falls inside a certain range. That could be someone’s age, a game score, or a sensor reading that has to stay within safe limits. In Java this comes down to comparing a value against two boundaries and deciding if it belongs between them. Comparisons are handled in specific ways by the language and the JVM, and edge cases can change the outcome in ways worth being aware of. I publish free articles like this daily, if you want to support my work and get access to exclusive content and weekly recaps, consider subscribing to my Substack. How Range Checks Work in Java Range checks sit on top of Java’s comparison system, where every check is distilled into a boolean outcome. What looks like a short expression in source code is actually a sequence of JVM instructions that load values, perform the comparison, and branch depending on the result. These operators are designed to make intent clear to the developer while still compiling down to highly efficient instructions at runtime. Looking at them this way turns a range check from a surface-level condition into a predictable process that explains why certain inputs pass and others don’t. Comparison Operators in Action At the heart of every range check are comparison operators. These include <, <=, >, >=, ==, and !=. Each operator compares two values and produces a boolean outcome. That result can either control a flow of execution in a branch, be stored as a variable, or be combined with other conditions. int speed = 65;boolean highwayLimit = speed <= 70;System.out.println(highwayLimit); // true int speed = 65 boolean highwayLimit = 70 // true The operator <= compares the current speed with the posted limit. The compiler turns this into bytecode instructions such as if_icmple, which evaluates the integer comparison and chooses the next instruction path based on the outcome. Comparisons also work beyond simple constants. They can involve variables, method calls, or arithmetic results. int value = 45;int min = 20;int max = 60;boolean result = value >= min && value <= max;System.out.println(result); // true int value = 45 int min = 20 int max = 60 boolean result = // true The JVM doesn’t treat these any differently from literals. Each operand is loaded onto the stack and the comparison triggers a conditional branch. When a boolean value needs to be stored, the compiler emits code that writes 1 or 0 accordingly. That efficiency is part of why these checks perform well even when called repeatedly in loops or large datasets. Inclusive vs Exclusive Boundaries Boundaries set the rules for what counts as being inside the range. Inclusive boundaries accept equality at the edges, while exclusive boundaries reject it. The choice depends on the logic being applied to the data. int temperature = 100;// Inclusiveboolean safeInclusive = temperature >= 0 && temperature <= 100; System.out.println(safeInclusive); // true// Exclusiveboolean safeExclusive = temperature > 0 && temperature < 100; System.out.println(safeExclusive); // false int temperature = 100 // Inclusive boolean safeInclusive = 0 100 // true // Exclusive boolean safeExclusive = 0 100 // false For inclusive checks, the edges are part of the valid region. A temperature of exactly 100 is considered safe here. In the exclusive check, that same value falls outside because equality is not allowed. This control is what lets Java range checks adapt to a wide set of contexts, such as validating user input, setting thresholds, or evaluating loop conditions. It’s also possible to combine inclusive and exclusive checks in a single condition when the problem requires it. A grading system might give full marks at 100 but exclude zero as a passing value. int score = 0;boolean passing = score > 0 && score <= 100;System.out.println(passing); // false int score = 0 boolean passing = 0 100 // false For int values the compiler uses the family of if_icmp branch instructions. For long, float, and double, it emits a compare (lcmp, fcmpl/fcmpg, or dcmpl/dcmpg) followed by a conditional branch. The outcome is always a boolean, and when conditions are joined with &&, short-circuit logic makes sure the second comparison is skipped if the first already fails. Floating Point Ranges Floating point comparisons carry an extra layer of detail because they follow IEEE 754 rules. Unlike integers, floats and doubles can represent NaN (Not a Number), positive and negative infinity, and very small fractional values. Each of these cases affects how a range check behaves. NaN is the most striking case, because any comparison with it will return false. double reading = Double.NaN;boolean check = reading >= 0.0 && reading <= 100.0;System.out.println(check); // false double reading = boolean check =0.0100.0 // false Even though the boundaries appear reasonable, the result is false because NaN refuses to be equal to, greater than, or less than any number. This is a design decision in the IEEE standard, and Java follows it exactly through bytecode instructions like dcmpl and dcmpg. Something to consider as well is infinity values. Comparisons involving positive infinity only pass when checking upper bounds that also include infinity itself, and negative infinity only passes lower bounds of the same kind. double high = Double.POSITIVE_INFINITY;System.out.println(high <= Double.POSITIVE_INFINITY); // trueSystem.out.println(high < 1_000_000); // falsedouble low = Double.NEGATIVE_INFINITY;System.out.println(low >= Double.NEGATIVE_INFINITY); // trueSystem.out.println(low > -1_000_000); // false double high = // true 1_000_000 // false double low = // true 1_000_000 // false Floating point precision also shapes range checks at fine levels. Very small differences between decimal values can produce unexpected outcomes if equality is part of the condition. A double holding 0.1 + 0.2 does not exactly equal 0.3, so checks that rely on precise matches should be replaced with tolerance-based comparisons. double value = 0.1 + 0.2;System.out.println(value == 0.3); // false double value =0.10.20.3 // false This doesn’t mean range checks with floating point numbers are unreliable, but it does mean they carry rules that integers don’t. The JVM faithfully applies the IEEE rules, and developers need to be mindful of how NaN, infinity, and floating point precision interact with boundaries. Different Ways to Verify a Range Range checks in Java can be written in more than one form. Some are direct and easy to follow, while others bring in methods or broader APIs that provide different trade-offs. The choice depends on context, readability, and the type of data at hand. What all of them share is the same end result, a boolean that tells if a number falls between two boundaries. Direct Condition Checks The most common way to test a range is by chaining two comparisons together with the logical && operator. This style is clear, fast, and doesn’t call anything extra. It works for integers, floating point values, and even characters, since characters in Java are numeric values underneath. int n = 18;if (n >= 10 && n <= 25) { System.out.println("Within range");} int n = 18 if 10 25 "Within range" That line checks the lower boundary, then the upper boundary, and both must be true. The compiler turns this into bytecode instructions that run in sequence, skipping the second check if the first one already fails. Get Alexander Obregon’s stories in your inbox Join Medium for free to get updates from this writer. It’s also common to assign the result to a variable or return it directly from a method. char letter = 'h';boolean lowercase = letter >= 'a' && letter <= 'z';System.out.println(lowercase); // true char letter = 'h' boolean lowercase = 'a' 'z' // true A char in Java is a 16-bit unsigned UTF-16 code unit, so comparisons with >= and <= work like numeric comparisons. Math Methods for Bounds Another way is to rely on Math methods. These let you clamp values inside boundaries rather than checking them directly. It can be useful if you need both the boolean check and the adjusted value later. int x = 35;int clamped = Math.max(10, Math.min(x, 30));System.out.println(clamped); // 30 int x = 35 int clamped = 10 30 // 30 That forces the value down to the top limit, then up to the bottom limit. If the final value matches the original, then it was inside the range. int y = 15;boolean valid = Math.max(0, Math.min(y, 20)) == y;System.out.println(valid); // true int y = 15 boolean valid = 0 20 // true This form is less direct than chained comparisons, as each Math call adds stack operations in the bytecode. Performance usually isn’t a concern for a few checks, but if you’re processing large sets of values, direct comparisons are often lighter. Range Checks with Streams When working with collections, range logic often gets paired with filtering. Streams in Java make these kinds of bulk checks concise and readable. List<Integer> numbers = Arrays.asList(5, 12, 19, 25, 32);List<Integer> valid = numbers.stream() .filter(num -> num >= 10 && num <= 20) .toList();System.out.println(valid); // [12, 19] 5 12 19 25 32 10 20 // [12, 19] The range predicate is applied to every element, and only those that pass both checks move forward in the pipeline. Streams also combine well with other operations. A filter can feed into aggregation or transformation while still only keeping values inside the range. List<Double> readings = Arrays.asList(3.4, 7.1, 10.5, 15.2, 22.8);double avg = readings.stream() .filter(r -> r >= 5.0 && r <= 20.0) .mapToDouble(Double::doubleValue) .average() .orElse(Double.NaN);System.out.println(avg); // 10.933...3.47.110.515.222.8 double avg =5.020.0 // 10.933... Here the average is based only on values between the limits, which avoids skew from data that falls outside. A range filter can also be combined with counting or checks for the presence of valid values. This keeps the range logic clear while producing different outcomes depending on what the data requires. List<Integer> scores = Arrays.asList(45, 72, 88, 95, 102, 110);// Count how many scores fall inside the passing rangelong passingCount = scores.stream() .filter(s -> s >= 50 && s <= 100) .count();System.out.println(passingCount); // 3 45 72 88 95 102 110 // Count how many scores fall inside the passing range long passingCount = 50 100 // 3 This filter makes sure that only scores between 50 and 100 are included, and count() tells how many match. Something like this works well when you don’t need the actual elements, just a summary of how many values lie within the boundaries. Handling Edge Cases Most checks behave as expected when the boundaries are correct, but some conditions call for extra attention. A common issue is when the lower and upper boundaries are given in the wrong order. int lower = 50;int upper = 10;int value = 25;if (lower > upper) { int tmp = lower; lower = upper; upper = tmp;}boolean inRange = value >= lower && value <= upper;System.out.println(inRange); // true int lower = 50 int upper = 10 int value = 25 if int tmp = boolean inRange = // true Swapping the values makes sure the logic holds even when inputs arrive reversed. Integer overflow is another case. Some developers shorten range checks into an arithmetic form like (value - low) (value - high) <= 0. On paper this looks correct, but with values near Integer.MAX_VALUE or Integer.MIN_VALUE the subtraction can wrap around and return a wrong result. int low = Integer.MAX_VALUE - 2;int high = Integer.MAX_VALUE;int test = Integer.MAX_VALUE;boolean badCheck = (test - low) (test - high) <= 0; // overflow riskboolean safeCheck = test >= low && test <= high; // correct int low = 2 int high = int test = boolean badCheck = 0 // overflow risk boolean safeCheck = // correct Direct comparisons are safe from this problem, since the JVM’s integer comparison instructions don’t overflow. Conclusion Range checks in Java rely on the mechanics of comparison operators that the JVM turns into fast bytecode instructions. Each method, whether it’s a direct boolean condition, a clamp through Math, or a filter with streams, ends up producing the same binary outcome of true or false. What makes these checks worth studying is how different inputs behave when passed through them. Integers respond predictably to boundary logic, but floating point numbers bring in special cases like NaN, infinity, and rounding quirks. Boundary order and overflow also play a part, shaping how reliable the result is in practice. Java Language Specification on Expressions Java SE Documentation for Math Java SE Documentation for Stream Java SE Documentation for Double Thanks for reading! If you found this helpful, highlighting, clapping, or leaving a comment really helps me out. Java JVM Programming Software Development Learning To Code Some rights reserved ## Written by Alexander Obregon 26K followers ·15 following I post daily about programming topics and share what I learn as I go. For recaps, exclusive content, and to support me: No responses yet Write a response What are your thoughts? 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https://www.tutorialspoint.com/step-response-and-impulse-response-of-series-rc-circuit-using-laplace-transform
Data Structure Networking RDBMS Operating System Java MS Excel iOS HTML CSS Android Python C Programming C++ C# MongoDB MySQL Javascript PHP Selected Reading UPSC IAS Exams Notes Developer's Best Practices Questions and Answers Effective Resume Writing HR Interview Questions Computer Glossary Who is Who Step Response and Impulse Response of Series RC Circuit using Laplace Transform Signals and SystemsElectronics & ElectricalDigital Electronics An electric circuit consisting of a resistance (R) and a capacitor (C), connected in series, is shown in Figure-1. Consider the switch (S) is closed at t=0t=0. Step Response of Series RC Circuit Using Laplace Transform To obtain the step response of the series RC circuit, the applied input is given by, x(t)=Vu(t) x(t)=Vu(t) By applying KVL to the circuit, the following equation describing the series RC circuit is obtained − Vu(t)=Ri(t)+1C∫t−∞i(t)dt Vu(t)=Ri(t)+1C∫t−∞i(t)dt This equation can be written as, Vu(t)=Ri(t)+1C∫0−∞i(t)dt+1C∫t0i(t)dt Vu(t)=Ri(t)+1C∫0−∞i(t)dt+1C∫t0i(t)dt Taking the Laplace transform on both sides, i.e., L[Vu(t)]=L[Ri(t)]+L[1C∫0−∞i(t)dt]+L[1C∫t0i(t)dt] L[Vu(t)]=L[Ri(t)]+L[1C∫0−∞i(t)dt]+L[1C∫t0i(t)dt] ⇒Vs=RI(s)+1C[I(s)s]+1C[q(0+)s] ⇒Vs=RI(s)+1C[I(s)s]+1C[q(0+)s] Where, q(0+)q(0+) is the charge on the capacitor at t=(0+)t=(0+), i.e., it is the initial charge. If the initial conditions are neglected, then, Vs=RI(s)+I(s)sC Vs=RI(s)+I(s)sC ⇒Vs=I(s)(R+1sC) ⇒Vs=I(s)(R+1sC) Therefore, the current in the circuit is given by, I(s)=Vs(R+1sC)=Vs(sCsRC+1)=VCRC(s+1RC) I(s)=Vs(R+1sC)=Vs(sCsRC+1)=VCRC(s+1RC) ∴I(s)=VR[1(s+1RC)] Taking the inverse Laplace transform on both sides, we have, i(t)=VRe−(1/RC)t This is the step response of the series RC circuit. Impulse Response of Series RC Circuit Using Laplace Transform To obtain the impulse response of the series RC circuit (shown in Figure-1), the applied input is given by, x(t)=δ(t) Thus, the equation describing the system is given by, δ(t)=Ri(t)+1C∫t−∞i(t)dt=Ri(t)+1C∫0−∞i(t)dt+1C∫t0i(t)dt Taking the Laplace transform on both sides, we get, 1=RI(s)+1C[I(s)s]+1C[q(0+)s] Where, q(0+) is the initial charge on the capacitor and by neglecting the initial conditions, we get, 1=RI(s)+1C[I(s)s] ⇒(R+1sC)I(s)=1 Therefore, the current flowing in the circuit is I(s)=1(R+1sC)=1R(1+1sRC)=sRCR(sRC+1) ⇒I(s)=sC(sRC+1)=sCRC(s+1RC)=sR(s+1RC) Adding and subtracting (1/Rc) in the numerator on the RHS of the above equation, we get, I(s)=1R[(s+1RC)−(1RC)(s+1RC)] ⇒I(s)=1R{1−1RC[1(s+1RC)]} Now taking the inverse Laplace transform on both sides, we get, i(t)=1R[δ(t)−1RCe−(1/RC)tu(t)] This is the impulse response of the series RC circuit. Manish Kumar Saini Updated on: 2022-01-05T10:48:39+05:30 10K+ Views Related Articles Step Response and Impulse Response of Series RL Circuit using Laplace Transform Step Response of Series RLC Circuit using Laplace Transform Laplace Transform of Unit Impulse Function and Unit Step Function Signals and Systems – Symmetric Impulse Response of Linear-Phase System Circuit Analysis with Laplace Transform How to Calculate the Impulse Response in MATLAB? Z-Transform of Unit Impulse, Unit Step, and Unit Ramp Functions Difference between RL and RC Circuit Laplace Transform of Periodic Functions (Time Periodicity Property of Laplace Transform) Difference between Laplace Transform and Fourier Transform Difference between Z-Transform and Laplace Transform Relation between Laplace Transform and Fourier Transform Difference Between Stimulus and Response Events Laplace Transform of Sine and Cosine Functions Signals and Systems – Properties of Laplace Transform Kickstart Your Career Get certified by completing the course Get Started
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https://www.quora.com/What-is-an-algorithm-for-calculating-n-1-Mod-n-where-n-is-a-very-big-number-more-than-300-digit-Long-and-it-is-a-compound-number-compound-from-just-two-primes
Something went wrong. Wait a moment and try again. Number Theory Large Numbers Modular Mathematical Syst... Composite Numbers Computational Theory Modular Exponentiation Prime Numbers 5 What is an algorithm for calculating (n-1)! Mod n where n is a very big number more than 300 digit Long and it is a compound number (compound from just two primes)? Enrico Gregorio Associate professor in Algebra · Author has 18K answers and 15.7M answer views · 4y The answer is surprisingly easy: if n is prime, then (n−1)!≡(n−1)(modn) which is known as Wilson’s theorem. If n is composite and n>4, then (n−1)!≡0(modn). You easily find the proof of Wilson’s theorem. Here’s one that uses finite fields: in the field with n elements Z/nZ, every nonzero element, except and [−1]=[n−1], is different from its inverse, so you can pair up inverses in [(n−1)!]=…[n−2][n−1] and the product becomes [n−1]=[n−1] What if n is composite? Then we can write n=ab for some a,b>1. If we are able to find a≠b, then both factors appear in (n−1)!, yield The answer is surprisingly easy: if n is prime, then (n−1)!≡(n−1)(modn) which is known as Wilson’s theorem. If n is composite and n>4, then (n−1)!≡0(modn). You easily find the proof of Wilson’s theorem. Here’s one that uses finite fields: in the field with n elements Z/nZ, every nonzero element, except and [−1]=[n−1], is different from its inverse, so you can pair up inverses in [(n−1)!]=…[n−2][n−1] and the product becomes [n−1]=[n−1] What if n is composite? Then we can write n=ab for some a,b>1. If we are able to find a≠b, then both factors appear in (n−1)!, yielding that (n−1)! is divisible by ab=n. The only case in which this is not possible is when n=p2 for a prime p (check it). But, if p>2, in (n−1)! you find both p and 2p, so (n−1)! is divisible by 2p2, hence by n. The only exceptional cases are n=1 and n=4, but0!≡0(mod1) and 3!≡2(mod4). Amitabha Tripathi have more than a working knowledge of Z · Upvoted by Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 4.7K answers and 13.8M answer views · 5y Related Can we separate if the number n is prime or not by calculating (n-1)! (mod n)? Consider n>1. When n=p is prime, then (p−1)!≡−1(modp) by Wilson’s theorem. Suppose n is composite, and let d>1 be the smallest divisor of n. Then d=p is necessarily prime, and p≤np1 is also a divisor of n). So p, np appear in the set {1,2,3,…,n−1}. Unless p=np, their product (which is n) divides (n−1)!. So (n−1)!≡0(modn) with the possible except of the smallest divisor p of n satisfying p=np. In the exceptional case, n=p2. If p>2, then 2p<p2–1=n−1, and so both p and 2p appear in the set {1,2,3…,n−1}. T Consider n>1. When n=p is prime, then (p−1)!≡−1(modp) by Wilson’s theorem. Suppose n is composite, and let d>1 be the smallest divisor of n. Then d=p is necessarily prime, and p≤np1 is also a divisor of n). So p, np appear in the set {1,2,3,…,n−1}. Unless p=np, their product (which is n) divides (n−1)!. So (n−1)!≡0(modn) with the possible except of the smallest divisor p of n satisfying p=np. In the exceptional case, n=p2. If p>2, then 2p<p2–1=n−1, and so both and appear in the set . Their product is a multiple of . Hence, even for , . The only remaining case is , and in this case. For , we conclude that Sharvil Kesarwani Finished High School Maths at the end of Primary school. Casually learning uni. · Upvoted by Alon Amit , Lover of math. Also, Ph.D. and Michael Jørgensen , PhD in mathematics · Author has 76 answers and 331.9K answer views · 8y Related What is the result of: ? Where is the prime number. It was shown by H.P. Robinson and E. Potter in 1971 that in the book Mathematical Constants (Mathematical constants: H. P Robinson: Amazon.com: Books). This even has an OEIS index (A078437 - OEIS). However, a far more interesting, unsolved intrinsically related problem is proving the convergence/divergence of the below expression I would love to see an attempt on trying to prove it. Hope this helped. Related questions What's the number of primes between n! and (n+1)!? What is the algorithm to find a prime number from 1 to n? How do I get the number of primes less than factorial n (n!)? What is the easiest way to calculate N! When you know the value of N? What is an algorithm to find (1/n!) where n belongs to N? Jan van Delden MSc Math and still interested · Author has 4.8K answers and 6.5M answer views · 3y Related What algorithms perform the best for calculating a series of modulo operations for A mod b, where A is constant, b = [2..N], and N and A are large numbers? I guess it might be faster to avoid implementing a division algorithm if divisions are slow. Large integer division is well studied. Articles may be found for all kind of separate applications depending on things as the radix that is used (preventing a double conversion), whether parts of the procedure are processed in parallel, whether one actually needs a remainder or just the quotient. The underlying routines may also depend on size. I haven’t checked recently on current progress. There exist libraries that have good implementations of large integer division. That said, computing multiple residues for one fixed gives me the impression that you want to test whether or not Large integer division is well studied. Articles may be found for all kind of separate applications depending on things as the radix that is used (preventing a double conversion), whether parts of the procedure are processed in parallel, whether one actually needs a remainder or just the quotient. The underlying routines may also depend on size. I haven’t checked recently on current progress. There exist libraries that have good implementations of large integer division. That said, computing multiple residues for one fixed gives me the impression that you want to test whether or not is prime. In such a case I would consider to use a sieve (if is not too large) or use a composite test, like Millar Rabin, or maybe even a general purpose routine. This also might depend on the size of and whether it has a particular shape or not. All these test incorporate some modular arithmetic, but mostly with respect to a limited number of . If you first compute and subsequently your test of whether you have reached the end of the division process would certainly be simpler. There might be some algorithm that best suits your request, but it helps if you specify the word large and the actual goal that you wish to achieve. It also helps if I know what kind of math you can handle. Chances are that I only typed the obvious. 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. Assistant Bot · 1y To calculate math! \mod n[/math] for a large composite number [math]n[/math] that is the product of two primes, we can use properties of modular arithmetic and the properties of factorials. Here's a step-by-step algorithm to achieve this efficiently: Steps to Calculate math! \mod n[/math] Factor [math]n[/math]: Since [math]n[/math] is a compound number composed of two primes, factor [math]n[/math] as: [math]n = p \times q[/math] where [math]p[/math] and [math]q[/math] are prime numbers. Use Wilson's Theorem: Wilson's theorem states that for a prime [math]p[/math]: math! \equiv -1 \mod p[/math] We can use this theorem to find math! \mod p[/math] and math! \mod q[/math]. Calculate math! \mod p[/math]: Since [math]n = p \c[/math] To calculate math! \mod n[/math] for a large composite number [math]n[/math] that is the product of two primes, we can use properties of modular arithmetic and the properties of factorials. Here's a step-by-step algorithm to achieve this efficiently: Steps to Calculate math! \mod n[/math] Factor [math]n[/math]: Since [math]n[/math] is a compound number composed of two primes, factor [math]n[/math] as: [math]n = p \times q[/math] where [math]p[/math] and [math]q[/math] are prime numbers. Use Wilson's Theorem: Wilson's theorem states that for a prime [math]p[/math]: math! \equiv -1 \mod p[/math] We can use this theorem to find math! \mod p[/math] and math! \mod q[/math]. Calculate math! \mod p[/math]: Since [math]n = p \cdot q[/math], we have: math! = (p \cdot q - 1)! = (p-1)! \times (p \cdot q - p)! \times (p \cdot q - q)![/math] The term math! \mod p[/math] is: math! \equiv -1 \mod p[/math] The other terms can be calculated using the fact that math![/math] and math![/math] will be congruent to [math]0[/math] modulo [math]p[/math] and [math]q[/math] respectively because they will include [math]p[/math] or [math]q[/math] as factors. Calculate math! \mod q[/math]: Similarly, we can apply Wilson's theorem: math! \equiv -1 \mod q[/math] Again, the terms math! \mod p[/math] will yield [math]0[/math] because they will include [math]q[/math] as a factor. Combine Results Using the Chinese Remainder Theorem (CRT): Now we have: math! \equiv 0 \mod p[/math] math! \equiv 0 \mod q[/math] Since math![/math] is [math]0[/math] modulo both [math]p[/math] and [math]q[/math], we can conclude: math! \equiv 0 \mod n[/math] Conclusion For a composite number [math]n[/math] that is the product of two primes [math]p[/math] and [math]q[/math]: math! \mod n = 0[/math] This result holds for any composite number that is the product of two distinct primes, regardless of their size. Amitabha Tripathi have more than a working knowledge of Z · Upvoted by Horst H. von Brand , PhD Computer Science & Mathematics, Louisiana State University (1987) and Aditya Garg , M.Sc. Mathematics, Indian Institute of Technology, Delhi (2013) · Author has 4.7K answers and 13.8M answer views · 6y Related Let [math]f(n) = (n-1)!\,mod\,n[/math] , how can I prove or disprove [math]f(n) = \begin{cases} n-1 & \text{n is prime number}\ 0 & \text{n is nonprime number} \end{cases}[/math] ? You can disprove it by realizing that [math]f(4)=2[/math]. However, [math]f(p)=p-1[/math] for prime [math]p[/math] is Wilson's Theorem , and [math]f(1)=0[/math]. For [math]n>1[/math], [math]n[/math] composite, [math]n=ab[/math] with [math]1<a \le b<n[/math]. Unless [math]n=p^2[/math], with [math]p[/math] prime, we may choose [math]a[/math], [math]b[/math] distinct. If [math]n=ab[/math], with [math]1<a<b<n[/math], math![/math] is a product of [math]n-1[/math] positive integers which includes both [math]a[/math] and [math]b[/math]. Therefore [math]ab \mid (n-1)![/math], and so [math]f(n)=0[/math] in these cases. There remains the case where [math]n=p^2[/math], with [math]p[/math] prime. If [math]p>2[/math], then [math]2p2[/math]. So [math]f(n)=0[/math] whenever [math]n[/math] is composite, with the possible exception of [math]n=2^2[/math]. And You can disprove it by realizing that [math]f(4)=2[/math]. However, [math]f(p)=p-1[/math] for prime [math]p[/math] is Wilson's Theorem , and [math]f(1)=0[/math]. For [math]n>1[/math], [math]n[/math] composite, [math]n=ab[/math] with [math]1<a \le b<n[/math]. Unless [math]n=p^2[/math], with [math]p[/math] prime, we may choose [math]a[/math], [math]b[/math] distinct. If [math]n=ab[/math], with [math]1<a<b<n[/math], math![/math] is a product of [math]n-1[/math] positive integers which includes both [math]a[/math] and [math]b[/math]. Therefore [math]ab \mid (n-1)![/math], and so [math]f(n)=0[/math] in these cases. There remains the case where [math]n=p^2[/math], with [math]p[/math] prime. If [math]p>2[/math], then [math]2p2[/math]. So [math]f(n)=0[/math] whenever [math]n[/math] is composite, with the possible exception of [math]n=2^2[/math]. And we have seen that [math]f(4) \ne 0[/math]. [math]\blacksquare[/math] Related questions What is (n-1) /(n-1)! =? How do you calculate n! (mod m) where n is in the 1000s and m is a very large prime number eg. n=1000 m=10^9+7? Suppose S(n) is the sum of the first n factorial numbers, where n > 0. Without using a computer can you say whether S(n) is square free yet composite for any n? If so, what is an example of such an n? Can we separate if the number n is prime or not by calculating (n-1)! (mod n)? Given a number n, what is the probability that n is prime? Michael C Palau Author has 1.5K answers and 645.7K answer views · 4y This is a trapdoor function and even using 128 bit computing On a 64 bit CPU, it would take an extremely long time even if you had an efficient algorithm to do it. Unless you have access to a supercomputer, it would never get done Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Sponsored by NoBuzzZone This 2-Minute Ritual Will Keep Fruit Flies Out of Your Kitchen. Tired of sprays and vinegar traps? This chemical-free fix works 24/7—without the mess. Enrico Gregorio Associate professor in Algebra · Author has 18K answers and 15.7M answer views · 5y Related Can we separate if the number n is prime or not by calculating (n-1)! (mod n)? Yes, almost. Let [math]n[/math] be composite. Consider the minimum prime factor [math]p[/math] of [math]n[/math] and the largest exponent [math]k[/math] such that [math]p^k\mid n[/math]. If [math]p^k\ne n[/math], then [math]b=n/p^k>1[/math] and [math]\gcd(p^k,b)=1[/math]. Therefore [math]p^k\mid (n-1)![/math] and [math]b\mid (n-1)![/math], so that [math]n=p^kb\mid(n-1)![/math]. Hence, in this case, math!\equiv0\pmod{n}[/math]. It remains to show the case when [math]n=p^k[/math]. Since [math]n[/math] is composite by assumption, we have [math]k>1[/math]. If [math]kp<n-1[/math], we are done, because in math![/math] we can find [math]p,2p,\dots,kp[/math], so clearly [math]n=p^k\mid (n-1)![/math]. For what primes [math]p[/math] do we have [math]1+kp<p^k[/math] for all [math]k\ge2[/math]? The induction step works for every prime: if we assume that [math]p^k>1+kp[/math], then [math]p^{k+1}>p[/math] Yes, almost. Let [math]n[/math] be composite. Consider the minimum prime factor [math]p[/math] of [math]n[/math] and the largest exponent [math]k[/math] such that [math]p^k\mid n[/math]. If [math]p^k\ne n[/math], then [math]b=n/p^k>1[/math] and [math]\gcd(p^k,b)=1[/math]. Therefore [math]p^k\mid (n-1)![/math] and [math]b\mid (n-1)![/math], so that [math]n=p^kb\mid(n-1)![/math]. Hence, in this case, math!\equiv0\pmod{n}[/math]. It remains to show the case when [math]n=p^k[/math]. Since [math]n[/math] is composite by assumption, we have [math]k>1[/math]. If [math]kp<n-1[/math], we are done, because in math![/math] we can find [math]p,2p,\dots,kp[/math], so clearly [math]n=p^k\mid (n-1)![/math]. For what primes [math]p[/math] do we have [math]1+kp<p^k[/math] for all [math]k\ge2[/math]? The induction step works for every prime: if we assume that [math]p^k>1+kp[/math], then [math]p^{k+1}>p(1+kp)\ge 1+(k+1)p[/math] because the last inequality becomes [math]k(p^2-p)\ge1[/math] which is obviously true. When does the base step for [math]k=2[/math] fail? When [math]p^2\le1+2p[/math], that is, only for [math]p=2[/math], because the inequality [math]x^2- 2x-1>0[/math] is satisfied whenever [math]x>1+\sqrt{2}[/math]. However, for [math]p=2[/math], we do have [math]2^3>1+3\cdot2[/math], so for the prime [math]2[/math] the statement is true for every [math]k\ge3[/math]. Therefore If [math]n[/math] is prime, then math!\equiv-1\pmod{n}[/math]. If [math]n\ne4[/math] is not prime, then math!\equiv0\pmod{n}[/math]. Note that [math]n=4[/math] is indeed an exception, because math!=6\equiv2\pmod{4}[/math]. But it is the only one. Alberto Cid M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008) · Author has 2K answers and 3.8M answer views · 1y Related What is the relationship between Tn, Sn and Sn-1 where n is greater than 1 and n is an element of natural numbers? If you know what Tn and Sn refer to, and you know about that thing called addition, then it’s quite easy!!! Tn : the n-th term in a sequence, that is the element in the place number n. Example: Sequence 1², 2², 3², 4²… The T1 is 1, that's the first number The T3 is 3² = 9, that's the 3rd. Sn : the Sum of the elements up to n. Example : S1 = 1 S2 = 1+4 S3 = 1+4+9 S(n-1) or [math]S_{n-1} [/math]: is just the sum up to (n-1). Example: if n=3 then (n-1) = 2 then for n=3 : S(n-1) = S(3–1) = S(2), that is, [math]S_2[/math] From that it’s easy to see that: S3 = 1+4+9 (sum up to the 3rd) S2 = 1+4 (sum up to the 2nd) T3 = 9 (the 3rd element) If you know what Tn and Sn refer to, and you know about that thing called addition, then it’s quite easy!!! Tn : the n-th term in a sequence, that is the element in the place number n. Example: Sequence 1², 2², 3², 4²… The T1 is 1, that's the first number The T3 is 3² = 9, that's the 3rd. Sn : the Sum of the elements up to n. Example : S1 = 1 S2 = 1+4 S3 = 1+4+9 S(n-1) or [math]S_{n-1} [/math]: is just the sum up to (n-1). Example: if n=3 then (n-1) = 2 then for n=3 : S(n-1) = S(3–1) = S(2), that is, [math]S_2[/math] From that it’s easy to see that: S3 = 1+4+9 (sum up to the 3rd) S2 = 1+4 (sum up to the 2nd) T3 = 9 (the 3rd element) Then S3 = S2 + T3 The sum up to 3rd is the previous sum (up to 2nd) plus the 3rd element, which is called T3 Could you generalise that? Try it! … In general: The sum up to “n”, called Sn, is the previous sum, S(n-1), up to the (n+1)th element, plus the n-th element, which is called Tn: Sn = S(n-1) + Tn Equivalently, substracting S(n-1) from both sides: Sn - S(n-1) = Tn Sn = T1 + T2 + … + T(n-1) + Tn That's just the definition of Sn !!! S(n-1)= T1 + T2 + … + T(n-2) + T(n-1) So: Sn = [T1 + T2 + … + T(n-1)] + Tn The part in bold between the brackets is just S(n-1) … Sponsored by Adsy Access 100k+ quality websites ready to publish your articles. Choose exact sites by niche, traffic & DA/DR. Prices upfront. Most placements go live within days. Dana Jacobsen Software Engineer · Author has 313 answers and 1M answer views · Updated 7y Related How do you calculate n! (mod m) where n is in the 1000s and m is a very large prime number eg. n=1000 m=10^9+7? Your n is not particularly large, so we have many choices. We could do the naive way and calculate n! then modulo m. This takes less than a millisecond on a computer for your sizes, so it’s certainly practical for a few calls. Factorial is [math]O(n^2 \log n)[/math] the simple way, or [math]O(\log n M(n \log n))[/math] [[math]M(n)[/math] = complexity of multiplication of two n-digit numbers] with binary splitting, which is a fairly common technique. See among other source, Borwein’s “On the complexity of computing factorials” (1983). That doesn’t scale very well for our problem. We can do it the obvious way of doing n repeated modula Your n is not particularly large, so we have many choices. We could do the naive way and calculate n! then modulo m. This takes less than a millisecond on a computer for your sizes, so it’s certainly practical for a few calls. Factorial is [math]O(n^2 \log n)[/math] the simple way, or [math]O(\log n M(n \log n))[/math] [[math]M(n)[/math] = complexity of multiplication of two n-digit numbers] with binary splitting, which is a fairly common technique. See among other source, Borwein’s “On the complexity of computing factorials” (1983). That doesn’t scale very well for our problem. We can do it the obvious way of doing n repeated modular multiplications. This keeps the math easy (and on a computer, very fast if m is 64-bit). On my Macbook with Perl/ntheory’s new factorialmod(n,m) function it takes 36 microseconds for 6000! % (10^9+7). As I mentioned, n isn’t particularly large. We can do a few simple optimizations such as early exit if m > n (the answer is 0 since m evenly divides n!), possibly going backwards if m is prime (by Wilson’s theorem), checking for composite m and early exit with 0 if n exceeds the largest factor, and using Montgomery reduction. Note that once m is greater than 32-bit we need to be more clever than just “res = (resi) % m” because that will overflow. These very simple things don’t really change the O(n) asymptotic behavior (). For math! \mod (10^{18}+3)[/math] it takes 6 seconds on my Macbook, and it’s growing basically linearly. So n=10^14 is going to take a week. Ugh. There is another optimization that isn’t unreasonable, which is going through only the primes up to n. Up to n/2 we pull out repeated uses of the prime and do a modular exponentiation before the mulmod, so all the composites are accounted for. The downside is that Montgomery reduction in this setup has a lot more overhead so isn’t really useful. It’s a tradeoff whether the savings in only doing primes makes up for the linear 5x or so speedup that Montgomery math gives. If you aren’t doing Montgomery math then it certainly pays off. Note this combines just fine with the going-backward-via-Wilsons-Theorem optimization. But there is something much better! Fredrik Johannson has a blog article about an [math]O(n^{1/2})[/math] algorithm for factorials mod n. He links to his FLINT open source implementation (it’s part of the FLINT library). It uses “FFT-based polynomial multiplication and balanced subproduct trees.” Every time I see this I’m still amazed. n=10^14 would be something like 10 minutes. () I am ignoring the fact that mulmods aren’t really constant time, so the complexity is higher once we get large enough (32-bit or 64-bit depending on platform + cleverness in using asm). Ellis Cave 40+ years as an Electrical Engineer · Author has 7.9K answers and 4.3M answer views · 7y Related What is the result of: [math]\displaystyle\sum_{n=1}^{\infty}\tfrac {(-1)^{n-1}}{p_n}[/math] ? Where [math]p_n[/math] is the [math]n^{\text {th}}[/math] prime number. Using the J programming language: Write the J function representing the nth term in the series: ser =.3 :'(_1^y-1)%p:y-1' ser 1 0.5 ser 1 2 3 0.5 _0.333333 0.2 Add the terms up: +/ ser 1 2 3 0.366667 +/ser 1 2 3 4 5 6 7 0.296619 +/ ser 1+i.10 — compute the seies sum for n= 1->10 0.252983 +/ser 1+i.100 — n=1-> 100 0.268659 +/ser 1+i.1000 — n=1->1000 +/ser 1+i.10000 — n=1->10000 +/ser 1+i.100000 +/ser 1+i.1000000 0.269606 So the series converges on 0.269606 David Goldstein Software Engineer · Author has 272 answers and 1.3M answer views · 12y Related What is the fastest algorithm for finding [math]a^{n!}[/math] modulo [math]p[/math] where [math]p[/math] is a prime number? well, first off, [math]a^{p-1} = 1 \mod{p}[/math], so really this reduces to finding [math]n! \pmod{p-1}[/math]. Once this result is computed to some number in [math]{0, 1, \ldots p-2}[/math], the result can be computed via exponentiation by squaring (idea is to compute [math]a^{2^k} \pmod{p}[/math] by repeated squaring, and then multiply the subset of these terms together to get the desired power of [math]a[/math]) We'll focus on computing [math]n! \pmod{p-1}[/math] from here on. Case 0: [math]p=2[/math]. Then (trivially) [math]a^{n!} = a \mod 2[/math]. Case 1: If [math]n \ge p-1[/math], obviously [math]n! = 0 \pmod{p-1}[/math]. Case 2: [math]p=3[/math]. Then [math]n! \mod 2[/math] is trivial to compute (0 unless n=1) Case 3: [math]p-1[/math] isn't prime (th well, first off, [math]a^{p-1} = 1 \mod{p}[/math], so really this reduces to finding [math]n! \pmod{p-1}[/math]. Once this result is computed to some number in [math]{0, 1, \ldots p-2}[/math], the result can be computed via exponentiation by squaring (idea is to compute [math]a^{2^k} \pmod{p}[/math] by repeated squaring, and then multiply the subset of these terms together to get the desired power of [math]a[/math]) We'll focus on computing [math]n! \pmod{p-1}[/math] from here on. Case 0: [math]p=2[/math]. Then (trivially) [math]a^{n!} = a \mod 2[/math]. Case 1: If [math]n \ge p-1[/math], obviously [math]n! = 0 \pmod{p-1}[/math]. Case 2: [math]p=3[/math]. Then [math]n! \mod 2[/math] is trivial to compute (0 unless n=1) Case 3: [math]p-1[/math] isn't prime (the only consecutive primes are 2 and 3). Which means that for larger values of [math]n[/math], [math]n![/math] will contain two factors which multiple to [math]p-1[/math]. More formally: If there exists [math]k \le n[/math] such that [math]k | p-1[/math], math/k \le n[/math] and [math]p-1 \neq k^2[/math], then [math]n! = 0 \pmod{p-1}[/math]. Now this can't be applied when - n is too small for any factors of p-1 to qualify as potential k. This basically implies that [math]n < (p-1)/4[/math] ([math]p-1[/math] is even, so math/2[/math] might be prime), with tighter bounds more likely. - [math]p-1 = q^2[/math] for some prime [math]q[/math]. Considering that [math]p-1[/math] is even, we must have [math]q=2[/math] and [math]p=5[/math]. Which is easy to compute, since there are only 4 values of n not caught by case 1. Summary: the hard case is when p is a very large prime, and n is much smaller than p - namely, [math]n < (p-1)/4[/math] - and (p-1)/2 turns out to be hard to factor, i.e. if it is the product of two large primes. Jaron Walker High School Student · 8y Related Why does n(n+1)/2 work with finding the number of steps needed for n blocks? Let’s take the fourth triangular number as an explanation. Now work backwards. A diagram of the ten dots in a right angle look like: Add a second triangle of the same size to create a rectangle. Notice how the rectangle is a 45 rectangle, or a 4(4+1) rectangle. [math]n(n+1)/2[/math] works because it is taking the area of the rectangle and dividing it by two. Let’s take the fourth triangular number as an explanation. [math]n(n+1)/2[/math] [math]=4(4+1)/2[/math] [math]=4(5)/2[/math] [math]=20/2[/math] [math]=10[/math] Now work backwards. A diagram of the ten dots in a right angle look like: Add a second triangle of the same size to create a rectangle. Notice how the rectangle is a 45 rectangle, or a 4(4+1) rectangle. [math]n(n+1)/2[/math] works because it is taking the area of the rectangle and dividing it by two. Related questions What's the number of primes between n! and (n+1)!? What is the algorithm to find a prime number from 1 to n? How do I get the number of primes less than factorial n (n!)? What is the easiest way to calculate N! When you know the value of N? What is an algorithm to find (1/n!) where n belongs to N? What is (n-1) /(n-1)! =? How do you calculate n! (mod m) where n is in the 1000s and m is a very large prime number eg. n=1000 m=10^9+7? Suppose S(n) is the sum of the first n factorial numbers, where n > 0. Without using a computer can you say whether S(n) is square free yet composite for any n? If so, what is an example of such an n? Can we separate if the number n is prime or not by calculating (n-1)! (mod n)? Given a number n, what is the probability that n is prime? How do you calculate n! /(n-2)! X 2! Becomes n(n-1)? What is the significance of having a prime number modulo n where ? How do I calculate the sum: n mod 1 + n mod 2 + n mod 3 + … + n mod m? Assume (N) is a natural number where there are one or two primes between (N), and (2N-1), what are all values of (N)? What is n/(n!) 1/n? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://math.stackexchange.com/questions/3720983/value-of-n-for-which-the-function-xn-sin-frac1x-is-continuous-at-x
continuity - Value of $n$ for which the function $x^n \sin {\frac{1}{x}}$ is continuous at $x=0$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Value of n n for which the function x n sin 1 x x n sin⁡1 x is continuous at x=0 x=0 Ask Question Asked 5 years, 3 months ago Modified5 years, 3 months ago Viewed 1k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. The question is as follows, Determine the values of n n for which the function, f(x)={x n sin(1 x)0,x≠0,x=0 f(x)={x n sin⁡(1 x),x≠0 0,x=0 is continuous at x=0 x=0 The way I tried to solve it is by using inequalities, starting with, −1≤sin(1 x)≤1−1≤sin⁡(1 x)≤1 −x n≤x n sin(1 x)≤x n−x n≤x n sin⁡(1 x)≤x n lim x→0−x n≤lim x→0 f(x)≤lim x→0 x n lim x→0−x n≤lim x→0 f(x)≤lim x→0 x n Which gives us, continuity ∀n∀n This seems right but I'm not sure, could someone confirm it and also is there a better method? functions continuity Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Jun 15, 2020 at 17:26 Prakhar NagpalPrakhar Nagpal 1,360 12 12 silver badges 25 25 bronze badges 8 2 Well done! Your answer is correct. The function is continuous for all n∈N n∈N General Grievous –General Grievous 2020-06-15 17:28:12 +00:00 Commented Jun 15, 2020 at 17:28 3 what if n=0 n=0?Angina Seng –Angina Seng 2020-06-15 17:28:39 +00:00 Commented Jun 15, 2020 at 17:28 An interesting extension is to answer the same question, but instead of asking about continuity, consider whether the 1st derivative exists. What about the 2nd derivative? n-th derivative?Benjamin Wang –Benjamin Wang 2020-06-15 17:29:52 +00:00 Commented Jun 15, 2020 at 17:29 @BenjaminWang I will try that too :)Prakhar Nagpal –Prakhar Nagpal 2020-06-15 17:32:50 +00:00 Commented Jun 15, 2020 at 17:32 @AnginaSeng I'm not sure since that would be an indeterminant form right? (0 0 0 0)Prakhar Nagpal –Prakhar Nagpal 2020-06-15 17:33:41 +00:00 Commented Jun 15, 2020 at 17:33 |Show 3 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Let f n(x)=x n sin(1/x)f n(x)=x n sin⁡(1/x). About the easy cases first, this function is continuous when x x is away from 0 0 by as a composition of continuous functions, and as you say, when n>0 n>0, |f n(x)|≤|x|n→x→0 0|f n(x)|≤|x|n→x→0 0, so the function is also continuous in 0 0. Now if n=0 n=0. Take the sequence x k=1 2 π k+π/2 x k=1 2 π k+π/2 Then x k→k→∞0 x k→k→∞0, but f(x k)=sin(2 π k+π/2)=sin(π/2)=1 f(x k)=sin⁡(2 π k+π/2)=sin⁡(π/2)=1 does not converge to 0 0, so the function is discontinuous in 0 0 by the sequential definition of the limit (the same is true by the way if n<0 n<0). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 15, 2020 at 18:28 LL 3.14LL 3.14 14.2k 2 2 gold badges 17 17 silver badges 54 54 bronze badges 3 Where do you see any (x k)n(x k)n? The only case when I take x k x k is when n=0 n=0, so x n=1 x n=1 for any x>0 x>0 (so if you prefer, the sequence (x k)0=1(x k)0=1 so it converges to 1 1, was that your problem?)LL 3.14 –LL 3.14 2020-06-16 16:21:15 +00:00 Commented Jun 16, 2020 at 16:21 Yes, I deleted my comment, I came to the same conclusion. Although I am wondering now, how did you come with the expression for x k x k? After all we could take x k x k as 1 2 π k+π/6 1 2 π k+π/6 Prakhar Nagpal –Prakhar Nagpal 2020-06-16 19:54:20 +00:00 Commented Jun 16, 2020 at 19:54 Yes, of course, you can take a lot of different sequences that will converges to whatever value between −1−1 and 1 1. I just chose 1 1 to take a simple case, since you just need to find one sequence along which the function is not converging to 0 0.LL 3.14 –LL 3.14 2020-06-16 20:54:51 +00:00 Commented Jun 16, 2020 at 20:54 Add a comment| This answer is useful 1 Save this answer. Show activity on this post. Define f n(x)=x n sin 1 x f n(x)=x n sin⁡1 x, x≠0 x≠0. For any n>0 n>0, setting f n(0)=0 f n(0)=0, one obtains a continuous function since |f n(x)|≤x n−→−x→0 0|f n(x)|≤x n→x→0 0. If n≤0 n≤0, the discontinuity at x=0 x=0 is not removable as can can see by choosing sequences x n=2 π(2 n+1)x n=2 π(2 n+1) and y n=1 n π y n=1 n π. Both sequences converge to 0 0 but f(y n)=0 f(y n)=0 while f(y n)f(y n) is highly oscillatory. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 15, 2020 at 21:20 MJD 67.8k 44 44 gold badges 309 309 silver badges 621 621 bronze badges answered Jun 15, 2020 at 19:07 MittensMittens 46.6k 6 6 gold badges 60 60 silver badges 115 115 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 functions continuity See similar questions with these tags. 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home + KNOB & TUBE WIRING + LIGHTING, EXTERIOR - home + LIGHTING, INTERIOR - home + LOW VOLTAGEWIRING + METERS & BASES + MOTOR REPAIR - home + MOTOR WIRE SIZE + MULTI-WIRE CIRCUITS + NOISES, ELECTRICAL + OLD HOUSE ELECTRIC- home + OUTLET, WIRE - home + PANEL- home + RELAY SWITCHES + SAFETY + SERVICE ENTRY- home + SPLICE + THERMAL IMAGING + TURN BACK ON + ZINSCO SYLVANIA ENERGY + + AIR CHANGE RATE + AIR LEAKS - home + AIR LEAKS RETURN DUCTS + AIR LEAKS SUPPLY DUCTS + BASEMENT HEAT LOSS + BIO-FUEL + BLOWER DOORS + DUCT SYSTEM + ENERGY AUDIT + ENERGY RETROFIT + ENERGY SAVINGS PRIORITY + ENERGY USE MONITOR + HEAT COST SAVINGS + HEAT LOSS INDICATORS + HIGH MASS TRADEOFFS + ROOF COLOR + R U & K VALUE + SEER RATING + SOLAR ENERGY + TIMERS + VENTILATION, HEAT COST + WATER HEATER TIMER + WIND ENERGY + WINDOW EFFICIENCY ENVIRONMENT + + AIR POLLUTANTS + ALLERGENS + + ALLERGEN TESTS + ARSENIC HAZARDS + ASBESTOS HAZARDS + ASBESTOS IDENTIFICATION - home + ASBESTOS in THIS MATERIAL? + ASBESTOS LIST of PRODUCTS + ASBESTOS PHOTO GUIDE + BACKUP, SEPTIC-SEWAGE + BACTERIA, MOLD, POLLEN + BANNED ASBESTOS PRODUCTS + BEDBUGS + BIOLOGICAL POLLUTANTS + CEILING TILE ASBESTOS ID + CELL PHONE RADIATION + CHINESE DRYWALL + DISINFECTANTS, SANITIZERS, SEALANTS + ELECTROMAGNETIC FIELDS + FIBERGLASS HAZARDS + FIBERGLASS INSULATION MOLD + FIBERGLASS SHEDDING + FIBERGLASS CONTAMINANTS + FLOOR TILE ASBESTOS + FORMALDEHYDE HAZARDS + GAS DETECTION + HAZARD vs RISK + HOUSE DUST + INDOOR AIR QUALITY IAQ + MOLD CONTAMINATION + MOLD / ENVIRONMENT EXPERT + MORGELLONS SYNDROME + MYCOTOXIN EFFECTS + NOISE DIAGNOSIS + ODOR DIAGNOSIS + PESTICIDE EXPOSURE + POPCORN CEILING ASBESTOS + SEWAGE CONTAMINATION EXTERIOR + + BRICK WALL WEEP HOLES + DECK & PORCH CONSTRUCT - home + DOORS, EXTERIOR + FIBER CEMENT SIDING - home + FLASHING on BUILDINGS - home + PAINT FAILURE + LIGHTNING PROTECTION + PAINT FAILURE - home + RAMPS, ACCESS - home + SHEATHING, FIBERBOARD + SLIP TRIP & FALL HAZARDS + STAIR CONSTRUCTION - home + STAIR DIMENSIONS + STUCCO WALL METHODS + WINDOWS & DOORS, AGE, TYPES HEAT + + AGE of A/C & HEAT PUMPS + AGE of HEATER, BOILER, FURNACE + AIR FILTERS f- home + AIR HANDLER / BLOWER UNIT - home + AQUASTAT CONTROL - home + BACKDRAFTING + BACKUP HEAT for HEAT PUMPS + BANGING HEAT SYSTEM NOISES + BANGING HEAT ZONE VALVES + BANGING PIPES RADIATORS + BUZZING NOISE + BASEBOARD HEAT REPAIR - home + BLOWER FAN + BOILERS - home + CAD CELL RELAY + CHECK VALVES + CIRCULATOR PUMPS- home + CLEARANCE DISTANCES + COMBUSTION AIR + CONDENSING BOILERS/FURNACES + CONTROLS & SWITCHES + CONVECTOR HEATERS + DAMPERS & DRAFT REGULATORS + DATA TAGS + DIAGNOSE & FIX A/C / HEAT PUMP + DIAGNOSE & FIX BOILER - home + DIAGNOSE & FIX FURNACE - home + DIRECT VENT / SIDE WALL VENT + DRAFT REGULATORS / HOODS, GAS + DRAFT MEASUREMENT + DRAFT REGULATOR + DUCT SYSTEM - home + ELECTRIC HEAT - home + EVAPORATIVE COOLING SYSTEM + EXPANSION TANK, BOILER - home + FAN, AIR HANDLER BLOWER - home + FAN LIMIT SWITCH - home + FILTERS, AIR + FILTERS, OIL + FIRE SAFETY CONTROLS + FIREPLACES & HEARTHS - home + FLUE SIZE + FURNACE CONTROLS + FURNACES, HEATING - home + GAS BURNER FLAME & NOISE + GAS BURNER PILOT LIGHT + GEOTHERMAL HEAT + HEAT PUMP REPAIR - home + HEAT LOSS DIAGNOSIS-BOILERS + HEAT LOSS DIAGNOSIS-FURNACES + HEAT WON'T TURN OFF + HEAT WON'T TURN ON + HEATING COST SAVINGS + HEATING OIL- home + HEATING SYSTEM NOISE + HUMMING NOISE + LIFE EXPECTANCY A/C / HEAT PUMP + LIFE EXPECTANCY FAN / WALL CONVECTOR + LIFE EXPECTANCY FURNACE + LOW VOLTAGE WIRING + MANUALS + MINI SPLIT A/C & HEAT PUMPS + MOBILE HOME HEAT + NO HEAT - BOILER + NO HEAT - FURNACE + OIL STORAGE TANKS + OPERATING TEMPERATURES + PORTABLE ELECTRIC HEATER + RADIANT HEAT + STEAM HEAT + THERMOSTATS - home + THERMOSTAT WIRING + ZONE VALVES INSPECTION + + ADVANCED METHODS + CARPENTER ANTS + CARPENTER BEES + DISASTER INSPECTION- home + DUST SAMPLING + FEAR-O-METER: Dan's 3 D's SET REPAIR PRIORITIES + FIBER & HAIR IDENTIFICATION + FIBERGLASS PARTICLE + FIRE OFF-GASSING + FORENSIC INVESTIGATION + GAS TEST PROCEDURES + HISTORIC & OLD BUILDINGS + HOUSE DUST ANALYSIS + INSECT INFESTATION - home + LIGHT, GUIDE to FORENSIC USE + LIGHT, UV BLACK LIGHT USES + MICROSCOPY + STRUCTURAL DAMAGE PROBING + TERMITE DAMAGE + THERMAL EXPANSION INDOOR AIR + + AIRBORNE MOLD LEVEL + AIRBORNE PARTICLE ANALYSIS + ALLERGEN TESTS + ANIMAL ALLERGENS DANDER + CARBON DIOXIDE + CARBON MONOXIDE + CARPETING + CAT DANDER + COMBUSTION GASES + DUST SAMPLING + EMERGENCY RESPONSE, IAQ, GAS, MOLD + FIBERGLASS - home + HUMIDITY + IAQ & HOUSE TIGHTNESS + INDOOR AIR HAZARDS + ODORS GASES SMELLS- home + VENTILATION INSULATION + + ATTIC + BASEMENT + FIBERGLASS + FRAMING DETAILS + HOT ROOF PROBLEMS + INSULATION AIR & HEAT LEAKS + INSULATION CHOICES + INSULATION FACT SHEET- DOE + INSULATION GREENHOUSE + INSULATION ID + INSULATION LOCATION + INSULATION MOLD + INSULATION R-VALUES + POLYSTYRENE FOAM + RIGID FOAM + UFFI UREA FORMALDEHYDE FOAM INTERIOR + + ASBESTOS in DRYWALL + BATH & KITCHEN DESIGN - home + CABINETS & COUNTERTOPS - home + CARPETING - home + CARPET STAIN ID + CEILING STAIN DIAGNOSIS + CERAMIC TILE FLOOR, WALL + CONDENSATION + COUNTERTOPS + DRYWALL FIBERBOARD PLASTER- home + EFFLORESCENCE WHITE DEPOSIT + FIBERBOARD- home + FIREPLACES & HEARTHS + FLOOR, CONCRETE SLAB + FLOOR TYPES & DEFECTS - home + INTERIOR FINISHES + KITCHEN DESIGN + MOISTURE CONTROL + PAINT FAILURE - home + PLASTER METHODS + RESILIENT SHEET FLOORING - home + SHEATHING, FIBERBOARD + SHEET FLOORING ID + SLIP TRIP & FALL + STAIR CONSTRUCTION + STAIN DIAGNOSIS + STUCCO WALL METHODS + THERMAL TRACKING + TILED SURFACES + TRIM, INTERIOR + WALL FINISHES + WOOD STOVE OPERATION - home + WOOD FLOOR DAMAGE MOBILE HOME + + BUYERS ADVICE + CODES & MANUALS + COMBUSTION AIR SAFETY + CONNECTIONS, MULTI-WIDE + COOLING SYSTEM + CRAWL SPACES + CROSSOVER CONNECTORS + DATA TAGS & LABELS + DEMOLISH REMOVE MOVE + ELECTRICAL POWER LOST + ELECTRICAL SYSTEM + EMERGENCY EGRESS WINDOWS + ENERGY ZONES + EXTERIOR DEFECTS + FLICKERING LIGHTS + FOUNDATIONS + GFCI DIAGNOSIS + HEALTH DEPARTMENT HELP + HEATING SYSTEM + INSPECTIONS + INSULATION & VENTILATION + INTERIOR DEFECTS + LEAKS + MODULAR CONSTRUCTION + MOLD CONTAMINATION + PIERS + PLUMBING + ROOF + SAFETY + SKIRTING + STABILIZING & TIE DOWNS + STRUCTURE + TEMPORARY OFFICE TRAILER + WALL DEFECTS + WATER HEATERS + WIND RATINGS + WINTERIZE MOLD + + ACTION GUIDE + AIRBORNE MOLD COUNT - home + AIRBORNE PARTICLE LEVEL- home + ASPERGILLOSIS + ATTIC MOISTURE or MOLD + BLACK MOLD, HARMLESS + BLEACHING MOLD + BOOK / DOCUMENT MOLD + CABINET MOLD + CACTUS FUNGI / MOLD + CAR MOLD CONTAMINATION + CARPET MOLD / ODOR TESTS + CAR MOLD CONTAMINATION + CEILING STAIN DIAGNOSIS + DIRT FLOOR MOLD + DRYWALL MOLD + DUST / MOLD SAMPLING + EFFLORESCENCE & WHITE DEPOSITS + EMERGENCY RESPONSE + FEAR of MOLD - MYCOPHOBIA + FIBERBOARD SHEATHING MOLD + FIBERGLASS INSULATION MOLD + FIND MOLD, ESSENTIAL STEPS + FOXING STAINS + HARD TO SEE MOLD, SPOTTING + HIDDEN MOLD + HUMIDITY CONTROL & TARGETS + LIGHT, USE TO FIND MOLD + MERULIPORIA FUNGUS + MILDEW + MOBILE HOME MOLD + MODULAR HOME MOLD + MOLD A COMPLETE GUIDE - home + MOLD SAFETY ADVICE for TENANTS + MOLD CLEANUP + MOLD AGE + MOLD APPEARANCE + MOLD CLEARANCE INSPECTION + MOLD COUNT NUMBERS + MOLD CULTURE SAMPLING + MOLD DETECTION + MOLD DOCTOR + MOLD ENVIRONMENTAL EXPERTS + MOLD EXPOSURE STANDARDS + MOLD FREQUENCY + MOLD INVESTIGATION PROCEDURE + MOLD ODORS, MUSTY SMELLS + MOLD PREVENTION - home + MOLD RELATED ILLNESS + MOLD SANITIZER, SPRAY, BIOCIDE + MOLD TEST PROCEDURES + MVOCs & MOLDY MUSTY ODORS + MYCOTOXIN EFFECTS + OZONE TREATMENT WARNING NOISE + + BANGING BOOMING NOISES - home + ELECTRICAL SYSTEM NOISE + FAN NOISES + HEATING SYSTEM NOISE + HVAC SYSTEM NOISE + NOISE CONTROL for ROOFS + PLUMBING SYSTEM NOISE - home + RELAY SWITCH NOISE + ROOF IMPACT NOISE + ROOF NOISE TRANSMISSION - home + SOUND CONTROL + TEMPERATURE CHANGE & ROOF NOISE + WATER HAMMER NOISE ODOR + + AIR CONDITIONING + ANIMAL or URINE + CAR ODORS, ANIMALS + FLOOR DRAIN / TRAP + METHANE & SEWER GAS + MOLD ODORS + MVOCs MOLDY MUSTY + ODOR CONTROL for SEPTIC + ODORS FROM HEATING SYSTEMS + ODORS, PLUMBING SYSTEM + ODORS, SEPTIC or SEWER + ODOR SENSITIVITIY + OZONE MOLD / ODOR TREATMENT + PLUMBING SYSTEM - home + SMELL PATCH FIND ODOR SOURCE + URINE ODOR SOURCE + WATER ODOR CURE PLUMBING + + AIR DISCHARGE at FAUCETS + CHECK VALVES + CLEARANCE DISTANCES + CLOGGED DRAIN REPAIR + COMPOSTING TOILETS + DISPOSABLE WET WIPE CLOGS + DRAIN CLEANOUTS + FLOOR DRAIN / TRAP ODORS + GAS TANKS & PIPING + OIL TANKS & PIPING + PLUMBING TRAPS + PLUMBING VENTS - home + SEPTIC SYSTEMS + SEWAGE PUMPS - home + TANKLESS COIL HOT WATER + TOILETS - home + WATER HEATERS, ELECTRIC - home + WATER PIPE CLOG + WATER PRESSURE DIAGNOSE + WATER PRESSURE IMPROVE + WATER PRESSURE LOSS- home + WATER PUMPS & WELLS + WATER SHUTOFF VALVE + WATER SOFTENERS - home + WATER PIPING - home + WATER TANK - home + WINTERIZE A BUILDING ROOF + + AGE + ASBESTOS & FIBER CEMENT - home + ASPHALT SHINGLES - home + CLAY TILE - home + CLEANING + COLOR + CONCRETE + CONTRACTOR, CHOOSE + CORRUGATED + DEBRIS STAINING + DISPUTE RESOLUTION + EPDM, RUBBER, PVC + EXTRACTIVE BLEEDING SHINGLES + FELT UNDERLAYMENT - home + FIBER CEMENT ROOFING - home + FIBERBOARD & FIBER-WOOD + FIRE RATINGS + FLASHING on BUILDINGS - home + FLAT ROOF LEAKS + HAIL DAMAGE + ICE DAM + INSPECTION + LEAD ROOFING & FLASHING + LEAK REPAIR - home + LOW SLOPE - home + MATERIALS, AGE, TYPES + MEMBRANE & SINGLE PLY + METAL- home + PLASTIC ROOFING TYPES + PVC, EPDM, RUBBER MEMBRANE + ROLL ROOFING, ASPHALT & SBS + RUBBER SHINGLES SLATES + SBS ROOFING ROLL & BUR ROOFS + SEALANTS & MASTICS + SHINGLE STORAGE + SLATE - home + SLOPE CALCULATIONS + STAINS - home + STANDARDS + STONE ROOF + THATCH ROOF + TILE, CLAY - home + TILES, CONCRETE + VENTILATION - home + WALKABLE + WARRANTIES + WHITE STAINS - home + WIND DAMAGE + WIND NOISES + WIND DAMAGE RESISTANT + WOOD SHAKE & SHINGLE - home + WORKMANSHIP & DAMAGE SEPTIC + + AEROBIC ATUs - home + AGE of SEPTIC SYSTEM + BACKUP PREVENTION + BIOMAT FORMATION & SEPTIC LIFE + BOD WASTEWATER TEST + CAMERAS, SEWER / SEPTIC + CARE - home + CESSPOOLS + CHAMBER SEPTIC SYSTEMS + CLEARANCE DISTANCES + CLOGGED DRAIN REPAIR + CLOGGED DRAIN FIELD + CODES - home + COMMERCIAL SEPTIC + COMPONENT LOCATIONS - home + D-BOX INSTALL REPAIR + DESIGN ALTERNATIVES - home + DESIGN BASICS - home + DESIGN PREVENT FLOOD DAMAGE + DIFFICULT SEPTIC SITE + DISPERSAL METHODS + DISPOSAL vs TREATMENT + DRAINFIELD TEST - home + DRIVING OVER SEPTIC + DRYWELL + EFFLUENT DISTRIBUTION + EFFLUENT RETENTION TIME + FAILURE SIGNS + FILTERS + FLOODED SYSTEM REPAIR + FREEZE PROTECTION + FREEZE-UP SOLUTIONS + FROZEN AEROBIC SEPTIC + GARBAGE DISPOSAL vs SEPTICS + GRAVELLESS SEPTIC + GREYWATER SYSTEMS + HOOT AEROBIC SEPTIC + HOME BUYERS GUIDE + HOME SELLERS GUIDE + HOW SEPTIC SYSTEMS WORK + INSPECT & TEST - home + INSPECT & TEST LAWS + LAUNDROMAT WASTEWATER + LIFE EXPECTANCY + LOADING & DYE TEST - home + LOW COST SYSTEMS + MAINTENANCE - home + MEDIA FILTER SYSTEMS - home + ODOR CONTROL + ODORS, SEWER GAS + PLANTS OVER SEPTIC SYSTEMS + PUMPING the SEPTIC TANK + PUMPS + REPAIR - home + SAFETY + SANDY SOIL SYSTEMS + SEEPAGE PITS + SEWAGE BACKUP + SEWAGE CONTAMINANTS + SEWAGE TREATMENT PLANTS + SEWER CONNECTION? - home + TANKS - home + TANK CLEANING + TANK COVERS + TANK DEPTH + TANK, HOW TO FIND + TANK PUMPING + TANK PUMPING SCHEDULE + TANK TEES STRUCTURE + + ADOBE CONSTRUCTION + BASEMENT WATERPROOFING + BLOCK FOUNDATION / WALLS - home + BRICK FOUNDATIONS & WALLS - home + BUILDING DAMAGE REPAIR + BULGED vs. LEANING FOUNDATIONS + CARPENTER ANTS + CHIMNEY REPAIR - home + CLAY HOLLOW TILE + CLEARANCE DISTANCES - topic home + COLUMNS & POSTS, DEFECTS + CONCRETE FOUNDATION, WALL, SLAB + CONTROL JOINT CRACKS + CRAWL SPACES + DAMAGE ASSESSMENT + DECK & PORCH - home + EARTHQUAKE, FLOOD, DISASTER + FIBERBOARD SHEATHING - home + FIRE ESCAPES & STAIRS + FIRE STOPPING + FOUNDATION TYPES + FOUNDATION CRACKS & DAMAGE- home + FOUNDATION REPAIR - home + FRAMING AGE, SIZE, SPACING, TYPES + FRAMING CONNECTORS + FRAMING DAMAGE REPAIR - home + FRAMING FASTENERS + FRAMING SQUARE - home + FRAMING TABLES, SPANS + HOUSEWRAP + INSECT DAMAGE - home + LOG HOME REPAIR + MOBILE HOMES + MODULAR CONSTRUCTION + MOISTURE CONTROL + PANELIZED CONSTRUCTION + PORCH CONSTRUCTION + ROOF BEND SAG COLLAPSE + ROOF FRAMING TIES & BEAMS + ROOF TRUSS UPLIFT + ROT TYPES + SHIPPING CONTAINER HOUSING + SINKHOLES / SUBSIDENCES + STAIRS, RAILINGS, LANDINGS, RAMPS - home + STRESS SKIN / SIPS + STRUCTURAL DAMAGE PROBING + TERMITE INSPECTION & DAMAGE + TIMBER FRAMING, ROT + TRUSSES, FLOOR & ROOF + TRUSS UPLIFT, ROOF + WATER ENTRY, LEAKS + WOOD STRUCTURE ASSESSMENT VENTILATE + + ATTIC VENTILATION + BALANCED VENTILATION + BATHROOM VENT CODES - home + BATHROOM VENT DUCT ROUTING + KITCHEN VENT DESIGN + HIP ROOF VENTILATION + HOT ROOF PROBLEMS + ICE DAM PREVENTION + ROOF ICE DAM LEAKS + ROOF VENTILATION CODE- home + SKYLIGHT VENTILATION + SOFFIT VENTILATION WATER + + AIR VOLUME CONTROLS + ARTESIAN WELLS, SPOOLS + CHECK VALVES + CHEMICAL CONTAMINANTS + CISTERNS + FOOT VALVES + GREYWATER SYSTEMS - home + NO WATER PRESSURE + ODORS + PUMP PRESSURE CONTROL + PUMPS + SPRINGS + SWIMMING WATER + WATER CONSERVATION- home + WATER CONTAMINATION - home + WATER FILTERS + WATER METERS + WATER ODORS + WATER POLLUTANTS + WATER PRESSURE BOOSTER + WATER PRESSURE LOSS - home + WATER PRESSURE REDUCER / REGULATOR + WATER PUMP PRESSURE CONTROL + WATER PUMP PRIMING + WATER PUMP REPAIR - home + WATER TANK REPAIR - home + WATER TESTS - home + WATER TREATMENT - home + WELLS CISTERNS & SPRINGS - home CONSULT ESTIMATE CONTACT | ADVERTISEMENT ADVERTISEMENT Table of Coefficient of Expansion of Building MaterialsCLTE Coefficient of Linear Thermal Expansion for Common Building Materials POST a QUESTION or COMMENT about the thermal expansion or contraction of building materials as temperatures vary Thermal coefficient of expansion of building materials: Here we provide a Table of Coefficient of Thermal Expansion of Building Materials - what is the linear expansion of glass, metal, wood, masonry or plastic in response to temperature changes. We include a discussion of the definition of thermal coefficient of expansion, how thermal expansion causes trouble in building materials, and how to use the data in the thermal expansion table to calculate changes in material size as temperatures change. InspectAPedia tolerates no conflicts of interest. We have no relationship with advertisers, products, or services discussed at this website. - Daniel Friedman, Publisher/Editor/Author - See WHO ARE WE? Table of Coefficients of Thermal Expansion of Common Building Materials How do building materials change in dimension in response to temperature changes, sun, shade, ice, snow?How do the dimensions of different building materials vary with temperature, heat, cold? Table of Coefficient of Expansion of Building Materials. Sketch at page top and accompanying text are reprinted/adapted/excerpted with permission from Solar Age Magazine - editor Steven Bliss. Photo: vinyl siding being installed in very cold temperatures in Two Harbors, Minnesota. [Click to enlarge any image] Definition of CoTE: Coefficient of Thermal Expansion CoTE or Coefficient of Thermal Expansion gives change per unit dimension of material in response to change (increase or decrease) in the material's temperature. Definition of CLTE: Coefficient of Linear Theral Expansion The coefficient of linear thermal expansion (CLTE) of any material is the change of a material's dimension per unit change in temperature. - special thanks to reader Eric, P.E. The coefficient of thermal expansion or CLTE, Coefficient of Linear Thermal Expansion The linear expansion of a heated solid (or liquid) is measured by α = the coefficient of linear expansion, The coefficient of thermal expansion is defined such that α measures the percentage change in the length of the material per degree of temperature change. Be careful in comparing the coefficient of expansion of different materials from different reference sources for thermal expansion coefficients as various references quote α in degrees C, others in degrees F. Usually people use the term CLTE or Coefficient of Linear Thermal Expansion as expansion along the long dimension of a material is most likely to be greater and thus of greater concern in construction. CLTE may be expressed as a fraction or as a power of ten as you'll see in the table below. The following simple formula for the coefficient of thermal linear expansion in a building material is written to measure the percentage change in length per degree of temperature change: α = (Change in Length / Original Length) / Change in Temperature One can write similar formulas to calculate the coefficient of thermal expansion of a material in area (applicable, for example in thermal splitting of asphalt roof shingles ) or to calculate the coefficient of thermal expansion of volume. But because so many building material failures and leaks derive from cracks or openings due to thermal expansion of materials in length, that is our focus here. | Comparison of Coefficients of Linear Temperature Expansion (CLTE) of Common Building Materials | | CLTE Ordered by Name of Material | Ordered by Coefficient of Thermal [Temperature] Expansion | | Material | Coefficient of Expansion in inches of expansion per inch of material per degree F. | Material | Coefficient of Expansion in inches of expansion per inch of material per degree F. | | ABS plastic | 0.0000170 (glass fiber-reinforced) | Acrylic | 0.0001300 (extruded) | | ABS plastics | 0.0000410 | Polyethylene | 0.0001110 | | | | Vinyl2 (siding & trim) | 0.000045 or 4.5×10 −5 in/in/F 1, 2 | | | | PVC (cellular) | 0.000045 or 4.5 x 10-5 in/in/F 1 | | Acrylic | 0.0001300 (extruded) | Polycarbonates | 0.0000440 | | Acrylic | 0.0000410 (sheet cast) | ABS plastics | 0.0000410 | | Aluminum | 0.0000123 - 0.0000129 or 0.000029 (other sources) 0.000117 to 0.111137 or 11.7 to 13.7 x 10-5 in/in/F 1 α (ppm) = 29 (0–600 °C) α = coefficient of linear expansion (°C−1) | Acrylic | 0.0000410 (sheet cast) | | Aluminum Alloys | 0.0000117 or 11.7 x 10-6 in/in/F Note that "alloys" is not defined and could vary significantly | | | | Brass | 0.0000104 - 190 | Epoxy | 0.0000310 | | Brick | 0.011811 - 0.0629921 /degF vs 0.0000031 (brick masonry) 11, 13, 14, 15 5 - 7 x 10-6 per degC 18 best citation 13 | Ice | 0.0000280 (effects of freezing water) | | Cast iron | 0.0000058 | ABS plastic | 0.0000170 (glass fiber-reinforced) | | Cast iron | 0.0000060 (gray cast iron) | Zinc | 0.0000165 | | Cement | 0.0000060 | Lead | 0.0000151 | | Clay tile | 0.0000033 | Aluminum | 0.0000123 - 0.0000129 | | Concrete | 0.0000080 (Concrete structure = 0.0000055)14 0.000004 to 0.000007 or 0.4 to 0.7 x 10-5 in/in/F 1 | Brass | 0.0000104 - 190 | | Copper | 0.0000093 | Copper | 0.0000093 | | Epoxy | 0.0000310 | Concrete | 0.0000080 (Concrete structure = 0.0000055) | | Glass, hard | 0.0000033or 0.0000040 or 0.4 x 10-5 in/in/F 1 | Iron, pure | 0.0000067 | | Glass, plate | 0.00000500 | Steel | 0.0000063 - 0.0000073 (also Iron, forged) | | Glass, Pyrex | 0.0000022 | Cast iron | 0.0000060 (gray cast iron) | | Granite | 0.0000044 (also Limestone, Marble) | Cement | 0.0000060 | | Ice | 0.0000280 (effects of freezing water) | Cast iron | 0.0000058 | | Iron, pure | 0.0000067 | Glass, plate | 0.0000050 | | Lead | 0.0000151 | Granite | 0.0000044 (also Limestone, Marble) | | Masonry | 0.0000026 - 0.0000050 | Nylon | 0.00000447 (molding & extruding compound) | | Mortar | 0.0000041 - 0.0000075 | Mortar | 0.0000041 - 0.0000075 | | Nylon | 0.0000447 (molding & extruding compound) | Clay tile | 0.0000033 | | Polycarbonates | 0.0000440 | Glass, hard | 0.0000033 | | Polyethylene | 0.0001110 | Brick | 0.0000031 (brick masonry) | | Polystyrene | 4.16 x 10-5 in/in/F | | | | Polyurethane | 11.0 x 10-4 in/in/F (Density Dependent) 1 | | | | PVC1(rigid) | 0.000035 or 3.5x10,-5 in/in/F 1 | | | | PVC (cellular) | 0.000045 or 4.5 x 10-5 in/in/F 1 | | | | Stainless Steel SS 301SS 303, 304, 305 & SS 308SS 309SS 316 & SS 317SS 321 & SS 347SS 403SS 420 & 430FSS 446SS 501 & SS 502 (varies by alloy/composition) | 16.9 ppm/°C 17.3 14.9 16.0 16.7 9.9 10.45 10.6 11.15 | | | | Steel | 0.0000063 - 0.0000073 (also Iron, forged) 0.000008 or 0.8 x 10-5 in/in/F 1 | Wood,Oak | 0.0000030 (across grain) | | Vinyl2 (siding & trim) | 0.000045 or 4.5×10 −5 in/in/F 1, 2 | | | | Water | see HOT WATER PRESSURE EXPANSION RATE | | | | Wood,Oak | 0.0000030 (across grain) | Wood, Pine | 0.0000028 | | Wood, Oak | 0.0000027 (parallel to grain) | Wood, Oak | 0.0000027 (parallel to grain) | | Wood, Pine | 0.0000028 or in other sources0.000004 or 0.4 to 0.5 x 10-5 parallel to the grain and 0.000034 to 0.000058or 3.4 to 5.8 x 10 -5 at right angles to the fiber direction or grain | Masonry | 0.0000026 - 0.0000050 | | Zinc | 0.0000165 | Glass, Pyrex | 0.0000022 | Notes to the table above ASTM. 1997. Standard methods for testing small clears pecimens of timber. ASTM D143. West Conshohocken,PA: American Society for Testing and Materials. Kollman, F.F.P.; Côté, W.A., Jr. 1968. Principles ofwood science and technology I—solid wood. New York,Springer–Verlag New York, Inc. Panshin, A.J.; deZeeuw, C. 1980. Textbook of woodtechnology. New York: McGraw–Hill. Vol. 1, 4th ed Special thanks to Bob Fankhauser Email: blueboxconst@hevanet.com, a retired engineer / professional handyman and Habitat for Humanity volunteer who offered comments, suggestions, additions for vinyl CLTE (Coefficient of Linear Thermal Expansion), CPVC, PVC, cellular PVC, and vinyl (25 Feb 20-16) as well as helpful discussion concerning the wide variation in coefficients of expansion of materials given by various sources. Stainless steel coefficients of expansion: University of Chicago, LINEAR THERMAL EXPANSION COEFFICIENTS of METALS & ALLOYS, Table 17-1, [PDF] retrieved 2017.11.13, original source: Smith, Thomas L., AIA, CRC, METAL ROOFING: 'FIXING' for THERMAL MOVEMENT [PDF] NRCA, National Roofing Contractors Association Weatherwax, R.C.; Stamm, A.J. THE COEFFICIENTS OF THERMAL EXPANSION OF WOOD AND WOOD PRODUCTS [PDF] (1956) Transactionsof American Society of Mechanical Engineers. 69(44):421–432 Simpson, William, Anton TenWolde, WOOD HANDBOOK, PHYSICAL PROPERTIES of WOOD - Chapter 3 Physical Properties & Moisture RElations of Wood [PDF] p. 3-21, "Coefficients of Expansion" in Forest Products Laboratory. 1999. Wood handbook—Wood as an engineering material.Gen. Tech. Rep. FPL–GTR–113. Madison, WI: U.S. Department of Agriculture, Forest Service,Forest Products Laboratory. 463 p. USFPS, retrieved 2020/10/23 original source: Skaar, C. 1988. Wood–water relations. New York:Springer–Verlag. New York, Inc. See Notes 1 & 2 and details now found at VINYL SIDING COEFFICIENT of LINEAR EXPANSION Engineering Toolbox website, provides a more extensive table of coefficients of linear expansion at engineeringtoolbox.com/linear-expansion-coefficients-d_95.htmlretrieved de-novo 2023/07/24 Western Washington University Thermal Expansion is described and defined in a clear article that also gives both linear and volumetric coefficients of thermal expansion at 20 degC. for a variety of materials at ac.wwu.edu/~vawter/PhysicsNet/Topics/Thermal/ThermExpan.htmlWeb: www.wwu.edu Ross, Culbertson W., THERMAL EXPANSION of CLAY BUILDING BRICKS [PDF] (1941) NIST, National Institute of Standards & Technology, retrieved 2023/07/24, original source: nvlpubs.nist.gov/nistpubs/jres/27/jresv27n2p197_A1b.pdf Abstract: The coefficients of thermal expansion of 139 bricks were measured over the range -10° to +400 C (140 to 1040 F). These bricks included 1 sample of sand-lime, 9 of fire-clay, and 61 of clay and shale bricks. The clay and shale bricks represented a wide range in properties and included samples from various districts in the United States. The coefficients of 87 percent of the clay and shale bricks were between 5 and 7 millionths per 0 C (2.8 to 3.9 per 0 F). The average coefficient of the clay bricks was 6.0 (3.3), of the shale bricks 6.1 (3.4), and of the fireclay bricks 3.9 millionths per 0 C (2.2 per 0 F). No relation was observed between the thermal expansion and the other physical properties of the bricks. Excerpts: The change in length of building bricks due to temperature changes is one of the factors sometimes blamed for cracking of brick-masonry walls. The coefficients of thermal expansion ranged from 3.7 to 13.6 millionths per 0 C (2.1 to 7.6 per 0 F), with an average of 6.3 (3.5). Kidger, Peter, Dickinson, Chris, CLAY vs BRICK MOVEMENT of MASONRY [PDF] (2016) Clay Brick Association of Southern Africa, Web: claybrick.org Addresses movement of masonry due to changes in temperature or moisture or both. AZO Materials, BRICK WALLS - Expansion and Expansion Gaps - (2022) retrieved 2023/07/24, original source: azom.com/article.aspx?ArticleID=1327 Excerpts: The coefficient of expansion or ‘e’ value of clay bricks when tested in accordance with AS/NZS 4456.11 can vary from a very low 0.3mm/m (millimetres per metre) to over 1.6mm/m. Most general purpose bricks have a coefficient of expansion in the range 0.5 to 1.0mm/m over 15 years. Clay bricks have a once off, permanent expansion after manufacture due to the firing process that extracts all moisture from the brick. This expansion is between 0 and 0.2%. CLAY BRICK: Thermal expansion and contraction is minimal throughout the year. Depending on the clay mixture and firing process, the coefficient of linear thermal movement is 4-8. We think the authors mean 4-8 ? per ˚C - Ed. details needed. CONCRETE BRICK: Moisture movement is exacerbated by a daily cycle of thermal expansion and contraction. Depending on the type of aggregate and proportions, the coefficient of linear thermal movement is 7-14. CONCRETE BLOCK: The movement coefficient of concrete blocks is similar to concrete bricks, but because blocks are larger the magnitude of movement is greater. Large blocks show cracks and plaster crazing more than smaller concrete units. NCMA, CONCRETE MASONRY MANUAL [PDF] 7th Ed (2005) National Concrete Manufacturers Association, P O Box 168 Halfway House 1685 / 18 McIntyre St 18 McIntyre St, Glenlily, Cape Town, 7500 South Africa, Tel: +27 11 805-6742 Facsimile: + 27 11 315-4683 e-mail: cma@mweb.co.za Web site: www.cma.org.za Seventh Edition – revised 2005. Illustrations: John L Betts MIA Editor: J W Lane ISBN: 0-9584142-7-0 SABS, SOUTH AFRICAN NATIONAL STANDARD, Part K: WALLS [PDF] SABS Standards Division 1 Dr Lategan Road Groenkloof Image Private Bag X191 Pretoria 0001 Tel: +27 12 428 7911 Fax: +27 12 344 1568 www.sabs.co.za - retrieved 2023/07/24, original source: Excerpts from Table 19: Burnt Clay moisture expansion: < .05 to 0.20 % NOTE 1 SANS 227 contains a test procedure to establish the moisture expansion of burnt clay bricks. Readers can see from these building material coefficients of thermal expansion (also called coefficient of linear temperature expansion) that assembling a building component that uses multiple materials requires methods that allow for these differences in the degree of expansion as temperatures change. Failing to permit movement of abutting or connected building materials whose rate of thermal expansion varies significantly will lead to separation, cracks, leaks, or damage in many instances. Examples of the problems caused by differences in thermal expansion of building materials are particularly seen in windows and skylights. See SLOPED GLAZING DETAILS At SKYLIGHT LEAK DIAGNOSIS & REPAIR we include an example of failure of roof flashing cement that has lost its ability to tolerate thermal expansion and contraction on the building. As we discuss at CRACKS & THERMAL SPLITTING in FIBERGLASS SHINGLES, we have not found a source defining the coefficient of thermal expansion of asphalt roof shingles - CONTACT Us if you can provide that information. Calculation Examples using the Thermal Coefficient of Expansion of Materials How much does aluminum siding expand when heated by rising ambient temperatures or sunlight? While an accurate answer would include the effect of the siding color on thermal gain from direct exposure to sunlight along with the angle of sunlight hitting the siding and the duration of sun exposure, what follows is a "ballpark" calculation ignoring those niceties. The coefficient of linear expansion of Aluminium is about 22 x (10-6) m/mC or 22 x 0.000001 per unit-length per degree C of temperature rise. Or re-stating, aluminum siding will expand along its length from 21 to about 24 micrometers (about 22 um) per meter of length for each 1°C increase in its temperature. What would be the thermal expansion of aluminum siding of a single segment of 20 ft (6.1m) if we heat it from 41°F up to 81°F? if the temperature increases by 25C (say from 5 C to 30C = 41F to 86F) the increase in length in the siding segment will be 22 x 20ft x 25C x (10-6) = 0.011ft of expansion That doesn’t look like much. But wait! Let's convert that to inches: 0.11 ft x 12 (inches per foot) = 0.132 inches. (If we've got this right) That's enough to make a popping or banging noise if the aluminum siding is installed too-tightly or is constrained between vertical abutting trim. Sanity Check: I looked at George Metaxas commenton Quorum quora.com 4 October 2020. George, a retired EE, wrote The coefficient of linear expansion of Aluminium is about 22x(10^-6) m/mC. As the piece is 3.6m long, for a temperature increase of 55C, the elongation will be: 22 x 3.6 x 55 x (10^-6) = 4356 x10^-6 m = 4.35 mm. 4.33 mm, for you inch-people, is 0.17 inches across just 12 ft. of length. ... ADVERTISEMENT Reader Comments, Questions & Answers About The Article Above Below you will find questions and answers previously posted on this page at its page bottom reader comment box. Reader Q&A - also see RECOMMENDED ARTICLES & FAQs On 2023-05-10 by InspectApedia Editor (mod) - correction to the table of coefficients of expansion of materials, arranged by value @Mike, Thank you SO much, really, for taking the time to write and offer a correction. We've edited and re-arranged the table as per the error that you pointed out. On 2023-05-10 by Mike - sorted table of thermal coefficients should put vinyl and pvc near the top The sorted table of thermal coefficients is not correctly sorted, vinyl and pvc should be number 3 and 4 but without converting from engineering notation to decimal it is not immediately obvious unless you deal with engineering notation daily. On 2021-11-04 y Inspectapedia Com Moderator (mod) - diagnosing bang & boom noises that might be due to thermal changes @Yihang, Please help us out by posting your questions just once in one place - we have thousands of readers we try to serve- tripling our workload makes it hard for us. You posted and we replied at On 2021-11-03 by Yihang Hi, I lived in an apartment in Auckland Central for one year. Everything was fine before. But 3 weeks ago I started to hear strange thudding/thumping noise almost every morning from about 10am to 1pm. Sometimes I will also hear the noise at 4pm. The thudding noises seem to occur randomly throughout the time period, so two thumps, one is less louder and one is louder and then several minutes later another two thumps. There is also another strange phenomenon is that, sometimes, within a very short time interval, say 30 seconds, there are about 5 thudding noises, like boom, boom, boom , boom, boom. It never happens at evening(but this is what I am afraid for). What is actually going on?I lived in the top floor of the apartment and the thudding noises sometimes seem to coming from the ceiling but other times seem to coming from the floor where my downstairs neighbour promised me so many times that they did not make any of such noises! On 2020-10-23 by danjoefriedman (mod) - coefficient of expansion of engineered wood. Joe Thanks for the question about the coefficient of expansion of engineered wood. Composite wood beams, like other wood products, will expand and contract with changes in moisture (and less so with temperature) across the grain - across the width and thickness of the beam more than along the length (parallel to the grain). The coefficient of thermal expansion of pine and spruce parallel to the wood-grain is about 0.4 to 0.5 x 10-5 and perpendicular to the fiber direction 3.4 to 5.8 x 10 -5. Other sources like that used in my table above give Pine as 0.0000028 /inch as the Coefficient of Expansion in inches of expansion per inch of material per degree F. So let's calculate the "worst case" or most expansion that occurs across the grain using Pine: a 2x12 is roughly 11.25" deep (not really 12" unless it's rough cut) -40C is about -40F too. 30C is about + 86F So we have a total F-degree change of 40+86 = 126 degrees F (11.25 inches of material x 0.0000028 expansion per degree) x 126 degrees = about 0.004" across the grain or using another source 0.4 to 0.5 x 10-5 expansion along the grain (parallel to the grain or in length) as a "worst case" for lengthening, we have .000005 x (total beam length not given) x 126 degrees = ?? length change Maybe if you post a sketch I'll understand better; Interestingly, dry wood reacts less than damp wood to temperature variations. I have not found stable estimates of the coefficients of expansion of wood in response to moisture alone or at a fixed temperature and am researching that. On 2020-10-22 by Joe Tworek - thermal expansion/contraction in an engineered wood beam I am building a shop mezzanine, the main beam will be a composite made out of 3, 2X12’s held together with 1/2 in carriage bolts spaced 4 in vertically every 3 ft the entire length of the beam. I don’t want the beam to contact the shop walls. The 2x12’s are a soft lumber probably Spruce, I live in Alberta, Canada where the temp can vary from -40C in the winter to 30C in the summer. How much thermal expansion of the beam do you expect based on this. Thank You. On 2020-07-19 by danjoefriedman (mod) - different coefficient of expansion for aluminum alloys. Roger thanks so much for taking the time to give a different coefficient of expansion for aluminum alloys. I can but speculate that there can be rather different LCTEs depending on just which alloy is being tested. One may also get a different coefficient of expansion depending on the exact conditions, such as how aluminum wire of a particular alloy is employed, the use of antioxidents or thermal grease. Xu, Ling, Miaocao Wang, Yang Zhou, Zhengfang Qian, and Sheng Liu. "Effect of silicone gel on the reliability of heavy aluminum wire bond for power module during thermal cycling test." In 2016 IEEE 66th Electronic Components and Technology Conference (ECTC), pp. 1005-1010. IEEE, 2016. Ganesh, V. V., P. K. Tan, and M. Gupta. "Development and characterization of an aluminum alloy containing interconnected-wires as reinforcement." Journal of alloys and compounds 315, no. 1-2 (2001): 203-210. One may also get a different LCTE depending on the measurement method used, as we see in some research articles. Otte, H. M., W. G. Montague, and D. O. Welch. "X‐Ray Diffractometer Determination of the Thermal Expansion Coefficient of Aluminum near Room Temperature." Journal of Applied Physics 34, no. 10 (1963): 3149-3150. And of course how the aluminum is being employed Thomas, R. "Stress-induced deformation of aluminum metallization in plastic molded semiconductor devices." IEEE Transactions on Components, Hybrids, and Manufacturing Technology 8, no. 4 (1985): 427-434. Let's do some further research on sources. On 2020-07-19 by Roger - the LCTE of aluminum alloys is approximately 11 x 10^-6 /F rather than 11 x 10^-5 /F. The Engineering Toolbox (and others) suggest that the LCTE of aluminum alloys is approximately 11 x 10^-6 /F rather than 11 x 10^-5 /F. Any comments? Question: find and fix banging & knocking noises from pillars & roof of this Florida Patio After endlessly searching online I happened across your site. I have been dealing with this problem for over a year. I'm in Florida and I had a patio screen room added to my home last August. It's constructed entirely of aluminum except for the roof which is an unknown material insulated with some sort of styrofoam. I attached a picture. [Click to enlarge any image] My problem is that I am hearing intermittent banging and knocking noises coming from the pillars and the areas of roof. It sounds as if something is hitting it. I have inspected for everything including birds, insects, lizards, squirrels etc. It is NOT any sort of animal. What I do find interesting is that I do seem to hear it more frequently when there are slightly breezy conditions but I still hear it whether it's breezy or not. Nothing is hanging or loose. I contacted the builder and told him about it. He came out about 4-5 months ago to inspect it. He checked it from top to bottom and couldn't find anything wrong with it. He said it could possibly be expanding and releasing from temperature changes but he can't be sure. But literally minutes after he left, it began knocking and banging again. Obviously he can fix what isn't present. I am trying to figure out HOW I can capture these sounds as evidence of what is happening here daily! All I can seem to find online are voice activated devices. This clearly isn't a voice. I am desperate to fix this issue as these knocking sounds interfere with my peace and enjoyment. Can you help me figure this out? This enclosure cost me $13,000 and I can't even enjoy it! It's extremely aggravating.Thank you and I hope to hear from you soon. - Anonymoys by private email 2020/10/17 Moderator reply: The most-common explanation is just what your builder gave, Thermal expansion and contraction, is probably the culprit, anon. Those noises can appear as banging, knocking, or clicks, squeaks, or tapping sounds as metal or other components of your Florida patio move against one another during temperature changes, also affected by wind and changes in humidity, even barometric pressure. The conceptual explanation is that different materials expand and contract at different rates - as illustrated in the article above on this page on coefficients of thermal expansion. So they move differently and so move against one another to produce sound. Reader follow-up: So there's nothing that can be done? I feel I should have been advised of this BEFORE construction. I never would have agreed to this Moderator reply: I did not say that Dayna But I'm reluctant to pretend to diagnose and repair sounds without any more information; Of course there is something to be done, and in many cases the repair is economically reasonable. Where we get into trouble fixing thermal expansion/contraction noises are where large, expensive materials have been "permanently" installed such as in some metal roof systems, as we discuss at TEMPERATURE CHANGE & ROOF NOISE. In that case we still have the option of insulating under the roof to reduce noise transmission into the building interior. But for your Florida patio, with aluminum-wrappd components, the solution may be far more reasonable. First identify the actual moving parts and confirm what's going on; often an adjustment in fasteners or use of sealants can solve the problem. POPPING SNAPPING NOISE DIAGNOSIS is the topic home page where we organize all of our diagnostic and repair articles on this. Ask your roofer to be prepared to come by quickly the next time there's a prolonged session of noises from your patio, and also see if you can yourself by careful observation, perhaps aided by a mechanic's stethoscope, pinpoint the location of at least one connection that's noisy. If I were doing this I'd first look outside at the components that are getting the most heat from sun exposure. On 2020-03-18 by (mod) - coefficient of expansion for marble Re-posting Q&AJayantibhai What thermal expansion of composite marble dimensions 20ft by 30 ft at 40degree centigrade?This Q&A were posted originally at FORENSIC INVESTIGATION of BUILDINGS Moderator reply: The thermal expansion of marble is about the same as granite - The thermal expansion of marble is about the same as granite - Coefficient of Expansion in inches of expansion per inch of material per degree F. Granite = 0.0000044 (also Limestone, Marble) Use your browser's on-page search function such as Ctrl-F to search the page above for "marble" to find more details. On 2019-12-29 by (Mod)- engineer tunes-up table of coefficients of expansion of materials Thak you so much for taking the time to comment, Eric; we will edit the article above accordingly and will keep your comments with this page. On 2019-12-29 by Eric Suggestions from a licensed professional engineer:1. "Definition of CLTE: The coefficient of linear thermal expansion (CLTE) of any material is the change in the material’s length [and to a much lesser degree the width or thickness] per unit change in temperature.CLTE applies to the length, width, and thickness equally, not "too a much lesser degree". If the width and thickness are smaller, the expansion will be smaller, however the ratio remains the same. I'd suggest replacing with: "Definition of CLTE: The coefficient of linear thermal expansion (CLTE) of any material is the change of a material's dimension per unit change in temperature."2. "The coefficient of thermal expansion is defined such that α measures the percentage change in the length of the material per degree of temperature change."The CoTE does not indicate percentage change, it indicates change per unit dimension of material. Suggestion to take out "percentage".Also, the table should list all numbers in the same format (i.e., you have "0.00000234", "16.9 ppm", "4.5 x 10(-5)", and also show (sometimes) the units in the table, even including C for a table that everything else is in F. The units are already given in the column header, so suggestion to take out units, convert those that are in C to F, and format all numbers to be in a single format. On 2019-05-19 - by (mod) - dimensional changes in a wood framed house HermanYou ask an interesting question but we need to be rather clear, and perhaps more narrow in our question and answer since even a traditional wood frame house has plenty of other materials involved in its construction besides wood. The coefficient of expansion for pine is 0.0000028 inches per inch of material per degree F. You could add up the widths of all of the framing members whose width or narrow dimension is along the same axis as a house exterior wall - but not all of those would change the building width. For example, even if a wall stud shrinks or expands in the middle of a wall, it's not likely to make a measurable change in overall building width - it simply expands and contracts inside its adjoining joist bays. So you'd look only at framing members whose position is such that their thickness determines the overall wall width or length. That's probably pretty-few: corner posts, intermediate posts not abutting an open joist bay. That might be just three or four 2x4 or 2x6 widths (depending on how the house is framed) - So take a look at the framing plan for your house, add up the pertinent widths along the long dimension, and then plug those into the table above, probably assuming the wood is pine. You'll also have to make some assumptions about moisture levels and perhaps tightness of framing and framing connectors. I suspect that other materials such as siding expand and contract more than the frame - considerably more - which is why installation instructions call for appropriate gaps. On 2019-05-06 by Herman If I had a traditional wood frame house, say 25' x 35' how many inches would it shrink and grow from -10°F to 100° F? ... Continue reading at THERMAL EXPANSION of HOT WATER or select a topic from the closely-related articles below, or see the complete ARTICLE INDEX. Or see THERMAL EXPANSION of MATERIALS FAQs - questions & answers posted originally at this page Or see these Recommended Articles FORENSIC INVESTIGATION of BUILDINGS FROST HEAVES, FOUNDATION, SLAB THERMAL EXPANSION of MATERIALS ALUMINUM WIRING HAZARDS & REPAIRS BRICK WALL THERMAL EXPANSION CRACKS BRICK WALL WOOD NAILER INSERTS CHIMNEY CRACK DETECTION & DIAGNOSIS FIBERGLASS SHINGLE CRACKS FORENSIC INVESTIGATION of BUILDINGS HOT WATER PRESSURE EXPANSION RATE SLOPED GLAZING DETAILS SPONTANEOUS GLASS BREAKAGE NiS TEMPERATURE CHANGE & ROOF NOISE TEMPERATURE CHANGE RELATED NOISES THERMAL EXPANSION CRACKS in BRICK THERMAL EXPANSION of HOT WATER THERMAL EXPANSION TPR VALVE LEAKS VINYL SIDING COEFFICIENT of LINEAR EXPANSION THERMAL IMAGING, THERMOGRAPHY THERMAL MASS in BUILDINGS TRUSS UPLIFT, ROOF Suggested citation for this web page THERMAL EXPANSION of MATERIALS at InspectApedia.com - online encyclopedia of building & environmental inspection, testing, diagnosis, repair, & problem prevention advice. Share this article Or see this INDEX to RELATED ARTICLES: ARTICLE INDEX to EXTERIORS of BUILDINGS Or use the SEARCH BOX found below to Ask a Question or Search InspectApedia Ask a Question or Search InspectApedia Try the search box just below, or if you prefer, post a question or comment in the Comments box below and we will respond promptly. Search the InspectApedia website Note: appearance of your Comment below may be delayed: if your comment contains an image, photograph, web link, or text that looks to the software as if it might be a web link, your posting will appear after it has been approved by a moderator. Apologies for the delay. Only one image can be added per comment but you can post as many comments, and therefore images, as you like. You will not receive a notification when a response to your question has been posted. Please bookmark this page to make it easy for you to check back for our response. Comment Form is loading comments... IF above you see "Comment Form is loading comments..." then COMMENT BOX - countable.ca / bawkbox.com IS NOT WORKING. In any case you are welcome to send an email directly to us at InspectApedia.com at editor@inspectApedia.com We'll reply to you directly. Please help us help you by noting, in your email, the URL of the InspectApedia page where you wanted to comment. Citations & References In addition to any citations in the article above, a full list is available on request. Bob Fankhauser <blueboxconst@hevanet.com>, 503 206 9824 Cell, a retired engineer / professional handyman and Habitat for Humanity volunteer who offered comments, suggestions, additions for vinyl CLTE (Coefficient of Linear Thermal Expansion), CPVC, PVC, cellular PVC, and vinyl (25 Feb 20-16) as well as helpful discussion concerning the wide variation in coefficients of expansion of materials given by various sources. Building Pathology, Deterioration, Diagnostics, and Intervention, Samuel Y. Harris, P.E., AIA, Esq., ISBN 0-471-33172-4, John Wiley & Sons, 2001 [General building science-DF - Particularly useful text Certainteed Weatherboard fiber cement siding and trim products - see certainteed.com/ or see certainteed.com/resources/sidingandtrimspecsheet.pdf In addition to citations & references found in this article, see the research citations given at the end of the related articles found at our suggestedCONTINUE READING or RECOMMENDED ARTICLES. Carson, Dunlop & Associates Ltd., 120 Carlton Street Suite 407, Toronto ON M5A 4K2. Tel: (416) 964-9415 1-800-268-7070 Email: info@carsondunlop.com. Alan Carson is a past president of ASHI, the American Society of Home Inspectors. Thanks to Alan Carson and Bob Dunlop, for permission for InspectAPedia to use text excerpts from The HOME REFERENCE BOOK - the Encyclopedia of Homes and to use illustrations from The ILLUSTRATED HOME . Carson Dunlop Associates provides extensive home inspection education and report writing material. 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Sign Up Sign In English English العربية English English العربية My Wallet Sign Up Sign In My Wallet My Classes My Messages My Reports Lesson Explainer: Intervals Mathematics • Second Year of Preparatory School Join Nagwa Classes Attend live Mathematics sessions on Nagwa Classes to learn more about this topic from an expert teacher! Check Available Classes Next Session: 09/21/2025 09/21/2025 09/21/2025 undefined • 02:00:00pm Number of Seats: One-to-One Class Try This In this explainer, we will learn how to determine limited and unlimited intervals. An interval is a way of describing a subset of the real numbers between two given values. For example, we can describe the set of positive numbers, , that are less than 2 as an inequality: . The equivalent set in interval notation is , so saying means that it is between 0 and 2. These represent the same set of numbers: those above 0 and less than 2. We call this a limited or bounded set since all of the values in the set are between two numbers, and we call the values of 0 and 2 the endpoints of this interval. We can also represent this set by using a number line; we want to include all of the values between 0 and 2 but not the endpoints. We do this by using open circles to show that 0 and 2 are not included, as shown. Before we formally define interval notation, we should note that sometimes we want to include the endpoints in our set. To do this, we introduce the notation , which is equivalent to the inequality . In other words, it is the set of all values greater than or equal to 0 and less than or equal to 2. When both endpoints are included in the interval, we call it a closed interval; if neither endpoint is included, then we say it is an open interval. We can also represent this interval on a number line. This time, we used closed circles to represent that the endpoints are included in the set as shown. It is worth noting that we can include a single endpoint and exclude the other by using each type of parenthesis. For example, the interval would be equivalent to the inequality and we could represent this interval on a number line as shown. We are now ready to define interval notation formally, as follows. Definition: Limited Intervals For real numbers and , where , we can define the following limited intervals: : this is called the open interval from to . This interval contains all of the values between and but does not include or . : this is called the closed interval from to . This interval contains all of the values between and including both or . We can also define the following two half-open (or half-closed) intervals: : this interval contains all of the values between and but does not include ; however, it does contain . : this interval contains all of the values between and but does not include ; however, it does contain . This definition allows for a few interesting cases. First, we can consider the interval ; this set would include all real numbers that satisfy the inequality , which is just the number . Hence, . Second, we can consider the interval ; this would be all real numbers that satisfy the inequality . Since there are no numbers that are both greater than and less than , this set has no members. Hence, . Third, if , then these intervals are all empty. For example, the interval would include all real numbers greater than 2 that are also less than 1, which is not possible, so the set is empty. Let’s now see an example of describing a given interval notation on a number line. Example 1: Identifying the Number Line That Represents a Given Interval Which of the following figures represents the interval ? Answer We recall that the interval represents all of the numbers between and 0. It is a half-open interval and in particular includes the endpoint 0 but does not include . More formally, we have To represent this set on a number line, we want an open circle at to represent that is not included in the set and a closed circle at 0 to show that 0 is a member of the set. We then highlight all of the numbers between and 0 with a line giving us the following. We can see that this is option b. Let’s now see an example of using the definitions of interval notation to determine whether a number is an element of a given interval. Example 2: Determining Whether a Number Belongs to an Interval Which of the following is true? Answer We first recall that is the open interval from to ; this means that it includes all of the numbers between these two values but not the endpoints themselves. In particular, we can write this as follows: To determine whether 5 is a member of this set, we need to determine whether 5 is greater than and less than . We can do this in a few different ways; one method is to write out the decimal expansion of the radicals. We have We can then see that . Hence, . Thus far, we have only dealt with limited intervals. However, we can describe unlimited subsets of the real numbers by using inequalities. For example, the positive numbers can be written as . This is an unlimited (or unbounded) set, since there is no greatest element. We can represent this set on a number line as follows. We use an open circle at 0 to show that this element is not included in the set and we can add an arrow to show that this interval extends infinitely in the positive direction. This is not the only way we can represent this interval. We can write this in interval notation by introducing the symbol , called infinity. Infinity is not a real number; however, we can think of it as being bigger than any real number. In other words, it is not finite. This allows us to represent the positive numbers in the interval notation . We can describe this concept and the idea of unlimited intervals as follows. Definition: Unlimited Intervals If an interval extends forever in either direction of the number line, we call it an unlimited interval. We use the symbol (read infinity) to represent something bigger than any real number. We can also use to represent something smaller than any real number. For any real number , we have the following intervals: : this is the set of all values greater than or equal to . : this is the set of all values greater than . : this is the set of all values less than or equal to . : this is the set of all values less than . The introduction of the symbol for infinity does introduce a few key concepts that are important to note. First, we cannot have an interval that is closed on the side of infinity. The reason for this is that , so we cannot include this in our intervals, since we are only interested in subsets of the real numbers. Second, we can represent as an interval. The set of real numbers can be written as , since this interval includes all real numbers. Let’s now see an example of representing a given set as an unlimited interval. Example 3: Expressing a Given Set in Interval Notation Express the following set using interval notation . Answer By looking at the set notation for , we see that its elements must be real numbers and they must satisfy the inequality . We can write this in interval notation by using 2 and infinity as its endpoints; we note that we want to include 2 in the interval and we should not include infinity. This gives us . In our next example, we will determine which statement is true of a given unlimited interval. Example 4: Identifying the Relation between a Number and a Set Which of the following is true? Answer We first recall that the notation means that is a subset of . For this notation to be valid, both and must be sets. However, in the question, we have and we know that is not a set, so option C is not a correct statement. A very similar statement is true for option D. Although is not a subset of , we still cannot use the notation , since this requires to be a set. Since it is not a set, we cannot compare the two objects in this way, so the answer is not D. The remaining options raise the question of whether is a member of this interval. To determine this, let’s recall what is meant by . It is the set of all real numbers greater than 5; we can write this as We know that , and this is greater than 5. Hence, , which is option A. Since intervals are sets, we can perform all of the set operations on intervals. For example, we can take the union of intervals to represent the real numbers in either interval; we can take the intersection of intervals to represent the real numbers in both intervals; we can take the set difference of two intervals to remove elements of an interval from another interval; we can take the complement of an interval to represent all of the real numbers not in the interval. There are several techniques for evaluating these set operations on intervals. One method is to directly work with the inequalities. However, it can be useful to work these examples through with a number line. For example, let’s evaluate . We can do this by sketching both intervals on a number line. This gives us the following. To determine their union, we want any number in either interval. We can see that this includes any number greater than or equal to that is less than 5. We can then write this as . Let’s now see an example of determining an expression for a set represented graphically on a number line. Example 5: Identifying the Interval Represented on a Given Number Line Which of the following expressions represents the set shown on the number line? Answer There are many different ways of representing the set given on the number line. For example, we can see that the numbers greater than and less than or equal to 1 are the only ones not included in this set. We can then recall that the complement of a set means all of the elements that are not in the set. We note that the elements not in this set are the ones in the interval . Hence, the complement of this set, , is the one represented on the number line. However, this is not one of the given options. Instead, we note that we can think of the complement as removing elements from the entire set of real numbers. We have that since both sets are all the real numbers that are not greater than and less than or equal to 1. We can also write this set as This highlights why option E is incorrect: the set in option E includes 1; however, 1 is not an element on the given number line. We can see that the correct answer is option D, . In our previous example, we showed a useful result comparing the complement of an interval and the difference with the set . Definition: Complement of an Interval If is an interval, then . This is of course true for any subset of , not just intervals, but it is useful to commit this result to memory. In our next example, we will simplify an expression of the intersection of two given intervals. Example 6: Finding the Intersection of Two Intervals Given that and , find . Answer We first recall that the intersection of two sets is the set of elements in both sets. Therefore, to find , we want to find all real numbers that are in both intervals, and . We will find these numbers by sketching both intervals on a number line. First, : we can sketch this on a number line by drawing a closed circle at to show that this endpoint is included in the set and an open circle at 3 to show that this is an element of the interval. We get the following. We can follow the same process for . We can see that all of the interval is contained within . In other words, . This means that their intersection is the entire subset: . Hence, . Let’s finish by recapping some of the important points from this explainer. Key Points For real numbers and , where , we can define the following limited intervals: : this is called the open interval from to . this is called the closed interval from to . : this is a half-open (or half-closed) interval from to . : this is a half-open (or half-closed) interval from to . If an interval extends forever in either direction of the number line, we call it an unlimited interval. We use the symbol (read infinity) to represent something bigger than any real number. We can also use to represent something smaller than any real number. For any real number , we have the following unlimited intervals: : this is the set of all values greater than or equal to . : this is the set of all values greater than . : this is the set of all values less than or equal to . : this is the set of all values less than . We can take the union, intersection, difference, and complement of intervals. If is an interval, then . Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Interactive Sessions Chat & Messaging Realistic Exam Questions Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
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https://stackoverflow.com/questions/68043558/python-logic-puzzle-relationships
Skip to main content 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 python logic puzzle relationships Ask Question Asked Modified 3 years, 1 month ago Viewed 460 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Attempting to use Logic to solve the following in python: ``` from kanren import run, eq, membero, var, conde from kanren.constraints import neq, isinstanceo rules = (eq, (var(), var(), var(), var()), people), (membero, (4, x, ('Steve' , var() , 'blue' , var() ), people)), (membero, (4, x, ( var() , 'cat' , var() , 'Canada' ), people)), (membero, (4, x, ('Matthew', var() , var() , 'USA' ), people)), (membero, (4, x, ( var() , var() , 'black', 'Australia'), people)), (membero, (4, x, ('Jack' , 'cat' , var() , var() ), people)), (membero, (4, x, ('Alfred' , var() , var() , 'Australia'), people)), (membero, (4, x, ( var() , 'dog' , var() , 'France' ), people)), (membero, (4, x, ( var() , 'rabbit', var() , var() ), people)) ``` results in: ``` (, (4, ~_2, (~_2422, 'rabbit', ~_2423, ~_2424), ~_809)) ``` which keeps me from running ``` solutions = run(0, people, rules) ``` Can you help me with a solution? Thank you. python logic member zebra-puzzle minikanren Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Jun 19, 2021 at 4:49 Will Ness 71.4k1010 gold badges105105 silver badges192192 bronze badges asked Jun 19, 2021 at 4:15 Lynn WilburnLynn Wilburn 3111 bronze badge 3 just shooting in the dark here (looking at the opening page at pypi.org/project/miniKanren), maybe you need to add x = var() somewhere. – Will Ness Commented Jun 19, 2021 at 4:57 or maybe try removing the first two arguments to membero. in the examples that I could find membero is called with only two arguments. – Will Ness Commented Jun 19, 2021 at 8:00 1 I wrote a zebra puzzle solver that you can find here: puzzle-solvers.readthedocs.io/en/latest. Not Kanren, but it works – Mad Physicist Commented Jan 10, 2022 at 10:46 Add a comment | 1 Answer 1 Reset to default This answer is useful 0 Save this answer. Show activity on this post. Not sure what version of kanren you were using but it sure doesn't work that way under the latest one. This seemed to produce the result: ``` people = var() things = lall(eq((var(), var(), var(), var()), people), membero(('Steve', var(), 'blue', var()), people), membero((var(), 'cat', var(), 'Canada'), people), membero(('Matthew', var(), var(), 'USA'), people), membero((var(), var(), 'black', 'Australia'), people), membero(('Jack', 'cat', var(), var()), people), membero(('Alfred', var(), var(), 'Australia'), people), membero((var(), 'dog', var(), 'France'), people), membero((var(), 'rabbit', var(), var()), people) solutions = run(0, people, things) ``` It results in many solutions, but I think they're just permutations of each other. Maybe it can be made tighter like this: ``` people = var() things = lall(eq((('Steve', var(), var(), var()), ('Matthew', var(), var(), var()), ('Jack', var(), var(), var()), ('Alfred', var(), var(), var())), people), membero(('Steve', var(), 'blue', var()), people), membero((var(), 'cat', var(), 'Canada'), people), membero(('Matthew', var(), var(), 'USA'), people), membero((var(), var(), 'black', 'Australia'), people), membero(('Jack', 'cat', var(), var()), people), membero(('Alfred', var(), var(), 'Australia'), people), membero((var(), 'dog', var(), 'France'), people), membero((var(), 'rabbit', var(), var()), people) ) solutions = run(0, people, things) ``` This still resulted in two solutions, both of which look identical to me. ``` (('Alfred', ~_16, 'black', 'Australia'), ('Jack', 'cat', ~_14, 'Canada'), ('Matthew', 'rabbit', ~_21, 'USA'), ('Steve', 'dog', 'blue', 'France')) ``` Of course, only mentioning three animals and two colors, that's all you're gonna get. I can see where you might have gotten the syntax in the original question: was it perhaps here? I wish I could get that puzzle going, but nothing I do works for it... Update Had a very helpful exchange with the developers at kanren. Check it out Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications edited Jun 30, 2022 at 0:15 answered Jun 24, 2022 at 3:27 rschwiebrschwieb 78622 gold badges1616 silver badges4242 bronze badges Add a comment | Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions python logic member zebra-puzzle minikanren See similar questions with these tags. 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https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem?srsltid=AfmBOopf_GdyMeK-HqguJpWsg3zLSKU68f5YIzyfyuvuMgVik1ev1Qae
Art of Problem Solving Shoelace Theorem - 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 Shoelace Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Shoelace Theorem The Shoelace Theorem is a nifty formula for finding the area of a simple polygon given the coordinates of its vertices. Contents [hide] 1 Theorem 2 Other Forms 3 Proof 1 3.1 Proof of claim 1: 3.2 Proof: 4 Proof 2 5 Proof 3 6 Problems 6.1 Introductory 6.2 Exploratory 7 External Links Theorem Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is You can also go counterclockwise order, as long as you find the absolute value of the answer. The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes. Other Forms This can also be written in form of a summation or in terms of determinants as which is useful in the variant of the Shoelace theorem. Note here that and . The formula may also be considered a special case of Green's Theorem where and so . Proof 1 Claim 1: The area of a triangle with coordinates , , and is . Proof of claim 1: Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product . ) Proof: We will proceed with induction. By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for . We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get Hence as claimed. ~ShreyJ Proof 2 Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the Theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result Proof 3 This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids. The proof is in this book: (The only thing that needs to be slightly modified is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.) Problems Introductory In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of . Exploratory Observe that is the area of a triangle with vertices and is the volume of a tetrahedron with vertices . Does a similar formula hold for Dimensional triangles for any ? If so how can we use this to derive the D Shoelace Formula? External Links A good explanation and exploration into why the theorem works by James Tanton: Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: AOPS Retrieved from " Categories: Geometry Theorems Mathematics 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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https://artofproblemsolving.com/wiki/index.php/2019_AMC_10B_Problems/Problem_9?srsltid=AfmBOor2Vg7usGogYzcAuQg0XuHIUqkdEsnBHaK3C19Ayj8eOqtRYUBQ
Art of Problem Solving 2019 AMC 10B Problems/Problem 9 - 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 2019 AMC 10B Problems/Problem 9 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2019 AMC 10B Problems/Problem 9 Contents 1 Problem 2 Solution 1 3 Solution 2 4 Solution 3 (Formal) 5 Solution 4 6 Video Solution 7 Video Solution 8 See Also Problem The function is defined by for all real numbers , where denotes the greatest integer less than or equal to the real number . What is the range of ? Solution 1 There are four cases we need to consider here. Case 1: is a positive integer. Without loss of generality, assume . Then . Case 2: is a positive fraction. Without loss of generality, assume . Then . Case 3: is a negative integer. Without loss of generality, assume . Then . Case 4: is a negative fraction. Without loss of generality, assume . Then . Thus the range of the function is . ~IronicNinja Solution 2 It is easily verified that when is an integer, is zero. We therefore need only to consider the case when is not an integer. When is positive, , so When is negative, let be composed of integer part and fractional part (both ): Thus, the range of x is . Note: One could solve the case of as a negative non-integer in this way: Solution 3 (Formal) Let {} denote the fractional part of ; for example, {}, and {}. Then for , {} and for , {}. Now we can rewrite , breaking the expression up based on whether or . For , the above expression is equal to {} {}{} . For , the expression is equal to {} {} {}. Therefore the only two possible values for , and thus the range of the function, is . ~KingRavi Solution 4 We have 2 cases: either is positive or is negative. Case 1 - x is positive: Let , where is a positive integer and is a positive real number between 0 and 1. We have and , so the possible value of if is positive is . Case 2 - x is negative: Let , where is a positive integer and is a positive real number between 0 and 1. We have and and , so the possible values of if is negative are and Hence, the possible values of are and , so the answer is . ~azc1027 Video Solution ~Education, the Study of Everything Video Solution ~savannahsolver See Also 2019 AMC 10B (Problems • Answer Key • Resources) Preceded by Problem 8Followed by Problem 10 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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https://www.teacherspayteachers.com/Product/Properties-of-Logarithms-CST-ACT-SAT-Test-Practice-643526
Properties of Logarithms - CST ACT SAT Test Practice Description Properties of Logarithms - CST ACT SAT Test Practice Reviews Questions & Answers
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https://www.ncbi.nlm.nih.gov/books/NBK551528/
An official website of the United States government 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 in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Gastroparesis Anil Kumar Reddy Reddivari; Parth Mehta. Author Information and Affiliations Authors Anil Kumar Reddy Reddivari1; Parth Mehta. Affiliations 1 University of Illinois Last Update: February 28, 2024. Continuing Education Activity Gastroparesis is a condition of delayed gastric emptying, lasting at least 3 months in the absence of mechanical obstruction. The illness is defined by symptoms such as nausea, vomiting, bloating, early satiety, and abdominal pain. Delayed or ineffective gastric emptying results from abnormalities in gastrointestinal motor function, a complex sequence of events involving the parasympathetic and sympathetic nervous systems, gastric smooth muscle cells, pacemaker cells within the stomach and intestine, and the pyloric sphincter. Patients with gastroparesis have alterations in mucosal immune cell infiltration and cytokine expression. More than 50% of cases of gastroparesis are idiopathic; diabetic, postsurgical, and postinfectious causes are common. The increasing incidence of gastroparesis has been attributed to the higher prevalence of diabetes, obesity, and causative medications. Clinically, many patients have symptoms overlapping with functional dyspepsia, and the two disorders commonly co-occur; diagnostic testing is important to document the presence of gastroparesis. This activity reviews the epidemiology, etiology, pathogenesis, evaluation, and management of gastroparesis and highlights the important role of the interprofessional healthcare team in improving outcomes for patients with this gastrointestinal disease process. Objectives: Identify patients at risk for gastroparesis based on their clinical history. Select efficacious diagnostic modalities for patients with suspected gastroparesis. Apply holistic treatment approaches to patients with gastroparesis. Develop and implement effective interprofessional team strategies to improve outcomes and quality of life for patients with gastroparesis. Access free multiple choice questions on this topic. Introduction Gastroparesis is a condition of delayed gastric emptying in the absence of mechanical obstruction, lasting for at least 3 months. Gastroparesis causes symptoms such as nausea, vomiting, bloating, early satiety, and abdominal pain. Delayed or ineffective gastric emptying occurs due to abnormal gastrointestinal motor function, a complex sequence of events involving the parasympathetic and sympathetic nervous systems, gastric smooth muscle cells, pacemaker cells within the stomach and intestine, and the pyloric sphincter. Patients with gastroparesis also show alterations in mucosal immune cell infiltration and cytokine expression. The most common etiology of gastroparesis is idiopathic, followed by diabetic, postsurgical, and postinfectious causes. The incidence of gastroparesis has been increasing over the last few decades, attributed to increases in diabetes, obesity, and causative medications. Clinically, many patients have symptoms overlapping with functional dyspepsia, and the two disorders commonly co-occur; diagnostic testing is essential to document the presence of gastroparesis. Treating gastroparesis is limited by medication tolerance and decreasing effectiveness; dietary modification is important. Etiology Gastroparesis is most frequently idiopathic. Other common etiologies include diabetic, postsurgical, and postinfectious causes. Various neurologic disorders can also affect gastrointestinal motility by altering the parasympathetic or sympathetic nerve supply to the gastrointestinal (GI) tract. Etiologic categories of gastroparesis include but are not limited to: Idiopathic: the etiology is approximately 50% of cases Diabetes mellitus (DM): diabetic gastroparesis is more frequent and more severe in patients with type 1 DM but is seen in patients with type 2 DM Postsurgical: iatrogenic vagal nerve injury and altered stomach anatomy affect gastric emptying Viral infections: particularly Norwalk virus, rotavirus, and Epstein-Barr virus Rheumatological diseases: amyloid neuropathy and systemic sclerosis (scleroderma) Autoimmune: may benefit from plasmapheresis of autoantibodies Neurological conditions Traumatic spinal cord injury Medications: opioids, cyclosporine, phenothiazines, dopamine agonists, octreotide, alpha-2-adrenergic agonists, tricyclic antidepressants, calcium channel blockers, GLP-1 agonists or analogs, lithium, and progesterone. Patients with idiopathic gastroparesis are more likely to be women, White, and have symptoms of abdominal pain and early satiety than patients with diabetic gastroparesis. Similar to diabetic gastroparesis, episodes can be episodic or cyclical. Patients with diabetes at increased risk of developing gastroparesis usually have a combination of the following: long duration of the disease, poorly controlled blood sugars, and other autonomic neuropathic complications. Patients with type 1 diabetes are more likely to have gastroparesis than those with type 2, and up to 50% of patients with diabetes and suboptimal glucose control will have gastroparesis. However, optimal glucose control is not always associated with improved gastric motility, especially in the short term. Gastroparesis can also affect glucose control, and hyperglycemia decreases gastric motility, even in nondiabetic patients. Gastroparesis is common following surgery involving the esophagus, stomach, duodenum, or pancreas and procedures that risk injury to the vagal nerve. Postsurgical gastroparesis may develop immediately or months to years following the procedure. Surgical procedures associated with postoperative gastroparesis include Nissen fundoplication, gastrectomy, pancreatic cancer cryoablation, and pancreatoduodenectomy. Gastroparesis has been reported in gynecological surgeries performed in the lower abdominopelvic cavity. Uncomplicated fundoplication and sleeve gastrectomy can prolong gastric emptying time due to reduced stomach capacity, as can fluid-filled balloon bariatric surgery. Postinfectious is common and typically self-limited; full recovery is usual. Norwalk virus and rotavirus are the most commonly identified etiologic viral pathogens. However, Epstein-Barr virus, cytomegalovirus, and Varicella Zoster virus have also been implicated in gastroparesis due to their potential to cause long-term autonomic dysfunction. Epidemiology The epidemiology of gastroparesis is difficult to estimate due to the relatively weak correlation of symptoms with gastric emptying, resulting in a high rate of underdiagnosis. In one of the largest population-based studies, 3604 cases of possible gastroparesis were evaluated, of which 83 patients classified as definite gastroparesis and 127 patients as probable gastroparesis. A 2022 systematic review of 13 studies from 1994 to 2019 found a prevalence of gastroparesis in the general population between 13.8 to 267.7 per 100,000 adults and an incidence of 1.9 to 6.3 per 100,000 person-years. The incidence of gastroparesis in women is quadruple that in men and increases with increasing age in both sexes. The overall incidence of hospitalizations for gastroparesis increased by 300% between 1997 to 2013. Pathophysiology The initial step for digesting solids is gastric expansion of the fundus and proximal stomach, followed by antral contraction. Peristaltic waves cause ingested food to be triturated into smaller particles. Coordinated antral motor function is crucial for grinding, mixing, and gastric emptying of solids; abnormal motor function results in accumulated indigestible food and eventual bezoar formation. Once a particular force is reached, pyloric contraction allows particles smaller than 2 mm to proceed to the duodenum; larger particles are returned to the stomach body to allow further trituration. Decreases in antral contraction and pyloric relaxation are the primary mechanisms of gastroparesis. Antral hypomotility is defined as fewer than one contraction per minute in the first postprandial hour. Neuromuscular control is exerted at several levels, including intrinsic and extrinsic neurons, smooth muscle cells, and interstitial cells of Cajal. Interactions among smooth muscle cells, interstitial cells of Cajal located between the circular and longitudinal muscle layers, and fibroblast-like PGFRα-positive cells regulate the contraction and relaxation involved in peristalsis. The enteric nervous system is complex and comprises over 100 million neurons. Interstitial cells of Cajal also have a pacemaker function to generate and propagate gastric and intestinal slow waves; they also initiate pyloric contraction and relaxation. In gastroparesis, decreased nitrous oxide levels may contribute to increased pyloric tone. Increased localized inflammatory cell infiltration is notable, and one possibility is that gastric stasis is associated with macrophage and CD8 T-cell infiltration of the lamina propria, an infiltration unseen in patients with functional dyspepsia but no gastroparesis. Both the parasympathetic and sympathetic nervous systems regulate gastric motility. The vagal nerve is essential, innervating over 70% of myenteric nerves. Vagus nerve motor pathways control muscle contraction of the stomach, and sensory pathways relay information to coordinate the forward propulsion of gastric contents. Idiopathic and diabetic gastroparesis are associated with antral hypomotility. In contrast, infiltrative disorders like systemic sclerosis are associated with a normal number of antral contractions but a decreased contraction strength (amplitude). Higher postprandial gastric volume is also associated with delayed gastric emptying. In postvagotomy and post-gastric resection patients, phasic contractility and the accommodation response of the stomach are impaired. Histopathology Histopathologically, gastroparesis is associated with a reduced number of autonomic ganglia and inflammatory changes resulting in abnormal gastric myoelectric motor functions. A study in patients with refractory gastroparesis evaluated gastric neuromuscular histopathology revealed lymphocytic infiltration of the myenteric plexus, fibrosis in the inner circular and outer longitudinal layers, and decreased interstitial cells of Cajal. Gastric biopsies of patients with idiopathic and diabetic gastroparesis show loss of Cajal cells and enteric neurons and increased inflammatory infiltrates. History and Physical History Gastroparesis is a heterogeneous disorder. Symptoms do not always correlate with the degree of delayed gastric emptying, and some patients positive for gastroparesis on diagnostic testing are asymptomatic. Nausea and vomiting are the most common clinical symptoms and correlate with the degree of gastric emptying delay. Other symptoms include postprandial fullness and bloating; weight loss is common in severe and persistent cases. Vomitus may contain food ingested several hours prior. There are similarities and differences in the clinical characteristics of diabetic and idiopathic gastroparesis. Patients with idiopathic gastroparesis report more symptoms of early satiety, postprandial fullness, and abdominal pain as compared with patients with diabetic gastroparesis. Patients with diabetic gastroparesis have more severe nausea and vomiting. Even though abdominal pain is a frequent complaint in patients with gastroparesis, it is rarely the predominant symptom. In patients whose principal symptom is abdominal pain, etiologies other than gastroparesis merit consideration. Functional dyspepsia has many overlapping symptoms with gastroparesis and can occur in the same cohort of patients. Gastric emptying studies are often needed to differentiate the two conditions. Symptoms of gastroparesis may also be quantifiable via the gastroparesis cardinal symptom index, a questionnaire based on the severity of symptoms of postprandial fullness: early satiety, nausea, vomiting, and bloating. Physical Examination The physical examination of a patient with gastroparesis may reveal epigastric distention or tenderness but no guarding or rigidity suggestive of peritonitis. Patients may have signs of the underlying disorder causing gastroparesis, such as a parkinsonian tremor, diminished sensation in diabetic neuropathy, or surgical scars in cases of postsurgical gastroparesis. Evaluation Gastroparesis should be suspected in patients with chronic nausea, vomiting, early satiety, postprandial fullness, abdominal pain, or bloating. After obtaining a thorough history and performing a physical examination, most patients in whom gastroparesis is suspected will undergo basic laboratory testing with a metabolic panel, liver function testing, complete blood count, and lipase. Abdominal imaging often involves computed tomography (CT) scans or magnetic resonance imaging (MRI) to rule out other causes of pain and vomiting. Imaging can help exclude a mechanical obstruction, such as from malignancy. Upper GI endoscopy can also assist in ruling out obstructive etiology. Scintigraphic gastric emptying studies (GES) are the gold standard for detecting gastroparesis. A solid or liquid radioisotope-containing meal may be used to measure gastric emptying. Based on the scintigraphic gastric retention of contents after 4 postprandial hours, gastroparesis is graded as mild (< 15% retention), moderate (15% to 35% retention), and severe (> 35% retention). Some centers perform testing after 2 postprandial hours; 4-hour testing is more accurate. Adding a liquid GES study can further increase sensitivity in patients with normal gastric emptying of solids. The gastric emptying breath test (GEBT) is also used for evaluating gastroparesis. In this test, a stable isotope, 13C, is incorporated into a substrate, usually spirulina, and measured from the patient's breath by spectroscopy. This test is noninvasive and avoids radiation exposure. Correlation with scintigraphy is estimated at 73% to 95%. The use of GEBT is limited in patients with liver or lung disease. The wireless motility capsule (WMC) is another FDA-approved method for evaluating gastroparesis. The utility of WMC is limited by cost, availability, and a poor correlation with gastric emptying of food; the capsule is an indigestible solid. The WMC has an estimated sensitivity of 59% to 86% and a specificity of 64% to 81% for gastroparesis. Gastric ultrasonography has been utilized to assess antral wall motion, image transpyloric flow patterns, and estimate gastric emptying based on measurements of the gastric antrum. Gastric ultrasonography is limited by the lack of image detail, the need for an experienced technician, and its impracticality for prolonged examinations. However, ultrasonography is noninvasive and does not involve radiation, so it is sometimes used in children and pregnant patients. Treatment / Management A holistic approach is recommended when treating patients with gastroparesis. Treatment should comprise assessment of nutritional status, measures to correct fluid, electrolyte, and nutritional deficiencies, relief of symptoms of gastroparesis, improvement of gastric emptying, and treatment of the underlying cause to prevent disease progression. Dietary Modifications Given the importance of dietary modification, formal nutrition and dietetic consults are recommended. Small meals limit emptying time and can alleviate symptoms. Patients may need to eat 4 or 5 times daily to compensate for the small meal size. Meals composed of low-fat, low-fiber, and small-particle foods are also recommended. Liquids are more easily absorbed than solids. Carbonated drinks release carbon dioxide that aggravates gastric distention. Tobacco and alcohol decrease antral contractility and impair gastric emptying. Despite the prevalence of obesity in this cohort, malnutrition is common, and vitamin supplements are often required. If oral intake is inadequate, enteral nutrition via jejunostomy tube should be considered. Parenteral nutrition is rarely needed. Pharmacological Treatment Antiemetics such as prochlorperazine (5 to 10 mg up to 3 times daily), diphenhydramine (12.5 to 25 mg up to 4 times daily), and ondansetron (4 or 8 mg up to 3 times daily) can provide symptomatic relief in gastroparesis. The prokinetic medication metoclopramide is the only drug approved by the United States Food and Drug Administration (FDA) for gastroparesis and is usually administered 15 minutes before meals. The initial dosage is 5 mg 3 times daily. The total dose may be increased to 40 mg daily; this dose requires close monitoring for adverse reactions. The duration of treatment approved by the FDA is 12 weeks due to the possibility of severe side effects, including anxiety, restlessness, dystonia, hyperprolactinemia, and QT prolongation. Metoclopramide has a black box warning for extrapyramidal symptoms like dystonia and tardive dyskinesia, which can occur in up to 1% of patients. Other off-label medications aimed at symptom relief include domperidone, centrally-acting antidepressants, and the serotonin type 4 (5-HT4) partial agonist tegaserod. Domperidone is a dopamine-D2 antagonist that works similarly to metoclopramide to improve gastric emptying and decrease nausea and vomiting. The FDA restricts domperidone use in the United States; it is currently under review as an investigational drug. The initial dosage of domperidone is usually 10mg 3 times daily but can be increased to 20 mg per dose. Significant adverse effects of domperidone include cardiac arrhythmias and hyperprolactinemia; drug reactions are prevalent. Monitoring the QTc interval in patients taking domperidone is recommended; avoid or discontinue use if the QTc exceeds 450 ms. Patients who cannot tolerate metoclopramide or domperidone may benefit from erythromycin, a macrolide antibiotic and motilin agonist that stimulates high-amplitude gastric fundal peristaltic contractions, thereby improving symptoms during acute exacerbations. Erythromycin can be given intravenously (3 g/kg every 8 hours) in acute cases and orally (250 to 500 mg 3 times daily) for maintenance. Patients exhibit tachyphylaxis, a decrease in response quickly due to desensitization, to erythromycin; its use is restricted to 4 weeks of continuous administration. Common side effects of erythromycin include GI toxicity, ototoxicity, bacterial resistance, and QTc prolongation. Erythromycin is the best-studied of this medication class, but clarithromycin and azithromycin have similar therapeutic and adverse effects. Nonpharmacological Therapies Gastric electrical stimulation (GES) decreases the frequency of vomiting and the need for nutrition supplementation. In patients with symptoms refractory to pharmacologic measures, GES is considered a compassionate treatment approved for diabetic and idiopathic gastroparesis. Patients with diabetic gastroparesis respond more effectively than other groups. Implantation of a GES stimulator is a surgical procedure via laparotomy or laparoscopic approach. The device consists of a pair of leads implanted in the muscularis propria 10 cm proximal to the pylorus, then connected to a pulse generator. Risks of GES placement include infection, lead migration, and perforation. The devices require battery replacement every 10 years. Other less studied interventions include venting gastrostomy, feeding jejunostomy, partial gastrectomy, pyloroplasty, and intrapyloric botulinum toxin injection; botulinum toxin has not been shown effective in randomized control trials. Investigational Therapies Multiple prospective treatments for gastroparesis are under investigation. Granisetron, a 5-HT3 antagonist administered as a transdermal patch, appears to improve gastroparesis-related nausea and vomiting. Similarly, a randomized control trial of aprepitant, a neurokinin-1 receptor antagonist, showed symptom improvement in gastroparesis patients. Cannabinoids, such as dronabinol, have been considered in the treatment of gastroparesis symptoms but have not been studied in clinical trials. However, cannabinoids can also delay gastric emptying. Acupuncture has proven effective for diabetic gastroparesis in randomized control trials. Other investigational therapies undergoing evaluation include elamorelin, a pentapeptide ghrelin-receptor agonist; the 5HT-4 receptor agonists prucalopride and velusetrag; and metoclopramide nasal spray. Differential Diagnosis The clinician should consider a wide differential diagnosis list in patients with symptoms of nausea and vomiting. The following conditions can present similarly to gastroparesis: Functional dyspepsia: very similar symptoms to gastroparesis, and a functional test is often required to differentiate these processes. Gastric outlet obstruction: a similar presentation but with underlying mechanical obstruction. Cyclical vomiting syndrome: recurrent episodes of intractable nausea and vomiting lasting hours to days. Rumination syndrome: a behavioral disorder that consists of daily, effortless regurgitation of undigested food after ingestion of a meal. In contrast to gastroparesis, it is not preceded by nausea or vomiting. Cannabinoid hyperemesis syndrome: prolonged cannabis use can cause episodic vomiting, accompanied by a tendency for hot showers to aid symptom relief; symptoms improve with the cessation of cannabis. Psychiatric disorders: including anxiety neurosis, anorexia nervosa, or bulimia, could have persistent upper GI symptoms confused with gastroparesis. Prognosis Patients with gastroparesis may have a relatively poor quality of life due to persistent symptoms and poor nutrition. In a study of 20 patients with diabetes followed for 12 years, gastric emptying symptoms were reasonably stable. In another study of 86 patients with diabetes, after adjusting for comorbidities, gastroparesis symptoms did not correlate with mortality. In contrast, results from a study done at a tertiary center revealed that 7% of gastroparesis patients died, and 22% of patients needed long-term enteral or parenteral nutrition after 6 years of follow-up. Controlled community studies for the prognosis of gastroparesis are needed, and data from tertiary care hospitals may not accurately represent the disease outcomes in the general population. Postviral gastroparesis often improves within 12 months. Complications The incidence of complications from gastroparesis will vary according to underlying etiology. Patients with longstanding gastroparesis may have the following complications: Severe protein-calorie malnourishment Bezoar formation requiring surgical resection Mallory-Weiss tears Procedure-related complications Recurrent hospitalizations Significant healthcare burden. Deterrence and Patient Education Systematic education targeting different members of the healthcare team can ensure quality care for patients with gastroparesis. All people with diabetes and other comorbidities that correlate with gastroparesis should receive education about symptom diagnosis. Patients should have detailed education about diet and lifestyle modifications, including eating multiple small meals and avoiding alcohol, smoking, and carbonated beverages. Due to the chronicity of symptoms and the lack of definitive treatment options, gastroparesis impairs the quality of life of patients. Pearls and Other Issues Gastroparesis is a common condition causing significant morbidity. The most common etiologies of gastroparesis are idiopathic and diabetic. If associated with significant weight loss, consider eating disorders as an etiology. Pathology is highly linked to autonomic dysfunction of the sympathetic and parasympathetic autonomic nervous systems and decreased interstitial cells of Cajal, which are the "pacemaker" cells of the GI tract. Medications, especially opioid medications, are increasingly linked to the incidence of gastroparesis. Diabetic gastroparesis is especially linked to antrum and pyloric dysfunction and increased pyloric contraction. Scintigraphy is the gold standard for imaging gastric emptying and should be performed over 4 hours. If scintigraphy is unavailable, the gastric emptying breast test (GEBT) or wireless motility capsule can be used. The wireless motility capsule has significantly less sensitivity and specificity when compared to scintigraphy. Metoclopramide is the only FDA-approved medication for gastroparesis, but its use is limited by significant side effects, including a black box warning for tardive dyskinesia. Many medications are used off-label for symptomatic treatment. Dietary modifications should be advised. All patients with this condition should be offered consultation with a nutritionist. Enhancing Healthcare Team Outcomes An interprofessional team approach is essential to improve the morbidity associated with gastroparesis. Nutritionists are essential to managing gastroparesis effectively. Many patients with gastroparesis are undernourished despite the prevalence of obesity in this cohort. Small meals composed of low-fat, low-fiber, and small-particle foods are recommended. Pharmacists should educate the patient on medications that can help improve gastric motility while minimizing adverse effects. The diabetes educator should inform patients with diabetic gastroparesis about blood glucose control. Patients should be encouraged to eat a healthy diet and maintain hydration. Any food that triggers gastroparesis should be avoided. Healthy body weight and regular exercise are also recommended. Primary care practitioners are likely the first health care professionals to encounter patients with symptoms of gastroparesis and should be educated regarding further evaluation and treatments. Ideally, patients should receive care from gastroenterologists specializing in gastroparesis. These examples demonstrate how an interprofessional team approach to gastroparesis can improve patient outcomes. Patient and healthcare provider education plays a critical role in improving accurate diagnosis and evidence-based management of gastroparesis. Review Questions Access free multiple choice questions on this topic. Click here for a simplified version. Comment on this article. References 1. : Camilleri M, Sanders KM. Gastroparesis. Gastroenterology. 2022 Jan;162(1):68-87.e1. [PMC free article: PMC8678360] [PubMed: 34717924] 2. : Lacy BE, Parkman HP, Camilleri M. Chronic nausea and vomiting: evaluation and treatment. Am J Gastroenterol. 2018 May;113(5):647-659. 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[PubMed: 10097936] : Disclosure: Anil Kumar Reddy Reddivari declares no relevant financial relationships with ineligible companies. : Disclosure: Parth Mehta declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK551528PMID: 31855372 PubReader Print View Cite this Page Reddivari AKR, Mehta P. Gastroparesis. [Updated 2024 Feb 28]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. 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Using a naive Bayesian approach to identify academic risk based on multiple sources: A conceptual replication Carly Oddleifson1,Stephen Kilgus2,David A Klingbeil2,Alexander D Latham2,Jessica S Kim2,Ishan N Vengurlekar2 Affiliations Expand Affiliations 1 Department of Educational Psychology, University of Wisconsin-Madison, United States. Electronic address: oddleifson@wisc.edu. 2 Department of Educational Psychology, University of Wisconsin-Madison, United States. PMID: 39710436 DOI: 10.1016/j.jsp.2024.101397 Item in Clipboard Using a naive Bayesian approach to identify academic risk based on multiple sources: A conceptual replication Carly Oddleifson et al. J Sch Psychol.2025 Feb. Show details Display options Display options Format J Sch Psychol Actions Search in PubMed Search in NLM Catalog Add to Search . 2025 Feb:108:101397. doi: 10.1016/j.jsp.2024.101397. Epub 2024 Dec 4. Authors Carly Oddleifson1,Stephen Kilgus2,David A Klingbeil2,Alexander D Latham2,Jessica S Kim2,Ishan N Vengurlekar2 Affiliations 1 Department of Educational Psychology, University of Wisconsin-Madison, United States. Electronic address: oddleifson@wisc.edu. 2 Department of Educational Psychology, University of Wisconsin-Madison, United States. PMID: 39710436 DOI: 10.1016/j.jsp.2024.101397 Item in Clipboard Cite Display options Display options Format Abstract The purpose of this study was to conduct a conceptual replication of Pendergast et al.'s (2018) study that examined the diagnostic accuracy of a nomogram procedure, also known as a naive Bayesian approach. The specific naive Bayesian approach combined academic and social-emotional and behavioral (SEB) screening data to predict student performance on a state end-of-year achievement test. Study data were collected in a large suburban school district in the Midwest across 2 school years and 19 elementary schools. Participants included 5753 students in Grades 3-5. Academic screening data included aimswebPlus reading and math composite scores. SEB screening data included Academic Behavior subscale scores from the Social, Academic, and Emotional Behavior Risk Screener. Criterion scores were derived from the Missouri Assessment Program (MAP) tests of English Language Arts and Mathematics. The performance of each individual screener was compared to the naive Bayesian approach that integrated pre-test probability information (i.e., district-wide base rates of risk derived from prior year MAP test scores), academic screening scores, and SEB screening scores. Post-test probability scores were then evaluated using a threshold model (VanDerHeyden, 2013) to determine the percentage of students within the sample that could be differentiated in terms of ruling in or ruling out intervention versus those who remained undifferentiated (as indicated by the need for additional assessment to determine risk status). Results indicated that the naive Bayesian approach tended to perform similarly to individual aimswebPlus measures, with all approaches yielding a large percentage (65%-87%) of undifferentiated students when predicting proficient performance. Overall, the results indicated that we likely failed to replicate the findings of the original study. Limitations and future directions for research are discussed. Keywords: Evidence-based assessment; MTSS; Nomogram; Replication; Screening. Copyright © 2024 Society for the Study of School Psychology. Published by Elsevier Ltd. All rights reserved. PubMed Disclaimer Similar articles Education support services for improving school engagement and academic performance of children and adolescents with a chronic health condition.Barnett T, Tollit M, Ratnapalan S, Sawyer SM, Kelaher M.Barnett T, et al.Cochrane Database Syst Rev. 2023 Feb 8;2(2):CD011538. doi: 10.1002/14651858.CD011538.pub2.Cochrane Database Syst Rev. 2023.PMID: 36752365 Free PMC article. Measures implemented in the school setting to contain the COVID-19 pandemic.Krishnaratne S, Littlecott H, Sell K, Burns J, Rabe JE, Stratil JM, Litwin T, Kreutz C, Coenen M, Geffert K, Boger AH, Movsisyan A, Kratzer S, Klinger C, Wabnitz K, Strahwald B, Verboom B, Rehfuess E, Biallas RL, Jung-Sievers C, Voss S, Pfadenhauer LM.Krishnaratne S, et al.Cochrane Database Syst Rev. 2022 Jan 17;1(1):CD015029. doi: 10.1002/14651858.CD015029.Cochrane Database Syst Rev. 2022.Update in: Cochrane Database Syst Rev. 2024 May 2;5:CD015029. doi: 10.1002/14651858.CD015029.pub2.PMID: 35037252 Free PMC article.Updated. Comparison of Two Modern Survival Prediction Tools, SORG-MLA and METSSS, in Patients With Symptomatic Long-bone Metastases Who Underwent Local Treatment With Surgery Followed by Radiotherapy and With Radiotherapy Alone.Lee CC, Chen CW, Yen HK, Lin YP, Lai CY, Wang JL, Groot OQ, Janssen SJ, Schwab JH, Hsu FM, Lin WH.Lee CC, et al.Clin Orthop Relat Res. 2024 Dec 1;482(12):2193-2208. doi: 10.1097/CORR.0000000000003185. Epub 2024 Jul 23.Clin Orthop Relat Res. 2024.PMID: 39051924 Sexual Harassment and Prevention Training.Cedeno R, Bohlen J.Cedeno R, et al.2024 Mar 29. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.2024 Mar 29. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan–.PMID: 36508513 Free Books & Documents. Accounting for learning environments in academic screening.Lekwa AJ, Reddy LA, Kettler RJ, Van Norman ER.Lekwa AJ, et al.J Sch Psychol. 2025 Feb;108:101403. doi: 10.1016/j.jsp.2024.101403. Epub 2024 Nov 29.J Sch Psychol. 2025.PMID: 39710438 See all similar articles References Publication types Research Support, U.S. Gov't, Non-P.H.S. Actions Search in PubMed Search in MeSH Add to Search MeSH terms Academic Performance Actions Search in PubMed Search in MeSH Add to Search Academic Success Actions Search in PubMed Search in MeSH Add to Search Bayes Theorem Actions Search in PubMed Search in MeSH Add to Search Child Actions Search in PubMed Search in MeSH Add to Search Educational Measurement / methods Actions Search in PubMed Search in MeSH Add to Search Female Actions Search in PubMed Search in MeSH Add to Search Humans Actions Search in PubMed Search in MeSH Add to Search Male Actions Search in PubMed Search in MeSH Add to Search Risk Assessment / methods Actions Search in PubMed Search in MeSH Add to Search Schools Actions Search in PubMed Search in MeSH Add to Search Students / psychology Actions Search in PubMed Search in MeSH Add to Search Related information MedGen Grants and funding R305B200026/US Department of Education R305A210019/US Department of Edcuation LinkOut - more resources Miscellaneous NCI CPTAC Assay Portal [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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https://www.youtube.com/watch?v=1i_ei-Nwooc
CONVERTING PERCENTS TO DECIMALS & DECIMALS TO PERCENTS (BEGINNER): 6th Grade Common Core Math Blooming With Ms.B 16600 subscribers 12 likes Description 657 views Posted: 9 Jan 2024 Hi y'all! We're at it again with a 5th/6th-grade math skill of converting percents to decimals and decimals to percents & it's easier than you could ever imagine! California Math Chapter 2. Music by Bensound Transcript: Intro hey friends what's up okay today we are going over another sixth grade standard moving percentages to decimals and decimals to percentages let's go okay so Converting Percent to Decimal if I have a percent and I want to change that to a decimal all I have to do is start at the percent and jump two times to the left add your decimal and you no longer need that percent but if you want to go the other way where you have your decimal and you want to turn it into a percent you go ahead and jump two spaces to the back where the percentage goes add your percent and you no longer need your decimal so percent you jump two spaces to the front where your decimal goes change decimal to percent you jump two spaces to the back where the percent goes let me show you with some Converting Percent to Decimal Examples numbers so if I have something that is 25% and I want to make that a decimal all I have to do is jump one two spaces to the front and now my decimal is 2 five but what if I have a number like three percent well I would still start at the percent jump one jump two add your decimal but I don't have a number right here so I need to add a zero meaning our decimal is going to be 0.3 let's try another one say I have a bigger number where I have 200 and 43% and I want to make that a decimal I started the % I jump one I jump two that's where my decimal goes so my new decimal will be 2.43 that'll be my decimal now if I have a percent that is smaller than 1% say I have 03 per. we can still do that I follow the same rules but instead of starting at the percent this time I have a decimal to work with so I'm going to jump one jump two to the left my new decimal will be there now I have no spaces there so I need to add zeros I did two jumps so I need to add two zeros so my decimal will be .03 what if I have1 16% well in that case I started my decimal cuz I already have one I jump one two spaces to the left I don't have numbers there so I need to add zeros and take away my old decimal so my new decimal will be 0.16 now let's go the reverse let's say I have 76 and I want to change that to a percent so I'm going to start at my decimal and percents are in the back so that's where I'm going so I jump one I jump two add my percent get rid of my decimal so I have 76% let's say I I have 36 I want to change that to a percent percents are at the back so I'm going to jump towards the back 1 2 add my percent get rid of my decimal so I'm going to have 36% now what if I have a bigger number let's say I have 2.14 but I want to make that a percent I start with the decimal and I go back towards the percentage area where I go 1 two that is now a percent and I have 24% and yes even though it is greater than 100 you can definitely have percentages that are greater than 100 that will be a whole other lesson so let's do another one that is big let's say I have 54.3 that is okay I'm going to go ahead I want to make a percent so I'm jumping towards the end where the percent goes I am jumping one I'm jumping two add my percent but I don't have a number here so I need to add a zero that means that is 5,430 Outro per. and that's it for today thank you guys for joining me I'll see you in the next one [Music] bye
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https://testbook.com/question-answer/the-steady-flow-energy-equationq-mh2--5e19cb59f60d5d2d6cf1855c
[Solved] The steady flow energy equation: Q = m(h2 – h1) is app Get Started ExamsSuperCoachingTest SeriesSkill Academy More Pass Skill Academy Free Live Classes Free Live Tests & Quizzes Previous Year Papers Doubts Practice Refer & Earn All Exams Our Selections Careers English Hindi Home Thermodynamics First Law of Thermodynamics Steady Flow Energy Equation Question Download Solution PDF The steady flow energy equation: Q = m(h 2 – h 1) is applicable for? This question was previously asked in SSC JE ME Previous Paper 8 (Held on: 27 Sep 2019 Morning) Attempt Online View all SSC JE ME Papers > Nozzle Turbine Compressor Boiler Answer (Detailed Solution Below) Option 4 : Boiler Crack SSC JE Mechanical (CBT 1) Selection Kit 2025 with India's Super Teachers FREE Demo Classes Available Explore Supercoaching For FREE Free Tests View all Free tests > Free Thermodynamics for All AE/JE ME Exams Mock Test 6.1 K Users 20 Questions 20 Marks 20 Mins Start Now Detailed Solution Download Solution PDF Explanation: Steady Flow Energy Equation (S.F.E.E.) m(h 1+v 1 2 2+g z 1)+Q˙=m(h 2+v 2 2 2+g z 2)+W˙ Now, Boiler: Ẇ = 0, no work is done by the boiler, Change in kinetic and potential energy is neglected (ΔK.E = ΔP.E = 0) ∴ Q̇= ṁ(h 2 - h 1) Nozzle: Q̇= 0, as the nozzle is perfectly insulated. Ẇ= 0, no work is done by the nozzle, v 1<<< v 2 and potential energy change is neglected ∴v 2=2(h 1−h 2) Turbine: Q = 0, for the adiabatic or perfectly insulated turbine, Change in kinetic and potential energy is neglected ∴ Ẇ= ṁ(h 1 – h 2) Compressor: Q̇= 0, for the adiabatic or perfectly insulated turbine, Change in kinetic and potential energy is neglected(ΔK.E = ΔP.E = 0) ∴Ẇ = ṁ(h 2 - h 1) Points to remember: DeviceS.F.E.E Nozzlev 2=2(h 1−h 2) TurbineẆ = ṁ(h 1 – h 2) CompressorẆ = ṁ(h 2 - h 1) BoilerQ = ṁ(h 2 - h 1) Download Solution PDFShare on Whatsapp Latest SSC JE ME Updates Last updated on Aug 5, 2025 -> SSC JE vacancies 2025 for Mechanical Engineering have been released. A total of 327vacancy are available for SSC JE ME. ->SSC JE ME Notification 2025has been released on June 30. -> The SSC JE Mechanical engineering application form was available from June 30 to July 21. -> SSC JE 2025 CBT 1 exam for Mechanical Engineering will be conducted from October 27 to 31. -> SSC JE exam to recruit Junior Engineers in different disciplines under various departments of the Central Government. -> The selection process of the candidates for the SSC Junior Engineer post consists of Paper I, Paper II, Document Verification, and Medical Examination. -> Candidates who will get selected will get a salary range between Rs. 35,400/- to Rs. 1,12,400/-. -> Candidates must refer to the SSC JE Previous Year Papers and SSC JE Civil Mock Test, SSC JE Electrical Mock Test, and SSC JE Mechanical Mock Test to understand the type of questions coming in the examination. India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank Mock Tests & Quizzes Get Started for Free Trusted by 7.6 Crore+ Students More Steady Flow Energy Equation Questions Q1.Dry saturated steam enters a frictionless adiabatic nozzle with negligible velocity at a temperature of 300 °C [h₁ = 2751 kJ/kg]. It is expanded to a pressure of 5 MPa isentropically [h₂ = 2651 kJ/kg]. What will be the exit velocity of steam? Q2.A steam turbine receives steam steadily at 10 bar with an enthalpy of 3000 kJ/kg and discharges at 1 bar with an enthalpy of 2700 kJ/kg. The work output is 250 kJ/kg. The changes in kinetic and potential energies are negligible. The heat transfer from the turbine casing to the surroundings is equal to Q3.If h1 and h2 are enthalpies at the inlet and the outlet, respectively and C1 and C2 are inlet and outlet velocities, respectively, then which of the following statements is INCORRECT about the equation (c 2 2−c 1 2 2=h 1−h 2)for steam nozzles? Q4.In a steady-flow air compressor, air enters at a speed of 5 m/s with a pressure of 1 bar and leaves at a speed of 7.5 m/s with a pressure of 7 bar. If the inlet specific volume is 0.5 m3/kg and the outlet specific volume is 0.15 m3/kg, what is the ratio of the inlet pipe diameter to the outlet pipe diameter? Q5.Air enters a nozzle with negligible velocity. The enthalpy of air decreases by 180 kJ/kg. What will be the exit velocity of air? Nozzle is horizontal and heat losses are negligible. Q6.In a steady flow system, a substance flows at a rate of 5 kg/s. The following data is given for the entry and exit of the system. At entry: Pressure = 600 kPa, Velocity = 300 m/s, Internal energy = 2000 kJ/kg, Specific volume = 0.4 m3/kg. At exit : Pressure = 200 kPa, Velocity = 300 m/s, Internal energy = 1600 kJ/kg, Specific volume = 1.0 m3/kg. The system is considered to be adiabatic. Neglect the change in potential energy of the system. What will be the power (developed/consumed) by the system? Q7.Find the efficiency of a pump (rated 400 W) that can lift 500 kg of water by 30 m in 10 minutes. (Use g = 10m/s2) Q8.When the compressor and motor operate on the same shaft and are enclosed in a common casing, then it is known as ____. Q9.A fully insulated and completely empty tank is being filled with some fluid with negligible fluid velocity. Which of the following is correct? Q10.A roots blower compresses 0.08 m3 air from 1 bar to 1.5 bar per revolution and ideal work input is found to be 3438.89 Nm, then roots efficiency is: More First Law of Thermodynamics Questions Q1.The sum of internal energy and product of pressure and volume is known as Q2.The change in internal energy of a closed system undergoing a cycle Q3.For a closed system undergoing a process (not a cycle), the first law is expressed as: Q4.What is the power of 2 tonne refrigeration system? Latent heat of ice may be taken as 336 kJ/kg and specific heat capacity of water as 4 kJ/kgK. Q5.For water, the latent heat of freezing is 334 kJ/kg and the specific heat capacity averages 4.19 kJ/kgK. The quantity of heat to be removed from 1 kg of water at 30°C in order to turn it into ice at 0°C is: Q6.A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa the pressure remaining constant throughout the process. There is a transfer of 37.6 KJ of heat from the gas during the process. The internal energy of gas changes by Q7.An object of mass 5 kg falls from rest through a vertical distance of 20 m and gains a velocity of 10 m/s. The work done by the resistance of the air on the object will be Q8.For a closed system undergoing a thermodynamic cycle, the first law states which of the following? Q9.Despite having the highest possible efficiency for Carnot cycle, it is not suitable for a practical engine using a gaseous working fluid as: Q10.For a heat engine cycle, which of the following relation is always true? [Q = heat transfer, W = Work transfer] Crack All AE & JE Exams with India's Super Teachers Rahul Mishra Testbook Mohit Kasanya Testbook Explore Supercoaching For FREE Suggested Test Series View All > SSC Reasoning PYP Mock Test Series (20k+ Questions) 1712 Total Tests with 1 Free Tests Start Free Test SSC GK PYP Mock Test Series (20k+ Questions) 1791 Total Tests with 1 Free Tests Start Free Test Suggested Exams UPPSC AE NLC Graduate Executive Trainee SSC JE ME UPPSC AE Important Links More Thermodynamics Questions Q1.The sum of internal energy and product of pressure and volume is known as Q2.When the pressure of a given mass of liquid at a temperature of 303 K is increased from 3 MPa to 3.5 MPa the relative density of liquid increases from 0.5 to 0.501. What is the average bulk modulus of liquid over the given pressure range Q3.An isothermal process is governed by Q4.An adiabatic boundary is one which Q5.If cyclic integral of dQ/T > 0, the cycle is Q6.In a quasi-static process, irreversibilities are: Q7.Which of the following processes is always quasi-static? Q8.The TdS equation can be expressed as: Q9.The change in internal energy of a closed system undergoing a cycle Q10.A reversible power cycle operates with temperature limits of 800 K and 300 K. If it takes 560 KJ of heat, what would be the unavailable work? 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https://link.springer.com/article/10.1007/s44337-024-00139-y
Advertisement Prevalence and management of ICU delirium—Time to think outside the box You have full access to this open access article 1937 Accesses 1 Citation 1 Altmetric Explore all metrics Abstract Delirium presents a significant management challenge in the intensive care unit (ICU) setting of various hospital systems and leads to significant morbidity and mortality. Significant portion of psychiatry consultation involves management of agitation associated with this condition. While it is acknowledged that a definitive consensus on the pharmacological management of delirium is lacking, resulting in notable frustration among care teams and families, this review aims to aggregate existing evidence and present a comprehensive overview of the current landscape and potential gaps in literature. This narrative review seeks to condense the disparate research and work done on the medical and non-medical management of delirium in the ICU. Similar content being viewed by others Monitoring for Delirium in Critically Ill Adults Delirium in Intensive Care ICU Delirium: Diagnosis, Risk Factors, and Management Explore related subjects Avoid common mistakes on your manuscript. 1 Introduction Delirium is a critical issue in ICU management across various healthcare systems, resulting in high rates of morbidity and mortality. While the community prevalence of delirium is at approximately 1–2%, a prospective cohort study on the population above the age of 65 reported the prevalence of delirium at 55.4% with 35.4% of these cases going missed by the primary team. Hyperactive delirium in this study was found to be the commonest type of delirium [17 .")]. Hypoactive delirium is frequently misdiagnosed as depression or catatonia and is often associated with a poorer outcome. Studies on the effects of delirium on length of hospital stay, morbidity, and mortality, show that patient stay, likelihood of institutionalization and mortality are inevitably increased. Zhang et al. investigate the prevalence of delirium in intensive care unit (ICU) patients and its association with adverse clinical outcomes through a meta-analysis of clinical observational studies. Sixteen studies involving 5,891 patients were included, revealing that delirious patients had significantly higher mortality rates (odds ratio of 3.22) and experienced more complications (odds ratio of 6.5). Additionally, these patients were more likely to be discharged to skilled nursing facilities (odds ratio of 2.59). The analysis also showed that delirious patients had longer stays in the ICU (weighted mean difference of 7.32 days) and the hospital (weighted mean difference of 6.53 days), as well as extended durations of mechanical ventilation (weighted mean difference of 7.22 days). In conclusion, delirium in critically ill patients is associated with higher mortality rates, increased complications, and longer hospital and ICU stays [23 .")]. The risk factors for delirium are variable. In a study on prevalence, risk factors and outcomes of delirium, of 229 enrolled patients, 50 met criteria for delirium. Abnormal sodium levels, illness severity, dementia, temperature dysregulation, psychoactive drug use, and azotemia were seen to be risk factors for delirium and patients with 3 or more risk factors simultaneously had a 60% likelihood to go into a delirious state . A systematic review identified key risk factors for delirium in critically ill adults. After reviewing 33 studies (70% high quality), strong evidence supported the following risk factors: age, dementia, hypertension, emergency surgery or trauma before ICU admission, mechanical ventilation, metabolic acidosis, coma, and high APACHE II score. Gender was not associated with delirium, and dexmedetomidine use was linked to a lower delirium prevalence. Moderate evidence indicated multiple organ failure as a risk factor. These findings inform prevention strategies and future research . Medications are a leading cause of delirium and agents commonly linked to delirium include anticholinergics, sedatives, analgesics, H2 receptor antagonists, and antiparkinsonian drugs. Drug-drug and drug-disease interactions may also cause delirium in patients in situations when a single drug alone could be well tolerated. Delirium can occur despite ‘therapeutic’ serum concentrations. Major challenges faced by healthcare teams involves excluding underlying medical conditions. The goal is to aggressively treat underlying causes and keep risk of morbidity and mortality low while delirium abates . Other independent risk factors for delirium include pre-existing cognitive impairment; poor functional status; polypharmacy; urinary catheterization; dehydration; and electrolytes derangements. Furthermore, 56.9% of patients with delirium secondary to these causes continued to have delirium upon discharge from the hospital [17 .")]. Often delirium is secondary to unrecognized vitamin deficiencies, particularly thiamine deficiency, with delirium occurring in 6/19 (32%) patients with normal thiamine status and 13/17 (76%) thiamine-deficient patients. other possible causes for delirium were present in all cases both of patients with normal thiamine status and of patients with abnormal thiamine status . In the ICU setting, delirium has a prevalence of approximately 14 to 28%, continuing to rise to upwards of 70 to 87% in the case of patients on mechanical ventilation [79 .")]. Post-operative rates of delirium range from approximately 15 to 53% of older adults [8/ ")]. Factors increasing risk of delirium in patients admitted to the ICU included Age (odds ratio of 4.1) hypertension (OR = 3.3) COPD (OR = 3.5 steroids (OR = 2.8) with sedation presenting the highest risk for onset of delirium (OR = 5.4) [97 .")]. The presence of delirium has important prognostic implications; in mechanically ventilated patients it is associated with a 2.5-fold increase in short-term mortality and a 3.2-fold increase in 6 month mortality . 2 Methods 2.1 Study design This study was designed as a narrative review to provide a comprehensive analysis of existing literature on ICU delirium and its management. The aim was to synthesize findings from a broad range of sources and present an interpretative summary of the current understanding, key advancements, and evolving management strategies for ICU delirium. 2.2 Literature search strategy A systematic search of electronic databases, including PubMed, Scopus, and Google Scholar, was conducted to identify relevant studies. The search covered literature published between 1990 and 2024. Keywords used in the search included “ICU Delirium,” “Pharmacological management of ICU Delirium,” “Non-pharmacological management of ICU Delirium,” “Management of ICU Delirium,” “Guidelines for ICU Delirium management,” and “Recent advances in ICU Delirium management.” Additional studies were identified through screening UpToDate for related topics and from reference lists of key articles. 2.3 Inclusion and exclusion criteria Articles were selected based on their relevance to ICU delirium and its management, with a focus on publications in peer-reviewed journals. Only studies written in English were included. Exclusion criteria were limited to non-peer-reviewed articles and gray literature (e.g., conference abstracts or dissertations). Studies from a broad time frame were selected to provide a holistic overview of the evolution of ICU delirium management, highlighting historical context when relevant. 2.4 Data extraction and synthesis Data from selected studies were extracted systematically. The extracted information included study design, sample size, key outcomes, interventions (pharmacological and non-pharmacological), and limitations. Thematic analysis was performed to identify recurring trends, gaps in current knowledge, and areas requiring further investigation. No formal meta-analysis was conducted due to heterogeneity in study designs and outcome measures. 2.5 Quality assessment Although formal quality assessments were not conducted, emphasis was placed on studies with clear methodologies, well-defined patient populations, and clinically meaningful outcomes. High-quality studies, particularly those involving randomized controlled trials and comprehensive reviews, were prioritized. 2.6 Ethical considerations Since this study only involved reviewing and analyzing previously published literature, no ethical approval was required. 3 Managing delirium in the ICU It is important to have early detection and management strategies present and available because delirium has shown to increase hospital stay by around 5–7 days and increase costs to patients and hospitals by approximately $5000-$6000 . It is also critical to note that up to 1/3rd of cases of delirium can be prevented if early screening and management of risk factors is appropriately done . The management of delirium is primarily by non-pharmacological and pharmacological methods. 4 Non-pharmacological strategies Given the numerous risk factors for delirium and its correlation with ailments such low electrolyte levels, polypharmacy, prior delirium, etc., the non-pharmacological management of delirium may be more crucial than the pharmacological management of delirium. Many electronic medical systems now incorporate order sets to implement these non-pharmacological interventions. The best therapies for preventing delirium are those that involve multiple components. Early patient mobilization, volume replacement, treating visual or auditory impairments, and discontinuing non-essential medications are some examples of these strategies. The assessment of bowel and bladder functions, staff education, geriatric consulting, nutritional status, sleep cycle, pain management, and supplemental oxygen are critical components. Targeting infections, avoiding catheterization, and medication reconciliation are other crucial measures . For those at risk of delirium, avoiding new prescriptions for benzodiazepines, opioids, dihydropyridines, and antihistamines can prevent delirium setting in [13, 14]. In light of this, some post-acute care (PAC) facilities have introduced a new nurse-led model for the management of newly admitted patients with delirium. The model consists of four modules that are based on best practices as described by the peer-reviewed literature: Uniform screening for delirium symptoms and indicators upon PAC unit admission. Evaluation and management of potential delirium contributors and causes. Preventing and treating typical consequences from delirium. The recovery of the patient's cognitive and self-care abilities. The Delirium Room, a 4-bed unit offering extensive 24 h nursing care, is another model approach for managing patients who have unpreventable delirium or who are already delirious at admission to the ICU. It has also been used for the treatment of geriatric patients both independently and as a crucial component of a 22-bed acute care unit . A study evaluating the impact of an interdisciplinary intervention program led by nurses on delirium's incidence, course (severity and duration), cognitive functioning, functional rehabilitation, mortality, and length of stay in older hip-fracture patients revealed a significant reduction in both the severity and frequency of delirium episodes if targeted interventions are implemented. The nurse led interdisciplinary intervention program consisted of education of nursing staff, systematic cognitive screening, consultative services by a delirium resource nurse, a geriatric nurse specialist, or a psychogeriatrician, and use of a scheduled pain protocol. Patients were monitored for signs of delirium, as measured by the Confusion Assessment Method (CAM). Severity of delirium was assessed using a variant of the CAM. Cognitive and functional status were measured by the Mini-Mental State Examination (MMSE) . As delirium is an acute state of confusion, reduction in patient stimulation is also beneficial for early recovery of patients, this is corroborated by a study in which earplugs during the night lowered the incidence of confusion in the studied intensive care patients . Marcantonio et al. studied the effectiveness of geriatric consultation in reducing delirium post-hip fractures. A geriatrician making daily visits for the duration of the hospitalization and making targeted recommendations based on a structured protocol reduced delirium by over one third and reduced the incidence of severe delirium by over one half . It was also observed that the risk-factor intervention strategies led to significant reductions in the number and duration of episodes of delirium in hospitalized older patients as compared to usual care in a study where specific risk factors were monitored and managed daily and compared to usual care, with delirium, the primary outcome, being assessed daily until discharge. The intervention consisted of standardized protocols for the management of six risk factors for delirium: cognitive impairment, sleep deprivation, immobility, visual impairment, hearing impairment, and dehydration . There was a benefit found to using family members for non-pharmacological delirium prevention when compared to standard management, according to a randomized controlled trial that involved 287 hospitalized patients who developed delirium and either received non-pharmacological intervention (144 patients) or standard management (143 patients) . In another randomized control trial of 174 patients with delirium brought to the hospital for treatment of delirium, mean age of patients being 83 years old and 31% with previous dementia, group which received multicomponent geriatric care had faster alleviation of delirium and improve cognition, however there was no significant improvement in terms of hard endpoints of prognosis, with 60.9% of the intervention group and 64.9% of the control group being either deceased or permanently institutionalized on 1-year follow up . In addition, virtual reality (VR) stimulation, administered via a head-mounted display in ICU settings, offers promising benefits for patients, particularly in reducing the incidence of delirium through its relaxing effects. The intervention involves immersive 360-degree videos, such as nature scenes or urban parks, played with noise-cancelling headphones [232 .")]. Notably, VR-based cognitive stimulation in critically ill patients has shown safety and elicited no negative reactions. Additionally, VR therapy demonstrates potential in alleviating psychophysiological stress, pain, and anxiety, while enhancing cognitive and attentional capacities. Healthcare providers reported the headset's comfort and a strong sense of presence, unanimously agreeing on its availability for patients, with minimal side effects observed in patient [24:748–53. .")]. The application of machine learning (ML) and artificial intelligence (AI) is in predicting delirium and agitation. Recent studies have shown that large language models and other AI techniques can predict delirium by analyzing patient-specific factors. For instance, one study explored the use of machine learning (ML) to predict postoperative delirium by integrating clinical and sleep-related data, including polysomnography (PSG) and sleep-disorder questionnaires. Analyzing 912 patients who underwent surgery, the study identified key delirium risk factors, such as midazolam use, prolonged surgery duration, hypoalbuminemia, fewer REM episodes, and higher daytime sleepiness. Using a gradient-boosting algorithm, the model achieved an AUC of 0.84 when combining clinical and PSG data. The results demonstrate that ML models can effectively identify patients at high risk of delirium, allowing for closer monitoring and potentially improved outcomes [255 .")]. Similarly, another study aimed to develop a machine learning model to predict sepsis-associated delirium (SAD) in ICU patients using data from the MIMIC-IV and eICU-CRD databases. Various machine learning models, including logistic regression, support vector machines, decision trees, random forests, and XGBoost, were compared for their predictive performance. The XGBoost model demonstrated the best results, with an AUROC of 0.793 in internal validation and 0.701 in external validation. The model proved effective in predicting SAD earlier than traditional delirium assessments, including for patients who are challenging to assess through conventional methods, highlighting its potential for improving early intervention in critical care settings [264 .")]. Furthermore, Zhang et al. employed an ensemble machine learning model to predict agitation in mechanically ventilated patients under light sedation. This model combined multiple algorithms and accurately predicted agitation within 24 h based on variables such as body temperature, ventilation, and sedation levels [271 .")]. We acknowledge that the current use of AI and large language models in managing delirium is largely predictive, focusing on identifying high-risk patients based on specific clinical factors during ICU admissions. An ICU delirium-prediction model, as defined, incorporates at least one clinical factor measured during a hospital stay to estimate the risk of delirium. This predictive capability can guide clinicians in targeting high-risk patients and developing early intervention strategies. As demonstrated in a systematic review by Matthew et al., ICU delirium-prediction models developed over the past five years vary significantly in performance. The review analyzed 23 models, noting that their effectiveness—measured by AUC, specificity, and sensitivity—ranged widely. In their review of prediction models, they noted that many existing models are static and fail to account for the dynamic nature of ICU patients. They suggested that the limitations in current models could be mitigated by employing machine learning, which can adapt to changing patient conditions and provide more actionable, real-time predictions. It is essential to pursue further research into more dynamic AI-driven models that can adapt to patient changes and offer practical guidance to clinicians. Additionally, more robust reviews, similar to that of Matthew et al., are necessary to evaluate the broader applicability of AI and language models in predicting and managing ICU delirium. Expanding the generalization and utility of such algorithms remains a critical step toward improving their effectiveness in real-world settings . Non-pharmacological management is vital for preventing delirium, including early mobilization, addressing visual or auditory impairments, and minimizing non-essential medications. Nurse-led models and specialized units like the Delirium Room offer comprehensive care. Interdisciplinary interventions, such as systematic screening and geriatric consultation, reduce delirium severity and frequency. Simple measures like reducing patient stimulation with earplugs can aid recovery. However, a gap in the literature exists regarding the absence of a step-wise or structured flowchart highlighting all non-pharmacological interventions for delirium patients. This gap hampers the standardization of care strategies. Further research should focus on developing such guidelines to enhance delirium management. Another gap also exists in assessing long-term outcomes and evaluating interventions' impact on mortality and institutionalization rates, as seen in trials showing improved cognition but no significant change in prognosis. These gaps should be addressed to enhance delirium management strategies Table 1. 5 Pharmacological treatment 5.1 Haloperidol The use of antipsychotics to manage agitation associated with delirium is not free of controversies. Traditionally, the drug of choice for treatment of delirium in the ICU has been the antipsychotic haloperidol, however, recent research has shown that low-dose haloperidol prophylactic treatment demonstrated no efficacy in reducing the incidence of postoperative delirium. Various studies have shown that Haloperidol has a positive effect on the severity and duration of delirium, Haloperidol reduced the number of days’ patients stayed in the hospital, and the therapy was well tolerated . This is corroborated by another study on critically ill patients after having undergone non cardiac surgery who were given 0.1 mg/hr IV haloperidol for 12 h showing a reduction in incidence of 8% to 15%as compared to 23%in the control group with an increased mean time to onset and significantly reduced ICU stay . Patients on haloperidol have however been observed to experience over sedation and QTc prolongation and patients receiving ongoing therapy should be on telemetry on receiving regular EKG monitoring . 5.2 Dexmedetomidine The current mainstay drug for the treatment of delirium in the ICU setting is the sedative Dexmedetomidine which is a highly selective alpha-2 receptor agonist. In a study conducted to observe the effects of Dexmedetomidine on outcomes of cardiac surgery in geriatric patients, results suggested perioperative use of Dexmedetomidine was associated with decreases in hospital and operative mortality in patients following cardiac surgery. It also reduced incidences of both postoperative stroke and delirium . Comparing with other drugs Dexmedetomidine reduces risk of postoperative ventricular tachycardia and delirium but may increase risk of bradycardia. There was also reduced risk of atrial fibrillation, shorter length of ICU stays and hospitalization but increased risk of hypotension with Dexmedetomidine and as such should not be started on hypotensive or bradycardic patients. Other studies which have been conducted have similar results with incidence of delirium and tachycardia being significantly reduced when it is administered at a dosage of 0.1 mg/kg post operatively as compared to a patient group which is given only placebo . There has also been research done on the timing of administration of Dexmedetomidine and its potential method of action showing that when it is given 1 µg/kg bolus followed by 0.2–0.7 µg/kg/hour maintenance infusion from the beginning till the end of surgery the incidence of delirium and levels of IL-6 were much lower both 1 h and 24 h after administration as compared to just giving bolus 15 min prior to end surgery or giving normal saline before end of surgery . 5.3 Guanfacine A retrospective study underscores guanfacine's efficacy as a well-tolerated medication for managing delirium, even in medically and critically ill patients, with rare cardiovascular adverse events reported. Patients treated with guanfacine demonstrated reduced acute sedative use for behavioural agitation and received fewer opioids, enhancing their participation in PT/OT activities . Moreover, future prospective, randomized, placebo-controlled studies are warranted to further explore guanfacine's potential in delirium management. In another case series, guanfacine's role in managing hyperactive delirium-associated agitation is advocated, highlighting its sedative properties without significant cardiovascular effects compared to clonidine and dexmedetomidine. Combining guanfacine with sedatives may effectively reduce sedative dosage while ensuring agitation control, emphasizing the need for formal investigations into its efficacy alongside other alpha-2 agonists in delirium prevention and management [37:83–8. .")]. 5.4 Cholinesterase inhibitors Donepezil, a cholinesterase inhibitor, has also been found to be effective in reducing the hospital stay for delirium. In a double blind, placebo controlled, parallel group randomized trial was undertaken. Donepezil was well tolerated with no serious adverse events. Donepezil did not significantly reduce the incidence of delirium. However, mean length of hospital stay was 9.9 days for the donepezil group vs 12.1 days in the placebo group , However, this is controversial as other studies on Donepezil have found no benefit to its administration over placebo . There have been studies on the use of donepezil for non-ICU delirium, as well as its role in reducing delirium in patients with dementia when administered before hospitalization. However, more trials are needed to determine the definite role of this drug and its specific indication in managing delirium in the ICU setting. Lieberman et al. investigated whether pre-hospital use of donepezil, a cholinesterase inhibitor, improves outcomes for dementia patients during critical illness, focusing on its effect on delirium. Data from the MIMIC-III database included 2734 ICU patients with dementia, 447 of whom had taken donepezil before hospitalization. After adjusting for confounders, results showed that patients on donepezil had significantly lower in-hospital and 90 day mortality, shorter ICU stays, and reduced mechanical ventilation duration. Donepezil's benefits were associated with reduced delirium. The study suggests that donepezil may improve critical illness outcomes in dementia patients by mitigating delirium, and future research should explore its preventive role in delirium [407 .")]. Another study aimed to evaluate the effectiveness and safety of cholinesterase inhibitors for treating delirium in non-ICU settings. The researchers included one randomized controlled trial with 15 participants (mean age 82.5 years), diagnosed with delirium using the Confusion Assessment Method. The study compared rivastigmine (a cholinesterase inhibitor) with a placebo. Results showed no significant differences between the two groups in terms of the duration of delirium, adverse events, use of rescue medications, mortality, or early withdrawal from the study. Due to the small sample size, there was limited evidence on other outcomes such as the severity of delirium, cognitive impairment, and hospitalization length. The quality of evidence was low. The authors concluded that there is insufficient evidence to support or refute the use of cholinesterase inhibitors for treating delirium in non-ICU settings. More trials are needed to draw definitive conclusions . 5.5 Second generation antipsychotics A systematic review and meta-analysis conducted by Hirota et al. aimed to evaluate the efficacy and tolerability of antipsychotics in preventing postoperative delirium in surgical patients. Six randomized controlled trials involving 1689 patients were included, comparing prophylactic use of antipsychotics (haloperidol, olanzapine, and risperidone) to placebo. The analysis showed that antipsychotics significantly reduced the occurrence of delirium (RR = 0.50, NNT = 7). Specifically, second-generation antipsychotics, such as olanzapine and risperidone, were more effective than placebo (RR = 0.36, NNT = 4), while haloperidol did not demonstrate a clear benefit. There were no significant differences between the groups in terms of delirium severity, discontinuation rates, or adverse events. The study concluded that second-generation antipsychotics are more effective than placebo in preventing delirium, but they do not reduce its severity in patients who develop it . Preoperative sublingual olanzapine has also been seen to significantly reduce rates of delirium in patients undergoing joint replacement surgery . 5.6 Gabapentin Anticonvulsants such as Gabapentin have also been found to reduce delirium, in a clinical trial, postoperative delirium occurred in 5/12 patients (42%) who received placebo vs 0/9 patients who received gabapentin, p = 0.045. The reduction in delirium appears to be secondary to the opioid-sparing effect of gabapentin. Another study published in 2017 investigated whether perioperative gabapentin could reduce postoperative delirium in older patients undergoing noncardiac surgery. It was a randomized, placebo-controlled, double-blinded trial involving 697 patients (mean age 72 years). Patients received either gabapentin (900 mg) or placebo preoperatively and for the first three postoperative days. The primary outcome was the incidence of postoperative delirium, measured using the Confusion Assessment Method. Secondary outcomes included postoperative pain, opioid use, and hospital length of stay. Results showed that the incidence of delirium did not significantly differ between the gabapentin (24%) and placebo (20.8%) groups. While gabapentin was associated with slightly reduced opioid use, it did not impact delirium or hospital stay duration. The study concluded that perioperative gabapentin reduced opioid consumption but did not prevent postoperative delirium . 5.7 Valproic acid In the ICU, analgesics, sedatives, and antipsychotics are frequently utilized for managing agitation and delirium, yet their effectiveness is constrained by adverse reactions. Valproic acid (VPA) emerges as a viable alternative. Quinn et al. conducted a single-center retrospective study and examined valproic acid prescribing practices in ICU patients of their institution for agitation or delirium. Out of 80 patients, 55% experienced delirium resolution, with 69% achieving resolution with valproic acid alone versus 44% with valproic acid plus antipsychotics. The most common regimen was 250 mg orally 3 times daily. Delirium incidence significantly decreased from valproic acid day 0 to day 3 (93% to 68%). While valproic acid appears frequently prescribed and effective, larger studies are needed to confirm its role in ICU delirium management [463 .")]. Another retrospective cohort study compared the efficacy and safety of valproic acid (VPA) to other antipsychotics for ICU delirium and agitation. One hundred eight patients were analyzed, with no significant differences found in delirium-free days, agitation-free days, mortality, ventilation duration, or ICU length of stay between the VPA and control groups. While VPA showed similar efficacy, adverse effects were noted [476 .")]. 5.8 Meperidine The usage of meperidine increases normeperidine, the toxic metabolite which increases the confusion state and eventually leads to delirium. So, for postoperative analgesia, it is recommended to use Morphine instead. The decreased use of meperidine worldwide has decreased the incidence of delirium . 5.9 Melatonin and Melatonin agonists A new group of drugs has also been recently found to be effective in the treatment of delirium, this being melatonin and melatonin agonists like Ramelteon. In a multicenter rater blinded randomized placebo control trial. The eligibility criteria of this trial included age:65–89, newly admitted and being able to take medicine orally. Ramelteon was associated with lower risk of delirium (3% vs 32%). Ramelteon administered nightly may hence provide protection against delirium. The study supports a possible pathogenic role of melatonin neurotransmission in delirium . Melatonin was successful in decreasing postoperative delirium when used preoperatively (control group: 32.65%, group taking melatonin: 9.43%) and in treating more than half of patients (58.06%) developed postoperative delirium when used for three postoperative nights . In a randomized double blinded placebo-controlled study in an internal medicine service in a tertiary care center in London and Ontario, Canada. 145 individuals aged 65 or above were randomized to receive melatonin 0.5 mg every night for 14 days until discharge. Melatonin was found to be associated with lower risk of delirium (12% vs 31%) with an odds ratio of 0.19 and 95% confidence interval . 5.10 Electroconvulsive therapy In a study described by Nielsen et al. 5 severe, protracted delirium cases in the ICU were treated with electroconvulsive therapy (ECT) after conventional medical therapy failed. ECT effectively controlled agitation, anxiety, and discomfort in all patients except one who experienced posttraumatic amnesia. Despite controversy, ECT is deemed safe and efficient for various psychiatric conditions, including delirium. Given the high mortality and cognitive decline associated with ICU delirium, ECT emerges as a valuable option when medical treatment fails, aiding in ventilator weaning and avoiding prolonged sedation . A systematic review examined 10 studies to evaluate the efficacy and safety of modified electroconvulsive therapy (ECT) for delirium. The studies comprised six case reports, three case series, and one quasi-experimental study. However, the quality of evidence was generally low, with only one quasi-experimental study rated as fair quality. Notably, most cases (33 out of 40) involved delirium stemming from substance withdrawal. While ECT was reported to contribute positively to delirium treatment across cases, objective improvement measures were only available for a subset. Adverse events, such as transient confusion and memory deficits, were noted but deemed reversible. Overall, the evidence remains insufficient to firmly advocate for ECT as a standard delirium treatment, highlighting the need for further research with more rigorous methodologies and diverse delirium populations [53 to treat delirium. Acta Psychiatr Scand. 2023;147(5):403–19. .")]. 5.11 Thiamine Thiamine supplementation has emerged as a potential therapeutic approach for managing ICU delirium, a common complication in critically ill patients. Several studies suggest that thiamine deficiency may contribute to the etiology of ICU delirium [54, 55:785–90. .")]. A meta-analysis by Sedhai et al. highlighted that thiamine supplementation significantly reduced the odds of developing ICU delirium by 42% in randomized controlled trials (RCTs), though it did not demonstrate a consistent mortality benefit . Similarly, Dichtwald et al. examined the effect of thiamine on delirium in ICU patients in a RCT. Comparing two groups—one receiving thiamine and one not—researchers found that despite worse disease severity and longer ICU stays, the group given thiamine had fewer delirium events. However, there was no difference in mortality rates between the two groups. The findings suggest that thiamine may reduce delirium in critically ill patients but does not affect overall survival [57:200–4.")]. Despite these promising results in the context of delirium events, the effect of thiamine on long-term outcomes, especially in patients with septic shock remains uncertain. Thiamine administration in patient of septic shock has gained a lot of light in recent times; since sepsis and septic shock forms a huge burden of critically ill patients in icu, studies investigating its impact on various parameters of septic shock patients re published. For example, a study investigated the effect of early vitamin C and thiamine administration on ICU delirium-free days in patients with septic shock. Conducted as a single-center, retrospective study, it compared 89 matched pairs of patients who received vitamin C and thiamine versus those who did not. The results showed no significant difference in delirium-free days between the treatment and control groups, with both groups having similar median delirium-free days. Additionally, there were no differences in secondary outcomes, such as delirium incidence or 28 day mortality. These findings remained consistent across subgroup analyses [58:193. .")]. Furthermore, a secondary analysis of the VICTAS trial by Williams Roberson et al. investigated the long-term cognitive, psychological, and functional outcomes in sepsis patients treated with intravenous vitamin C, thiamine, and hydrocortisone compared to a placebo group. Conducted from August 2018 to July 2019, it included 213 adult patients assessed 6 months after randomization. The findings revealed that the intervention group had lower immediate memory scores and higher odds of developing posttraumatic stress disorder (PTSD) while being less likely to receive mental health care. No significant differences were observed in other cognitive, psychological, or functional outcomes. Overall, the results indicate that the combination therapy did not improve and may have worsened outcomes in sepsis survivors at 6 months, challenging the belief that early antioxidant and anti-inflammatory treatment can prevent long-term impairments associated with sepsis [59: e230380. .")]. Given the variability in findings, particularly concerning long-term cognitive and functional outcomes, it is critical to recognize that thiamine supplementation may have both benefits and potential harms, acting as a "double-edged sword." Before thiamine enters mainline management for ICU delirium, more robust systematic reviews and large-scale, prospective trials are needed. Such investigations should clarify its role, especially concerning mortality, ICU stay, and long-term cognitive outcomes. Although initial evidence supports thiamine's role in reducing ICU delirium, its impact on other critical illness parameters is still inconclusive, necessitating cautious optimism Table 2. 6 Conclusion In conclusion, the management of ICU delirium necessitates a comprehensive, multicomponent non-pharmacological approach as the first-line intervention. Key strategies such as effective communication, patient reorientation, reassurance, and the establishment of a supportive care environment are essential for mitigating the impact of delirium. While there are currently no drugs approved specifically for the treatment of delirium, various medications are employed clinically to provide symptomatic relief; our emphasis must remain on these non-pharmacological interventions. Moreover, there is significant scope for further research in this area. Future studies should focus on developing advanced AI algorithms capable of uniformly predicting ICU delirium, thereby enhancing early identification and intervention. Additionally, larger-scale studies and trials investigating the efficacy of specific pharmacological treatments could contribute valuable insights to the field. Addressing these research gaps will be crucial for improving delirium management and patient outcomes in critical care settings. Pharmacological interventions, particularly dexmedetomidine, may be beneficial in select cases but should be used judiciously and in conjunction with non-pharmacological strategies. Ongoing research into novel pharmacological agents, such as melatonin agonists, holds promise for further optimizing delirium management and improving patient outcomes. In conclusion, a comprehensive approach combining non-pharmacological and pharmacological interventions tailored to individual patient needs is essential for effectively managing delirium in the ICU setting. By implementing evidence-based strategies and staying abreast of recent advances, healthcare providers can strive to mitigate the impact of delirium on patient morbidity, mortality, and healthcare utilization. Data availability No datasets were generated or analysed during the current study. References Al Farsi RS, Al Alawi AM, Al Huraizi AR, Al-Saadi T, Al-Hamadani N, Al Zeedy K, Al-Maqbali JS. 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Article PubMed PubMed Central Google Scholar Download references Author information Authors and Affiliations SSG Hospital/Medical College Baroda, The Maharaja Sayajirao University (Primary), 51, Green Park Society, Bholav, Bharuch, 392001, India Nritya Trivedi University of Missouri-Columbia, Columbia, Missouri, USA Ashish Sarangi The Maharaja Sayajirao University, Vadodara, Gujarat, India Nritya Trivedi, Husen Boriwala, Vraj Bhatt & Sunidhi Rohatgi Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Contributions Literature review: N.T., A.S., H.B., V.B., and S.R.; Drafting the manuscript: N.T., A.S., H.B., V.B., and S.R.; Preparing the tables: N.T., A.S., H.B., V.B., and S.R.; Reviewed the manuscript: N.T., A.S., H.B., V.B., and S.R.; Editing the manuscript: N.T., A.S., H.B., V.B., and S.R. 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Learn more: PMC Disclaimer | PMC Copyright Notice Living Rev Relativ . 2012 Jul 4;15(1):8. doi: 10.12942/lrr-2012-8 Search in PMC Search in PubMed View in NLM Catalog Add to search Binary Neutron Star Mergers Joshua A Faber Joshua A Faber 1 Center for Computational Relativity and Gravitation and School of Mathematical Sciences, Rochester Institute of Technology, 85 Lomb Memorial Drive, Rochester, NY 14623 USA Find articles by Joshua A Faber 1,✉, Frederic A Rasio Frederic A Rasio 2 Center for Interdisciplinary Exploration and Research in Astrophysics, and Department of Physics & Astronomy, Northwestern University, 2145 Sheridan Road, Evanston, IL 60208 USA Find articles by Frederic A Rasio 2 Author information Article notes Copyright and License information 1 Center for Computational Relativity and Gravitation and School of Mathematical Sciences, Rochester Institute of Technology, 85 Lomb Memorial Drive, Rochester, NY 14623 USA 2 Center for Interdisciplinary Exploration and Research in Astrophysics, and Department of Physics & Astronomy, Northwestern University, 2145 Sheridan Road, Evanston, IL 60208 USA ✉ Corresponding author. Accepted 2012 May 22; Issue date 2012. © The Author(s) 2012 PMC Copyright notice PMCID: PMC5255524 PMID: 28163622 Abstract We review the current status of studies of the coalescence of binary neutron star systems. We begin with a discussion of the formation channels of merging binaries and we discuss the most recent theoretical predictions for merger rates. Next, we turn to the quasi-equilibrium formalisms that are used to study binaries prior to the merger phase and to generate initial data for fully dynamical simulations. The quasi-equilibrium approximation has played a key role in developing our understanding of the physics of binary coalescence and, in particular, of the orbital instability processes that can drive binaries to merger at the end of their lifetimes. We then turn to the numerical techniques used in dynamical simulations, including relativistic formalisms, (magneto-)hydrodynamics, gravitational-wave extraction techniques, and nuclear microphysics treatments. This is followed by a summary of the simulations performed across the field to date, including the most recent results from both fully relativistic and microphysically detailed simulations. Finally, we discuss the likely directions for the field as we transition from the first to the second generation of gravitational-wave interferometers and while supercomputers reach the petascale frontier. Introduction Binaries composed of neutron stars (NSs) and black holes (BHs) have long been of interest to astrophysicists. They provide many important constraints for models of massive star evolution and compact object formation, and are among the leading potential sources for detection by gravitational-wave (GW) observatories. While it remains uncertain whether mergers of compact binaries are an important contributor to the production of r-process elements, they are now thought to be the leading candidate to explain short-duration, hard-spectrum gamma-ray bursts (often abbreviated to “short-hard” GRBs, or merely SGRBs). The first neutron star-neutron star (NS-NS) binary to be observed was PSR B1913+16, in which a radio pulsar was found to be in close orbit around another NS . In the decades since its discovery, the decay of the orbit of PSR B1913+16 at exactly the rate predicted by Einstein’s general theory of relativity (see, e.g., [306, 325]) has provided strong indirect evidence that gravitational radiation exists and is indeed correctly described by general relativity (GR). This measurement led to the 1993 Nobel Prize in physics for Hulse and Taylor. According to the lowest-order dissipative contribution from GR, which arises at the 2.5PN level (post-Newtonian; where the digit indicates the expansion order in [υ/c]2 in the Taylor expansion term), and assuming that both NSs may be approximated as point masses, a circular binary orbit decays at a rate da/dt = −a/τ GW where a is the binary separation and the gravitational radiation merger timescale τ GW is given by 1 where M 1, M 2, and M ≡ M 1 + M 2 are the individual NS masses and the total mass of the binary, μ = M 1 M 2 /M is the reduced mass, q = M 2/M 1 is the binary mass ratio, and we assume geometrized units where G = c =1 (as we do throughout this paper, unless otherwise noted). The timescale for an elliptical orbit is shorter, and it can be shown that eccentricity is reduced over time by GW emission, leading to a circularization of orbits as they decay. A quick integration shows that the time until merger is given by τ merge = τ GW/4. The luminosity of such systems in gravitational radiation is 2 which, at the end of a binary’s lifetime, when the components have approached to within a few NS radii of each other, is comparable to the luminosity of all the visible matter in the universe (∼ 10 53 erg/s). The resulting strain amplitude observed at a distance D from the source (assumed to be oriented face-on) is given approximately by 3 at a characteristic frequency 4 The first measurement that will likely be made with direct GW observations is the orbital decay rate, with the period evolving (for the circular case) according to the relation 5 where T is the orbital period and ω the angular frequency, and thus the “chirp mass,” 6 is likely to be the easiest parameter to determine from GW observations. Several NS-NS systems are now known, including PSR J0737-3039 , a binary consisting of two observed pulsars, which allows for the prospect of even more stringent tests of GR . Even with the handful of observed sources to date, one may use this sample to place empirical limits on the expected rate of NS-NS mergers and to constrain the many parameters that enter into population synthesis calculations . With regard to the former, the very short merger timescale for J0737, τ merge = 85 Myr, makes it especially important for estimating the overall rate of NS-NS mergers since it is a priori very unlikely to detect a system with such a short lifetime. Although black hole-neutron star (BH-NS) binaries are expected to form through the same processes as NS-NS binaries, none has been detected to date. This is generally thought to reflect their lower probability of detection in current surveys, in addition to intrinsically smaller numbers compared to NS-NS systems . BH-NS systems are an expected byproduct of binary stellar evolution, and properties of the population may be inferred from population synthesis studies calibrated to the observed NS-NS sample (see, e.g., ). In this review, we will summarize the current state of research on relativistic mergers, beginning in Section 2 with a description of the astrophysical processes that produce merging binaries and the expected parameters of these systems. The phases of the merger are briefly described in Section 3. In Section 4, we discuss the numerical techniques used to generate quasi-equilibrium (QE) sequences of NS-NS configurations, and we summarize the QE calculations that have been performed. These sequences yield a lot of information about NS physics, particularly with regard to the nuclear matter equation of state (EOS). They also serve as initial data for dynamical merger calculations, which we discuss next, focusing in turn on the numerical hydrodynamics techniques used to compute mergers and the large body of results that has been generated, in Sections 5 and 6, respectively. We pay particular attention to how numerical studies have taken steps toward answering a number of questions about the expected GW and electromagnetic (EM) emission from merging binaries, and we discuss briefly the possibility that they may be the progenitors of SGRBs and a source of r-process elements. We close with conclusions and a look to the future in Section 7. While most of this review focuses on NS-NS mergers, many of the methods used to study NS-NS binaries are also used to evolve BH-NS binaries, and it has become clear that both merger types may produce similar observational signatures as well. For a review focusing on BH-NS merger calculations, we encourage the reader to consult the recent work by Shibata and Taniguchi . Evolutionary Channels and Population Estimates Merging NS-NS and BH-NS binaries, i.e., those for which the merger timescale is smaller than the Hubble time, are typically formed through similar evolutionary channels in stellar field populations of galaxies (both may also be formed through dynamical processes in the high-density cores of some star clusters, but the overall populations are smaller and more poorly constrained; see for a review). It is difficult to describe the evolutionary pathways that form NS-NS binaries without discussing BH-NS binaries as well, and it is important to note that the joint distribution of parameters such as merger rates and component masses that we could derive from simultaneous GW and EM observations will constrain the underlying physics of binary stellar evolution much more tightly than observing either source alone. Population synthesis calculations for both merging NS-NS and BH-NS binaries typically favor the standard channel in which the first-born compact object goes through a common-envelope (CE) phase, although other models have been proposed, including recent ones where the progenitor binary is assumed to have very nearly-equal mass components that leave the main sequence and enter a CE phase prior to either undergoing a supernova [42, 54]. Simulations of this latter process have shown that close NS-NS systems could indeed be produced by twin giant stars with core masses ≳ 0.15 M⊙, though twin main sequence stars typically merge during the contact phase . In the standard channel (see, e.g., [44, 178], and Figure 1 for an illustration of the process), the progenitor system is a high-mass binary (with both stars of mass M ≳ 8–10 M⊙ to ensure a pair of supernovae). The more massive primary evolves over just a few million years before it leaves the main sequence, passes through its giant phase, and undergoes a Type Ib, Ic, or II supernova, leaving behind what will become the heavier compact object (CO): the BH in a BH-NS binary or the more massive NS in an NS-NS binary. The secondary then evolves off the main sequence in turn, triggering a CE phase when it reaches the giant phase and overflows its Roche lobe. Dynamical friction shrinks the binary separation dramatically, until sufficient energy is released to expel the envelope. Without this step, binaries would remain too wide to merge through the emission of GWs within a Hubble time. Eventually, the exposed, Helium-rich core of the secondary undergoes a supernova, either unbinding the system or leaving behind a tight binary, depending on the magnitude and orientation of the supernova kick. Figure 1. Open in a new tab Cartoon showing standard formation channels for close NS-NS binaries through binary stellar evolution. Image reproduced from . This evolutionary pathway has important effects on the physical parameters of NS-NS and BH-NS binaries, leading to preferred regions in phase space. The primary, which can accrete some matter during the CE phase, or during an episode of stable mass transfer from the companion Helium star, should be spun up to rapid rotation (see for a review). In NS-NS binaries, we expect that this process will also reduce the magnetic field of the primary down to levels seen in “recycled” pulsars, typically up to four orders of magnitude lower than for young pulsars [180, 73]. The secondary NS, which never undergoes accretion, is likely to spin down rapidly from its nascent value, but is likelier to maintain a stronger magnetic field. While this evolutionary scenario has been well studied for several decades, many aspects remain highly uncertain. In particular: The CE efficiency, which helps to determine the expected range of binary separations and the mass of the primary compact object after its accretion phase, remains very poorly constrained [229, 37, 78, 338]. If too much matter is accreted by the NS, it may undergo accretion-induced collapse to a BH , though the growing body of observed NS-NS systems does help place constraints on the allowed range of accretion-related parameters. The exact relation between a star’s initial mass and the eventual compact object mass is better understood, but significant theoretical uncertainties remain, and the relation is sensitive to the metallicity (often unknown), mainly through the effects of mass loss in stellar winds, and to the details of the explosion itself [336, 205]. The maximum allowed NS mass will affect whether the primary remains a NS or undergoes accretion-induced collapse to a BH; its value is dependent upon the as-yet undetermined nuclear matter EOS. At present, the strongest limit is set by the binary millisecond pulsar PSR J1614-2230, for which a mass of M NS = 1.97 ± 0.04 M⊙ was determined by Shapiro time delay measurements . Determining the NS EOS from GW observations may eventually provide stronger constraints [237, 221, 184], including a determination of whether supernova remnants are indeed classical hadronic NS or instead have cores consisting of some form of strange quark matter or other elementary particle condensate [223, 119, 230, 120, 23, 5]. The supernova kick velocity distribution is only partially understood, leading to uncertainties as to which systems will become unbound after the second explosion [133, 324, 219, 152]. Given all these uncertainties, it is reassuring that most estimates of the NS-NS and BH-NS merger rate, expressed either as a rate of mergers per Myr per “Milky Way equivalent galaxy” or as a predicted detection rate for LIGO (the Laser Interferometer Gravitational-Wave Observatory) and Virgo (see Section 5.5 below), agree to within 1–2 orders of magnitude, which is comparable to the typical uncertainties that remain once all possible sources of error are folded into a population synthesis model. In Table 1, we show the predicted detection rates of NS-NS and BH-NS mergers for both the first generation LIGO detectors (“LIGO”), which ran at essentially their design specifications , and the Advanced LIGO (“AdLIGO”) configuration due to go online in 2015 . We note that the methods used to generate these results varied widely. In , the authors used the observed parameters of close binary pulsar systems to estimate the Galactic NS-NS merger rate empirically (such results do not constrain the BH-NS merger rate). In [198, 128], the two groups independently estimated the binary merger rate from the observed statistics of SGRBs. In these cases, one does not get an independent prediction for the NS-NS and BH-NS merger rate, but rather some linear combination of the two. In both cases, the authors estimated that, if NS-NS and BH-NS mergers are roughly equal contributors to the observed SGRB sample, LIGO will detect about an order of magnitude more BH-NS mergers since their higher mass allows them to be seen over a much larger volume of the Universe. As they both noted, should either type of system dominate the SGRB sample, we would expect a doubling of LIGO detections for that class, and lose our ability to constrain the rate of the other using this method. Many population synthesis models have attempted to understand binary evolution within our galaxy by starting from a basic parameter survey of the various assumptions made about CE evolution, supernova kick distributions, and other free parameters. In [323, 79], population synthesis models are normalized by estimates of the star formation history of the Milky Way. In [140, 218], parameter choices are judged based on their ability to reproduce the observed Galactic binary pulsar sample, which allows posterior probabilities to be applied to each model in a Bayesian framework. A review by the LIGO collaboration of this issue may be found in . Table 1. Estimated initial and advanced LIGO rates for BH-NS and NS-NS mergers from population synthesis calculations and other methods. The methods used are, in order, empirical constraints from the observed sample of binary pulsars (‘Empirical’), constraints on the combined NS-NS/BH-NS merger rate assuming that they are the progenitors of short-hard gamma-ray bursts (‘SGRBs’), population synthesis models calibrated to the star formation rate in the Milky Way (‘Pop. Synth. — SFR’), and population synthesis calibrated against the observed Galactic binary pulsar sample (‘Pop. Synth. — NS-NS’). We note that observations of binary pulsars do not yield constraints for BH-NS binaries. SGRB observations may produce constraints on NS-NS merger rates, BH-NS, merger rates, or both, depending on which sources are the true progenitors, but this remains unclear. Therefore, the table quotes results assuming a roughly equal split between the two. The official review of these results and their implications by the LIGO/Virgo Scientific Collaborations may be found in . | Author | NS-NS | BH-NS | Method | --- --- | | | LIGO | AdLIGO | LIGO | AdLIGO | | | Kim et al. | 5e-3 | 27 | | | Empirical | | Nakar et al. | | ∼ 2 | | ∼ 20.0 | SGRBs | | Guetta & Stella | 7.0e-3 | 22 | 7.0e-2 | 220 | SGRBs | | Voss & Tauris | 6.0e-4 | 2.0 | 1.2e-3 | 4.0 | Pop. Synth. — SFR | | de Freitas Pacheco et al. | 8.0e-4 | 6.0 | | | Pop. Synth. — SFR | | Kalogera et al. | 1.0e-2 | 35 | 4.0e-3 | 20 | Pop. Synth. — NS-NS | | O’Shaughnessy et al. | 1.0e-2 | 10 | 1.0e-2 | 10 | Pop. Synth. — NS-NS | Open in a new tab Should the next generation of GW interferometers begin to detect a statistically significant number of merger events including NSs, it should be possible to constrain several astrophysical parameters describing binary evolution much more accurately. These include The relative numbers of BH-NS and NS-NS mergers: Interferometric detections are sensitive to a binary’s “chirp mass” (see Eq. 6), and to the binary mass ratio as well [57, 11, 321, 320] if the signal-to-noise ratio is sufficiently high. Even if the merger signal takes place at frequencies too high to fall within the LIGO band, it should still be possible in most cases to determine whether the primary’s mass exceeds the maximum mass of a NS. The mass ratio probability distribution for merging binary systems: If both binary components’ masses are determined, we will be able to constrain both the BH mass distribution in merging binaries and the NS mass ratio distribution. Knowledge of the former would determine, e.g., whether the current low estimates for the mass accreted onto the primary CO core during the CE phase are correct , as this model predicts that BH masses in close BH-NS binaries should cluster narrowly around M BH = 10 M⊙. Previous calculations assuming larger accreted masses typically favored mass ratios closer to unity, since the primaries often began as NSs and underwent “accretion-induced collapse” to a BH during the CE phase. The NS-NS mass distribution will allow for tests of whether “twins”, i.e., systems whose zero-age main sequence (or “ZAMS”) masses are so close that they both leave the main sequence before either undergoes a supernova, play a significant role in the merging NS-NS population . While Advanced LIGO or another interferometer will likely be required to make the first direct observations of NS-NS mergers and their immediate aftermath, it is possible that more than just the high-energy prompt emission from mergers may be observable using EM telescopes. Although the particular candidate source they identified resulted from a pointing error , Nakar and Piran suggest that the mass ejection from mergers should yield an observable radio afterglow , although the afterglows may be too faint to be seen by current telescopes at the observed distances of existing localized SGRBs . While such outbursts could also result from a supernova, the luminosity required would be an order of magnitude larger than those previously observed. Given the length and timescales characterizing radio bursts, no NS-NS simulation has been able to address the model directly, but it certainly seems plausible that the time-variable magnetic fields within a stable hypermassive remnant could generate enough EM energy to power the resulting radio burst . If mergers produce sufficiently large ejecta masses, M ej ≳ 10−3 M⊙, r-process nuclear reactions may produce a “kilonova” afterglow one day after a merger with a V-band optical luminosity ν Ł ν ≈ 10 41 erg/s (roughly 1000 times brighter than a classical nova) . These potential EM observations of mergers are likely to spur further research into the amount and velocity of merger ejecta, which could then be coupled to a larger-scale astrophysical simulation of the potential optical and radio afterglows. Stages of a Binary Merger The qualitative evolution of NS-NS mergers, or indeed any compact binary merger, has long been understood, and may be divided roughly into inspiral, merger, and ringdown phases, each of which presents a distinct set of challenges for numerical modeling and detection. As a visual aid, we include a cartoon summary in Figure 2, originally intended to describe black hole-black hole (BH-BH) mergers, and attributed to Kip Thorne. Drawn before the advent of the supercomputer simulations it envisions, we note that merger waveforms for all compact binary mergers have proven to be much smoother and simpler than shown here. To adapt it to NS-NS mergers, we note that NSs are generally assumed to be essentially non-spinning, and that the “ringdown” phase may describe either a newly formed BH or a NS that survives against gravitational collapse. Figure 2. Open in a new tab Cartoon picture of a compact binary coalescence, drawn for a BH-BH merger but applicable to NS-NS mergers as well (although NSs are generally assumed to be non-spinning). Image adapted from Kip Thorne. Summarizing the evolution of the system through the three phases: After a pair of supernovae yields a relatively tight NS-NS binary, the orbital separation decays over long timescales through GW emission, a phase that takes up virtually all of the lifetime of the binary except the last few milliseconds. During the inspiral phase, binary systems may be accurately described by QE formalisms, up until the point where the gravitational radiation timescale becomes comparable to the dynamical timescale. The evolution in time is well-described by PN expansions, currently including all terms to 3PN , though small deviations can arise because of finite-size effects, especially at smaller separations (see Eqs. 1–6 for the lowest-order 2.5PN expressions for circular inspirals). Once the binary separation becomes no more than a few times the radii of the two NSs, binaries rapidly become unstable. The stars plunge together, following the onset of dynamical instability, and enter the merger phase, requiring full GR simulations to understand the complicated hydrodynamics that ensues. According to all simulations to date, if the NS masses are nearly equal, the merger resembles a slow collision, while if the primary is substantially more massive than the secondary the latter will be tidally disrupted during the plunge and will essentially accrete onto the primary. Since the NSs are most likely irrotational just prior to merging, there is a substantial velocity discontinuity at the surface of contact, leading to rapid production of vortices. Meanwhile, some fraction of the mass may be lost through the outer Lagrange points of the system to form a disk around the central remnant. This phase yields the maximum GW amplitude predicted by numerical simulations, but with a signal much simpler and more quasi-periodic than in the original cartoon version. GW emission during the merger encodes important information about the NS EOS, particularly the GW frequency f cut at which the binary orbit becomes unstable (see Eq. 4) resulting in a characteristic cutoff in GW emission at those frequencies. Meanwhile, the merger itself may generate the thermal energy that eventually powers a SGRB, which occurs when the neutrinos and anti-neutrinos produced by shock-heated material annihilate around the remnant to produce high-energy photons. Finally, the system will eventually settle into a new, dynamically stable configuration through a phase of ringdown, with a particular form for the GW signal that depends on the remnant’s mass and rotational profile. If the remnant is massive enough, it will be gravitationally unstable and collapse promptly to form a spinning BH. Otherwise it must fall into one of three classes depending on its total mass. Should the remnant mass be less than the maximum mass M iso supported by the nuclear matter EOS for an isolated, non-rotating configuration, it will clearly remain stable forever. Instead a remnant that is “supramassive”, i.e., with a mass above the isolated stationary mass limit but below that allowed for a uniformly rotating NS (typically ≲ 1.2 M iso, with very weak dependence on the EOS; see, e.g., [147, 70, 71], and references therein) may become unstable. Supramassive remnants are stable against gravitational collapse unless angular momentum losses, either via pulsar-type emission or magnetic coupling to the outer disk, can drive the angular velocity below the critical value for stability. If the remnant has a mass above the supramassive limit, it may fall into the hypermassive regime, where it is supported against gravitational collapse by rapid differential rotation. Hypermassive NS (HMNS) remnants can have significantly larger masses, depending on the EOS (see, e.g., [31, 275, 86, 293, 114]), and will survive for timescales much longer than the dynamical time, undergoing a wide variety of oscillation modes. They can emit GWs should a triaxial configuration yield a significant quadrupole moment, and potentially eject mass into a disk around the remnant. Eventually, some combination of radiation reaction and magnetic and viscous dissipation will dampen the differential rotation and lead the HMNS to collapse to a spinning BH, again with the possibility that it may be surrounded by a massive disk that could eventually accrete. The energy release during HMNS collapse provides the possibility for a “delayed” SGRB, in which the peak GW emission occurs during the initial merger event, but the gamma-ray emission, powered by the collapse of the HMNS to a BH, occurs significantly later. Most calculations indicate that a geometrically thick, lower-density, gravitationally bound disk of material will surround whatever remnant is formed. Such disks, which are geometrically thick, are widely referred to as “tori” throughout the literature, though there is no clear distinction between the two terms, and we will use “disk” throughout this paper to describe generically the bound material outside a central merger remnant. Such disks are expected to heat up significantly, and much of the material will eventually accrete onto the central remnant, possibly yielding observable EM emission. Given the low densities and relatively axisymmetric configuration expected, disks are not significant GW emitters. There may be gravitationally unbound outflow from mergers as well, though dynamical simulations neither confirm nor deny this possibility yet. Such outflows, which can be the sites of exotic nuclear reactions, are frequently discussed in the context of r-process element formation, but their inherently low densities make them difficult phenomena to model numerically with high accuracy. Comparison to BH-NS mergers Since this three-phase picture is applicable to BH-NS mergers as well, it is worthwhile to compare the two merger processes at a qualitative level to understand the key similarities and differences. Inspiral for BH-NS mergers is also well-described by PN expansions up until shortly before the merger, but the parameter space is fundamentally different. First, since BHs are heavier than NSs, the dynamics can be quite different. Also, since BHs may be rapidly-spinning (i.e., have dimensionless spin angular momenta as large as J/M 2 ∼ 1), spin-orbit couplings can play a very important role in the orbital dynamics of the binary, imprinting a large number of oscillation modes into the GW signal (see, e.g., [126, 57]). From a practical standpoint, the onset of instability in BH-NS mergers should be easier to detect for LIGO, Virgo, and other interferometers, since the larger mass of BH-NS binaries implies that instability occurs at lower GW frequencies (see Eq. 4, noting that the separation a at which mass-transfer begins scales roughly proportionally to the BH mass). The onset of instability of a BH-NS binary is determined by the interplay of the binary mass-ratio, NS compactness, and BH spin, with the first of these playing the largest role (see Figures 13–15 of and the summary in ). In general, systems with high BH masses and/or more compact NSs tend to reach a minimum in the binding energy as the radius increases, leading to a dynamical orbital instability that occurs near the classical innermost stable circular orbit (ISCO). In these cases, the NS plunges toward the BH before the onset of tidal disruption, and is typically “swallowed whole”. leaving behind almost no material to form a disk. The GW emission from such systems is sharply curtailed after the merger event, yielding only a low-amplitude, high-frequency, rapidly-decaying “ringdown” signal (see, e.g., ). Numerical calculations have shown that even in borderline cases between dynamical instability and mass-shedding the NS is essentially swallowed whole, especially in cases where the BH in either non-spinning or spinning in the retrograde direction, which pushes the ISCO out to larger radii (see, e.g., [290, 289, 283, 91, 94]). Figure 13. Open in a new tab Evolution of the density in a NS-NS merger, with magnetic field lines superposed. The first panel shows the binary shortly after contact, while the second shows the short-lived HMNS remnant shortly before it collapses. In the latter two panels, a BH has already formed, and the disk around it winds up the magnetic field to a poloidal geometry of extremely large strength, ∼ 10 15 G, with an half-opening angle of 30°, consistent with theoretical SGRB models. Image reproduced by permission from Figure 1 of , copyright by AAS. Figure 15. Open in a new tab Effective strain at a distance of 100 Mpc shown as a function of the GW frequency (solid red curve) for the same four merger calculations depicted in Figure 14. Post-merger quasi-periodic oscillations are seen as broad peaks in the GW spectrum at frequencies f GW = 2–4 kHz. The blue curve shows the Taylor T4 result, which represents a particular method of deducing the signal from a 3PN evolution. The thick green dashed curve and orange dot-dashed curves depict the sensitivities of the second-generation Advanced LIGO and LCGT (Large Scale Cryogenic Gravitational Wave Telescope) detectors, respectively, while the maroon dashed curve shows the sensitivity of a hypothetical third-generation Einstein Telescope. Image reproduced by permission from Figures 5 and 6 of , copyright by APS. A richer set of phenomena occurs when the BH-NS mass ratio is closer to unity, the NS is less compact, the BH has a prograde spin direction, or more generally, some combination of those factors. In such cases, the NS will reach the mass-shedding limit prior to dynamical instability, and be tidally disrupted. Unlike what was described in semi-analytic Newtonian models (see, e.g., [68, 228, 77]) and seen in some early Newtonian and quasi-relativistic simulations (see, e.g., [165, 166, 138], stable mass transfer, in which angular momentum transfer via mass-shedding halts the inspiral, has never been seen in full GR calculations, nor even in approximate GR models (see the discussion in ). Even so, unstable mass transfer can produce a substantial disk around the BH, though in every GR simulation to data the substantial majority of the NS matter is accreted promptly by the BH (see for a detailed summary of all current results). The exact structure of the disk and its projected lifetime depend on the binary system parameters, with the binary mass ratio and spin both important in determining the disk mass and the BH spin orientation critical for determining both the disk’s vertical structure and its thermodynamic state given the shock heating that occurs during the NS disruption. In general, the mass and temperature of the post-merger disks are comparable to those seen in some NS-NS mergers, and inasmuch as either is a plausible SGRB progenitor candidate then both need to be viewed as such. To date, no calculation performed in full GR has found any bound and self-gravitating NS remnant left over after the merger, including both NS cores that survive the initial onset of mass transfer by recoiling outward (predicted for cases in which stable mass transfer was thought possible, as noted above) or those in which bound objects form through fragmentation of the circum-BH disk. Motivated by observations of wide-ranging timescales for X-ray flares in both long and short GRBs , the latter channel has been suggested to occur in collapsars and mentioned in the context of mergers , possibly on longer timescales than current simulations permit. Even so, there is no analogue to the HMNS state that may result from NS-NS mergers, nor any mechanism for a delayed SGRB as provided by HMNS collapse. Qualitative numerical results Constructing QE sequences for a given set of NS parameters requires sophisticated numerical schemes, but not supercomputer-scale resources, as we discuss in Section 4 below, focusing first on the numerical techniques used to construct QE binary data in GR, and the astrophysical information contained in the GW emission during the inspiral phase. Merger and ringdown, on the other hand, typically require large-scale numerical simulations, including some of the largest calculations performed at major supercomputer centers, as we discuss in detail Section 5 and 6 below. To illustrate the various physical processes that occur during NS-NS mergers, we show the evolution of three different NS-NS merger simulations in Figures 3, 4, and 5, taken from Figures 4–6 of . In Figure 3, we see the merger of two equal-mass NSs, each of mass M NS = 1.4 M⊙, described by the APR (Akmal-Pandharipande-Ravenhall) EOS . In the second panel, clear evidence of “tidal lags” is visible shortly after first contact, leading to a slightly off-center collision pattern. By the third panel, an ellipsoidal HMNS has been formed, surrounded by a disk of material of lower density, which gradually relaxes to form a more equilibrated HMNS by the final panel. In Figure 4, we see a merger of two slightly heavier equal-mass NSs with M NS = 1.5 M⊙. In this case, the deeper gravitational potential limits the amount of mass that goes into the disk, and once a BH is formed (with a horizon indicated by the dashed blue circle in the final panel) it accretes virtually all of the rest mass initially present in the two NSs, with only 0.004% of the total remaining outside the horizon. Figure 3. Open in a new tab Isodensity contours and velocity profile in the equatorial plane for a merger of two equal-mass NSs with M ns = 1.4 M⊙ assumed to follow the APR model for the NS EOS. The hypermassive merger remnant survives until the end of the numerical simulation. Image reproduced by permission from Figure 4 of , copyright by APS. Figure 4. Open in a new tab Isodensity contours and velocity profile in the equatorial plane for a merger of two equal-mass NSs with M ns = 1.5 M⊙ assumed to follow the APR model for the NS EOS. With a higher mass than the remnant shown in Figure 3, the remnant depicted here collapses promptly to form a BH, its horizon shown by the dashed blue circle, absorbing all but 0.004% of the total rest mass from the original system. Image reproduced by permission from Figure 5 of , copyright by APS. Figure 5. Open in a new tab Isodensity contours and velocity profile in the equatorial plane for a merger of two unequal-mass NSs with M 1 = 1.3 M⊙ and M 2 = 1.6 M⊙, with both assumed to follow the APR model for the NS EOS. In unequal-mass mergers, the lower mass NS is tidally disrupted during the merger, forming a disk-like structure around the heavier NS. In this case, the total mass of the remnant is sufficiently high for prompt collapse to a BH, but 0.85% of the total mass remains outside the BH horizon at the end of the simulation, which is substantially larger than for equal-mass mergers with prompt collapse (see Figure 4). Image reproduced by permission from Figure 6 of , copyright by APS. Figure 6. Open in a new tab Dimensionless binding energy E b/M 0 vs. dimensionless orbital frequency M 0 Ω, where M 0 is the total ADM (Arnowitt-Deser-Misner) mass of the two components at infinite separation, for two QE NS-NS sequences that assume a piecewise polytropic NS EOS. The equal-mass case assumes M NS = 1.35 M⊙ for both NSs, while the unequal-mass case assumes M 1 = 1.15 M⊙ and M 2 = 1.55 M⊙. The thick curves are the numerical results, while the thin curves show the results from the 3PN approximation. The lack of any minimum suggests that instability for these configurations occurs at the onset of mass shedding, and not through a secular orbital instability. Image reproduced by permission from Figure 16 of , copyright by AAS. In Figure 5, we see the merger of an unequal-mass binary, with masses M 1 = 1.3 M⊙ and M 2 = 1.6 M⊙. In this case, the heavier NS partially disrupts the lighter NS prior to merger, leading to the secondary NS being accreted onto the primary. In this case, a much more massive disk is formed and, even after a BH forms in the center of the remnant, a substantial amount of matter, representing 0.85% of the total mass, remains outside the horizon. Later accretion of this material could potentially release the energy required to power a SGRB. Initial Data and Quasi-Equilibrium Results Overview While dynamical calculations are required to understand the GW and EM emission from BH-NS and NS-NS mergers, some of the main qualitative features of the signals may be derived directly from QE sequences. From the variation of total system energy with binary angular velocity along a given sequence, it is possible to construct an approximate GW energy spectrum dE GW/df immediately from QE results, essentially by performing a numerical derivative (see Figure 6). Doing so for a number of different sequences makes it possible to identify key frequencies where tidal effects may become measurable and to identify these with binary parameters such as the system mass ratio and NS radius. Similarly, since QE sequences should indicate whether a binary begins to shed mass prior to passage through the ISCO (see Figure 7), one may be able to classify observed signals into mass-shedding and non-mass-shedding events, and to use the critical point dividing those cases to help constrain the NS EOS. Single-parameter estimates have been derived for NS-NS binaries using QE sequences (and for BH-NS binaries using QE and dynamical calculations ). NS-NS binaries typically approach instability at frequencies f GW ≳ 1 kHz, where laser shot noise is severely degrading the sensitivity of an interferometer detector. To observe ISCO-related effects with higher signal-to-noise, it may be necessary to operate GW observatories using narrow-band signal recycling modes, in which the sensitivity in a narrow range of frequencies is enhanced at the cost of lower sensitivity to broadband signals . Figure 7. Open in a new tab Mass-shedding indicator vs. orbital frequency M 0 Ω, where h is the fluid enthalpy and the derivative is measured at the NS surface in the equatorial plane toward the companion and toward the pole in the direction of the angular momentum vector, for a series of QE NS-NS sequences assuming equal-mass components. Here, χ = 1 corresponds to a spherical NS, while χ = 0 indicates the onset of mass shedding. More massive NSs are more compact, and thus able to reach smaller separations and higher angular frequencies before mass shedding gets underway. Image reproduced by permission from Figure 19 of , copyright by AAS. It is important to note that, while the potential parameter space for NS EOS models is still very large, a much smaller set may serve to classify models for comparison with the first generation of GW detections. Indeed, by breaking up the EOS into piecewise polytropic segments, one may use as few as four parameters to roughly approximate all known EOS models, including standard nuclear models as well as models with kaon or other condensates . To illustrate this, we show in Figure 8 four different QE models for NS-NS configurations with different EOS, taken from ; all have M 1 = 1.15 M⊙ and M 2 = 1.55 M⊙, and they correspond to the closest separation for which the QE code still finds a convergent result. Figure 8. Open in a new tab Isodensity contours for QE models of NS-NS binaries. In each case, the two NSs have masses M 1 = 1.15 M⊙ (left) and M 2 = 1.55 M⊙ (right), and the center-of-mass separation is as small as the QE numerical methods allow while able to find a convergent result. The models assume different EOS, resulting in different central concentrations and tidal deformations. Image reproduced by permission from Figures 9–12 of , copyright by AAS. The inspiral of NS-NS binaries may yield complementary information about the NS structure beyond what can be gleaned from QE studies of tidal disruption. NSs have a wide variety of oscillation modes, including f-modes, g-modes, and r-modes, any of which may be excited by resonances with the orbital frequency as the latter sweeps upward. Should a particular oscillation mode be excited resonantly, it can then serve briefly as an energy sink for the system, potentially changing the phase evolution of the binary. For example, in a rapidly spinning NS, excitation of the m =1 r-mode can be significant, yielding a change of over 100 radians for the pre-merger GW signal phase in the case of a millisecond spin period . For NS-NS mergers in the field, this would require one of the NSs to be a young pulsar that has not yet spun down significantly, which is unlikely because of the difficulty in obtaining such an extremely small binary separation after the second supernova. Other modes, such as the l = 2 f-mode, may be excited in less extreme circumstances, also yielding information about NS structure parameters . Quasi-equilibrium formalisms It has long been known that the GW emission from eccentric binaries is very efficient at radiating away angular momentum relative to the radiated energy ; as a result, the orbital eccentricity decreases as a binary inspirals, so that orbits should be very nearly circular long before they enter the detection range of ground-based interferometers. The only exception could be from a dynamical capture process that would create a binary with a significant eccentricity and very small orbital separation. Such eccentric binaries have been predicted to form in the nuclear cluster of our galaxy (see, e.g., ) and in core-collapsed globular clusters [127, 167]. However, at present, no formalism exists to construct initial data for such systems, besides superposing the individual components with sufficiently large initial separations to minimize constraint violations. Using this circularity of primordial binaries as a starting point, one may use the constraint equations of GR, along with an assumption of quasi-circularity, to derive sets of elliptic equations describing compact binary configurations. For both QE and dynamical calculations, most groups typically make use of the Arnowitt-Deser-Misner (ADM) 3+1 splitting of the metric , which foliates the metric into a set of three-dimensional hypersurfaces by introducing a time coordinate. The resulting form of the metric, which is completely general, is written 7 where α is known as the lapse function, β i the shift vector, and γ ij the spatial three-metric intrinsic to the hypersurface. We are following the standard relativistic notation here where Greek indices correspond to four-dimensional quantities and Latin indices to spatial three-dimensional quantities. Thus, the shift vector is a 3-vector, raised and lowered with the spatial 3-metric γ ij rather than the spacetime 4-metric g μν. To simplify matters, one typically defines a conformal factor ψ that factors out the determinant of the 3-metric, such that 8 introducing the conformal 3-metric with unit determinant. While the 3-metric is a fundamental component of the geometric structure of the spacetime, the lapse function and shift vector are gauge quantities that simply reflect our choice of coordinates. Thus, while one often determines the lapse and shift in order to construct a appropriately “stationary” solution in the relevant coordinates between neighboring time slices, their values are often replaced to initialize dynamical runs with more convenient choices and thus different assumptions about coordinate evolution in time. The field equations of general relativity take the deceptively simple form 9 where G μν is the Einstein tensor, R μν and R the Ricci curvature tensor and the curvature scalar, and T μν the stress-energy tensor that accounts for the presence of matter, electromagnetic fields, and other physical effects that contribute to the mass-energy of the spacetime. Since GR is a second-order formulation, valid initial data must include not only the metric but also its first time derivative. It generally proves most convenient to introduce the time derivative of the metric after subtracting away the Lie derivative with respect to the shift, yielding a quantity known as the extrinsic curvature, K ij: 10 Both the 3-metric and extrinsic curvature are symmetric tensors with six free parameters. For systems containing NSs, one must consider the effects of nuclear matter through its presence in the stress-energy tensor T μν. It is common to assume that the matter has the EOS describing a perfect, isotropic fluid, for which the stress energy tensor is given by 11 where ρ 0, ε, P and u μ are the fluid’s rest-mass density, specific internal energy, pressure, and 4-velocity, respectively. Many calculations further assume that the NS EOS is described by an adiabatic polytrope, for which 12 where Γ is the adiabatic index of the gas and k a constant, though a number of models designed to incorporate nuclear physics and/or strange matter condensates have also been constructed and studied (see Sections 4.4 and 6 below). The problem in constructing initial data is not so much producing solutions that are self-consistent within GR, but rather to specify a sufficient number of assumptions to fully constrain a solution. Indeed, there are only four constraints imposed by the equations of GR, known as the Hamiltonian and momentum constraints. The Hamiltonian constraint is found by projecting Einstein’s equations twice along the direction defined by a normal observer, and describes the way stress-energy leads to curvature in the metric (see, e.g., for a thorough review): 13 where R is the scalar curvature of the 4-metric, is the trace of the extrinsic curvature, and 14 is the total energy density seen by a normal observer. The third term indicates that the total energy density is found by projecting the stress-energy tensor in the direction of the unit-length timelike normal vector n, whose components are given by 15 In the final expression h ≡ 1 + ε + P/ρ 0 is the specific enthalpy of the fluid, and the combination Γ n ≡ α u 0 represents the Lorentz factor of the matter seen by an inertial observer. The notation here makes use of the standard summation convention, in which repeated indices are summed over. Projecting Einstein’s equations in the space and time directions leads to the vectorial momentum constraint 16 where D i represents a three-dimensional covariant derivative and j i ≡ ρ 0 h Γ n u i is the total momentum seen by a normal observer. The Conformal Thin Sandwich formalism In order to specify all the free variables that remain once the Hamiltonian and momentum constraints are satisfied, two different techniques have been widely employed throughout the numerical relativity community. One, known as the Conformal Transverse-Traceless (CTT) decomposition, underlies the Bowen-York solution for black holes with known spin and/or linear momentum that is widely used in the “moving puncture” approach. To date, however, the CTT formalism has not been used to generate NS-NS initial data, and we refer readers to [284, 63] for descriptions of the CTT formalism applied to BH-NS and BH-BH initial data, respectively. To date, most groups have used the Conformal Thin Sandwich (CTS) formalism to generate QE NS-NS data (see for a review, [13, 137] for the initial steps in the formulation, and [326, 327, 333, 69] for derivations of the form in which it is typically used today). One first specifies that the conformal 3-metric is spatially flat, i.e., , where δ ij is the Kronecker delta function. Under this assumption, the only remaining parameter defining the spatial metric is the conformal factor ψ, which serves the role of a gravitational potential. Indeed, in the limit of weak sources, it is linearly related to the standard Newtonian potential. Next, one specifies that there exists a helical Killing vector, so that, as the configuration advances forward in time, all quantities remain unchanged when properly rotated at constant angular velocity in the azimuthal direction. This is sufficient to fix all but the trace of the extrinsic curvature, with the other components forced to satisfy the relation 17 The trace of the extrinsic curvature K remains a free parameter in this approach. While one may choose arbitrary prescriptions to fix it, most implementations choose a maximal slicing of the spatial hypersurfaces by setting K = ∂ t K = 0. Under these assumptions the Hamiltonian and momentum constraints, along with the trace of the Einstein equations, yield five elliptic equations for the lapse, shift vector, and conformal factor: 18 19 20 where 21 is the trace of the stress-energy tensor projected twice in the spatial direction. While these five equations are linked and the right-hand sides are nonlinear, they are amenable to solution using iterative methods. Boundary conditions are set by assuming asymptotic flatness: at large radii, the metric takes on the Minkowski form so α → 1, ψ → 1, and . We note that a purely corotating shift term yields zero when we apply the left-hand side of Eq. 20, so we may subtract it away and solve the equation with a boundary condition of zero instead. The breakdown in Eqs. 18, 19, and 20 is not unique. The Meudon group [125, 124], to pick one example, has often chosen to define ν ≡ ln α and β ≡ ln (αψ 2 ), and replace Eqs. 18 and 19 with the equivalent pair This approach is sufficient to define the field component of the configuration, but one still needs to solve for the matter quantities as well. One starts by assuming that there is a known prescription for reconstructing the density, internal energy, and pressure from the enthalpy h. Next, one has to assume some model for the spin of the NS. While corotation is often a simpler choice, since the velocity field of the matter is zero in the corotating frame, the more physically reasonable condition is irrotational flow. Indeed a realistic NS viscosity is never sufficiently large to tidally lock the NS to its companion during inspiral [45, 146]. If we define the co-momentum vector w i = hu i, irrotational flow implies the vanishing of its curl: 22 which allows us to define a velocity potential Ψ such that w ≡ ∇Ψ. Using these quantities, one may write down the integrated Euler equation 23 where the 3-velocity U i of the fluid with respect to an Eulerian observer is given by 24 and the orbital 3-velocity with respect to the same observer by . For details on the ways in which one may construct an elliptic equation for the velocity potential, we refer to the derivation in . To date, all QE sequences and dynamical runs in the literature have assumed that NSs are either irrotational or synchronized, but it is possible to construct the equations for arbitrary NS spins so long as they are aligned [29, 309]. While suggestions are also given there on how to construct QE sequences with intermediate spins using the new formalism, none have yet appeared in the literature. Similarly, a formalism to add magnetic fields self-consistently to QE sequences has been constructed , as current dynamical simulations typically begin from data assuming either zero magnetic fields or those that only contribute via magnetic pressure. Other formalisms The primary drawback of the CTS system is the lack of generality in assuming the spatial metric to be conformally flat, which introduces several problems. The Kerr metric, for example, is known not to be conformally flat, and conformally flat attempts to model Kerr BHs inevitably include spurious GW content. The same problem affects binary initial data: in order to achieve a configuration that is instantaneously time-symmetric, one actually introduces spurious gravitational radiation into the system, which can affect both the measured parameters of the initial system as well as any resulting evolution. Other numerical formalisms to specify initial data configurations in GR have been derived using different assumptions. Usui and collaborators derived an elliptic set of equations by allowing the azimuthal component of the 3-metric to independently vary from the radial and longitudinal components [319, 318], finding good agreement with the other methods discussed above. A number of techniques have been developed to construct helically symmetric spacetimes in which one actually solves Einstein’s equations to evaluate the non-conformally-flat component of the metric, which are typically referred to as “waveless” or “WL” formalisms [260, 50, 291]. In terms of the fundamental variables, rather than specifying the components of the conformal spatial metric by ansatz, one specifies instead the time derivative of the extrinsic curvature using a physically motivated prescription. These methods are designed to match the proper asymptotic behavior of the metric at large distances, and may be combined with techniques designed to enforce helical symmetry of the metric and gauge in the near zone the near zone helical symmetry, or “NHS” formalism) to produce a global solution [315, 334, 316]. QE sequences generated using this formalism have shown that the resulting conformal metric is indeed non-flat, with deviations of approximately 1% for the metric components, and similar differences in the system’s binding energy when compared to equivalent CTS results. They suggest that underestimates in the quadrupole deformations of NS prior to merger may result in total phase accumulation errors of a full cycle, especially for more compact NS models. QE formalisms reflect the assumption that binaries will be very nearly circular, since GW emission acting over very long timescales damps orbital eccentricity to negligible values for primordial NS-NS binaries between their formation and final merger. Binaries formed by tidal capture and other dynamical processes, which may be created with much smaller initial separations, are more likely to maintain significant eccentricities all the way to merger (see, e.g., for a discussion of such processes for BH-BH binaries) and it has been suggested based on simple analytical models that such mergers, likely occurring in or near dense star clusters, may account for a significant fraction of the observed SGRB sample . However, more detailed modeling is required to work out accurate estimates of merger rates given the complex interplay between dynamics and binary star evolution that determines the evolution of dense star clusters, and given the large uncertainties in the distributions of star cluster properties in galaxies throughout the universe. No initial data have ever been constructed in full GR for merging NS-NS binaries with eccentric orbits since the systems are then highly time-dependent, while the calculations performed to evolve them generally use a superposition of two stationary NS configurations . Numerical implementations There are a number of numerical techniques that have been used to solve these elliptic systems. The first calculations of NS-NS QE sequences, in both cases for synchronized binaries, were performed by Wilson, Mathews, and Marronetti [327, 328, 189] and Baumgarte et al. [25, 26]. The former used a finite differencing scheme, and centered different quantities at cells, vertices, and faces in order to construct a system of equations that was solved using fast matrix inversion techniques, while the latter used a Cartesian multigrid scheme, restricted to an octant to increase computational efficiency. After a formalism for evaluating QE irrotational NS-NS sequences was developed [51, 313], some of the first results were obtained by Uryū and Eriguchi, who developed a finite-differencing code in spherical coordinates allowing for the solution of relativistic NS-NS binaries using Green’s functions[313, 317]. Their method extended the self-consistent field (SCF) work of , which had previously been applied to axisymmetric configurations. Irrotational configurations were also generated by Marronetti et al. , using the same finite difference scheme as found in the work on synchronized binaries. The most widely used direct grid-based solver in numerical relativity is the Bam_Elliptic solver , which solves elliptic equations on single rectangular grids or multigrid configurations. It is included within the Cactus code, which is widely used in 3-D numerical relativity . In particular it has been used to initiate a number of single and binary BH simulations, including one of the original breakthrough binary puncture works . Lagrangian methods, typically based on smoothed particle hydrodynamics (SPH) [181, 118, 194]) have been used to generate both synchronized and irrotational configurations for PN [10, 99, 101, 100] and conformally flat (CF) [211, 210, 97, 212, 207, 209, 208, 34, 35, 33] calculations of NS-NS mergers, but they have not yet been extended to fully GR calculations, in part because of the difficulties in evolving the global spacetime metric. The most widely used data for numerical calculations are those generated by the Meudon group (see Section 4.4 below for details on their calculations and for a detailed description of their methods). The code they developed, Lorene , uses multidomain spectral methods to solve elliptic equations (while the code has been used primarily for relativistic stellar and binary configurations, it can be used as a more general solver). Around each star, one creates a set of nested, contiguous grids, with points arrayed in the radial and angular directions. The innermost grid has spheroidal geometry, and the surrounding grids are annular. The outermost grid may be allowed to extend to spatial infinity through a compactification transformation of the radial coordinate. To solve elliptic equations for various field quantities, one breaks each into a sum of two components, each of whose source terms are concentrated in one NS or the other. Similarly, the source terms themselves are split into two pieces, ideally, so each component is well-described by spheroidal spectral coefficients centered around each star. Using the spectral expansion, one may pass values from one star to the other and then recalculate spectral coefficients for the other grid configuration. This scheme has several efficiency advantages over direct grid-based methods, which helps to explain its popularity. First, the domain geometry may be chosen to fit to a NS surface, which eliminates Gibbs phenomenon-related errors and allows for exponential convergence with respect to the number of grid points, rather than the geometric convergence that characterizes finite difference-based grid codes. Second, the use of spectral methods requires much less computer memory than grid-based codes, and, as a result, Lorene is a serial code that can run easily on any off-the-shelf PC, rather than requiring a supercomputer platform. Quasi-equilibrium and pre-merger simulations NS-NS binaries may be well approximated by QE configurations up until they reach separations comparable to the sizes of the binary components themselves, that latter phase lasting a fraction of a second after an inspiral of millions of years or more. The eventual merger will occur after the binary undergoes one of two possible orbital instabilities. If the total binary energy and angular momenta reach a minimum at some separation, which defines the ISCO, the binary becomes dynamically unstable and plunges toward merger. Alternately, if the NS fills its Roche lobe (typically the lower density NS) mass will transfer onto the primary and the secondary will be tidally disrupted. The parameters of some NS-NS systems could technically allow for stable mass transfer, in which mass loss from a lighter object to a heavier one leads to a widening of the binary separation. This does occur for some binaries containing white dwarfs, but every dynamical calculation to date using full GR or even approximate GR has found that the rapid inspiral rate leads to inevitably unstable mass transfer and the prompt merger of a binary. Many of the results later confirmed using relativistic QE sequences were originally derived in Newtonian and PN calculations, particularly as explicit extensions of Chandrasekhar’s body of work (see ). Chandrasekhar’s studies of incompressible fluids were first extended to compressible binaries by Lai, Rasio, and Shapiro [156, 155, 158, 157, 159], who used an energy variational method with an ellipsoidal treatment for polytropic NSs. They established, among other results, the magnitude of the rapid inspiral velocity near the dynamical stability limit , the existence of a critical polytropic index (n ≈ 2) separating binary sequences undergoing the two different terminal instabilities , the role played by the NS spin and viscosity and magnitude of finite-size effects in relation to 1PN terms [158, 157], and the development of tidal lag angles as the binary approaches merger . They also determined that for most reasonable EOS models and nonextreme mass ratios, as would pertain to NS-NS mergers, an energy minimum is inevitably reached before the onset of mass transfer through Roche lobe overflow. The general results found in those works were later confirmed by , who used a SCF technique [131, 132], finding similar locations for instability points as a function of the adiabatic index of polytropes, but a small positive offset in the radius at which instability occurred. Similar results were also found by [311, 312], but with a slight modification in the total system energy and decrease in the orbital frequency at the onset of instability. The first PN ellipsoidal treatments were developed by Shibata and collaborators using self-consistent fields [270, 269, 279, 281, 299] and by Lombardi, Rasio, and Shapiro . Both groups found that the nonlinear gravitational effects imply a decrease in the orbital separation (increase in the orbital frequency) at the instability point for more compact NS. This result reflects a fairly universal principle in relativistic binary simulations: as gravitational formalisms incorporate more relativistic effects, moving from Newtonian gravity to 1PN and on to CF approximations and finally full GR, the strength of the gravitational interaction inevitably becomes stronger. The effects seen in fully dynamical calculations will be discussed in Section 6, below. The first fully relativistic CTS QE data for synchronized NS-NS binaries were constructed by Baumgarte et al. [26, 25], using a grid-based elliptic solver. Their results demonstrated that the maximum allowed mass of NSs in close binaries was larger than that of isolated NSs with the same (polytropic) EOS, clearly disfavoring the “star-crushing” scenario that had been suggested by [327, 187] using a similar CTS formalism (but see also the error in these latter works addressed in , discussed in Section 6.3 below). Baumgarte et al. also identified how varying the NS radius affects the ISCO frequency, and thus might be constrained by GW observations. Using a multigrid method, Miller et al. showed that while conformal flatness remained valid until relatively near the ISCO, the assumption of syncronized rotation broke down much earlier. Usui et al. used the Green’s function approach with a slightly different formalism to compute relativistic sequences and determined that the CTS conditions were valid up until extremely relativistic binaries were considered. The first relativistic models of physically realistic irrotational NS-NS binaries were constructed by the Meudon group using the Lorene multi-domain pseudo-spectral method code. Since then, the Meudon group and collaborators have constructed a wide array of NS-NS initial data, including polytropic NS models [125, 303, 304], as well as physically motivated NS EOS models or quark matter condensates . Irrotational models have also been constructed by Uryū and collaborators [313, 317] for use in dynamical calculations, and nuclear/quark matter configurations have been generated by Oechslin and collaborators [212, 209]. A large compilation of QE CTS sequences constructed using physically motivated EOS models including FPS (Friedman-Pandharipande) , SLy (Skyrme Lyon) , and APR models, along with piecewise polytropes designed to model more general potential cases (see ), was published in . The most extensive set of results calculated using the waveless/near-zone helical symmetry condition appear in , with equal-mass NS-NS binary models constructed for the FPS, Sly, and APR EOS in addition to Γ = 3 polytropes. Results spanning all of these QE techniques are summarized in Table 2. Table 2. A summary of various studies focusing on QE sequences of NS-NS binaries. Please refer to Section 6 for a discussion of papers that focus on dynamical simulations instead. Gravitational schemes include Newtonian gravity (‘Newt.’), lowest-order post-Newtonian theory (‘PN’), conformal thin sandwich (‘CTS’) including modified forms of the spatial metric (‘Mod. CTS’), and waveless/near-zone helical symmetry techniques. Numerical methods include ellipsoidal formalisms (‘Ellips.’), self-consistent fields (‘SCF’), numerical grids (‘Grid’), multigrids, and multipatch, Green’s function techniques (‘Green’s’), spectral methods (‘Spectral’), or SPH relaxation (‘SPH’). With regard to EOS models, ‘WD’ refers to the exact white dwarf EOS assuming a cold degenerate electron gas . The ‘Physical’ EOS models include the FPS , SLy , and APR nuclear EOS models, along with their parameterized approximations and other physically motivated models. The compactness refers to the value for a NS in isolation before it is placed in a binary, and plays no role in Newtonian physics. The mass ratio q = M 2/M 1 is defined to be less than unity, and ‘spin’ refers to either synchronized or irrotational configurations. | Author | Ref. | Grav. | Method | EOS | Compact ness | Mass ratio | Spin | --- --- --- --- | | Lai | | Newt. | Ellips. | | N/A | 1.0 | Syn. | | Lai | | Newt. | Ellips. | | N/A | 1.0 | Syn./Irr. | | Lai | | Newt. | Ellips. | | N/A | 0.2–1.0 | Syn./Irr. | | New | | Newt. | SCF | | N/A | 1.0 | Syn. | | Uryū | | Newt. | SCF | | N/A | 1.0 | Irr. | | Shibata | | PN | Grid | Γ = 2 | | 1.0 | Syn. | | Shibata | | PN | Grid | Γ = 3 | | 1.0 | Syn. | | Shibata | | PN | Ellips. | | | 1.0 | Syn. | | Lombardi | | PN | Ellips. | Γ = 2,3 | | 1.0 | Syn./Irr. | | Baumgarte | | CTS | Multigrid | Γ = 2 | | 1.0 | Syn. | | Usui | | Mod. CTS | Green’s | Γ = 2,3,∞ | | 1.0 | Syn. | | Uryū | | CTS | Green’s | Γ = 2 | | 1.0 | Syn./Irr. | | Uryū | | CTS | Green’s | | | 1.0 | Irr. | | Bonazzola | | CTS | Spectral | Γ = 2 | | 1.0 | Syn/Irr. | | Taniguchi | | CTS | Spectral | Γ = 2 | | 0.9–1.0 | Syn./Irr. | | Taniguchi | | CTS | Spectral | Γ = 1.8,2.25,2.5 | | 0.83–1.0 | Syn./Irr. | | Miller | | CTS | Multigrid | Γ = 2 | | 1.0 | Syn. | | Bejger | | CTS | Spectral | Phys. | | 1.0 | Irr. | | Limousin | | CTS | Spectral | Quark | | 1.0 | Syn./Irr. | | Oechslin | | CTS | SPH | Quark | | 1.0 | Irr. | | Taniguchi | | CTS | Spectral | Physical | | 0.7–1.0 | Irr. | | Uryū | | WL/NHS | Multipatch | Γ = 3, Physical | | 1.0 | Irr. | Open in a new tab Dynamical Calculations: Numerical Techniques Overview: General relativistic (magneto-)hydrodynamics and microphysical treatments NS-NS binaries are highly relativistic systems, numerous groups now run codes that evolve both GR metric fields and fluids self-consistently, with some groups also incorporating an ideal magnetohydrodynamic evolution scheme that assumes infinite conductivity. The codes that evolve the GR hydrodynamics or magnetohydrodynamics (GRHD and GRMHD, respectively) equations are many and varied, incorporating different spatial meshes, relativistic formalisms, and numerical techniques, and we will summarize the leading variants here. All full GR codes now make use of the significant insight gained from BH-BH merger calculations, but much work on these systems predates the three 2005 “breakthrough” papers by Pretorius , Goddard , and the group then at UT Brownsville (now at RIT) , with the first successful NS-NS merger calculations announced already in 1999 . A list of the groups that have performed NS-NS merger calculations using full GR is presented below; note that many of these groups have also performed BH-NS simulations, as discussed in the review by Shibata and Taniguchi . Of the full GR codes used to evolve NS-NS binaries, almost all are grid-based and make use of some form of adaptive mesh refinement. The one exception is the SpEC code developed by the SXS collaboration, formed originally by Caltech and Cornell, which has used a hybrid spectral-method field solver with grid-based hydrodynamics. Most make use of the BSSN formalism for evolving Einstein’s equations (see Section 5.2.1 below), while the HAD code uses the alternate Generalized Harmonic Gauge (GHG) approach. This technique is also used by the SXS collaboration and the Princeton group, who have both performed simulations of merging BH-NS binaries (see Section 6.6) but have yet to report any results on NS-NS mergers. Three groups have reported results for NS-NS mergers including MHD (HAD, Whisky, and UIUC), while the KT (Kyoto/Tokyo) group has reported magnetized evolutions of HMNS remnants (see and references therein for a discussion of their work and that of other numerical relativity groups), but have yet to use that code for a NS-NS merger calculation. While full GR codes were being developed to study NS-NS binaries, a parallel and rather independent track developed to study detailed microphysical effects in binary mergers using approximate relativistic schemes. This includes codes like that developed by the MPG group that accurately track the production of neutrinos and antineutrinos and their annihilation during a merger, as well as post-processing routines that use extensive nuclear chains to track the production of rare high-atomic number r-process elements in merger ejecta . Meanwhile, the Bremen group’s SPH code includes variable-temperature physically motivated equations of state and magnetohydrodynamics , and has been used with a multi-group flux-limited diffusion neutrino code to generate expected neutrino signatures from merger calculations . A summary of groups performing NS-NS merger calculations is presented in Table 3. Table 3. A summary of groups reporting NS-NS merger calculation results. The asterisk for the KT collaboration’s MHD column indicates that they have used an MHD-based code for other projects, but not yet for NS-NS merger simulations. Gravitational formalisms include full GR, assumed to be implemented using the BSSN decomposition except for the HAD collaborations’s GHG approach, the CF approximation, or Newtonian gravity. Microphysical treatments include physically motivated EOS models or quark-matter EOS and neutrino leakage schemes. | Abbrev. | Refs. | Grav. | MHD | Microphysics | --- --- | KT | [134, 144, 145, 265, 264, 287, 288] [285, 286, 282, 332] | GR | | Phys. EOS, ν-leakage | | HAD | [7, 6] | GR (GHG) | Y | N | | Whisky | [17, 18, 14, 15, 116, 117, 240, 241] | GR | Y | N | | UIUC | | GR | Y | N | | Jena | | GR | N | N | | MPG | [34, 35, 33, 32, 123, 209, 208, 296] | CF | N | Quark, Phys. EOS, ν-leakage | | Bremen | [82, 233, 247] | Newt | Y | Phys. EOS, ν-leakage | Open in a new tab GR numerical techniques GR formalisms and gauge choice There are two distinct schemes used in all binary merger calculations performed to date, the BSSN (Baumgarte-Shapiro-Shibata-Nakamura) [277, 27] and Generalized Harmonic formalisms. For general reviews of these formalisms, as well as other developments in numerical relativity, we refer readers to two recent books on numerical relativity [4, 30]. Here we only briefly summarize these two schemes. The BSSN formalism was adapted from the 3+1 ADM approach, with quantities defined as in Eqs. 7 and 8. While the original ADM scheme proved to be numerically unstable, defining the auxiliary quantities and treating these expressions as independent variables stabilized the system and allowed for long-term evolutions. While slight variants exist, the 19 evolved variables are typically either the conformal factor ψ or its logarithm ϕ, the conformal 3-metric, the trace K of the extrinsic curvature, the trace free extrinsic curvature A ij and the conformal connection functions The evolution equations themselves are given in Appendix A. The BSSN scheme was used in the binary merger calculations of the KT collaboration [287, 288, 285, 286], the first completely successful NS-NS calculations ever performed in full GR. Ever since the UTB/RIT and Goddard groups showed simultaneously that a careful choice of gauge allows BHs to be evolved in BSSN schemes without the need to excise the singularity, these “puncture gauges” have gained widespread hold, and have been used to evolve NS-NS binaries (and in some cases, BH-NS binaries) by the KT collaboration , UIUC , and Whisky . The generalized harmonic formalism, developed over about two decades from initial theoretical suggestions up to its current numerical implementation [112, 115, 232, 130, 231] was used to perform the first calculations of merging BH-BH binaries by Pretorius , and has since been applied to NS-NS binaries by the HAD collaboration and to BH-NS mergers by HAD , SXS [85, 84, 108], and the Princeton group [294, 88]. It involves introducing a set of auxiliary quantities denoted H μ representing the action of the wave operator on the spacetime coordinates themselves 25 which are treated as independent gauge variables whose evolution equation must be specified. Current first-order formulations [171, 6] evolve equations for the spacetime metric g μν along with its spatial derivative, Φ iμν = ∂ i g μν and projected time derivative Π μν = −n α ∂ α g μν, subject to consistency constraints on the spatial derivatives. The first BH-NS merger calculations in the GH formalism used a first-order reduction of the Einstein equations and specified the source functions to damp to zero exponentially in time, while the first binary NS merger work used a similar first-order reduction and chose harmonic coordinates with H μ = 0. In both formalisms, most groups employ grid-based finite differencing to evaluate spatial derivatives. While finite differencing operators may be easily written down to arbitrary orders of accuracy, there is a trade-off between the computational efficiency achievable by using smaller, second-order stencils and the higher accuracy that can be attained using larger, higher-order ones. For the moment, many groups are now moving to at least fourth-order accurate differencing techniques, with a high likelihood that at least the field sector of NS-NS merger calculations will soon be performed at comparable order to BH-BH calculations, at either sixth or eighth-order accuracy, if not higher. The main limitation to date involves the complexity of shock capturing using higher-order schemes, as we discuss in Section 5.2.3 below. Imposing physically realistic and accurate boundary conditions remains a difficult task for numerical codes. Ideally, one wishes to impose boundary conditions at large distances that preserve the GR constraints and yield a well-posed initial-boundary value problem. On physical grounds, the boundary should only permit outgoing waves, preventing the reflection of spurious waves back into the numerical grid. Otherwise, reflections may be avoided by placing the outer boundaries so far away from the matter that they remain causally disconnected from the merging objects for the full duration of the simulation. Building upon previous work (see, e.g., [113, 151, 12, 150, 256, 242], Winicour derived boundary conditions that satisfy all of the desired conditions for the generalized harmonic formalism. No such conditions have been derived for the full BSSN formalism, though progress has been made (see, e.g., [43, 129]) so that we may now construct well-posed boundary conditions in the weak-gravity linearized limit of BSSN and for related first-order gravitational formalisms . Grid decompositions While unigrid schemes are extremely convenient, they tend to be extremely inefficient, since one must resolve small-scale features in the central regions of a merger but also extend grids out to the point where the GW signal has taken on its roughly asymptotic form. Thus, nearly every code incorporates some means of focusing resolution on the high-density regions via some form of mesh refinement. A simple approach, for instance, is to use “fisheye” coordinates that represent a continuous radial deformation of a grid , a technique that had previously been used successfully, e.g., for BH-BH mergers [21, 62]. While fixed mesh refinement offers the chance for greater computational efficiency and accuracy, much current work focuses on adaptive mesh refinement, in which the level of refinement of the grid is allowed to evolve dynamically to react to the evolving binary configuration. Several codes, both public and private, now implement this functionality. The publicly available Carpet computational toolkit [263, 262], which is distributed to the community as part of the open source Einstein [90, 175] uses a “moving boxes” approach, and has been designed to be compatible with the widely used and publicly available Cactus framework. It has been successfully implemented by the Whisky group to perform NS-NS mergers, by UIUC for their BH-NS mergers , and a host of other groups for BH-BH mergers and additional problems. The KT code SACRA also implements an adaptive mesh refinement (AMR) scheme for NS-NS and BH-NS mergers , as does the most recent version of the HAD collaboration’s code , which is based on the publicly available infrastructure of the same name [169, 8], and the BAM code employed by the Jena group [308, 41, 122]. The Princeton group also has an AMR code, which has been used to perform BH-NS mergers to date [294, 88, 89] One drawback of employing rectangular grids is that memory costs scale like N 3, where N is the number of grid cells across a side, and total computational effort like N 4 once one accounts for the reduction in the timestep due to the Courant-Friedrich-Levy criterion. Since one does not necessarily need high angular resolution at large radii, there is great current interest in developing schemes that use some form of spheroidal grid, for which the memory scaling is merely ∝ N. A group at LSU has implemented a multi-patch method , in which space is broken up into a number of non-overlapping domains in such a way that the six outermost regions (projections of the faces of a cube onto spheres of constant radius), maintain constant angular resolution and thus produce linear dependence of the total number of grid points on the number of radial points. To date, it has been used primarily for vacuum spacetimes and hydrodynamics on a fixed background. The SXS collaboration, begun at Caltech and Cornell and now including members at CITA and Washington State, has used a spectral evolution code with multiple domains to evolve BH-NS binaries, which achieves the same scaling by expanding the metric fields in modes rather than in position space . Their first published results on NS-NS binaries are currently in preparation (see for a brief summary of work to date). While all of these grid techniques produce tremendous advantages in computational efficiency, each required careful study since deformations of a grid or the introduction of multiple domains can introduce inaccuracies and non-conservative effects. As an example, in AMR schemes, one must deal with the same reflection issues at refinement boundaries that are present at the physical boundaries of the grid, as discussed above, though the interior nature of the boundaries allows for additional techniques, such as tapered grid boundaries , to be used to minimize reflections there. The study of how to minimize spurious effects in these schemes continues, and will represent an important topic for years to come, especially as evolution schemes become more complicated by including more physical effects. Hydrodynamics, MHD, and high-resolution shock capturing Fluids cannot be treated in the same way as the spacetime metric because finite differencing operators do not return meaningful results when placed across discontinuities induced by shocks. Instead, GR(M)HD calculations must include some form of shock-capturing that accounts for these jumps. These are typically implemented in conservative schemes, in which the fluid variables are transformed from the standard “primitive” set , which includes the fluid density, pressure, and velocity (and magnetic field in MHD evolutions), into a new set so that the evolution equations may be written in the form 26 where the flux functions and source terms can be expressed in terms of the primitive variables but not their derivatives. These schemes allow one to evolve the resulting MHD set of equations so that numerical fluxes are conserved to numerical precision across cell walls as the fluid evolves in time. One widely used scheme, often referred to as the Valencia formulation , is described in Appendix B.. There are important mathematical reasons for casting the GRHD/GRMHD system in conservative form, primarily since the mathematical techniques describing Godunov methods may be called into play . In such methods, we assume that the evolution of the quantities may be expressed in the form 27 where the points have spatial coordinates x i ≡ x 0 + i Δ x and the timesteps satisfy t n = t 0 + n Δ t. The fluxes must be determined by solving the Riemann problem at each cell face (thus the half-integer indices), either exactly or approximately. It can be shown that solutions constructed this way, when convergent, must converge to a solution of the original problem, even when shocks are present . First one reconstructs the primitives from the conserved variables on both sides of an interface, using interpolation schemes designed to respect the presence of shocks. Simple schemes involving limiters yield second-order accuracy in general but first-order accuracy at shocks, while higher-order methods such as PPM (piecewise parabolic method) and essentially non-oscillatory (ENO) schemes such as CENO (central ENO) and WENO (weighted ENO) yield higher accuracy but at much higher computational cost. Once primitives are interpolated to the cell interfaces, flux terms are evaluated there and one solves the Riemann problem describing the evolution of two distinct fluid configurations placed on either side of a membrane (see for a description). While complete solutions of the Riemann problem are painstaking to evolve, a number of approximation techniques exist and do not reduce the order of accuracy of the code. Finally, one must take the conservative solution, now advanced forward in time, and recover the underlying primitive variables, a process that requires solving as many as eight simultaneous equations in the case of GRMHD or five for GRHD systems. A number of methods to do this have been widely studied , and simplifying techniques are known for specific cases (for the case of polytropic EOSs in GRHD evolutions, one need only invert a single non-analytic expression and the remaining variables can then be derived analytically). The inclusion of magnetic fields in hydrodynamic calculations adds another layer of complexity beyond shock capturing. Magnetic fields must be evolved in such a way that they remain divergence-free, much in the same way that relativistic evolutions must satisfy the Hamiltonian and momentum constraints. Brute force attempts to subtract away any spurious divergence often lead to instabilities, so more intricate schemes have been developed. “Divergence cleaning” schemes typically introduce a new field representing the magnetic field divergence and use parabolic/hyperbolic equations to damp the divergence away while moving it off the computational domain; the approach is relatively simple to implement but prone to small-scale numerical errors . “Constrained transport” schemes stagger the grids on which different physical terms are calculated to enforce the constraints (see, e.g., for a particular implementation), and have been applied widely to many different physical configurations. Recently, a new scheme in which the vector potential is used rather than the magnetic field was introduced by Etienne, Liu, and Shapiro [93, 95], and found to yield successful results for a variety of physical configurations including NSs and BHs. Microphysical numerical techniques Neutron star physics and equations of state One of the largest uncertainties in the input physics of NS-NS merger simulations is the true behavior of the nuclear matter EOS. To date, EM observations have yielded relatively weak constraints on the NS mass-radius relation, with the most precise simultaneous measurement of both as of now resulting from observations of Type 1 X-ray bursts from accreting NSs in three different sources . In each case, the NS mass was found to lie in the range 1.3 M⊙ ≲ M NS ≲ 2 M⊙ and the radius 8 km ≲ R NS ≲ 12 km, implying a NS compactness 28 A more stringent constraint on the NS EOS is provided by observations of the Shapiro time delay in the binary millisecond pulsar PSR J1614-2230, which was found to have a mass M NS = 1.97 ± 0.04 M⊙ , which would rule out extremely soft EOS models incapable of supporting such a massive NS against collapse. As we discuss in more detail below, GW observations are likely to eventually yield tighter constraints than our current EM-based ones, though BH-NS mergers, which can undergo stronger tidal disruptions than NS-NS mergers at frequencies closer to LIGO and other GW observatories’ maximum frequency sensitivity band, may prove to be more useful for the task than NS-NS mergers. Given the large theoretical uncertainties in describing the proper physical NS EOS, many groups have chosen the simplest possible parameterization: a polytrope (see Eq. 12). Under this choice, the enthalpy h takes the particularly simple form 29 Initial data are generally assumed to follow the relation 30 where K is constant across the fluid. In the presence of shocks, the value of K for a particular fluid element will increase with time. We note that the Whisky group [17, 18] uses the term “polytropic” to refer to simulations in which Eq. 30 is enforced throughout, which implies adiabatic evolution without shock heating, and use the term “ideal fluid” to describe an EOS that includes the effects of shock heating and enforces Eq. 29. Since the temperatures of NSs typically yield thermal energies per baryon substantially below the Fermi energy, one may treat nearly all NSs as effectively cold, except for the most recently born ones. During the merger process for NS-NS binaries, the matter will remain cold until the two NSs are tidally disrupted and a disk forms, at which point the thermal energy input and substantially reduced fluid densities require a temperature evolution model to properly model the underlying physics. In light of these results, some groups adopt a two-phase model for the NS EOS (see, e.g., ), where a cold, zero-temperature EOS, evaluated as a function of the density only, encodes as much information about as we possess about the NS EOS, and the hot phase depends on both the density and internal energy, typically in a polytropic way, 31 There are a number of physically motivated EOS models that have been implemented for merger simulations, whose exact properties vary depending on the assumptions of the underlying model. These include models for which the pressure is tabulated as a function of the density only: FPS , SLy , and APR ; as well as models including a temperature dependence: Shen [268, 267] and Lattimer-Swesty . A variety of models have been used to study the effects of quarks, kaons, and other condensates, which typically serve to soften the EOS, leading to reduced maximum masses and more compact NSs [223, 119, 230, 120, 23, 5]. Given the variance among even the physically motivated EOS models, it has proven useful to parameterize known EOS models with a much more restricted set of parameters. In a series of works, a Milwaukee/Tokyo collaboration determined that essentially all current EOS models could be fit using four parameters, so that their imprint on GW signal properties could be easily analyzed [237, 238, 184]. Their method assumes that the SLy EOS describes NS matter at low densities, and that the EOS at higher densities can be described by a piecewise polytropic fit with breaks at ρ = 10 14.7 and 10 15 g/cm 3. The four resulting parameters are P 1 = P (ρ = 10 14.7), the pressure at the first breakpoint density, which normalizes the overall density scale, as well as Γ 1, Γ 2, Γ 3, the adiabatic exponents in the three regions. Their results indicate that advanced LIGO should be able to determine the NS radius to approximately 1 km at an effective distance of 100 Mpc, which would place tight constraints on the value of P 1 in particular. Electromagnetic and neutrino signature modeling Motivated by the evidence that SGRBs frequently appear in galaxies with very low star formation rates [40, 109], astronomers have suggested that their progenitors are likely to be mergers of either NS-NS and/or BH-NS binaries. While soft-gamma repeaters (SGRs) have been confirmed as an SGRB source from observations of the system SGR 1806-20, they make up no more than approximately 15% of the total observed SGRB fraction according to the leading population estimates [164, 199]. There has been much interest in predicting the EM signatures of NS-NS and BH-NS mergers, along with the associated neutrino emission. The simplest models estimate a local radiation cooling rate for the matter but do not attempt to follow the paths of the photons and/or neutrinos after they are emitted, instead calculating the time-dependent luminosity assuming free streaming. Such models have been used in non-GR simulations of binary mergers going back more than a decade [253, 246], and recently such schemes have been used to perform full GR NS-NS mergers , including a self-consistent evolution of the electron fraction of the material Y e, rather than a passive advection approach. More complicated flux-limited diffusion schemes, in which the neutrino fluxes for given species and energies are given by explicit formulae that limit to the correct values for zero optical depth (free-streaming) and very large optical depth (diffusion), have been used as a post-processing tool to investigate the merger remnants in Newtonian NS-NS mergers , but have yet to be applied to full GR simulations. Finally, radiation transport schemes to evolve EM and neutrino fluxes passing through fluid configurations have been implemented in numerical GR codes [80, 103], but have yet to be used in binary merger simulations. GW signal modeling Measuring the GW signal from a dynamical merger calculation is a rather difficult task. One must determine, using a method unaffected by gauge effects, the perturbations at asymptotically large distances from a source by extrapolating various quantities measured at large but finite distances from the merger itself. In the early days of numerical merger simulations, most groups typically assumed Newtonian and/or quasi-Newtonian gravitation, for which there is no well-defined dynamical spacetime metric. GW signals were typically calculated using the quadrupole formalism, which technically only applies for slow-moving, non-relativistic sources (see for a thorough review of the theory). Temporarily reintroducing physical constants, the strains of the two polarizations for signals emitted in the z-direction are where r is the distance from the source to the observer and it the traceless quadrupole moment of the system. The energy and angular momentum loss rates of the system due to GW emission are given, respectively, by while only approximate, the quadrupole formulae do yield equations that are extremely straightforward to implement in both grid and particle-based codes using standard integration techniques. Quadrupole methods were adopted for later PN and CF simulations, again because the metric was assumed either to be static or artificially constrained in such a way that made self-consistent determination of the GW signal impossible. One important development from this period was the introduction of a simple method to calculate the GW energy spectrum dE/df from the GW time-series through Fourier transforming into the frequency domain . GW signals analyzed in the frequency domain allowed for direct comparison with the LIGO noise curve, making it much easier to determine approximate distances at which various GW sources would be detectable and the potential signal-to-noise ratio that would result from a template search. To constrain the nuclear matter EOS, one can examine where a GW merger spectrum deviates in a measurable way from the quadrupole point-mass form, 32 because of finite-size effects, and then link the deviation to the properties of the NS , as we show in Figure 9. Figure 9. Open in a new tab Approximate energy spectrum dE GW/df derived from QE sequences of equal-mass NS-NS binaries with isolated ADM masses M NS = 1.35 M⊙ and a Γ = 2 EOS, but varying compactnesses (denoted M/R here), originally described in . The diagonal lines show the energy spectrum corresponding to a point-mass binary, as well as values with 90%, 75%, and 50% of the power at a given frequency. Asterisks indicate the onset of mass-shedding, beyond which QE results are no longer valid. Image reproduced by permission from Figure 2 of , copyright by APS. Full GR dynamical calculations, in which the metric is evolved according to the Einstein equations, generally use one of two approaches to calculate the GW signal from the merger, if not both. The first method, developed first by by Regge and Wheeler and Zerilli and written down in a gauge-invariant way by Moncrief involves analyzing perturbations of the metric away from a Schwarzschild background. The second uses the Newman-Penrose formalism to calculate the Weyl scalar ψ 4, a contraction of the Weyl curvature tensor, to represent the outgoing wave content on a specially constructed null tetrad that may be calculated approximately . The two methods are complementary since they incorporate different metric information and require different numerical integrations to produce a GW time series. Regardless of the method used to calculate the GW signal, results are often presented by calculating the dominant s = −2 spin-weighted spherical harmonic mode. For circular binaries, the l = 2, m = 2 mode generally carries the most energy, followed by other harmonics; in cases where the components of the binary have nearly equal masses and the orbit is circular, the falloff is typically quite rapid, while extreme mass ratios can pump a significant amount of the total energy into other harmonics. For elliptical orbits, other modes can dominate the signal, e.g., a 3:1 ratio in power for the l = 2, m = 0 mode to the l = 2, m = 2 mode observed for high-ellipticity close orbits in . A thorough summary of both methods and their implementation may be found in . Dynamical Calculations NS-NS merger simulations address a broad set of questions, which can be roughly summarized as follows (note the same questions apply to BH-NS mergers as well): What is the final fate of the system, assuming a given set of initial parameters? Do we get a prompt collapse to a BH or the formation of a HMNS supported against collapse by differential rotation? Other outcomes are disfavored, at least for pre-merger NSs with masses M NS ≳ 1.4 M⊙ since the supramassive limit is at most 20% larger than that of a non-rotating NS [70, 71], and even for the stiffest EOS these values are typically less than 2.8 M⊙. What is the GW signal from the merger, and how does it inform us about the initial pre-merger parameters of the system? What fraction of the system mass is left in a disk around the central BH or HMNS? While deriving exact EM emission profiles from a hydrodynamical configuration remains a challenge for the future, minimum conditions that would allow for the energy release observed in SGRBs have been established based on scaling arguments. What is the neutrino and EM emission from the system, in both the time and energy domains? Obviously, the answer to this question and those that follow depend critically on the answers above. What role do B-fields play in the GW, EM, and neutrino emission, and how does that tie in with other models suspected of having the same disk/jet geometry and gamma-ray emission like active galactic nuclei, pre-main sequence stars, etc.? Do mergers produce a cosmologically significant quantity of r-process elements, or do those likely get produced by other astrophysical events instead? The influence of the gravitational formalism used in a numerical simulation on the answer one finds for the questions above differs item by item. Determining the final fate of a merging system is highly dependent on the gravitational formalism; NS-NS merger remnants only undergo collapse in quasi-relativistic and fully GR schemes. Moreover, orbital dynamics at separations comparable to the ISCO and even somewhat larger depend strongly on the gravitational scheme. In particular, mass loss rates into a disk are often suppressed by orders of magnitude in GR calculations when compared to CF simulations, and even more so in comparison to PN and Newtonian calculations. EM emission profiles from a disk are difficult to calculate accurately without the use of full GR for this reason. On the other hand, while GR is required to calculate the exact GW signal from a merger, even early Newtonian simulations predicted many of the qualitative GW emission features correctly, and PN and CF schemes yielded results with some degree of quantitative accuracy about the full wavetrain. B-fields have only begun to be explored, but it already seems clear that they will affect the hydrodynamical evolution primarily after the merger in cases where differential rotation in a HMNS or disk winds up magnetic field strengths up to energy equipartition levels, vastly stronger than those found in pre-merger NSs. For such configurations, non-relativistic calculations can often reproduce the basic physical scenario but full GR is required to properly understand the underlying dynamics. Finally, the production of r-process elements, which depends sensitively on the thermodynamic evolution of the merger, seems to generally disfavor binary mergers as a significant source of the observed stellar abundances since the temperature and thus the electron fraction of the fluid remains too small , regardless of the nature of the gravitational treatment used in the calculations. This picture may need to be revised if significant mass loss occurs from the hot accretion disk that forms around the central post-merger object, possibly due to energy release from the r-process itself, but numerical calculations do not currently predict sufficient mass loss to match observations . We will address each of these topics in greater detail in the sections below. Since the first NS-NS merger calculations, there have been two main directions for improvements: more accurate relativistic gravitation, resulting in the current codes that operate using a self-consistent fully GR approach, and the addition of microphysical effects, which now include treatments of magnetic fields and neutrino/EM radiation. Noting that several of the following developments overlapped in time, e.g., the first full GR simulations by Shibata and Uryū are coincident with the first PN SPH calculations, and predate the first CF SPH calculations, we consider in turn the original Newtonian calculations, those performed using approximate relativistic schemes, the calculations performed using full GR, and finally those that have included more advanced microphysical treatments. Quasi-equilibrium and semi-analytic methods vs fully dynamical results Before reviewing fully dynamical calculations of NS-NS mergers, it is worthwhile to ask how much information can already be deduced from QE calculations, which may be performed at much smaller computational cost, as well as from semi-analytic PN treatments and related approximate techniques. Clearly, the details of the merger and ringdown phases fall outside the QE regime, so only dynamical calculations can yield reliable information about the stability of remnants, properties of ejecta, or other processes that arise during the merger itself or in its aftermath. Thus, the primary point of comparison is the GW signal just prior to merger, which is also easier to detect (for first and second generation interferometers). The strength of QE calculations lies in their ability to model self-consistently finite-size effects not captured in PN treatments (which always assume two orbiting point masses). The increased tidal interaction between the objects typically results in a more rapid phase advance of the binary orbit, which is important for constructing template waveforms that cover the entire NS-NS inspiral, merger, and ringdown. While QE sequences potentially offer a wealth of information about well-separated binaries and can help fix the phase evolution of the inspiraling binary, they do have two weaknesses arising as the binary approaches the stability limit. First, most QE methods, including the CTS formalism described in Section 4.2.1, are time-symmetric, and assume that the NS possess a symmetry plane perpendicular to the direction of motion (i.e., a front-back symmetry whose axis is perpendicular to the orbital angular momentum and the binary separation vector). In reality, tidal lags develop prior to final plunge, with the innermost edge of each NS rotating forward and the outer edge backwards. This effect has been captured in analytic and semi-analytic approaches (see, e.g., for an early example), and is clearly seen in dynamical calculations (see Figure 3), but is not captured in CTS-based schemes (tidal lags also develop in BH-NS merger calculations when the BH has a non-zero spin, since this breaks the front-back symmetry; see for an example). A second weakness of QE methods is the treatment of the ISCO, particularly its importance as a characteristic point along an evolutionary sequence that, in theory, could encode information about the NS EOS. Simple estimates of the infall trajectory derived solely from QE sequences predict a very sudden and rapid infall near the ISCO, i.e., the point where the binding energy reaches a minimum along the sequence (see, e.g., the argument in ). However, this is clearly an oversimplification. In reality, binaries transition more gradually to the merger phase, and the inward plunge may occur significantly before reaching the formal ISCO; this in turns leads to more rapidly growing deviations from the QE approximation. Looking at the GW energy spectrum, one typically sees minor deviations from the point-mass predictions at frequencies below those characterizing the ISCO, but substantially more power at frequencies above it. Equivalently, the cutoff frequency for GW emission f cut, where the spectrum starts deviating strongly from the point-mass prediction, is usually higher than the QE frequency near the ISCO, f ISCO, while simple QE estimates assume these two frequencies to coincide. To date, most attempts to generate waveforms extended back to arbitrarily early starting points involve numerically matching PN signals, typically generated using the Taylor T4 approach , onto the early stages of numerically generated waveforms, with some form of maximum overlap method used to provide the most physical transition from one to the other. These approaches may be improved by adding tidal effects to the evolution, typically parameterized by the tidal Love numbers that describe how tidal gravity fields induce quadrupole deformations . Tidal effects can be placed into a relativistic framework [46, 74], which may be included within the effective one-body (EOB) formalism to produce high-accuracy waveforms . In the EOB approach , resummation methods are used to include higher-order PN effects, though some otherwise unfixed parameters need to be set by comparing to numerical simulations. Work is in its early stages to compare directly the GW spectra inferred from QE sequences of NS-NS binaries with those generated in numerical relativity simulations, but this comparison has been discussed at some length with regard to BH-NS mergers. Noting that NS-NS mergers generally correspond more closely to the BH-NS cases in which an ISCO is reached prior to the onset of tidal disruption, the KT collaboration [283, 276] concluded that the cutoff frequency marking significant deviations from PN point-mass behavior is roughly 30% higher than that marking emission near the classical ISCO for BH-NS systems (f cut ≃ 1.3 f ISCO). A more detailed study has now been performed comparing EOB methods to numerical evolutions. By comparing to long-term simulations of NS-NS mergers, Baiotti et al. find that EOB models may be tuned, via careful choices of their unfixed parameters, to reproduce the GW phases and amplitudes seen in NR evolutions up until the onset of merger. They further suggest that the EOB approach seems to cover a wider range of phase space than the Taylor T4 approach, presumable because of a more consistent representation of tidal effects, and offers the best route forward for construction of more accurate NS-NS inspiral templates. Early dynamical calculations The earliest NS-NS merger calculations were performed in Newtonian gravity, sometimes with the addition of lowest-order 2.5PN radiation reaction forces, and typically assumed that the NS EOS was polytropic. Both Eulerian grid codes [214, 196, 215, 197, 278, 255, 201, 298] and Lagrangian SPH codes [234, 235, 236, 76, 330, 331] were employed, and GW signals were derived under the assumptions of the quadrupole formalism. Configurations in Newtonian gravity cannot collapse, so a stable (possibly hypermassive) remnant was always formed. For polytropic EOS models with adiabatic indices larger than the classical minimum for production of a Jacobi ellipsoid, Γ ≳ 2.6, remnants were typically triaxial and maintained a significant-amplitude GW signal until the end of the simulation. For simulations using smaller values of Γ, remnants rapidly relaxed to spheroidal configurations, quickly damping away the resulting GW signal. Mass loss from the central remnant was often quite significant, with thick accretion disks or completely unbound material comprising up to 10–20% of the total system mass Mass loss was suppressed in numerical simulations by constructing irrotational, rather than synchronized, initial data. Irrotational flow is widely thought to be the more physically realistic case, since viscous forces are much too weak to synchronize a NS prior to merger [45, 146]. When irrotational NSs (which are counter-rotating in the corotating frame of the binary) first make contact, a vortex sheet forms. Since the low-density fluid layers at the contact surface are surrounded at first contact by the denser fluid layers located originally within each NS, the configuration is well understood to be Kelvin-Helmholtz unstable, resulting in rapid mixing through vortex production. Meanwhile, mass loss through the outer Lagrange points is hampered by the reduced rotational velocity along the outer halves of each NS. The GW emission from these mergers is composed of a “chirp,” increasing in frequency and amplitude as the NSs spiral inward, followed by a ringdown signal once the stars collide and merge. In [330, 331], a procedure to calculate the energy spectrum in the frequency band was laid out, with the resulting signal following the quadrupole, point-mass power-law form up to GW frequencies characterizing the beginning of the plunge. Above the plunge frequency, a sharp drop in the GW energy was seen, followed in some cases by spikes at kHz frequencies representing coherent emission during the ringdown phase. Approximate relativistic schemes The first steps toward approximating the effects of GR included the use of 1PN dynamics or the CF approximation. Using a formalism derived by Blanchet, Damour, and Schäfer , the 1PN equations of motion require the solution of eight Poisson-like equations in the form 33 where the source terms are compactly supported, and thus the fields ψ may be determined using the same techniques already in place to find the Newtonian potential. Adding in the lowest-order dissipative radiation reaction effects requires solution of a ninth Poisson equation for a reaction potential. The 1PN formalism was implemented in both grid-based and SPH codes [10, 99, 101, 100]. Unfortunately, physically realistic NSs are difficult to model using a PN expansion, since the characteristic NS compactness , leads to first order “corrections” that often rival Newtonian terms in magnitude. To deal with this problem Ayal et al. considered large (R ≈ 30 km), low-mass (< 1 M⊙) NSs, allowing them to study relativistic effects but making results more difficult to interpret for physically realistic mergers. In [99, 101, 100], a dual speed of light approach was used, in which all 1PN effects were scaled down by a constant factor to yield smaller quantities while Newtonian and radiation reaction terms were included at full-strength. Both SPH groups found that the GW signal in PN mergers is strongly modulated, whereas Newtonian merger calculations typically yielded smooth, either monotonically decreasing or nearly constant-amplitude ringdown signals. Even reduced 1PN effects were shown to suppress mass loss by a factor of 2–5 for initially synchronized cases, and disk formation was seen to be virtually non-existent for initially irrotational, equal-mass NSs with a stiff (Γ = 3 polytropic) EOS . Moving beyond the linearized regime, several groups explored the CF approximation, which incorporates many of the nonlinear effects of GR into an elliptic, rather than hyperbolic, evolution scheme. While nonlinear elliptic solvers are expensive computationally, they typically yield stable evolution schemes since field solutions are always calculated instantaneously from the given matter configuration. Summarized quickly, the CF approach involves solving the CTS field equations, Eqs. 18, 19, and 20, at every timestep, and evolving the matter configuration forward in time. The metric fields act like potentials, with various gradients appearing in the Euler and energy equations. While the CTS formalism remains the most widely used method to construct NS-NS (and BH-NS) initial data, it does not provide a completely consistent dynamical solution to the GR field equations. In particular, while it reproduces spherically symmetric configurations like the Schwarzschild solution exactly, it cannot describe more complicated configurations, including Kerr BHs. Moreover, because the CF approximation is time-symmetric, it also does not allow one to consistently predict the GW signal from a merging configuration. As a result, most dynamical calculations are performed by adding the lowest-order dissipative radiation reaction terms, either in the quadrupole limit or via the radiation reaction potential introduced in . The CTS equations themselves were originally written down in essentially complete form by Isenberg in the 1970s, but his paper was rejected and only published after a delay of nearly 30 years . In the intervening years, Wilson, Mathews, and Marronetti [327, 328, 188, 187] independently re-derived the entire formalism and used it to perform the first nonlinear calculations of NS-NS mergers (as a result, the formalism is often referred to as the “Wilson-Mathews” or “Isenberg-Wilson-Mathews” formalism). The key result in [327, 328, 188, 187] was the existence of a “collapse instability,” in which the deeper gravitational wells experienced by the NSs as they approached each other prior to merger could force one or both to collapse to BHs prior to the orbit itself becoming unstable. Unfortunately, their results were affected by an error, pointed out in , which meant that much of the observed compression was spurious. While their later calculations still found some increase in the central density as the NSs approached each other , these results have been contradicted by other QE sequence calculations (see, e.g., ). Furthermore, using a “CF-like” formalism in which the nonlinear source terms for the field equations are ignored, dynamical calculations demonstrated the maximum allowed mass for a NS actually increases in response to the growing tidal stress . The CF approach was adapted into a Lagrangian scheme for SPH calculations by the same groups that had investigated PN NS-NS mergers, with Oechslin, Rosswog, and Thielemann using a multigrid scheme and Faber, Grandclément, and Rasio a spectral solver based on the Lorene libraries . The effects of nonlinear gravity were immediately evident in both sets of calculations. In , NS-NS binaries consisting of initially synchronized NSs merged without appreciable mass loss, with no more that ∼ 10−4 of the total system mass ejected, strikingly different from previous Newtonian and PN simulations. When evolving initially irrotational systems, found no appreciable developments of “spiral arms” whatsoever, indicating a complete lack of mass loss through the outer Lagrange points. Both groups also found strong emission from remnants for a stiff EOS, as the triaxial merger remnant produced an extended period of strong ringdown emission. Neither set of calculations indicated that the remnant should collapse promptly to form a BH, but given the high spin of the remnant it was noted that conformal flatness would have already broken down for those systems. Full GR calculations A summary of full GR calculations of NS-NS mergers is presented in Table 4. The KT collaboration was responsible for the only full GR calculations of NS-NS mergers that predate the breakthrough calculations of numerically stable binary BH evolutions [231, 22, 61], which have since transformed the field of GR hydrodynamics and MHD in addition to vacuum relativistic evolutions (Miller et al. performed NS-NS inspiral calculations in full GR, but were not able to follow binaries through to merger). The first calculations of NS-NS mergers using a completely self-consistent treatment of GR were performed by Shibata and collaborators in the KT collaboration using a grid based code and the BSSN formalism . CTS initial data consisting of equal-mass NSs described by a Γ = 2 polytropic EOS were constructed via SCF techniques , for both synchronized and irrotational configurations. The hyperbolic system was evolved on a grid, with an approximate maximal slicing condition that results in a parabolic equation for the lapse and an approximate minimal distortion condition for the shift vector requiring the solution of an elliptic equation at every time step . The shift vector gauge condition was found to fail when BHs were produced in the merger remnant, a well-known problem that had long bedeviled simulations involving binary and even single BH evolutions, so modifications were introduced to extend the stability of the algorithm as far as possible. Among the key results from this early work was a clear differentiation between mergers of moderately low-compactness NSs , where the remnant collapsed promptly to a BH, and very low-compactness models, which yielded hypermassive remnants stabilized against gravitational collapse by differential rotation. Virtually all the NS matter was contained within the remnant for initially irrotational models, which served as evidence against equal-mass NSs mergers being a leading source of r-process elements in the universe through ejection. The lack of significant mass loss in equal-mass mergers, together with insignificant shock-heating of the material, also argued against the likelihood of such mergers as progenitors for SGRBs if the gamma-ray emission was assumed to be coincident with the GW burst; instead a delayed burst following the collapse of a HMNS to a BH appeared more likely. Table 4. A summary of Full GR NS-NS merger calculations. EOS models include polytropes, piecewise polytropes (PP), as well as physically motivated models including cold SLy , FPS , and APR models to which one adds an ideal-gas hot component to reflect shock heating, as well as the Shen [268, 267] finite temperature model and EOS that include Hyperonic contributions . “Co/Ir” indicates that both corotating and irrotational models were considered; “BHB” indicates that BH binary mergers were also presented, including both BH-BH and BH-NS types, “ν-leak” indicates a neutrino leakage scheme was included in the calculation, “GH” indicates calculations were performed using the GHG formalism rather than BSSN, “non-QE” indicates superposition initial data were used, including cases where eccentric configurations were studied (“Eccen.”); “MHD” indicates MHD was used to evolve the system. | Group | Ref. | NS EOS | Mass ratio | | notes | --- --- --- | | KT | | Γ=2 | 1 | 0.09–0.15 | Co/Ir | | Γ = 2, 2.25 | 0.89–1 | 0.1–0.17 | | | Γ=2 | 0.85–1 | 0.1–0.12 | | | SLy, FPS+Hot | 0.92–1 | 0.1–0.13 | | | SLy, APR+Hot | 0.64–1 | 0.11–0.13 | | | Γ = 2 | 0.85–1 | 0.14–0.16 | BHB | | APR+Hot | 0.8–1 | 0.14–0.18 | | | APR, SLy, FPS+Hot | 0.8–1.0 | 0.16–0.2 | | | Shen | 1 | 0.14–0.16 | ν-leak | | PP+hot | 1 | 0.12–0.17 | | | Shen, Hyp | 1.0 | 0.14–0.16 | ν-leak | | HAD | | Γ = 2 | 1.0 | 0.08 | GH, non-QE | | Γ = 2 | 1.0 | 0.08 | GH, non-QE, MHD | | Whisky | | Γ = 2 | 1.0 | 0.14–0.18 | | | Γ = 2 | 1.0 | 0.20 | | | Γ = 2 | 1.0 | 0.14–0.18 | MHD | | Γ = 2 | 1.0 | 0.14–0.18 | MHD | | Γ=2 | 0.70–1.0 | 0.09–0.17 | | [14, 15] | Γ = 2 | 1.0 | 0.12–0.14 | | | Γ = 2 | 1.0 | 0.18 | MHD | | UIUC | | Γ = 2 | 0.85–1 | 0.14–0.18 | MHD | | Jena | [308, 41] | Γ = 2 | 1.0 | 0.14 | | | Γ = 2 | 1.0 | 1.4 | Eccen. | Open in a new tab Later works, in particular a paper by Shibata, Taniguchi, and Uryū , introduced several new techniques to perform dynamical calculations that most codes at present still include in nearly the same or lightly modified form. These included the use of a high-resolution shock-capturing scheme for the hydrodynamics, as well as a Gamma-driver shift condition closely resembling the moving puncture gauge conditions that later proved instrumental in allowing for long-term BH evolution calculations. In the series of papers that followed their original calculations, the KT group established a number of results about NS-NS mergers that form the basis for much of our thinking about their hydrodynamic evolution: By varying the EOS model for the NS as well as the mass ratio, it was possible to constrain the binary parameters separating cases that form a HMNS rather than producing prompt collapse to a BH, and it was quickly determined that the total system mass as a proportion of the maximum allowed mass for an isolated NS is the key parameter, with only weak dependence on the binary mass ratio. For polytropic EOS models, the critical compactness values leading to prompt collapse for equal-mass binary mergers were found to be for Γ = 2 and for Γ = 2.25 . As a rough rule, collapse occurred for polytropic EOS when the total system rest-mass was at least 1.7 M max, where M max is the maximum mass of an isolated non-rotating NS for the given EOS. For physically motivated EOS models , the critical mass was significantly smaller; indeed, the critical NS mass was found to be ∼ 1.35 M max for the SLy EOS (i.e., collapse for M tot ≥ 2.7 M⊙ with M max = 2.04 M⊙) and ∼ 1.39 M max for the FPS EOS (collapse for M tot ≥ 2.5 M⊙ with M max = 1.8 M⊙). This was not a complete surprise, since for the physically motivated EOS the NS radius is nearly independent of the mass across much of the parameter space, limiting the ability of the HMNS to expand in response to the extra mass absorbed during the merger. The mass ratio was found to play a critical role in the evolution of the remnant/disk configuration, since unequal-mass cases are better characterized as disruptions of the smaller secondary followed by its accretion onto the primary, rather than a true merger between the two NSs. Disk masses from full GR calculations are generically smaller than those predicted from non-GR calculations. For polytropic EOS, disks contain approximately 4% of the total system mass for mass ratios q ≃ 0.8, varying roughly ∝ (1 − q) for a fixed total mass, with the disk mass decreasing for heavier binaries (and thus larger compactnesses) given the stronger gravity of the central remnant. Using the stiffer APR EOS , the dependence on the mass ratio was seen to be much steeper for a physical EOS than for polytropes, scaling like (1 − qp)p, where p ≃ 3–4 . With respect to GW emission, it was determined in that in low-mass cases in which a HMNS was formed, stiffer polytropic EOS models were able to support the development of a bar-mode instability, leading to transient spiral arm formation from the remnant and an extended period of strong GW emission, in the characteristic modulated form that results from differentially rotating ellipsoids (see, e.g., [160, 274, 259, 16, 183, 72]). Unequal-mass cases typically yielded one high-frequency peak at roughly GW ≈ 2 kHz corresponding to non-axisymmetric oscillations, and equal-mass cases yielded multiple peaks including those associated with quasi-radial oscillations as well . For physical EOS models , mass loss into a disk is reduced relative to the polytropic case given the higher compactness of the central region, and GW oscillation peaks, while very strong, occur at correspondingly higher frequencies. By contrast, prompt formation of a BH led to a ringdown signal with rapidly decreasing amplitude becoming negligible within a few dynamical times. The GW signals were evaluated under the gauge-dependent assumption of transverse tracelessness, and energy and angular momentum loss rates into each spherical harmonic mode were computed using the gauge-invariant Zerilli-Moncrief formalism [239, 335, 195] in much the same way that is used by some groups in numerical relativity today (many BH-BH and hydrodynamics simulations report GW signals derived from the alternate ψ 4 Weyl scalar formulation [202, 60], or use both methods). In , it was concluded that mergers of NSs with comparable masses made poor SGRB progenitor candidates, assuming prompt emission (because of the lack of energy available for neutrino annihilation), but that the energy budget in the HMNS case is orders of magnitude larger. Remarkably, this discussion from 2005 predates the first identifications of SGRBs with older populations, which greatly improved our theoretical understanding of compact object mergers as their likely progenitors. In the first work that followed the initial localizations of SGRBs, mergers of binaries with relatively small mass ratios, q ≈ 0.7, were seen to form sufficiently hot and massive disk to power a SGRB, albeit a relatively brief, low-luminosity one. It was suggested that the more likely SGRB progenitor is indeed a HMNS, since dissipative effects within the remnant can boost temperatures up to ∼ 10 11 K. Further approximate relativistic investigations of NS-NS mergers, along with BH-NS mergers, as potential SGRB sources quickly swept through the community after the initial localizations of SGRBs, with several groups using a wide variety of methods all concluding that mergers were plausible progenitors, but finding it extremely difficult to constrain the scenario in quantitative ways given the extremely complicated microphysics ultimately responsible for powering the burst (see, e.g., [207, 206] who investigated potential disk energies; , who modeled the fallback accretion phase onto a BH; and , who considered the thermodynamic and nuclear evolution of disks around newly-formed BHs produced by mergers). We will return to this topic below in light of recent GRMHD Simulations. In the past few years, five groups have reported results from NS-NS mergers in full GR; KT, HAD, Whisky, UIUC, and Jena. Much of the work of the HAD and Whisky groups, developers respectively of the code of those names, began at Louisiana State University (HAD) and the Albert Einstein Institute in Potsdam (Whisky), though both efforts now include several other collaborating institutions. Two other groups, the SXS collaboration that originated at Caltech and Cornell, and the Princeton group, have reported BH-NS merger results and are actively studying NS-NS mergers as well, but have yet to publish their initial papers about the latter. All of the current groups use AMR-based Eulerian grid codes, with four evolving Einstein’s equations using the BSSN formalism and the HAD collaboration making use of the GHG method instead. HAD, Whisky, and UIUC have all reported results about magnetized NS-NS mergers (the KT collaboration has used a GRMHD code to study the evolution of magnetized HMNS, but not complete NS-NS mergers). The KT collaboration has considered a wide range of EOS models, including finite-temperature physical models such as the Shen EOS, and have also implemented a neutrino leakage scheme, while all other results reported to date have assumed a Γ = 2 polytropic EOS model. Given the similarities of the various codes used to study NS-NS mergers, it is worthwhile to ask whether they do produce consistent results. A comparison paper between the Whisky code and the KT collaboration’s SACRA codes found that both codes performed well for conservative global quantities, with global extrema such as the maximum rest-mass density in agreement to within 1% and waveform amplitudes and frequencies differing by no more than 10% throughout a full simulation, and typically much less. Several of the the groups listed above have also been leaders in the field of BH-NS simulations: the KT, HAD, and UIUC groups have all presented BH-NS merger results, as have the SXS collaboration [85, 84, 108], and Princeton group [294, 88] (see for a thorough review). We discuss the current understanding of NS-NS mergers in light of all these calculations below. HMNS and BH remnant properties Using their newly developed SACRA code , the KT group , found that when a hybrid EOS is used to model the NS, in which the cold part is described by the APR EOS and the thermal component as a Γ = 2 ideal gas, the critical total binary mass for prompt collapse to a BH is M tot = 2.8–2.9 M⊙, independent of the initial binary mass ratio, a result consistent with previous explorations of other polytropic and physically motivated NS EOS models (see above). In all cases, the BH was formed with a spin parameter a ≈ 0.78 depending very weakly on the total system mass and mass ratio. They further classified the critical masses for a number of other physical EOS in , finding that binaries with total masses M tot ≲ 2.7 M⊙ should yield long-lived HMNSs (> 10 ms) and substantial disk masses with M disk> 0.04 M⊙ assuming that the current limit on the heaviest observed NS, M = 1.97 M⊙ is correct. In Figure 10, we show the final fate of the merger remnant as a function of the total pre-merger mass of the binary. “Type I” indicates a prompt collapse of the merger remnant to a BH, “Type II” a short-lived HMNS, which lasts for less than 5 ms after the merger until its collapse, and “Type III” a long-lived HMNS which survives for at least 5 ms. See for an explanation of the EOS used in each simulation. Figure 10. Open in a new tab Type of final remnant corresponding to different EOS models. The vertical axis shows the total mass of two NSs. The horizontal axis shows the EOSs together with the corresponding NS radii for M NS = 1.4 M⊙. Image reproduced by permission from Figure 3 of , copyright by APS. While all of the above results incorporated shock heating, the addition of both finite-temperature effects in the EOS and neutrino emission modifies the numerically determined critical masses separating HMNS formation from prompt collapse. Adding in a neutrino leakage scheme for a NS-NS merger performed using the relatively stiff finite-temperature Shen EOS, the KT collaboration reports in that HMNSs will form generically for binary masses ≲ 3.2 M⊙, not because they are centrifugally-supported but rather because they are pressure-supported, with a remnant temperature in the range 30–70 MeV. Since they are not supported by differential rotation, these HMNSs were predicted to be stable until neutrino cooling, with luminosities of ∼ 3–10 × 10 53 erg/s, can remove the pressure support. Even for cases where the physical effects of hyperons were included, which effectively soften the EOS and reduce the maximum allowed mass for an isolated NS to 1.8 M⊙, the KT collaboration still finds that thermal support can stabilize HMNS with masses up to 2.7 M⊙. Using a Carpet/Cactus-based hydrodynamics code called Whisky that works within the BSSN formalism (a version of which has been publicly released as GRHydro within the Einstein ), the Whisky collaboration has analyzed the dependence of disk masses on binary parameters in some detail. For mass ratios q = 0.7–1.0 , they found that bound disks with masses of up to 0.2 M⊙ can be formed, with the disk mass following the approximate form 34 where M max the maximum mass of a binary system for a given EOS (Γ = 2 ideal gas for these calculations), M is the maximum mass of an isolated non-rotating NS for the EOS, and M tot the mass of the binary, with all masses here defined as baryonic. The evolution of the total rest mass present in the computational domain for a number of simulations is shown in Figure 11. Figure 11. Open in a new tab Evolution of the total rest mass M tot of the remnant disk (outside the BH horizon) normalized to the initial value for NS-NS mergers using a Γ = 2 polytropic EOS with differing mass ratios and total masses. The order of magnitude of the mass fraction in the disk can be read off the logarithmic mass scale on the vertical axis. The curves referring to different models have been shifted in time to coincide at t coll. Image reproduced by permisison from Figure 5 of , copyright by IOP. Magnetized NS-NS mergers Using the HAD code described in that evolves the GHG system on an AMR-based grid with CENO reconstruction techniques, Anderson et al. performed the first study of magnetic effects in full GR NS-NS mergers . Beginning from spherical NSs with extremely strong poloidal magnetic fields (9.6 × 10 15 G, as is found in magnetars), their merger simulations showed that magnetic repulsion can delay merger by 1–2 orbits and lead to the formation of magnetically buoyant cavities at the trailing end of each NS as contact is made (see Figure 12), although the latter may be affected by the non-equilibrium initial data. Both effects would have been greatly reduced if more realistic magnetic fields strengths had been considered. Magnetic fields in the HMNS remnant, which can be amplified through dynamo effects regardless of their initial strengths, helped to distribute angular momentum outward via the magnetorotational instability (MRI), leading to a less differentially rotating velocity profile and a more axisymmetric remnant. The GW emission in the magnetized case was seen to occur at lower characteristic frequencies and amplitudes as a result. Figure 12. Open in a new tab Fluid density isocontours and magnetic field distribution (in a plane slightly above the equator) immediately after first contact for a magnetized merger simulation. The cavities at both trailing edges are attributed to magnetic pressure inducing buoyancy. Image reproduced by permission from Figure 1 of , copyright by APS. The UIUC group was among the first to produce fully self-consistent GRMHD results . Using a newly developed Cactus-based code, they performed the first studies of unequal-mass magnetized NS-NS mergers . Using poloidal, magnetar-level initial magnetic fields, Liu et al. found that magnetic effects are essentially negligible prior to merger, but can increase the mass in a disk around a newly formed BH moderately, from 1.3% to 1.8% of the total system mass for mass ratios of q = 0.85 and Γ = 2. They point out that MHD effects can efficiently channel outflows away from the system’s center after collapse , and may be important for the late-stage evolution of the system. In , the Whisky group performed simulations of magnetized mergers with field strengths ranging from 10 12 to 10 17 G. Agreeing with the UIUC work that magnetic field strengths would have essentially no effect on the GW emission during inspiral, they note that magnetic effects become significant for the HMNS, since differential rotation can amplify B-fields, with marked deviations in the GW spectrum appearing at frequencies of GW ≳ 2 kHz. They also point out that high-order MHD reconstruction schemes, such as third-order PPM, can produce significantly more accurate results that second-order limiter-based schemes. A follow-up paper showed that a plausible way to detect the effect of physically realistic magnetic fields on the GW signal from a merger was through a significant shortening of the timescale for a HMNS to collapse, though a third-generation GW detector could perhaps observe differences in the kHz emission of the HMNS as well. More recently, they have used very long-term simulations to focus attention on the magnetic field strength and geometry found after the remnant collapses to a BH . They find that the large, turbulent magnetic fields (B ∼ 10 12 G) present in the initial binary configuration are boosted exponentially in time up to a poloidal field of strength 10 15 G in the remnant disk, with the field lines maintaining a half-opening angle of 30° along the BH spin axis, a configuration thought to be extremely promising for producing a SGRB. The resulting evolution, shown in Figure 13, is perhaps the most definitive result indicating that NS-NS mergers should produce SGRBs for some plausible range of initial parameters. It is worth noting that all magnetized NS-NS merger calculations that have been attempted to date have made use of unphysically large magnetic fields. This is not merely a convenience designed to enhance the role of magnetic effects during the merger, though it does have that effect. Rather, magnetic fields are boosted in HMNS remnants by the MRI, whose fastest growing unstable mode depends roughly linearly on the Alfvén speed, and thus the magnetic field strength. In order to move to physically reasonable magnetic field values, one would have to resolve the HMNS at least a factor of 100 times better in each of three dimensions, which is beyond the capability of even the largest supercomputers at present, and likely will be for some time to come. GW emission In , the KT collaboration found a nearly linear relationship between the GW spectrum cutoff frequency f cut and the NS compactness, independent of the EOS, as well as a relationship between the disk mass and the width of the kHz hump seen in the GW energy spectrum. While f cut is a somewhat crude measure of the NS compactness, it occurs at substantially lower frequencies than any emission process associated with merger remnants, and thus is the parameter most likely to be accessible to GW observations with a second generation detector. The qualitative form of the high-frequency components of the GW spectrum is primarily determined by the type of remnant formed. In Figures 14 and 15, we show h(t) and , respectively, for four of the runs calculated by the KT collaboration and described in . Type I collapses are characterized by a rapid decrease in the GW amplitude immediately after the merger, yielding relatively low power at frequencies above the cutoff frequency. Type II and III mergers yield longer periods of GW emission after the merger, especially the latter, with the remnant oscillation modes leading to clear peaks at GW frequencies f GW = 2–4 khz that should someday be detectable by third generation detectors like the Einstein Telescope, or possibly even by advanced LIGO should the source be sufficiently close (D ≲ 20 Mpc) and the high-frequency peak of sufficiently high quality . Figure 14. Open in a new tab Dimensionless GW strain Dh/m 0, where D is the distance to the source and m 0 the total mass of the binary, versus time for four different NS-NS merger calculations. The different merger types become apparent in the post-merger GW signal, clearly indicating how BH formation rapidly drives the GW signal down to negligible amplitudes. Image reproduced by permission from Figures 5 and 6 of , copyright by APS. Using new multi-orbit simulations of NS-NS mergers, Baiotti et al. [14, 15] showed that the semi-analytic effective one-body (EOB) formalism severely underestimates high-order relativistic corrections even when lowest-order finite-size tidal effects were included. As a result, phase errors of almost a quarter of a radian can develop, although these may be virtually eliminated by introducing a second-order “next-to-next-to-leading order” (NNLO) correction term and fixing the coefficient to match numerical results. The excellent agreement between pre-merger numerical waveforms and the revised semi-analytic EOB approximant is shown in Figure 16. Figure 16. Open in a new tab Comparison between numerical waveforms, shown as a solid black line, and semi-analytic NNLO EOB waveforms, shown as a red dashed line (top panel). The top panels show the real parts of the EOB and numerical relativity waveforms, and the middle panels display the corresponding phase differences between waveforms generated with the two methods. There is excellent agreement between with the numerical waveform almost up to the time of the merger as shown by the match of the orbital frequencies (bottom panel). Image reproduced by permission from Figure 14 of , copyright by APS. Binary eccentricity The effects of binary eccentricity on NS-NS mergers was recently studied by the Jena group . Such systems, which would indicate dynamical formation processes rather than the long-term evolution of primordial binaries, evolve differently in several fundamental ways from binaries that merge from circular orbits. For nearly head-on collisions, they found prompt BH formation and negligible disk mass production, with only a single GW burst at frequencies comparable to the quasi-normal mode of the newly formed BH. For a collision in which mass transfer occurred at the first passage but two orbits were required to complete the merger and form a BH, a massive disk was formed, containing 8% of the total system mass even at time Δ t = 100 M tot ≈ 280 M⊙ after the formation of the BH. During that time, the black hole accreted an even larger amount of mass, representing over twice the mass of the remaining disk. Between the first close passage and the second, during which the two NS merged, the GW signal was seen to be quasi-periodic, and a a frequency comparable to the fundamental oscillation mode of the two NS, a result that was duplicated in a calculation for which the periastron fell outside the Roche limit and the eccentric binary survived for the full duration of the run, comprising several orbits. Simulations including microphysics In parallel to efforts in full GR, there has also been great progress in numerical simulations that include approximate relativistic treatments but a more detailed approach to microphysical issues. The first simulations to use a realistic EOS for NS-NS mergers were performed by Ruffert, Janka, and collaborators [253, 139, 254], who assumed the Lattimer-Swesty EOS for their Newtonian PPM-based Eulerian calculations. They were able to determine a physically meaningful temperature for NS-NS merger remnants of 30–50 MeV, an overall neutrino luminosity of roughly 10 53 erg/s for tens of milliseconds, and a corresponding annihilation rate of 2–5 × 10 50 erg/s given the computed annihilation efficiencies of a few parts in a thousand. This resulted in an energy loss of 2–4 × 10 49 erg over the lifetime of the remnant , a value later confirmed in multigrid simulations that replaced the newly formed HMNS by a Newtonian or quasi-relativistic BH surrounded by the bound material making up a disk . The temperatures in the resulting neutron-rich (Y e ≈ 0.05–0.2) remnant were thought to be encouraging for the production of r-process elements , although numerical resolution of the low-density ejecta limited the ability to make quantitatively accurate estimates of its exact chemical distribution. Further calculations, some of which involved unequal-mass binaries, indicated that the temperatures and electron fractions in the ejecta were likely not sufficient to produce solar abundances of r-process elements , with electron fractions in particular smaller than those set by hand in the r-process production model that appears in [111, 245]. More recently, it was suggested that the decompression of matter originally located in the inner crust of a NS and ejected during a merger has a nearly solar elemental distribution for heavy r-process elements (A> 140). This indicates that NS-NS mergers may be the source of the observed cosmic r-process elements should there be sufficient mass loss per merger event, M ej ∼ 3–5 × 10−5 M⊙, although these amounts have yet to be observed in full GR simulations which have often admittedly been performed using cruder microphysical treatments. In , Rosswog and Davies included a detailed neutrino leakage scheme in their calculations and also adopted the Shen EOS for several calculations, finding in a later paper that the gamma-ray energy release is roughly 10 48 erg, in line with previous results from other groups, but noting that the values would be significantly higher if temperatures in the remnant were higher, since the neutrino luminosity scales like a very high power of the temperature. These calculations also identified NS-NS mergers as likely SGRB candidates given the favorable geometry , and the possibility that the MRI in a HMNS remnant could dramatically boost magnetic fields on the sub-second timescales characterizing SGRBs . Rosswog and Liebendörfer found that electron antineutrinos dominate the emission, as had Ruffert and Janka , though the exact thermodynamic and nuclear profiles were found to be somewhat sensitive to the properties of the EOS model. More recently, using the VULCAN 2-dimensional multi-group flux-limited-diffusion radiation hydrodynamics code to evaluate slices taken from SPH calculations, Dessart et al. found that neutrino heating of the remnant material will eject roug hly 10−4 M⊙ from the system. Price and Rosswog [233, 247] performed the first MHD simulation of merging NS-NS binaries using an SPH code that included magnetic field effects, finding that the Kelvin-Helmholtz unstable vortices formed at the contact surface between the two NSs could boost magnetic fields rapidly up to ∼ 10 17 G. These results were not seen in GRMHD simulations, where gains in the magnetic field strength generated by dynamos were limited by the swamping of the vortex sheet at the surface of contact by rapidly infalling NS material that went on to form the eventual HMNS or BH . Longer-term simulations did note that shearing instabilities were able to support power-law, or perhaps even exponential, growth of the magnetic fields on longer timescales (∼ 10 s of ms), which augurs well for NS-NS mergers as the central engines of SGRBs . An effort to identify potential observational differences between NSs and COs with quark-matter interiors has been led by Oechslin and collaborators. Using an SPH code with CF gravity, Oechslin et al. [210, 212] considered mergers of NSs with quark cores described by the MIT bag model [67, 102, 142], which have significantly smaller maximum masses than traditional NSs. They found the hybrid nuclear-quark EOS yielded higher ISCO frequencies for NSs with masses ≳ 1.5 M⊙ and slightly larger GW oscillation frequencies for any resulting merger remnant compared to purely hadronic EOS. Bauswein et al. followed up this work by investigating whether “strangelets”, or small lumps of strange quark matter, would be ejected in sufficient amounts throughout the interstellar medium to begin the phase transition that would convert traditional hadronic NSs into strange stars. They determined that the total rate of strange matter ejection in NS-NS mergers could be as much as 10−8 M⊙ per year per galaxy or essentially zero depending on the parameters input into the MIT bag model, with the upper values clearly detectable by orbiting magnetic spectrometers such as the AMS-02 detector that was recently installed on the International Space Station [182, 148]. Further calculations concluded that the mergers of strange stars produce a much more tenuous halo than traditional NS mergers, more rapid formation of a BH, and higher frequency ringdown emission , as we show in Figure 17. Figure 17. Open in a new tab Evolution of a binary strange star merger performed using a CF SPH evolution. The “spiral arms” representing mass loss through the outer Lagrange points of the system are substantially narrower than those typically seen in CF calculations of NS-NS mergers with typical nuclear EOS models. Image reproduced by permission from Figure 4 of , copyright by APS. Oechslin, Janka, and Marek also analyzed a wide range of EOS models using their CF SPH code, finding that matter in spiral arms was typically cold and that the dynamics of the disk formed around a post-merger BH depends on the initial temperature assumed for the pre-merger NS . They also determined that the kHz GW emission peaks produced by HMNSs may help to constrain various parameters of the original NS EOS, especially its high-density behavior , with further updates to the prediction provided by Bauswein and Janka . Most recently, Stergioulas et al. studied the effect of nonlinear mode couplings in HMNS oscillations, leading to the prediction of a triplet peak of frequencies being present or low mass (M NS = 1.2–1.35 M⊙) systems in the kHz range. Comparison to BH-NS merger results While there is a history of Newtonian, quasi-relativistic, post-Newtonian, and CF gravitational formalisms being used to perform BH-NS merger simulations, their results are nearly always quantitatively, if not qualitatively, different than full GR simulations, and we focus here on the latter (see for a more thorough historical review). Most of the groups that have performed full GR NS-NS merger calculations have also published results on BH-NS mergers, including Whisky (for head-on collisions) , KT [290, 289, 283, 332, 276, 154, 153], UIUC [91, 94, 92], HAD , as well as the SXS collaboration [85, 84, 108, 107] and Princeton (for elliptical mergers) [294, 88]. Summarizing the results of these works, we get a rather coherent picture, which we describe below. The GW signal from BH-NS mergers is somewhat “cleaner” than that from NS-NS mergers, since the disruption of the NS and its accretion by the BH rapidly terminate the GW emission. In general, 3PN estimates model the signal well until tidal effects become important. The more compact the NS, the higher the dimensionless “cutoff frequency” M tot f cut at which the GW energy spectrum plummets, with direct plunges in which the NS is swallowed whole typically yielding excess power near the frequency maximum from the final pre-merger burst. For increasingly prograde BH spins, there is more excess power over the 3PN prediction at lower frequencies, but also a lower cutoff frequency marking the plunge (see the discussion in ). From an observational standpoint, the deviations from point-mass form become more visible for a higher mass BH-NS system, because frequencies scale characteristically like the inverse of the total mass. The distinction is particularly important for Advanced LIGO, as systems with BH ≳ 3 M⊙ typically yield cutoff frequencies within the advanced LIGO band at source distances of D ∼ 100 Mpc, while for lower-mass systems the cutoff occurs at or just above the upper end of the frequency band. This is significantly different than the situation for NS-NS mergers, in which the characteristic frequencies corresponding to the merger itself typically fall at frequencies above the advanced LIGO high-frequency sensitivity limit, and those corresponding to remnant oscillations in the range 2–4 kHz, which will prove a challenge even for third-generation GW detectors. Disk masses for BH-NS mergers were found to be extremely small in the first calculations, all performed using non-spinning BHs [290, 289, 91], but have since been corrected to larger values once more sophisticated grid-based schemes and atmosphere treatments were added to those codes. More recent results indicate disk masses for reasonable physical parameters can be as large as 0. 4 M⊙, for highly-spinning (a BH/M = 0.9) mergers , with values of 0.035–0.05 M⊙ characterizing non-spinning models with mass ratios q ≈ 1/5 . Mass loss into a disk is suppressed by misaligned spins, especially for highly-inclined BHs, so the aligned cases should currently be interpreted as upper limits for the disk mass when alignment is varied . Overall, disk masses for BH-NS merger remnants are comparable to those from NS-NS merger remnants, and may not be clearly distinguishable from them based solely on the emission properties of the disk. For BH-NS mergers with mass ratios q =1/3 and prograde spins of dimensionless magnitude a BH/M = 0.5, the disk parameters found after a run performed with the inclusion of a finite-temperature NS EOS indicated that the neutrino luminosity from the disk might be as high as 10 53 erg/s. While NS-NS merger simulations have led to predictions of neutrino luminosities a few times larger than this, the result does indicate that BH-NS mergers are also plausible SGRB progenitor candidates, possibly with lower characteristic luminosities than bursts resulting from NS-NS mergers. The role of magnetic fields in BH-NS mergers has only been investigated recently [66, 92], in simulations that apply an initially poloidal magnetic field to the NSs in the binary. Magnetic fields were found to have very little effect on the resulting GW signal and the mass accretion rate for the BH for physically reasonable magnetic field strengths, with visible divergences appearing only for B ∼ 10 17 G , which is not particularly surprising. Just as in NS-NS mergers, magnetic fields play very little role during inspiral, and unlike the case of NS-NS mergers there is no opportunity to boost fields at a vortex sheet that forms when the binary makes contact, nor in a HMNS via differential rotation. While the MRI may be important in determining the thermal evolution and mass accretion rate in a post-merger disk, such effects will likely be observable primarily on longer timescales. Just as full-GR NS-NS simulations do not indicate that such mergers are likely sources of the r-process elements we observe in the universe, BH-NS simulations in full GR make the same prediction: no detectable mass loss from the system whatsoever, at least in the calculations performed to date. The picture may change when even larger prograde spins are modeled, since this should lead to maximal disk production, or if more detailed microphysical treatments indicate that a significant wind can be generated from either a HMNS or BH disk and unbind astrophysically interesting amounts of material, but neither has been seen in the numerical results to date. As is seen in NS-NS mergers, the pericenter distance plays a critical role in the evolution of eccentric BH-NS mergers as well. Large disk masses containing up to 0.3 M⊙, with an unbound fraction of roughly 0.15 M⊙, can occur when the periastron separation is located just outside the classical ISCO, with GW signals taking on the characteristic zoom-whirl form predicted for elliptical orbits . In between pericenter passages, radial oscillations of the neutron star produce GW emission at frequencies corresponding to the f-mode for the NS as well . Summary and Likely Future Directions Returning to the questions posed in Section 6, we can now provide the current state of the field’s best answers, though this remains a very active area of research and new results will certainly continue to modify this picture. With regard to the final fate of the merger remnant, calculations using full GR are required, but the details of the microphysics do not seem to play a very strong role. It is now possible to determine whether or not a pair of NSs with given parameters and specified EOS will form a BH or HMNS promptly after merger, and to estimate whether a HMNS will collapse on a dynamical timescale or one of the longer dissipative timescales (see, e.g., ). For NS-NS binaries with sufficiently small masses, it is also possible to determine quickly whether the remnant mass is below the supramassive limit for which a NS is stabilized against collapse by uniform rotation alone, and thus would be unlikely to collapse, barring a significant amount of fallback accretion, unless pulsar emission or magnetic field coupling to the outer disk reduced the rotation rate below the critical value. This scenario likely applies only for mergers where the total system mass is relatively small M tot ≲ 2.5–2.6 M⊙ , even taking into account the current maximum observed NS mass of M NS = 1.97 M⊙ . Based on the wide arrays of EOS models already considered, it is entirely possible to infer the likely fate for sets of parameters and/or EOS models that have not yet been simulated, although no one has yet published a “master equation” that summarizes all of the current work into a single global form. While magnetic fields with realistic magnitudes are unlikely to affect the BH versus HMNS question [172, 117], finite-temperature effects might play a nontrivial role should NSs be sufficiently hot prior to merger (and see also ). In the end, by the time the second generation of GW detectors make the first observations of mergers, the high-frequency shot-noise cutoff will prove to be a bigger obstacle to determining the fate of the remnant than any numerical uncertainty. A schematic diagram showing the possible final fates for a NS-NS merger along with the potential EM emission (see Figure 21 of ) is shown in Figure 18. GW emission during merger is also well-understood, though there are a few gaps that need to be filled, with full GR again a vital requirement. While the PN inspiral signal prior to merger is very well understood, finite-size tidal effects introduce complications beyond those seen in BH-BH mergers, yet the longest calculations performed to date encompass fewer orbits prior to merger than the longest BH-BH runs . As noted in and elsewhere, longer calculations will likely appear over time, helping to refine the prediction for the NS-NS merger GW signal as the binary transitions from a PN phase into one that can only be simulated using full GR, and teasing out the NS physics encoded in the GW signal. It seems clear from the published work that the emission during the onset of the merger is well-understood, as is the very rapid decay that occurs once the remnant collapses to a BH, either promptly or following some delay. GW emission from HMNSs has been investigated widely, and there have been correlations established between properties of the initial binary and the late-stage high-frequency emission (see, e.g., [145, 117]), but given that magnetic fields, neutrino cooling, and other microphysical effects seem to be important, a great deal of work remains to be done. Perhaps more importantly, since HMNSs emit radiation at frequencies well beyond the shot-noise limit of even second-generation GW detectors, while the final inspiral occurs near peak sensitivity, it is likely that the first observations of NSs will constrain the nuclear EOS (or perhaps the quark matter EOS ) primarily via the detection of small finite-size effects during inspiral. Since QE calculations are computationally inexpensive compared to numerical merger simulations, there should be much more numerical data available about the inspiral stage than other phases of NS-NS mergers, which should help optimize the inferences to be drawn from future observations. Determining the mass of the thick disk that forms around a NS-NS merger remnant remains a very difficult challenge, since its density is much lower and harder to resolve using either grid-based or particle-based simulations. The parameterization given by Eq. 34 is generally consistent with the GR calculations of other groups (see, e.g., ), and seems to reflect a current consensus. It is also clear that disk masses around HMNSs (up to 0.2 M⊙) are significantly larger than those forming around prompt collapses, which are limited to about 0.05 M⊙. It is likely that several orders of magnitude more mass-energy are present in the remnant disk than is observed in EM radiation from a SGRB. Modeling the emission from the disk (and possibly a HMNS) remains extremely challenging. Neutrino leakage schemes have been applied in both approximate relativistic calculations [252, 246] and full GR , and a more complex flux-limited diffusion scheme has been applied to the former as a post-processing step , but there are no calculations that follow in detail the neutrinos as they flow outward, annihilate, and produce observable EM emission. At present, nuclear reactions are typically not followed in detail; rather, the electron fraction of the nuclear material, Y e, is evolved, and used to calculate neutrino emission and absorption rates. Magnetic fields, on the other hand, are starting to be much better understood. B-fields do seem to grow quite large through winding effects, even during the limited amount of physical time that can currently be modeled numerically [6, 172, 332, 241], with some calculations indicating exponential growth rates. The resulting geometries seem likely to produce the disk/jet structure observed throughout astrophysics when magnetized objects accrete material, which span scales from stellar BHs or pre-main sequence stars all the way up to active galactic nuclei . While recent numerical simulations have strengthened the case for NS-NS mergers as SGRB progenitors, full GR calculations have not generated much support for the same events yielding significant amounts of r-process elements. Noting the standard caveat that low-density ejecta are difficult to model, and that nuclear reactions are rarely treated self-consistently, there is still tension between CF calculations producing ejecta with the proper temperatures and masses to reproduce the observed cosmic r-process abundances (see, e.g., ), and full GR calculations that produce almost no measurable unbound material whatsoever. Figure 18. Open in a new tab Summary of potential outcomes from NS-NS mergers. Here, M thr is the threshold mass (given the EOS) for collapse of a HMNS to a BH, and Q m is the binary mass ratio. ‘Small’, ‘massive’, and ‘heavy’ disks imply total disk masses M disk ≪ 0.01 M⊙, 0.01 M⊙ ≲ M disk ≲ 0.03 M⊙, and M disk ≳ 0.05 M⊙, respectively. ‘B-field’ and ‘J-transport’ indicate potential mechanisms for the HMNS to eventually lose its differential rotation support and collapse: magnetic damping and angular momentum transport outward into the disk. Spheroids are likely formed only for the APR and other stiff EOS models that can support remnants with relatively low rotational kinetic energies against collapse. Image reproduced by permission from , copyright by APS. While NS-NS merger calculations have seen tremendous progress in the past decade, the future remains extremely exciting. Between the addition of more accurate and realistic physical treatments, the exploration of the full phase space of models, and the linking of numerical relativity to astrophysical observations and GW detection, there remain many unsolved problems that will be attacked over the course of the next decade and beyond. Acknowledgements J.A.F. acknowledges support from NASA under award 08-ATFP-0093 and from NSF Grant PHY-0903782. F.A.R. acknowledges support from NSF Grant PHY-0855592 and thanks the Aspen Center for Physics, supported by NSF Grant PHY-1066293, for hospitality while this work was being completed. We thank the referees for their careful reading and helpful suggestions. Field evolution equations In the BSSN evolution system, we define the following variables in terms of the standard ADM 4-metric g ij, 3-metric γ ij, and extrinsic curvature K ij: 35 36 37 38 39 The evolution system consists of 15 equations for the various field terms, 40 41 42 43 44 where the matter source terms contain various projections of the stress-energy tensor, defined through the relations 45 46 47 We have introduced the notation ∂ 0 = ∂ t − β j ∂ j. All quantities with a tilde involve the conformal 3-metric , which is used to raise and lower indices. In particular, and refer to the covariant derivative and the Christoffel symbols with respect to . Parentheses indicate symmetrization of indices, and the expression […]TF denotes the trace-free part of the expression inside the brackets. In the BSSN approach, the Ricci tensor is typically split into two pieces, whose respective contributions are given by 48 49 These equations must be supplemented with gauge conditions that determine the evolution of the lapse function α and shift vector β i. Noting that some groups introduce slight variants of these, the moving puncture gauge conditions that have become popular for all GR merger calculations involving BHs and NSs typically take the form 50 51 52 where B i is an intermediate quantity used to convert the second-order “Gamma-driver” shift condition into a pair of first-order equations, and η is a user-prescribed term used to control dissipation in the simulation. Note that it is possible to replace the three instances of ∂ t in the shift evolution equations 51 and 52 by the shift-advected time derivative (∂ t − β j ∂ j) without changing the stability or hyperbolicity properties of the evolution scheme; in both cases moving punctures translate smoothly across a grid over long time periods and both systems are strongly hyperbolic so long as the shift vector does not grow too large within the simulation domain . The generalized harmonic formulation involves recasting the Einstein field equations, Eq. 9 in the form 53 and, after some tensor algebra, rewriting the Ricci tensor in the form 54 The Christoffel coefficients are calculated from the full 4-metric, 55 and the gauge source terms H μ are defined in Eq. 25. Given well-posed initial data for the metric and its first time derivative (since the system is second-order in time according to Eq. 54), the evolution of the system may be treated by a first-order reduction that specifies the evolution of the four functions H μ along with the spacetime metric g μν, its projected time derivatives Π μ ν = −n> α ∂ α g μν, and its spatial derivatives 56 subject to a constraint specifying that the derivative terms Φ iμν remain consistent with the metric g μν in time. In practice, one typically introduces a constraint for the source functions, defining 57 and then modifies the evolution equation by appending a constraint damping term to the RHS of the stress energy-term (following [130, 231] 58 where n μ is the unit normal vector to the hypersurface (see Eq. 15). The gauge conditions used in the first successful simulations of merging BH binaries consisted of the set 59 with ξ 1, ξ 2, η constant parameters used to tune the evolution. The first one drives the coordinates toward the ADM form and the latter provides dissipation. The binary NS-NS merger work of chose harmonic coordinates with H μ = 0. GR Hydrodynamical and MHD equations In what follows, we will adopt the stress energy tensor of an ideal relativistic fluid, 60 where ρ, P, and u μ are the rest mass density, pressure, and fluid 4-velocity, respectively, and 61 is the relativistic specific enthalpy, with ϵ the specific internal energy of the fluid. The equations of ideal GR hydrodynamics may be derived from the local GR conservation laws of mass and energy-momentum: 62 where ∇μ denotes the covariant derivative with respect to the 4-metric, and J μ = ρu μ is the mass current. The 3-velocity υ i can be calculated in the form 63 where 64 is the Lorentz factor. The contravariant 4-velocity is then given by: 65 and the covariant 4-velocity is: 66 To cast the equations of GR hydrodynamics as a first-order hyperbolic flux-conservative system for the conserved variables D, S i, and τ, defined in terms of the primitive variables ρ, ϵ, υ i, we define 67 68 69 where γ is the determinant of γ ij. The evolution system then becomes 70 with 71 Here, and are the 4-Christoffel symbols. Magnetic fields may be included in the formalism, in the ideal MHD limit under which we assume infinite conductivity, by adding three new evolution equations and modifying those above to include magnetic stress-energy contributions of the form 72 where the magnetic field seen by a comoving observer, b μ is defined in terms of the dual Faraday tensor F νμ by the condition 73 where b 2 = b μ b μ represents twice the magnetic pressure. 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[DOI] [Google Scholar] Articles from Living Reviews in Relativity are provided here courtesy of Springer ACTIONS View on publisher site PDF (4.1 MB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction Evolutionary Channels and Population Estimates Stages of a Binary Merger Initial Data and Quasi-Equilibrium Results Dynamical Calculations: Numerical Techniques Dynamical Calculations Summary and Likely Future Directions Acknowledgements Contributor Information 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
11472
https://www.ncbi.nlm.nih.gov/books/NBK507850/
An official website of the United States government 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 in Account Logged in as:username Dashboard Publications Account settings Log out Access keys NCBI Homepage MyNCBI Homepage Main Content Main Navigation Browse Titles Advanced Help Disclaimer NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health. StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. StatPearls [Internet]. Show details Treasure Island (FL): StatPearls Publishing; 2025 Jan-. Biochemistry, Clotting Factors Walker Barmore; Tanvir Bajwa; Bracken Burns. Author Information and Affiliations Authors Walker Barmore1; Tanvir Bajwa2; Bracken Burns3. Affiliations 1 Mercer University School of Medicine 2 Midwestern University Chicago 3 East Tennessee State University (ETSU) Last Update: February 24, 2023. Introduction This topic analyzes the biochemistry of the coagulation cascade, specifically clotting factors and their biochemical interactions and roles among cell membranes, platelets, proteases, and cofactors. Other components involved in clot formation should be referenced, but the focus should be on clotting factors. The coagulation cascade is a well-studied and pertinent topic for health professionals to understand. Although this topic does not cover the coagulation cascade and its role in hemostasis as a simple chain of events, a brief overview should be included. A thorough examination of these biochemical interactions illuminates the coagulation cascade's underlying intricacies, enabling a seamless cohesive process. Fundamentals Hemostasis Clotting factors are arguably the crux and most essential components of hemostasis. Hemostasis is the body’s physiologic response to vascular endothelial injury, resulting in a series of processes that attempt to retain blood within the vascular system through clot formation. Hemostasis can be further divided into primary and secondary hemostasis. Primary hemostasis forms a soft platelet plug and involves vasoconstriction, platelet adhesion, activation, and aggregation. Secondary hemostasis is primarily defined as the formation of fibrinogen into fibrin, which ultimately evolves the soft platelet plug into a hard, insoluble fibrin clot. Within primary and secondary hemostasis, 3 coagulation pathways exist: intrinsic, extrinsic, and common. Pathways The intrinsic pathway responds to spontaneous, internal vascular endothelium damage, whereas the extrinsic pathway becomes activated secondary to external trauma. Both intrinsic and extrinsic pathways meet at a shared point to continue coagulation, the common pathway. Clotting factors in the intrinsic pathway include XII, XI, IX, and VIII. Clotting factors involved in the extrinsic pathway include factors VII and III. The common pathway includes clotting factors X, V, II, I, and XIII. Clotting factors can also be referred to outside of their Roman numeral designations. In the intrinsic pathway, factors XII, XI, IX, and VIII are the Hageman factor, plasma thromboplastin antecedent, Christmas factor, and antihemophilic factor A. In the extrinsic pathway, factors VII and III are stabilizing and tissue factors, respectively. The common pathway factors X, V, II, I, and XIII are the Stuart-Prower, proaccelerin, prothrombin, fibrinogen, and fibrin-stabilizing factors. Clotting factor IV is a calcium ion that plays an important role in all 3 pathways. Some clotting factors function as serine proteases, specifically factors II, VI, IX, and X. Cellular Level The overwhelming majority of clotting factors are manufactured principally in hepatocytes. Hepatocytes are responsible for providing the body with clotting factors XIII, XII, XI, X, IX, VII, V, II, and I. Clotting factors VIII (antihemophilic factor A) and III (tissue factor) originate from endothelial cells, whereas clotting factor IV (calcium ion) is freely available in plasma. Megakaryocytes produce the body’s platelets and contribute to factor V production. Mechanism Vascular injury results in the exposure of subendothelial collagen and von Willebrand factor (vWF). vWF is a glycoprotein that is the initial stationary foundation on which a clot forms. Subendothelial vWF, which is also present in the vasculature and acts to increase the half-life of VII, binds to glycoprotein Ib (GpIb) on platelets. This causes a conformational change on the platelet surface, exposing glycoprotein IIb/IIIa (GpIIb/IIIa). Due to the conformational change, circulating fibrinogen attaches to GpIIb/IIIa. A soft platelet plug has formed in hemostasis, and the importance of biochemical interactions of clotting factors arises. Membrane Binding In addition to the exposure of GpIIb/IIIa due to conformational change occurring on the platelet, phosphatidylserine also emerges on the platelet surface. Phosphatidylserine is a membrane phospholipid whose polar end has a negative charge and, as a result, provides an excellent surface for a calcium ion to bind. The interaction between negatively charged phosphatidylserine and calcium does not completely negate calcium’s positive charge. This allows for serine proteases to bind to the surface of the platelet membrane. This binding is possible due to the carboxylation of clotting factors II, VII, IX, and X. These clotting factors have a region called gamma-carboxyglutamic acid that undergoes vitamin-K dependent carboxylation via gamma-glutamyl carboxylase. The enzyme adds a negatively charged carboxyl group to glutamic acid residues, which calcium easily binds to. As a result, the clotting factors can adhere to the platelet surface as serine proteases. Intrinsic Pathway Proteases Factor XII activation is the first step of the intrinsic pathway. Its activation is induced via contact with subendothelial collagen in the presence of high molecular weight kininogen. The zymogen to enzyme activation was graphically denoted with the letter a, for example, XIIa. XIIa, in turn, activates XI into XIa, which leads to the activation of IXa. Our previous discussion of gamma-carboxylation and platelet membrane interaction becomes important at this point. Clotting factor IX is a serine protease within the intrinsic pathway. Although IXa is in its active form, IXa enzyme efficiency is abysmal without its essential cofactor, factor XIII. Proteolysis ensues once XIII and IXa are bound together (XIII-IXa) on the platelet membrane. Specifically, the serine protease cleaves certain C-terminal arginine residues in the zymogen, which results in its subsequent activation. From here, we can understand how VIII-IXa activates factor X into Xa and leads to the common pathway. Extrinsic Pathway P rotease Although the extrinsic pathway involves fewer steps than the common pathway, the role of serine proteases is just as important. When external insult occurs, clotting factor VII, along with its cofactor tissue thromboplastin, becomes an active protease and catalyzes X into Xa, which leads to the common pathway. Testing Prothrombin time measures coagulation throughout the extrinsic pathway and common pathway. A normal prothrombin time is between 11 to 15 seconds; however, this time may vary slightly in the healthcare setting. The international normalized ratio (INR) is used to mitigate the slight discrepancies in prothrombin time and is also the test of choice when a patient is on warfarin therapy. A therapeutic INR is usually considered between 2 to 3 (for most clinical situations requiring anticoagulation with warfarin). Partial thromboplastin time (PTT) measures coagulation throughout the intrinsic and common pathways. A normal PTT time is 25 to 40 seconds. PTT is the test of choice when monitoring a patient on unfractionated heparin. Routine PTT surveillance is not necessary for patients on low-molecular-weight heparin. Bleeding time measures platelet function and how well platelets can form a clot. Normal bleeding time is 2 to 7 minutes. Bleeding time is typically elevated in conditions of platelet dysfunction. Pathophysiology Here, commonly tested areas regarding the pathophysiology of clotting factors is discussed. Hemophilias Hemophilia A is an X-linked recessive coagulopathy that results in dysfunctional VIII. From our earlier discussion, we can see how dysfunctional VIII result in coagulopathy and prolonged PTT. Patients with this disorder often present with easy bruising, bleeding after dental procedures or from operations in general, and hemarthrosis. Hemophilia A can be treated with desmopressin and recombinant factor VIII. Desmopressin causes endothelial cells to release vWF, which stabilizes VIII. Hemophilia B, sometimes called Christmas disease, is an X-linked recessive coagulopathy resulting in dysfunction of IX. As with hemophilia A, hemophilia B also cause a prolonged PTT. The difference is hemophilia A is a cofactor deficiency while hemophilia B is a protease deficiency; therefore, desmopressin not be a good treatment option as these patients require recombinant factor IX. Hemophilia B present with the same symptoms as hemophilia A. It is important to note that hemophilia A and B have a normal prothrombin/INR. Von Willebrand Disease Von Willebrand disease (vWD) is the most commonly inherited coagulopathy. vWD can be differentiated from hemophilias in several ways. First, the vWF mode of inheritance is autosomal dominant. Secondly, vWD is a disease of platelet dysfunction and manifest primarily as mucosal membrane bleeding, such as epistaxis and prolonged menstrual cycles. Thirdly, bleeding time is normal in hemophilia, whereas it is prolonged in vWD. Since vWF increases the half-life of VIII, one can also expect to see a prolonged PTT in this disorder. Desmopressin can be used as a treatment option; however, certain subtypes of vWD do not warrant this treatment option. Vitamin K Deficiency Previously, we discussed the importance of vitamin K and its clotting factors II, VII, IX, X, protein C, and S. The effects of vitamin K deficiency can be observed in both the extrinsic and intrinsic pathways and directly measured via prothrombin time and PTT, which be prolonged. The etiology of vitamin K deficiency is extensive but commonly arises on test questions regarding patients with poor diet, pancreatic insufficiency, liver disease, intestinal flora imbalances, neonates, or mimicked by patients on warfarin therapy. Warfarin Vitamin K assists in the carboxylation of clotting factors II, VII, IX, X, protein C, and S. The enzyme responsible for gamma-carboxylation is vitamin K epoxide reductase, which is inhibited by warfarin. As mentioned previously, patients on warfarin have the coagulation status measured via INR. In emergency clinical settings, warfarin’s therapeutic effects are negated by the administration of fresh frozen plasma. In a less urgent clinical setting, patients may be administered vitamin K. Rarely; patients may experience warfarin-induced skin necrosis within the first few days of beginning warfarin. This is because protein C has the shortest half-life of vitamin K-dependent clotting factors; therefore, a patient enters a prothrombotic state. However, this rare complication is more common in patients with protein C deficiency. To help eliminate this complication, patients are often co-administered heparin while beginning warfarin therapy as heparin’s onset is immediate while warfarin’s onset takes 2 to 3 days. Clinical Significance By understanding the biochemistry of clotting factors, healthcare professionals can quickly identify probable causes of a patient's coagulopathy by examining a patient's coagulation studies. For elevations in prothrombine time/INR, we can focus on conditions such as liver disease, warfarin use, vitamin-K deficiency, and deficiencies in the extrinsic or common pathway. For elevations in PTT, we narrow our focus on more common causes such as hemophilias, unfractionated heparin use, vitamin-K deficiency, and vWF, with careful attention to bleeding time. It is important to remember that prolongations in prothrombin time and PTT could also be due to deficiencies in the common pathway. Still, the previous conditions and examples yield a higher probability of identifying the root of the coagulopathy. Review Questions Access free multiple choice questions on this topic. Comment on this article. References 1. : van der Meijden PEJ, Heemskerk JWM. Platelet biology and functions: new concepts and clinical perspectives. Nat Rev Cardiol. 2019 Mar;16(3):166-179. [PubMed: 30429532] 2. : Chinnaraj M, Planer W, Pozzi N. Structure of Coagulation Factor II: Molecular Mechanism of Thrombin Generation and Development of Next-Generation Anticoagulants. Front Med (Lausanne). 2018;5:281. [PMC free article: PMC6176116] [PubMed: 30333979] 3. : Tillman BF, Gruber A, McCarty OJT, Gailani D. Plasma contact factors as therapeutic targets. Blood Rev. 2018 Nov;32(6):433-448. [PMC free article: PMC6185818] [PubMed: 30075986] 4. : D'Alessandro E, Posma JJN, Spronk HMH, Ten Cate H. Tissue factor (:Factor VIIa) in the heart and vasculature: More than an envelope. Thromb Res. 2018 Aug;168:130-137. [PubMed: 30064684] 5. : Holthenrich A, Gerke V. Regulation of von-Willebrand Factor Secretion from Endothelial Cells by the Annexin A2-S100A10 Complex. Int J Mol Sci. 2018 Jun 13;19(6) [PMC free article: PMC6032327] [PubMed: 29899263] 6. : Swieringa F, Spronk HMH, Heemskerk JWM, van der Meijden PEJ. Integrating platelet and coagulation activation in fibrin clot formation. Res Pract Thromb Haemost. 2018 Jul;2(3):450-460. [PMC free article: PMC6046596] [PubMed: 30046749] 7. : Lippi G, Favaloro EJ. Laboratory hemostasis: from biology to the bench. Clin Chem Lab Med. 2018 Jun 27;56(7):1035-1045. [PubMed: 29455188] 8. : Montagnana M, Lippi G, Danese E. An Overview of Thrombophilia and Associated Laboratory Testing. Methods Mol Biol. 2017;1646:113-135. [PubMed: 28804823] 9. : Vinholt PJ, Hvas AM, Nybo M. An overview of platelet indices and methods for evaluating platelet function in thrombocytopenic patients. Eur J Haematol. 2014;92(5):367-76. [PubMed: 24400878] 10. : Cheung KL, Bouchard BA, Cushman M. Venous thromboembolism, factor VIII and chronic kidney disease. Thromb Res. 2018 Oct;170:10-19. [PubMed: 30081388] : Disclosure: Walker Barmore declares no relevant financial relationships with ineligible companies. : Disclosure: Tanvir Bajwa declares no relevant financial relationships with ineligible companies. : Disclosure: Bracken Burns declares no relevant financial relationships with ineligible companies. Copyright © 2025, StatPearls Publishing LLC. This book is distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International (CC BY-NC-ND 4.0) ( ), which permits others to distribute the work, provided that the article is not altered or used commercially. You are not required to obtain permission to distribute this article, provided that you credit the author and journal. Bookshelf ID: NBK507850PMID: 29939627 Share Views PubReader Print View Cite this Page Barmore W, Bajwa T, Burns B. Biochemistry, Clotting Factors. [Updated 2023 Feb 24]. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; 2025 Jan-. In this Page Introduction Fundamentals Cellular Level Mechanism Testing Pathophysiology Clinical Significance Review Questions References Related information PMC PubMed Central citations PubMed Links to PubMed Similar articles in PubMed Biochemistry, Antithrombin III.[StatPearls. 2025] Biochemistry, Antithrombin III. Hsu E, Moosavi L. StatPearls. 2025 Jan Review Clotting factors: Clinical biochemistry and their roles as plasma enzymes.[Adv Clin Chem. 2020] Review Clotting factors: Clinical biochemistry and their roles as plasma enzymes. Winter WE, Greene DN, Beal SG, Isom JA, Manning H, Wilkerson G, Harris N. Adv Clin Chem. 2020; 94:31-84. Epub 2019 Oct 18. Physiology, Factor V.[StatPearls. 2025] Physiology, Factor V. Lam W, Moosavi L. StatPearls. 2025 Jan Review How it all starts: Initiation of the clotting cascade.[Crit Rev Biochem Mol Biol. 2015] Review How it all starts: Initiation of the clotting cascade. Smith SA, Travers RJ, Morrissey JH. Crit Rev Biochem Mol Biol. 2015; 50(4):326-36. Epub 2015 May 28. Review New fundamentals in hemostasis.[Physiol Rev. 2013] Review New fundamentals in hemostasis. Versteeg HH, Heemskerk JW, Levi M, Reitsma PH. Physiol Rev. 2013 Jan; 93(1):327-58. See reviews...See all... Recent Activity Clear)Turn Off)Turn On) Biochemistry, Clotting Factors - StatPearls Biochemistry, Clotting Factors - StatPearls Your browsing activity is empty. Activity recording is turned off. Turn recording back on) See more... 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https://huggingface.co/datasets/inclusionAI/Ring-lite-distill-preview-sft-data/viewer/default/train?p=1
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11474
https://abm.memberclicks.net/assets/DOCUMENTS/PROTOCOLS/3-supplementation-protocol-english.pdf
ABM Protocol ABM Clinical Protocol #3: Supplementary Feedings in the Healthy Term Breastfed Neonate, Revised 2017 Ann Kellams, 1 Cadey Harrel, 2 Stephanie Omage, 3 Carrie Gregory, 4,5 Casey Rosen-Carole, 4,5 and the Academy of Breastfeeding Medicine A central goal of The Academy of Breastfeeding Medicine is the development of clinical protocols for managing common medical problems that may impact breastfeeding success. These protocols serve only as guidelines for the care of breastfeeding mothers and infants and do not delineate an exclusive course of treatment or serve as standards of medical care. Variations in treatment may be appropriate according to the needs of an individual patient. Definitions Used in This Protocol Exclusive breastfeeding: Feeding only breast milk (at the breast or own mothers’ expressed breast milk), no food or water except vitamins, minerals, and medications. Supplementary feedings: Additional fluids provided to a breastfed infant before 6 months (recommended du-ration of exclusive breastfeeding). These fluids may include donor human milk, infant formula, or other breast milk substitutes (e.g., glucose water). Complementary feedings: Solid or semisolid foods provided to an infant in addition to breastfeeding when breast milk alone is no longer sufficient to meet nutri-tional needs. Term infant: In this protocol ‘‘term infant’’ also includes early-term infants (gestational age 37–38 6/7 weeks). Background Given early opportunities to breastfeed, breastfeeding as-sistance, and instruction the vast majority of mothers and infants will successfully establish breastfeeding. Although some infants may not successfully latch and feed well during the first day (24 hours), most will successfully breastfeed with time, appropriate evaluation and support, with minimal intervention. Exclusive breastfeeding for the first 6 months is associated with the greatest protection against major health problems for both mothers and infants.1–3 Unfortunately, infant formula supplementation of healthy neonates in hos-pital is commonplace,4,5 despite widespread recommenda-tions to the contrary.6–8 Early supplementation with infant formula is associated with decreased exclusive breastfeeding rates in the first 6 months and an overall shorter duration of breastfeeding.9,10 Therefore, hospitals, healthcare facilities, and community organizations that promote breastfeeding are integral in improving the exclusivity and duration of breast-feeding.10 One way of achieving this is by following The Ten Steps to Successful Breastfeeding (the basis for the Baby-Friendly Hospital Initiative), both in the hospital and community. Newborn physiology Small quantities of colostrum are appropriate for the size of a newborn’s stomach,11–13 prevent hypoglycemia in a healthy, term, appropriate for gestational age infant,14,15 and are easy for an infant to manage as he/she learns to coordinate sucking, swallowing, and breathing. Healthy term infants also have sufficient body water to meet their metabolic needs, even in hot climates.16–18 Fluid necessary to replace insen-sible fluid loss is adequately provided by breast milk alone.7,18 Newborns lose weight because of physiologic di-uresis of extracellular fluid following transition from intra-uterine to extrauterine life and the passage of meconium. In a prospective cohort of mothers in a U.S. Baby-Friendly des-ignated hospital with optimal support of infant feeding, the mean weight loss of exclusively breastfed infants was 5.5%; notably, greater than 20% of healthy breastfed infants lost more than 7% of their birthweight.19 A study of over 160,000 healthy breastfed infants resulted in the creation of hour-specific nomograms for infant weight loss for exclusively breastfed newborns that showed differentially increased weight loss in those born by cesarean section than by vaginal birth. In this study, almost 5% of vaginally born infants and 1Department of Pediatrics, University of Virginia, Charlottesville, Virginia. 2Department of Family & Community Medicine, University of Arizona College of Medicine and Family Medicine Residency, Tucson, Arizona. 3Discipline of General Practice, The University of Queensland, Brisbane, Australia. Departments of 4Pediatrics and 5OBGYN, University of Rochester, Rochester, New York. BREASTFEEDING MEDICINE Volume 12, Number 3, 2017 ª Mary Ann Liebert, Inc. DOI: 10.1089/bfm.2017.29038.ajk 1 >10% of those born by cesarean section had lost ‡10% of their birth weight by 48 hours after birth. By 72 hours, >25% of infants born by cesarean section had lost ‡10% of their birth weight.20 Breastfed infants regain birth weight at an average of 8.3 days (95% confidence interval: 7.7–8.9 days) with 97.5% having regained their birth weight by 21 days.21 Infants should be followed closely to identify those who lie outside the predicted pattern, but the majority of those breastfed infants will not require supplementation. It should also be noted that excess newborn weight loss is correlated with positive maternal intrapartum fluid balance (received through intravenous fluids) and may not be directly indicative of breastfeeding success or failure.22,23 Early management of the new breastfeeding mother Some breastfeeding mothers question the adequacy of colostrum feedings and perceive that they have an insuffi-cient milk supply.24,25 These women may receive conflicting advice about the need for supplementation and would benefit from reassurance, assistance with breastfeeding technique, and education about the normal physiology of breastfeeding and infant behavior. Inappropriate supplementation may undermine a mother’s confidence in her ability to meet her infant’s nutritional needs26 and give inappropriate messages that may result in supplementation of breastfed infants at home.27 Introduction of infant formula or other supplements may decrease the feeding frequency of the infant, thereby decreasing the amount of breast stimulation a mother re-ceives, which results in a reduction of milk supply.28 Postpartum mothers with low confidence levels are very vulnerable to external influences, such as advice to offer breastfeeding infants supplementation of glucose water or infant formula. Well-meaning healthcare professionals may recommend supplementation as a means of protecting mothers from fatigue or distress, although this can conflict with their role in promoting breastfeeding.29–31 Several so-ciodemographic factors are associated with formula supple-mentation in the hospital, and vary geographically. It is important to recognize and address these factors in a cultur-ally sensitive manner. Inappropriate reasons for supplemen-tation and associated risks are multiple (Appendix Table A1). There are common clinical situations where evaluation and breastfeeding management may be necessary, but SUPPLE-MENTATION IS NOT INDICATED, including: 1. The healthy, term, appropriate for gestational age infant when the infant is feeding well, urinating and stooling adequately, weight loss is in the expected range, and bilirubin levels are not of concern (depending on ges-tational age, time since birth, and any risk factors).32 Newborns are normally sleepy after an initial alert period after birth (2 hours). They then have vari-able sleep–wake cycles, with an additional one or two wakeful periods in the next 10 hours whether fed or not.33 Careful attention to an infant’s early feeding cues, keeping the infant safely skin-to-skin with mother when she is awake, gently rousing the infant to attempt frequent breastfeeds, and teaching the mother hand expression of drops of colostrum,34 may be more appropriate than automatic supple-mentation after 6, 8, 12, or even 24 hours. Increased skin-on-skin time can encourage more frequent feeding. Ten percent weight loss is not an automatic mar-ker for the need for supplementation, but is an indicator for infant evaluation. 2. The infant who is fussy at night or constantly feeding for several hours Cluster feeding (several short feeds close together) is normal newborn behavior, but should warrant a feeding evaluation to observe the infant’s behavior at the breast35 and the comfort of the mother to en-sure that the infant is latched deeply and effectively. Some fussy infants are in pain that should be ad-dressed. 3. The tired or sleeping mother Some fatigue is normal for new mothers. However, rooming out for maternal fatigue does not improve mothers’ sleep time36 and has been shown to reduce breastfeeding exclusivity.37 Extreme fatigue should be evaluated for the safety of mother and baby to avoid falls and suffocation.38 Breastfeeding management that optimizes the infant feeding at the breast may make for a more satisfied infant AND allow the mother to get more rest. The following guidelines address strategies to prevent the need for supplementation (also see Appendix 2) as well as indi-cations for and methods of supplementation for the healthy, term (37- to 42-week), breastfed infant. Indications for supplementa-tion in term, healthy infants are few.7,39 Table 1 lists possible indications for the administration of supplemental feeds. In each case, the medical provider must decide if the clinical benefits outweigh the potential negative consequences of such feedings. Recommendations Step 1. Prevent the need for supplementation 1. There is mixed, but mainly positive, evidence about the role of antenatal education and in-hospital support on the rates of exclusive breastfeeding.40–42 (I) (Quality of evi-dence [levels of evidence I, II-1, II-2, II-3, and III] is based on the U.S. Preventive Services43 Task Force Appendix A Task Force Ratings and is noted in parentheses.) 2. All staff who care for postpartum women should be able to assist and assess breastfeeding infants, especially when other staff with expertise are not available. 3. Both mothers and healthcare professionals should be aware of the risks of unnecessary supplementation. 4. Healthy infants should be placed skin-to-skin with the mother, if she is awake and alert, immediately after birth to facilitate breastfeeding.7,44 (I) The delay in time between birth and initiation of the first breastfeed is a strong predictor of infant formula use and may affect future milk supply.10,45,46 (II-3, II-2, II-3) 5. It is ideal to have the mother and infant room-in 24 hours per day to respond to infant feeding cues, enhance opportunities for breastfeeding, and hence secretory activation (lactogenesis II).7,39,47,48 (III) 6. If mother–infant separation is unavoidable, milk supply is not well established, or milk transfer is inadequate, the mother needs instruction and encouragement to express her milk by hand or pump to stimulate milk 2 ABM PROTOCOL production and provide expressed milk for the in-fant.7,39,48,49 (I, III) This process should begin within 1 hour of birth.45 (II-2) Step 2. Address early indicators of the possible need for supplementation 1. The infant’s medical providers should be notified if the infant or mother meets any criteria for supplementa-tion, as listed in Table 1. 2. All infants must be formally evaluated for position, latch, and milk transfer before the provision of sup-plemental feedings. This evaluation should be under-taken by a healthcare provider with expertise in breastfeeding management, when available.7,48 Step 3. Determine whether supplementation is required and supplement with care 1. The status of the infant requiring supplementation should be determined and any decisions made on a case-by-case basis (guidelines in Table 1). 2. Hospitals should strongly consider formulating and instituting policies to require a medical provider’s order when supplements are medically indicated and informed consent of the mother when supplements are not medically indicated. It is the responsibility of the healthcare provider to fully inform parents of the benefits and risks of supplementation, document pa-rental decisions, and support the parents after they have made a decision.50,51 (III) 3. All supplemental feedings should be documented, in-cluding the content, volume, method, and medical indication or reason. 4. When supplementary feeding is medically necessary, the primary goals are to feed the infant and to optimize the maternal milk supply while determining the cause of low milk supply, poor feeding, or inadequate milk transfer. Supplementation should be performed in ways that help preserve breastfeeding such as limiting the volume to what is necessary for the normal new-born physiology, avoiding teats/artificial nipples,52 (I) stimulating the mother’s breasts with hand expression or pumping, and for the infant to continue to practice at the breast. 5. Optimally, mothers need to express milk frequently, usually once for each time the infant receives a Table 1. Possible Indications for Supplementation in Healthy, Term Infants (37–41 6/7 Weeks Gestational Age) 1. Infant indications a. Asymptomatic hypoglycemia, documented by laboratory blood glucose measurement (not bedside screening methods) that is unresponsive to appropriate frequent breastfeeding. Note that 40% dextrose gel applied to the side of the infant’s cheek is effective in increasing blood glucose levels in this scenario and improves the rate of exclusive breastfeeding after discharge with no evidence of adverse effects.78 Symptomatic infants or infants with glucose <1.4 mmol/L (<25 mg/dL) in the first 4 hours or <2.0 mmol/L (<35 mg/dL) after 4 hours should be treated with intravenous glucose.15 Breastfeeding should continue during intravenous glucose therapy. b. Signs or symptoms that may indicate inadequate milk intake: i. Clinical or laboratory evidence of significant dehydration (e.g., high sodium, poor feeding, lethargy, etc.) that is not improved after skilled assessment and proper management of breastfeeding.79 ii. Weight loss of ‡8–10% (day 5 [120 hours] or later), or weight loss greater than 75th percentile for age. 1. Although weight loss in the range of 8–10% may be within normal limits if all else is going well and the physical examination is normal, it is an indication for careful assessment and possible breastfeeding assistance. Weight loss in excess of this may be an indication of inadequate milk transfer or low milk production, but a thorough evaluation is required before automatically ordering supplementation.19,20,80 2. Weight loss nomograms for healthy newborns by hour of age can be found at: www.newbornweight.org20,80 iii. Delayed bowel movements, fewer than four stools on day 4 of life, or continued meconium stools on day 5 (120 hours).48,80 1. Elimination patterns for newborns for urine and stool should be tracked at least through to the onset of secretory activation. Even though there is a wide variation between infants, the patterns may be useful in determining adequacy of breastfeeding.81,82 II-2. Newborns with more bowel movements during the first 5 days following birth have less initial weight loss, earlier the transition to yellow stools, and earlier return to birth weight.83 c. Hyperbilirubinemia (see ABM Clinical Protocol #22: Guidelines for Management of Jaundice) i. Suboptimal intake jaundice of the newborn associated with poor breast milk intake despite appropriate intervention. This characteristically begins at 2–5 days and is marked by ongoing weight loss, limited stooling and voiding with uric acid crystals. ii. Breast milk jaundice when levels reach 340–425 lmol/L (20–25 mg/dL) in an otherwise thriving infant and where a diagnostic and/or therapeutic interruption of breastfeeding may be under consideration. First line diagnostic management should include laboratory evaluation, instead of interruption of breastfeeding. d. Macronutrient supplementation is indicated, such as for the rare infant with inborn errors of metabolism. 2. Maternal indications a. Delayed secretory activation (day 3–5 or later [72–120 hours] and inadequate intake by the infant).80 b. Primary glandular insufficiency (less than 5% of women—primary lactation failure), as evidenced by abnormal breast shape, poor breast growth during pregnancy, or minimal indications of secretory activation.84,85 c. Breast pathology or prior breast surgery resulting in poor milk production.84 d. Temporary cessation of breastfeeding due to certain medications (e.g., chemotherapy) or temporary separation of mother and baby without expressed breast milk available. e. Intolerable pain during feedings unrelieved by interventions. ABM PROTOCOL 3 supplement, or at least 8 times in 24 hours if the infant is not feeding at the breast. Breasts should be fully drained each time.53 (II-2) Maternal breast engorgement should be avoided as it will further compromise the milk supply and may lead to other complications.54 (III) 6. Criteria for stopping supplementation should be con-sidered from the time of the decision to supplement and should be discussed with the parents. Stopping supplementation can be a source of anxiety for parents and providers. Underlying factors should be addressed and mothers should be assisted with their milk supply, latch, and comfort with assessing the signs that their infant is adequately fed. It is important to closely follow up mother and infant. 7. When the decision to supplement is not medically indicated (Table 1), discussions with the mother should be documented by the nursing and/or medical staff followed by full support of her informed decision. Choice of Supplement 1. Expressed breast milk from the infant’s mother is the first choice for extra feeding for the breastfed in-fant.7,55 (III) Hand expression may elicit larger vol-umes than a breast pump in the first few days following birth and may increase overall milk sup-ply.56 Breast massage and/or compression along with expressing with a mechanical pump may also increase available milk.57 (II-3) 2. If the volume of the mother’s own colostrum/milk does not meet her infant’s feeding requirements and supplementation is required, donor human milk is preferable to other supplements.55 3. When donor human milk is not available or appropriate, protein hydrolysate formulas may be preferable to standard infant formula as they avoid exposure to intact cow’s milk proteins and reduce bilirubin levels more rapidly,58 (II-2) although recent data are less supportive of its role in preventing allergic disease.59 (I) The use of this type of formula may also convey the psychological message that the supplement is a temporary therapy, not a permanent inclusion of artificial feedings. 4. Supplementation with glucose water is not appropriate because it does not provide sufficient nutrition, does not reduce serum bilirubin,60,61 and might cause hypona-tremia. 5. The potential risks and benefits of other supplemental fluids, such as cow’s milk formulas, soy formulas, or protein hydrolysate formulas, must be considered along with the available resources of the family, the infant’s age, the amounts needed, and the potential impact on the establishment of breastfeeding. Volume of Supplemental Feeding 1. Several studies give us an idea of intakes at the breast over time. In most studies, the range of intake is wide, while formula-fed infants usually take in larger vol-umes than breastfed infants.62–66 (II-3) 2. Infants fed infant formula ad libitum commonly have much higher intakes than breastfed infants.65,66 (II-3) Acknowledging that ad libitum breastfeeding emulates evolutionary feeding and considering recent data on obesity in formula-fed infants, it appears that formula-fed infants may be overfed. 3. As there is no definitive research available, the amount of supplement given should reflect the normal amounts of colostrum available, the size of the infant’s stomach (which changes over time), and the age and size of the infant. Intake on day 2 postbirth is generally higher than day 1 in relation to infant’s demand.65 4. Based on the limited research available, suggested intakes for healthy, term infants are given in Table 2, although feedings should be based on infant cues. Methods of Providing Supplementary Feedings 1. When supplementary feedings are needed, there are a number of delivery methods from which to choose: a supplemental nursing device at the breast, cup feeding, spoon or dropper feeding, finger-feeding, syringe feed-ing, or bottle feeding.67 (III) 2. An optimal supplemental feeding device has not yet been identified, and may vary from one infant to an-other. No method is without potential risk or benefit.68 3. When selecting an alternative feeding method, clini-cians should consider several criteria: a. cost and availability b. ease of use and cleaning c. stress to the infant d. whether adequate milk volume can be fed in 20– 30 minutes e. whether anticipated use is short- or long-term f. maternal preference g. expertise of healthcare staff h. whether the method enhances development of breastfeeding skills. 4. There is no evidence that any of these methods are unsafe or that one is necessarily better than the other. There is some evidence that avoiding teats/artificial nipples for supplementation may help the infant return to exclusive breastfeeding20,52,69 (I); however, when hygiene is suboptimal, cup feeding is the recommended choice.55 Cup feeding also allows infants to control feeding pace68 (II-2). Cup feeding has been shown safe for both term and preterm infants and may help preserve breastfeeding duration among those who require mul-tiple supplemental feedings.52,70–72 (II-2, I, I, I, II-2) 5. If bottles are being used, pacing the feed may be beneficial, especially for preterm infants.73 (III) 6. Supplemental nursing systems have the advantages of supplying a supplement while simultaneously stimu-Table 2. Average Reported Intakes of Colostrum by Healthy, Term Breastfed Infants Time (hours) Intake (mL/feed) First 24 2–10 24–48 5–15 48–72 15–30 72–96 30–60 4 ABM PROTOCOL lating the breast to produce more milk, reinforcing the infant’s feeding at the breast, enabling the mother to have a breastfeeding experience, and encouraging skin-to-skin. However, mothers may find the systems awkward to use, difficult to clean, relatively expen-sive, requiring moderately complex learning, and the infant must be able to latch effectively.67 A simpler version, supplementing with a dropper, syringe, or feeding tube attached to the breast while the infant is feeding at breast, may be effective. 7. Bottle feeding is the most commonly used method of supplementation in more affluent regions of the world, but concerns have been raised because of distinct differences in tongue and jaw movements, and faster flow may result in higher (and unnecessary) volumes of feeds.67 Some experts have recommended a teat/ nipple with a wide base and slow flow to try to mimic breastfeeding and to avoid nipple confusion or pref-erence,68,74 (II-2), but little research has been done evaluating outcomes with different teats/nipples. Research Needs Research is necessary to establish evidence-based guide-lines on appropriate supplementation volumes for specific conditions and whether this varies for colostrum versus infant formula. Specific questions include the following: 1. Should the volume be independent of infant weight or a per kilogram volume? Should supplementation make up for cumulative losses? 2. Should feeding intervals or quantities be different for different types of delivery of supplementation (e.g., bottles, cup feeding)? 3. Are some methods (type and delivery mechanism) best for infants with certain conditions, ages, and available resources? Which methods interfere least with estab-lishing direct breastfeeding? Notes This protocol addresses the healthy, term newborn. For information regarding appropriate feeding and supplemen-tation for the late preterm infant (35–37 weeks), see ‘‘ABM Protocol #10: Breastfeeding the Late Preterm Infant’’75 and ‘‘Care and Management of the Late Preterm Infant Toolk-it.’’76 The World Health Organization broadened the annex of the ‘‘Global Criteria for the Baby Friendly Hospital In-itiative: Acceptable Medical Reasons for Supplementa-tion.’’77 to include acceptable reasons for use of breast milk substitutes in all infants. The handout (#4.5) is available at: www.who.int/nutrition/topics/BFHI_Revised_Section_4.pdf References 1. Kramer MS, Kakuma R. Optimal duration of exclusive breastfeeding. Cochrane Database Syst Rev 2012; CD003517. DOI: 10.1002/14651858.CD003517.pub2. 2. Smith HA, Becker GE. Early additional food and fluids for healthy breastfed full-term infants. Cochrane Database Syst Rev 2016;CD006462. DOI: 10.1002/14651858 .CD006462.pub4. 3. Victora CG, Bahl R, Barros AJD, et al. 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Pe ´rez-Escamilla R, Segura-Milla ´n S, Canahuati J, et al. Prelacteal feeds are negatively associated with breast-feeding outcomes in Honduras. J Nutr 1996;126:2765–2773. 99. Hale T, Rowe H. Medications and Mother’s Milk, 17th ed. New York: Springer Publishing Company, 2017. 100. Pease AS, Fleming PJ, Hauck FR, et al. Swaddling and the risk of Sudden Infant Death Syndrome: A meta-analysis. Pediatrics 2016;137:e20153275. 101. Bystrova K, Matthiesen A, Widstrom A, et al. The effect of Russian Maternity Home routines on breastfeeding and neonatal weight loss with special reference to swaddling. Early Hum Dev 2007;83:29–39. ABM PROTOCOL 7 102. Gagnon A, Leduc G, Waghorn K, et al. In-hospital for-mula supplementation of healthy breastfeeding newborns. J Hum Lact 2005;21:397–405. 103. Slaven S, Harvey D. Unlimited suckling time improves breastfeeding. Lancet 1981;1:392–393. 104. Geddes DT, Langton DB, Gollow I, et al. Frenulotomy for breastfeeding infants with ankyloglossia: Effect on milk removal and sucking mechanism as imaged by ultrasound. Pediatrics 2008;122:e188–e194. ABM protocols expire 5 years from the date of publication. Content of this protocol is up-to-date at the time of pub-lication. Evidence based revisions are made within 5 years or sooner if there are significant changes in the evidence. The 2009 edition of this protocol was authored by Nancy E. Wight and Robert Cordes. The Academy of Breastfeeding Medicine Protocol Committee: Wendy Brodribb, MBBS, PhD, FABM, Chairperson Larry Noble, MD, FABM, Translations Chairperson Nancy Brent, MD Maya Bunik, MD, MSPH, FABM Cadey Harrel, MD Ruth A. Lawrence, MD, FABM Kathleen A. Marinelli, MD, FABM Sarah Reece-Stremtan, MD Casey Rosen-Carole, MD, MPH, MSEd Tomoko Seo, MD, FABM Rose St. Fleur, MD Michal Young, MD For correspondence: abm@bfmed.org Table A1. Inappropriate Reasons for Supplementation in the Context of a Healthy Newborn and Mother, Responses, and Risks Concerns/inappropriate reasons Responses Risks of supplementation There is ‘‘no milk,’’5 belief that colostrum is insufficient until the milk ‘‘comes in’’ Mother and family should be educated about the benefits of colostrum including dispelling myths about the yellow color. Small amounts of colostrum are normal, physiologic, and appropriate for the term, healthy newborn. Can alter infant bowel flora and microbiome.86,87 Potentially sensitizes the infant to foreign proteins.88,89 Increases the risk of diarrhea and other infections, especially where hygiene is poor.3 Potentially disrupts the ‘‘supply-demand’’ cycle, leading to inadequate milk sup-ply and long-term supplementation. Supplementation is needed to prevent weight loss and dehydration in the postnatal period5 A certain amount of weight loss is normal in the first week of life and is due to diuresis of extracellular fluid and passage of meconium. Supplementation in the first few days may interfere with the normal frequency of breastfeeding. Supplementation with water or glucose water, increases the risk of jaundice,90 excessive weight loss,91 and longer hospital stays.92 Infant could become hypoglycemic Healthy, full-term infants do not develop symptomatic hypoglycemia as a result of suboptimal breastfeeding.15 Same risks as for weight loss/dehydration. Breastfeeding is related to jaundice in the postnatal period The more frequent the breastfeeding, the lower the bilirubin level.93,94 Same risks as for weight loss/dehydration. Bilirubin is a potent antioxidant95 and jaundice is normal in the newborn. Colostrum acts as a natural laxative helping to eliminate meconium that contains bilirubin.94 (continued) Appendix 8 ABM PROTOCOL Table A1. (Continued) Concerns/inappropriate reasons Responses Risks of supplementation Lack of time for counseling mother about exclusive breastfeeding when mothers request a supplement Train all staff in how to assist mothers with breastfeeding. Mothers may benefit from education about artificial feeds and/or how supplements may adversely affect subsequent breastfeeding.29 Time spent by healthcare professionals lis-tening to and talking with mothers is at least as important as other more active interventions (which may be viewed more as ‘‘real work’’ to them).29 If the supplement is infant formula, which is slow to empty from the stomach96 and often fed in larger amounts,66 the infant will breastfeed less frequently. Depending on the method of supplementation,52,74 or the number of supplements97 an infant may have difficulty returning to the breast. Feeds given before secretory activation and copious breast milk production (as opposed to supplementation) may be associated with delayed initiation of breastfeeding and negatively associated with exclusivity and duration of breastfeeding.98 Medications may be contraindi-cated with breastfeeding Accurate references are available to provid-ers (e.g., Medications and Mothers’ Milk 2017,99 LactMed on Toxnet website 15) Risk of decreasing breastfeeding duration or exclusivity.9,10 For most medical conditions, medication safe for breastfeeding mothers and babies is available Mother is too malnourished or sick to breastfeed or eats an inappropriate diet. Even malnourished mothers can breastfeed. Breast milk quality and quantity is only affected in extreme circumstances. Supplements are better given to the mother (with continued breastfeeding) than the infant. Risk of decreasing breastfeeding duration or exclusivity. Supplementation will quiet a fussy or unsettled infant5 Infants can be unsettled for many reasons. They may wish to cluster feed or simply need additional skin-to-skin time or holding.67 Filling (and often overfilling) the stomach with a supplement may make the infant sleep longer,96 missing important oppor-tunities to breastfeed, and demonstrating to the mother a short-term solution which may generate long-term health risks. Teaching other soothing techniques to new mothers such as breastfeeding, swaddling (but not if prone or side lying),100 swaying, singing, encouraging father or other relatives to assist. Caution should be taken to not ignore early feed-ing cues.101 Ensure comfortable, effective latch to max-imize signal to mother’s body and intake for the infant Risk of decreasing breastfeeding duration or exclusivity. Maternal engorgement due to decreased frequency of breastfeeding in the immediate postpartum period.54 Concern about the cause of frequent feeding and cluster feeding and other changes in infant behavior Periods when infants demand to breastfeed more are sometimes interpreted by moth-ers as insufficient milk. This may happen in later weeks but also in the second or third night (48–72 hours) postbirth. Changes in stooling patterns that often occur after 6–8 weeks of age can also be misinterpreted as insufficient milk. Anticipatory guidance for normal infant development and behavior is helpful. Risk of decreasing breastfeeding duration or exclusivity. Mothers need to rest or sleep Postpartum mothers are restless when separated from their infants and actually get less rest.29 Risk of decreasing breastfeeding duration or exclusivity. (continued) ABM PROTOCOL 9 Table A1. (Continued) Concerns/inappropriate reasons Responses Risks of supplementation Mothers lose the opportunity to learn their infants’ normal behavior and early feeding cues.48 Infants are at highest risk for receiving a supplement between 7 p.m. and 9 a.m.102 Sore nipples will improve if mother takes a break from breastfeeding5 Sore nipples are not a function of length of time breastfeeding. Position, latch, and sometimes individual anatomic variation (e.g., ankyloglossia) are more important.103 There is no evidence that limiting time at the breast will prevent sore nipples. The nipple should not be rubbed or com-pressed during breastfeeding even if the feedings are frequent or ‘‘clustered.’’104 Problem with latch not addressed. Risk of decreasing breastfeeding duration or cessation of breastfeeding. Risk of breast engorgement. 10 ABM PROTOCOL Appendix A2: Sample Maternity Care Infant Nutrition Algorithm
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https://www.youtube.com/watch?v=VzdF7nakkRY
Solving Trig Equation 14 + 7 cos 2x = 21 cos x Wyzant 5950 subscribers 1 likes Description 80 views Posted: 12 Nov 2023 View full question and answer details: Question: Find all values of x in the interval [0, 2𝜋] that satisfy the equation. (Enter your answers as a comma-separated list.) 14 + 7 cos(2x) = 21 cos(x) Answered By: Patrick F. Patient and Inspiring Math Teacher More information: See full answer: About: Wyzant Ask an Expert offers free answers to your toughest academic and professional questions from over 65,000 verified experts. It’s trusted by millions of students each month with the majority of questions receiving an answer within 1 hour of being asked. If you ever need more than just an answer, Wyzant also offers personalized 1-on-1 sessions with experts that will work with you to help you understand whatever you’re trying to learn. Ask your own question for free: Find a tutor for a 1-on-1 session: Subscribe to Wyzant on YouTube: Transcript: [Music] [Applause] okay so we're going to be solving a trig equation from zero to two Pi there's our equation and let's take a look okay first thing you might notice here is we can divide by seven so let's simplify it a bit okay now we're going to use the double angle formula identity in terms of cosine so that looks like that 2 cosine squ minus one we're going to move everything over to the left and you can see we have a quadratic equation okay can we factorize it yes we can okay so we need to make each factor zero the first one is zero if cosine X is a half second one is zero if cosine X is one so we need to look at each one of those cases let's look at the cosine xals a half first on the unit circle okay we're going to draw hypotenuse of one and we're going to rotate it to make a triangle where the coine is a half there's the hypotenuse and so the adjacent side should be a half and this C triangle should be familiar that angle there is going to be pi over three okay so there's the first quadrant answer the cosine is also positive in the fourth quadrant so we need to look at that so let's rotate to the fourth and make the same triangle adjacent side it's a half there's the angle that we're going to be counting all right so our angle we know is pi over 3 and so our answer is going to be 2 pi minus pi over 3 right we're Pi over3 short of a full circle okay so there's the two answers for that now we're going to look at cosine xal 1 easier to use the cosine graph as a reference so let's draw that there where is it one there and there so at zero and 2 pi and there are our four answers okay hope that was [Music] helpful
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https://www.khanacademy.org/math/geometry-tx/x790e3ac3e338c450:probability/x790e3ac3e338c450:area-based-probability/v/area-based-probability
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https://openstax.org/books/chemistry-2e/pages/17-4-potential-free-energy-and-equilibrium
Skip to ContentGo to accessibility pageKeyboard shortcuts menu Chemistry 2e 17.4 Potential, Free Energy, and Equilibrium Chemistry 2e17.4 Potential, Free Energy, and Equilibrium Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Explain the relations between potential, free energy change, and equilibrium constants Perform calculations involving the relations between cell potentials, free energy changes, and equilibrium Use the Nernst equation to determine cell potentials under nonstandard conditions So far in this chapter, the relationship between the cell potential and reaction spontaneity has been described, suggesting a link to the free energy change for the reaction (see chapter on thermodynamics). The interpretation of potentials as measures of oxidant strength was presented, bringing to mind similar measures of acid-base strength as reflected in equilibrium constants (see the chapter on acid-base equilibria). This section provides a summary of the relationships between potential and the related thermodynamic properties ΔG and K. E° and ΔG° The standard free energy change of a process, ΔG°, was defined in a previous chapter as the maximum work that could be performed by a system, wmax. In the case of a redox reaction taking place within a galvanic cell under standard state conditions, essentially all the work is associated with transferring the electrons from reductant-to-oxidant, welec: The work associated with transferring electrons is determined by the total amount of charge (coulombs) transferred and the cell potential: where n is the number of moles of electrons transferred, F is Faraday’s constant, and E°cell is the standard cell potential. The relation between free energy change and standard cell potential confirms the sign conventions and spontaneity criteria previously discussed for both of these properties: spontaneous redox reactions exhibit positive potentials and negative free energy changes. E° and K Combining a previously derived relation between ΔG° and K (see the chapter on thermodynamics) and the equation above relating ΔG° and E°cell yields the following: This equation indicates redox reactions with large (positive) standard cell potentials will proceed far towards completion, reaching equilibrium when the majority of reactant has been converted to product. A summary of the relations between E°, ΔG° and K is depicted in Figure 17.7, and a table correlating reaction spontaneity to values of these properties is provided in Table 17.2. Figure 17.7 Graphic depicting the relation between three important thermodynamic properties. | | | | | --- --- | | K | ΔG° | E°cell | | | > 1 | < 0 | > 0 | Reaction is spontaneous under standard conditions Products more abundant at equilibrium | | < 1 | > 0 | < 0 | Reaction is non-spontaneous under standard conditions Reactants more abundant at equilibrium | | = 1 | = 0 | = 0 | Reaction is at equilibrium under standard conditions Reactants and products equally abundant | Table 17.2 Example 17.6 Equilibrium Constants, Standard Cell Potentials, and Standard Free Energy Changes Use data from Appendix L to calculate the standard cell potential, standard free energy change, and equilibrium constant for the following reaction at 25 °C. Comment on the spontaneity of the forward reaction and the composition of an equilibrium mixture of reactants and products. Solution The reaction involves an oxidation-reduction reaction, so the standard cell potential can be calculated using the data in Appendix L. With n = 2, the equilibrium constant is then The standard free energy is then The reaction is spontaneous, as indicated by a negative free energy change and a positive cell potential. The K value is very large, indicating the reaction proceeds to near completion to yield an equilibrium mixture containing mostly products. Check Your Learning What is the standard free energy change and the equilibrium constant for the following reaction at room temperature? Is the reaction spontaneous? Answer: Spontaneous; n = 2; K = 6.8 109. Potentials at Nonstandard Conditions: The Nernst Equation Most of the redox processes that interest science and society do not occur under standard state conditions, and so the potentials of these systems under nonstandard conditions are a property worthy of attention. Having established the relationship between potential and free energy change in this section, the previously discussed relation between free energy change and reaction mixture composition can be used for this purpose. Notice the reaction quotient, Q, appears in this equation, making the free energy change dependent upon the composition of the reaction mixture. Substituting the equation relating free energy change to cell potential yields the Nernst equation: This equation describes how the potential of a redox system (such as a galvanic cell) varies from its standard state value, specifically, showing it to be a function of the number of electrons transferred, n, the temperature, T, and the reaction mixture composition as reflected in Q. A convenient form of the Nernst equation for most work is one in which values for the fundamental constants (R and F) and standard temperature (298) K), along with a factor converting from natural to base-10 logarithms, have been included: Example 17.7 Predicting Redox Spontaneity Under Nonstandard Conditions Use the Nernst equation to predict the spontaneity of the redox reaction shown below. Solution Collecting information from Appendix L and the problem, Notice the negative value of the standard cell potential indicates the process is not spontaneous under standard conditions. Substitution of the Nernst equation terms for the nonstandard conditions yields: The cell potential remains negative (slightly) under the specified conditions, and so the reaction remains nonspontaneous. Check Your Learning For the cell schematic below, identify values for n and Q, and calculate the cell potential, Ecell. Answer: n = 6; Q = 1440; Ecell = +1.97 V, spontaneous. A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. The example problem below illustrates the use of the Nernst equation in calculations involving concentration cells. Example 17.8 Concentration Cells What is the cell potential of the concentration cell described by Solution From the information given: Substituting into the Nernst equation, The positive value for cell potential indicates the overall cell reaction (see above) is spontaneous. This spontaneous reaction is one in which the zinc ion concentration in the cathode falls (it is reduced to elemental zinc) while that in the anode rises (it is produced by oxidation of the zinc anode). A greater driving force for zinc reduction is present in the cathode, where the zinc(II) ion concentration is greater (Ecathode > Eanode). Check Your Learning The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now? Answer: Ecell = 0.000 V; [Zn2+]cathode = [Zn2+]anode = 0.30 M PreviousNext Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Paul Flowers, Klaus Theopold, Richard Langley, William R. Robinson, PhD Publisher/website: OpenStax Book title: Chemistry 2e Publication date: Feb 14, 2019 Location: Houston, Texas Book URL: Section URL: © Jul 9, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
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https://2012books.lardbucket.org/books/beginning-algebra/s11-04-multiplying-and-dividing-radic.html
This is “Multiplying and Dividing Radical Expressions”, section 8.4 from the book Beginning Algebra (v. 1.0). For details on it (including licensing), click here. This book is licensed under a Creative Commons by-nc-sa 3.0 license. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz in an effort to preserve the availability of this book. Normally, the author and publisher would be credited here. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Additionally, per the publisher's request, their name has been removed in some passages. More information is available on this project's attribution page. For more information on the source of this book, or why it is available for free, please see the project's home page. You can browse or download additional books there. To download a .zip file containing this book to use offline, simply click here. Has this book helped you? Consider passing it on: Help Creative Commons Creative Commons supports free culture from music to education. Their licenses helped make this book available to you. Help a Public School DonorsChoose.org helps people like you help teachers fund their classroom projects, from art supplies to books to calculators. 8.4 Multiplying and Dividing Radical Expressions Learning Objectives Multiply radical expressions. Divide radical expressions. Rationalize the denominator. Multiplying Radical Expressions When multiplying radical expressions with the same index, we use the product rule for radicals. If a and b represent positive real numbers, Example 1: Multiply: . Solution: This problem is a product of two square roots. Apply the product rule for radicals and then simplify. Answer: Example 2: Multiply: . Solution: This problem is a product of cube roots. Apply the product rule for radicals and then simplify. Answer: Often there will be coefficients in front of the radicals. Example 3: Multiply: . Solution: Using the product rule for radicals and the fact that multiplication is commutative, we can multiply the coefficients and the radicands as follows. Typically, the first step involving the application of the commutative property is not shown. Answer: Example 4: Multiply: . Answer: Use the distributive property when multiplying rational expressions with more than one term. Example 5: Multiply: . Solution: Apply the distributive property and multiply each term by . Answer: Example 6: Multiply: . Solution: Apply the distributive property and then simplify the result. Answer: The process for multiplying radical expressions with multiple terms is the same process used when multiplying polynomials. Apply the distributive property, simplify each radical, and then combine like terms. Example 7: Multiply: . Solution: Begin by applying the distributive property. Answer: Example 8: Multiply: . Solution: Answer: Try this! Multiply: . Answer: Video Solution (click to see video) The expressions and are called conjugatesThe factors and are conjugates.. When multiplying conjugates, the sum of the products of the inner and outer terms results in 0. Example 9: Multiply: . Solution: Apply the distributive property and then combine like terms. Answer: −3 It is important to note that when multiplying conjugate radical expressions, we obtain a rational expression. This is true in general and is often used in our study of algebra. Therefore, for nonnegative real numbers a and b, we have the following property: Dividing Radical Expressions (Rationalizing the Denominator) To divide radical expressions with the same index, we use the quotient rule for radicals. If a and b represent nonnegative numbers, where , then we have Example 10: Divide: . Solution: In this case, we can see that 10 and 80 have common factors. If we apply the quotient rule for radicals and write it as a single square root, we will be able to reduce the fractional radicand. Answer: Example 11: Divide: . Solution: Answer: Example 12: Divide: . Solution: Answer: When the divisor of a radical expression contains a radical, it is a common practice to find an equivalent expression where the denominator is a rational number. Finding such an equivalent expression is called rationalizing the denominatorThe process of determining an equivalent radical expression with a rational denominator.. To do this, multiply the fraction by a special form of 1 so that the radicand in the denominator can be written with a power that matches the index. After doing this, simplify and eliminate the radical in the denominator. For example, Remember, to obtain an equivalent expression, you must multiply the numerator and denominator by the exact same nonzero factor. Example 13: Rationalize the denominator: . Solution: The goal is to find an equivalent expression without a radical in the denominator. In this example, multiply by 1 in the form . Answer: Example 14: Rationalize the denominator: . Solution: The radicand in the denominator determines the factors that you need to use to rationalize it. In this example, multiply by 1 in the form . Answer: Typically, we will find the need to reduce, or cancel, after rationalizing the denominator. Example 15: Rationalize the denominator: . Solution: In this example, we will multiply by 1 in the form . Notice that a and b do not cancel in this example. Do not cancel factors inside a radical with those that are outside. Answer: Try this! Rationalize the denominator: . Answer: Video Solution (click to see video) Up to this point, we have seen that multiplying a numerator and a denominator by a square root with the exact same radicand results in a rational denominator. In general, this is true only when the denominator contains a square root. However, this is not the case for a cube root. For example, Note that multiplying by the same factor in the denominator does not rationalize it. In this case, if we multiply by 1 in the form of , then we can write the radicand in the denominator as a power of 3. Simplifying the result then yields a rationalized denominator. For example, Therefore, to rationalize the denominator of radical expressions with one radical term in the denominator, begin by factoring the radicand of the denominator. The factors of this radicand and the index determine what we should multiply by. Multiply numerator and denominator by the nth root of factors that produce nth powers of all the factors in the radicand of the denominator. Example 16: Rationalize the denominator: . Solution: The radical in the denominator is equivalent to . To rationalize the denominator, it should be . To obtain this, we need one more factor of 5. Therefore, multiply by 1 in the form of . Answer: Example 17: Rationalize the denominator: . Solution: In this example, we will multiply by 1 in the form . Answer: Example 18: Rationalize the denominator: . Solution: In this example, we will multiply by 1 in the form . Answer: When two terms involving square roots appear in the denominator, we can rationalize it using a very special technique. This technique involves multiplying the numerator and the denominator of the fraction by the conjugate of the denominator. Recall that multiplying a radical expression by its conjugate produces a rational number. Example 19: Rationalize the denominator: . Solution: In this example, the conjugate of the denominator is . Therefore, multiply by 1 in the form . Answer: Notice that the terms involving the square root in the denominator are eliminated by multiplying by the conjugate. We can use the property to expedite the process of multiplying the expressions in the denominator. Example 20: Rationalize the denominator: . Solution: Multiply by 1 in the form . Answer: Example 21: Rationalize the denominator: . Solution: In this example, we will multiply by 1 in the form . Answer: Try this! Rationalize the denominator: . Answer: Video Solution (click to see video) Key Takeaways To multiply two single-term radical expressions, multiply the coefficients and multiply the radicands. If possible, simplify the result. Apply the distributive property when multiplying radical expressions with multiple terms. Then simplify and combine all like radicals. Multiplying a two-term radical expression involving square roots by its conjugate results in a rational expression. It is common practice to write radical expressions without radicals in the denominator. The process of finding such an equivalent expression is called rationalizing the denominator. If an expression has one term in the denominator involving a radical, then rationalize it by multiplying numerator and denominator by the nth root of factors of the radicand so that their powers equal the index. If a radical expression has two terms in the denominator involving square roots, then rationalize it by multiplying the numerator and denominator by its conjugate. Topic Exercises Part A: Multiplying Radical Expressions Multiply. (Assume all variables are nonnegative.) 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. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. What are the perimeter and area of a rectangle with length of centimeters and width of centimeters? What are the perimeter and area of a rectangle with length of centimeters and width of centimeters? If the base of a triangle measures meters and the height measures meters, then what is the area? If the base of a triangle measures meters and the height measures meters, then what is the area? Part B: Dividing Radical Expressions Divide. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. Rationalize the denominator. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. 103. 104. 105. 106. Part C: Discussion Research and discuss some of the reasons why it is a common practice to rationalize the denominator. Explain in your own words how to rationalize the denominator. Answers 1: 3: 5: 7 7: 9: 20 11: 13: 15: 17: 5 19: 21: 23: 16 25: 27: 29: 31: 33: 35: 37: 39: 41: 43: 45: 47: 10 49: 51: Perimeter: centimeters; area: square centimeters 53: 18 square meters 55: 5 57: 59: 61: 63: 65: 67: 69: 71: 73: 75: 77: 79: 81: 83: 85: 87: 89: 91: 93: 95: 97: 99: 101: 103: 105: Previous Section Table of Contents Next Section The factors and are conjugates. The process of determining an equivalent radical expression with a rational denominator.
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https://teachpastthepotholes.com/hands-on-activities-for-teaching-volume/?srsltid=AfmBOooZOkDkZPCfsmiumWhsozizwZ736fUHuAeVhniUBbO42oQQovEc
8 Hands On activities for Teaching Volume your Students will Love! - PAST THE POTHOLES Skip to content HOMEMenu Toggle Literacy Resources Math Resources Teacher Tools Assessment and Reporting ABOUT US SHOPMenu Toggle Shop CAD Shop USD INQUIRY-BASED MATH CURRICULUM REPORT CARD COMMENT BANKS MY ACCOUNTMenu Toggle My account Downloads Account details Search for:Search Search No products in the cart. No products in the cart. Continue Shopping PAST THE POTHOLES LITERACY MATH ASSESSMENT TEACHER TOOLS PAST THE POTHOLES Main Menu HOME Cart Math Goals Freebie Report Comments Math Bundles Shop CADMenu Toggle Shop USD My account 8 Hands On activities for Teaching Volume your Students will Love! byClaire Help a teacher out, share the love! Some links in this post may be affiliate links. If you click on them we may make a commission from qualifying purchases at absolutely no cost to you. Read our full Disclosure Policy here. Getting students engaged in their learning is easy with these hands-on activities for teaching volume! Teaching volume is such a fun unit but it can be challenging to come up with new activities as students do need a bit of practice. However, with a variety of hands-on volume activities for elementary students, it will never be wasted time. Volume is such a real-world concept that overlaps so many areas of our daily lives. Plus, basic number sense skills are entwined throughout so there’s so much opportunity for spiral review too! Fun Ways to Teach Volume Most of these activities to teach volume are open-ended or can be repeated. This means you won’t be constantly changing or creating new tasks. Perfect! Fill ‘Er Up! Start here! A foundational volume activity should be to use unit cubes to fill a variety of rectangular prisms to explore the concept of volume. Have students fill the prism, then dump out the cubes and try to recreate the same prism on their desk. This will really solidify their understanding of the concept. READ THIS FIRST: 6 Steps to Introduce Volume and Build a Solid Conceptual Understanding Volume Building Challenge Challenge your students to build prisms based on requirements you give. Have students work in pairs and add a timer for a fun bit of competition. Increase the level of challenge as you go. For example: Build a prism with a specific volume. Build a prism with a volume larger than _ or smaller than _. Build a prism with a specific volume and one given measurement. Build two different prisms with the same volume. Volume Scavenger Hunt I love being able to combine movement with math lessons! Even better when we get to go outside and enjoy a beautiful spring day. You can design this scavenger hunt in a variety of ways. Give students specific items (or categories of items) that they must find and measure the volume. For example, “Find the volume of a book”. Give students a list of tasks such as, “Find something with a volume that is greater than 48m³. Create a list of more general volume statements like, “Find something with a volume we would measure in cubic metres.” As you can see, this Volume Scavenger Hunt can be adapted to a variety of levels of difficulty and environments. Roll a Volume There are plenty of hands on activities for teaching volume that don’t require a ton of materials or preparation. Dice games are always a favourite and are perfect for easy, low-prep math centres! Roll-a-Volume is a spin on a popular area game in my class. Each student rolls 3 dice (or one die 3 times) to find the length, width and depth of their prism. They record these numbers on their paper then calculate the volume. All players must agree on the answers then the player with the largest volume wins a point. Play continues until a student gets 5 points. Mix and Match Volumes Looking for some really fun ways to teach volume? Channel your inner Bob Barker for some game show-esque fun! Preparation: Collect a few household boxes (cereal boxes, cracker boxes, shoe box, dictionary or textbook,…) and find the volume of each in whatever unit you choose. Record each of the volumes on large index cards that can be presented in front of the classroom. Play: Have students work in teams to try to match the box with the volume. Play it ‘Price it Right’ style and have each team take turns matching one box with one volume card or make it a race to match them all. Your choice! Volume Sort – image, dimensions, volume Create a set of cards with images of prisms, another set with the dimensions of those prisms and a third set with the actual volume of each prism. Have students work together to sort the 3 sets of cards to find the ones that match. This is another hands-on volume activity that is super easy to put in a math centre. You can decide the level of difficulty and the number of cards included to differentiate for your entire class or specific groups of students. Volume Task Cards Task cards are always an awesome versatile teaching activity. There are so many ways you can use task cards that go beyond just answering questions. I wrote a blog post filled with 18 different ways to use task cards so be sure to check that out. But some of my favourites are pasting them around the room for a type of scavenger hunt, games like Scoot or Bingo or as station rotations. Online Math Volume Games In today’s world, there are so many tools already available to us, we’d be foolish to keep re-creating the wheel and wasting our time! The internet is filled with interactive volume activities that students love. A few of my favourite online volume games and activities are… Kahoot is an interactive game that has plenty of quizzes already created and available for use. Or you can creat your own questions from scratch. I love Kahoot as a whole class in teams but it can also be done virtually or in small groups. MathGames.com – Practice questions with immediate feedback at different levels. NCTM Cubes – Students can fill boxes with cubes, rows of cubes or layers of cubes allowing for scaffolded practice. BBC Bitesize – What is Volume? and How to Calculate Volume include tutorials, examples, activities and quizzes for students. IXL.com – practice volume questions (I believe it allows 10 questions without an account). Tips for Hands On Activities for Teaching Volume I like to keep things simple and fuss-free but there are a few tips I’ve picked up over the years in the classroom. (Yep, learned the hard way!) Use fabric or flexible measuring tapes rather than rulers. Not only does their flexibility help make them more functional, you’ll thank me for the lack of ‘sword fights’ happening… You can probably pick some up at the dollar store or grab them on Amazon here. Provide a variety of tools to measure so students can explore different strategies and different units. Whenever students are moving around (like the scavenger hunt ideas above), use a classroom timer. It is so much easier to wrangle them back to you at the end! Print out grid paper for students to create their own nets/prisms. Make sure the grid size matches whatever unit cubes you have available. For example, if you’re using one inch cubes, print out 1 inch grid paper. Teaching Volume Activities Final Thoughts I hope this post has given you some ideas for teaching volume! Teaching it in the spring is the perfect time to get your kids up and moving with some hands-on learning. Not only because the weather’s getting nicer and they’re getting antsy, but because by now you’ve got your routines and expectations nailed. If you’d like more support with teaching volume in a hands-on, conceptual way, you might want to check out our complete Measurement Units for grades 4 – 8. 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11 Partial differentiation Contents 11 Partial differentiation# ``` import all python add-ons etc that will be needed later on %matplotlib inline import numpy as np import matplotlib.pyplot as plt from sympy import init_printing() # allows printing of SymPy results in typeset maths format plt.rcParams.update({'font.size': 14}) # set font size for plots ``` 11.1 Differentiating equations in more than one variable# When an equation depends on more than one variable, and perhaps, then it is possible to differentiate with respect to either or while keeping the other constant. The normal rules of differentiation apply; the only change is that the notation is a little different with a curly being used instead of a Roman and a subscript is used to indicate which variable is held constant. If the function will allow it, second and third partial derivatives can be calculated. Suppose that , then the and partial derivatives are The second partial derivatives are . As a second example is calculated where . Since it is indicated by the subscript that is a constant, differentiating gives . (i) Ideal gas# Differentiating the pressure in the ideal gas law to obtain the rate of change with respect to temperature is expressed as if the number of moles, , and volume are held constant. The result is Similarly at constant and , and constant and gives Calculating second derivatives has the same notation but, in this example, only is not zero. The other two derivatives are constants, . (ii) van der Waals gas# The van der Waals equation which, although empirical, describes real gases more accurately than does the ideal gas law. The equation is where is related to intermolecular interaction and describes the finite volume of a molecule. We shall find . Differentiating with respect to , and using the usual rules of differentiation with as a constant gives, Now find . Differentiating with respect to at constant is easier if the equation is expanded first as , then differentiating as a product produces, 11.2 Geometrical interpretation of partial derivatives# If a function can be differentiated with respect to either of two variables, and , then it represents a surface. The derivatives then represent the slope of the surface at any point but in a direction at a fixed or depending on which is held constant. The Sackur - Tetrode equation for the translational entropy of a perfect gas is where is the heat capacity at constant pressure , is a constant entropy, is the number of moles of the gas and the gas constant. The entropy vs temperature and pressure is shown as the surface in Fig. 33. Two partial derivatives are also shown, one at fixed parallel to the axis and one at fixed , parallel to the axis. These lines are parallel to the axes because there is no term in both and in the partial derivative equations and the partial derivative is the gradient of the lines shown at any point. Figure 33. Three-dimensional plot of the Sackur - Tetrode equation for one mole of an ideal monatomic gas where , with pressure in bar and in kelvin. The blue and red lines show how the partial derivatives vary with pressure or temperature when the other variable is held constant. The value of the partial derivative is the slope at any point. The liquefaction of gases was studied by Andrews before 1870 and he discovered the critical point using carbon dioxide. This gas was presumably used as it would be readily available from brewing. The van der Waals equation is nowadays commonly used to study non-ideal, i.e. real, gases. The equation is where is a measure of the attractive forces between molecules and accounts for their finite size. is here the molar volume, which is volume divided by number of moles; this is often represented by or . Fig. 34 shows the surface and shows gradients at three points with different variables held constant. The partial derivatives are also shown as lines, again note that the partial derivative is the slope at any point. The contour lines follow constant pressure where = constant. Fig. 35, shows the or isotherm plot, for Freon with the inflexion at the critical temperature of 385 K (black dot) is where the gradients, = 0. The data used to make the plot is that for freon, , with and and . The critical point values are ;. Figure 34. profile for a van der Waals gas. This figure and the next are calculated using data for freon, . The critical temperature is K where there is an inflexion point. Figure 35. plot at different temperatures separated by K. The parameters for freon, were used. The oscillating lines for pressures at temperatures lower than the critical point and inside the region bounded by the dotted line are not physically realistic. The straight solid lines show the path taken by the fluid in changing from a liquid to vapour. The curves on this graph have a gradient . The van der Waals equation provides a surprisingly good description of a real gas, considering that it has only two adjustable parameters, but does not reproduce the experimental data exactly. This is particularly true for the oscillatory nature of the curve below the critical point (385 K for freon), which is not seen experimentally. A horizontal tie line is drawn so that there is an equal area above and below the line (Maxwell equal area construction) and this is the path followed by the fluid. Three such lines are shown on the graph where the fluid has the same pressure and the phase change proceeds imperceptibly along this line from liquid to gas or vice versa. At a given temperature, horizontal pairs of points on the dotted line define the points where the pressure of the liquid and vapour are the same and where the area of the oscillating curve is the same above and below the straight line. The volume at each end of the tie line has to be found numerically by equating the pressure at either end and making the areas mentioned equal to zero, which is the integral The volumes and are found by solving these two equations simultaneously at each temperature. The integral for volume is . The dotted lines in fig 34 and 35 show the positions of the tie lines for a range of temperatures up to the critical point. The next short piece of code will calculate the volumes. ``` (nsolve is part of the sympy library) a = 10.78 # bar . (dm^3/mol)^2 b = 0.0998 # dm^3/mol R = 0.083145 # dm^3.bar.mol/k; 1 dm^3.bar == 100 joules p = lambda V,T : RT/(V-b) - a/V2 # bar, van der waals equation; V in dm^3/mol integrl = lambda V,T: RTln(V-b) + a/V # dm^3.bar integral pdV print('{:s}'.format( 'Temp V1 V2 dm^3/mol')) v1,v2 = symbols('v1 v2') # use sympy and nsolve, not fast but ok. define v1 and v2 for T0 in range(360,370,2): # choose some temperatures eqn1 = integrl(v2,T0)- integrl(v1,T0) - p(v2,T0)(v2-v1) eqn2 = p(v2,T0) - p(v1,T0) ans = nsolve(( eqn1,eqn2 ), (v1,v2), (0.15,0.8), verify=False ) # 0.15 and 0.8 are approx limits. print('{:d} {:f} {:f} '.format(T0, float(ans), float(ans) ) ) pass ``` ``` Temp V1 V2 dm^3/mol 360 0.196093 0.568952 362 0.198959 0.550496 ``` ``` 364 0.202042 0.532266 366 0.205378 0.514211 368 0.209012 0.496269 ``` 11.3 Mixed derivatives# The new feature of partial differentiation is that it is possible to differentiate with one variable then do so again with another. Using the first example, where , and differentiating with produces Next differentiating with produces which is the same result. The order of carrying out the differentiation is immaterial, and there is a mathematical theorem that guarantees this. In general for any well behaved function , This is also sometimes written in a more formal form. The left side of the equation tells us to differentiate with keeping constant, and then differentiate the result with , keeping constant, This is also frequently written as where the subscripts are assumed. In the ideal gas law, , differentiating with respect to at constant and (number of moles) produces Next, differentiating with respect to also at constant produces. Repeating the process but in the opposite order gives . 11.4 Chain rule# The chain rule can also be used with partial derivatives. If is some complicated expression in and the familiar function-of-function rule can be used to calculate for example, with held constant. The result is Notice the symmetry in the derivatives. A similar equation can be written with held constant if the differentiation were with respect to . Using this rule, if then where . The other derivative is 11.5 Use in Thermodynamics ( see also section 11.8 )# In thermodynamics, the chain rule is very often used to expand an expression in a new variable and this proves to be very useful. The reason for doing this is to change an unfamiliar derivative into two expressions each of which is related to something that can be measured. For example, suppose that the change in enthalpy with temperature at constant pressure is required this is , and is the constant pressure heat capacity , but, for the moment, assume that we do not know this. To find out what this derivative is, expand it in some other thermodynamic variable, entropy for example, then Notice again the ‘symmetry’ of the derivatives when making such an expansion. This procedure does not seem to make much sense until the following derivatives are looked up making the heat capacity at constant pressure. Had we chosen some other variable instead of , or for example, then although derivatives could be found these may not correspond to anything that could be measured. If the second derivatives (above) can be integrated and is used to experimentally determine entropy: at constant pressure, by measuring the heat capacity vs. temperature. 11.6 Reciprocal derivatives# Reciprocal derivatives follow the same rules as for normal differentiation: take the reciprocal and flip the derivative, with a function in and For example if then differentiating with respect to produces and differentiating with produces which proves (40). 11.7 Total derivatives# Total derivatives are probably used in thermodynamics more than elsewhere. It is common to see expressions such as The notation here is different to that used so far because , and exist as entities in themselves, rather than as a ratio, such as . In thermodynamics, one needs to know which variable depends upon another because this is rarely stated in the equations. An example is where pressure depends on and , and therefore temperature depends on and and depends on and ; . By definition, the internal energy of an ideal gas , depends only on temperature because there are no interactions between molecules. For example, does the entropy depend either on and , or and , or on other quantities? Not apparently an easy question to answer, and in fact any one thermodynamic quantity can depend upon any of the others, but is defined when it is a parameter of any other two; which two depends upon what problem you are trying to solve. There is almost too much choice and this can lead to confusion even though there are natural variables for each quantity and these are are generally used. A total differential can be derived in the same manner as was originally done for a simple derivative, such as ; we follow here McQuarrie & Simon (1997). Making a small change in pressure caused by a small change in temperature and volume produces and now subtract a small change in volume at temperature from the first term of this equation and add the same to the second, so adding zero. The result is Next, multiply both terms by unity, the first pair of terms with and the second pair with making In the first square brackets only changes; increasing to , the volume is unchanged at and in the second term, only changes. Now take the limits 0 and 0 to form the differential and again notice the symmetry in the expression. This derivation is quite general: if a function depends on variables and , i.e. , then it is always possible to write; 11.8 The ‘minus 1’ rule or Euler’s Chain Rule# Euler’s chain rule is a product of three derivatives that is very useful in thermodynamics and elsewhere. The pressure of a gas can be written as a function of temperature and pressure, . This can be expanded as a total derivative, equation (41), and if the pressure is kept constant then The two terms and can have any value but their ratio is fixed when and so when dividing by the notation should be changed to . Doing this produces Because is constant, by the ideal gas law, is only a function of temperature. Therefore as then Notice the ‘symmetry’ in the expression; each derivative involves all three parameters in a cyclic fashion. This type of equation is quite general; if a function has the form then In thermodynamics a function such as is usually called an equation of state; this means that and the Euler ‘minus 1’ equation (44) can be obtained in a different way using this last equation. Suppose that there is no change in pressure, which is the case in an isobaric process, then which is but this result is not very useful because it still contains the function . To remove this, the equation can be divided top and bottom by which produces which is equivalent to equation (44) but arrived at in a different way. 11.9 Partial Derivatives in Thermodynamics# In the study of thermodynamics, partial, mixed, and total derivatives are commonly used; this adds to the complexity of the subject, particularly if these appear to be ‘pulled out of fresh air’. Understanding how these relationships are produced really helps in understanding thermodynamics as it removes the mathematical burden, allowing the subject itself to be better understood. To this end, the parameters used in chemistry are and each one depends upon the others. Besides and , the other parameters used are the internal energy, the enthalpy, the Gibbs free energy and the Helmholtz free energy. When one quantity is to be determined, for instance the entropy , then it is defined provided it can be calculated as a function of any other two parameters. The two parameters chosen are normally determined by what can be measured experimentally; however, it can be shown that there are natural variables for each parameter; for the internal energy these are volume and entropy . The defining equations for , and in terms of natural variables are shown in table 1. When constructing equations involving thermodynamic derivatives as a rule of thumb, values are usually , and and values , and . As an aide memoir, the diagram below the table can be used to remember these equations. It is first necessary to remember the phrase ‘Good Pupils Have Studied Under Very Able Teachers’ with the first letter being the thermodynamic quantity. The rule is that to calculate or move to the two closest symbols and add + before the symbol if the move is to the right or up and negative otherwise. The other symbols are added according to the normal pairs. An example makes this clearer. To calculate start with then as variables are always in pairs , etc., then Table 1 Thermodynamic relationships and natural variables in a closed system where no matter enters or leaves. As an example of using Table 1 the partial derivatives with respect to entropy at constant pressure are calculated and then derivatives with temperature 11.10 Exact and non-exact differentials and state functions. Pfaffians.# In an adiabatic change, no heat enters or leaves the thermodynamic system. The first law therefore asserts that Now, if then and combining this with the first law produces This differential expression is in two variables and has the same form as the differential the mathematicians call a Pfaffian: where and are two functions of and . If the derivative of each of these two functions is now taken, but with respect to the other variable, only if and are then found to be equal is the differential said to be exact or perfect; The importance of this condition is understood only when equation (47) is integrated to find . If the differential is exact, then, provided exists, and is said to be integrable. The result of integrating is just the difference in at value of compared to that at i.e. the integral does not depend upon the path taken from the start to the end but just the values of the function at the start and end. In thermodynamics, any function with the same properties as is called a state function: enthalpy, entropy, temperature, and internal energy are examples of state functions. In contrast, heat and work, which depend on the path taken from start to end, are not state functions. The internal energy of a molecule consists of energy in rotational and vibrational levels. It does not matter how the energy gets into these levels, i.e. on the path that is taken to reach a certain amount of internal energy. For example, in an experiment, we might use radiation of appropriate frequencies, to excite a certain number of rotational levels first then vibrational ones later or vice versa; the result is that the same amount of internal energy is contained within the molecule. However, the amount of work done to achieve this need not be the same. Another state function is gravitational potential energy, which is the energy gained on climbing a hill. You will have the same potential energy whether you run up, walk up, or parachute there from a plane, but the amount of work done and heat generated to get you there, will be very different in each case. The equivalence between the derivatives in equation (48) may not be true in all cases and then the differential is not exact. The integration now depends upon the path taken from and the integration must be explicitly performed. Suppose that is an expression that depends on or , then it can be shown that the differential equation (47) can be made exact by dividing the expression by the factor , As an example, let us start with equation (46) or (47) and calculate (48) to see if the two derivatives are the same. If, in equation (46) we let differentiating with respect to produces and with These two derivatives are clearly not the same; for example, in an ideal gas the derivative but it should be zero if equation (48) is to be obeyed. Now do a similar calculation, but instead use equation (49), with and calculate , which is the entropy . Because depends upon and , by equation (49) a perfect differential should result; To check this, start with eqn. 48, calculate the derivatives again using the product rule with the term preceding and B that preceding in the last equation. The result is If the derivatives of and are equal, then with some rearranging the following relationship results which must be true if we believe that the original differential is exact. To confirm that this is the case, consider an ideal gas where the internal energy depends only on the temperature, then , and as , then . As a check, substituting into equation (50) gives which shows that equation (50) is correct and so is (49).
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[FREE] How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following - brainly.com Advertisement 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 +81,7k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +14,1k Ace exams faster, with practice that adapts to you Practice Worksheets +5,3k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified How many three-digit palindromes (numbers that read the same forward and backward) satisfy the following property: the sum of the hundreds digit, the units digit, and the product of the units and tens digits is eight more than the tens digit? 2 See answers Explain with Learning Companion NEW Asked by samuelgwonlol • 12/30/2017 0:00 / -- Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 31066818 people 31M 5.0 6 Upload your school material for a more relevant answer There are 3 three-digit palindromes satisfy the property. 242 , 313 and 404 Explanation Consider the provided information. It is given that the sum of three things -- the hundreds digit, the units digit, and the product of the units and tens digits -- is eight more than the tens digit. Let the number is "aba" (last and first digit will be same, As it is palindromes) Then we have: a + a + ba = b + 8 2a + ba - b = 8 Substitute a = 1 in 2a + ba - b = 8 2 + b - b = 8 2 ≠ 8 so 'a" can't be 1. Substitute a = 2 in 2a + ba - b = 8 4 + 2b - b = 8 b = 8- 4 b = 4 So, the first number is: 242 Substitute a = 3 in 2a + ba - b = 8 6 + 3b - b = 8 2b = 2 b = 1 So, the second number is: 313 Substitute a = 4 in 2a + ba - b = 8 8 + 4b - b = 8 3b = 8-8 b = 0 So, the third number is: 404 Substitute a = 5 in 2a + ba - b = 8 10 + 5b - b = 8 4b = -2 b should be whole number so no other numbers are possible. Hence, there are 3 three-digit palindromes satisfy the property. 242 , 313 and 404 Answered by FelisFelis •2.7K answers•31.1M people helped Thanks 6 5.0 (9 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 31066818 people 31M 5.0 6 Simple nature - Benjamin Crowell Mechanics - Benjamin Crowell Conceptual Physics - Benjamin Crowell Upload your school material for a more relevant answer There are three three-digit palindromes that satisfy the property: 242, 313, and 404. The relationship involves the hundreds and units digits being equal, and the tens digit fulfilling a specific equation related to a sum and product. Each potential value for the hundreds digit was tested to find valid solutions for the tens digit, leading to these three palindromes. Explanation To find three-digit palindromes that satisfy the given condition, we follow these steps: A three-digit palindrome is of the form 'aba', where 'a' is the hundreds and units digit, and 'b' is the tens digit. For example, if 'a' = 2 and 'b' = 4, the number would be 242. The condition states that the sum of the hundreds digit (a), the units digit (a), and the product of the tens digit (b) and the units digit (a) should be equal to the tens digit (b) plus 8. Mathematically, this gives us the equation: a+a+(b⋅a)=b+8 This simplifies to: 2 a+ab−b=8 Rearranging the equation, we get: 2 a+ab=b+8 Next, we can substitute values for 'a' (since it must be a digit between 1 and 9) and solve for 'b'. Substitute a=1: 2(1)+1 b=b+8 2+b=b+8 No solution. Substitute a=2: 2(2)+2 b=b+8 4+2 b=b+8 b=4 Thus one palindrome is 242. Substitute a=3: 2(3)+3 b=b+8 6+3 b=b+8 2 b=2 b=1 Thus another palindrome is 313. Substitute a=4: 2(4)+4 b=b+8 8+4 b=b+8 3 b=0 b=0 Thus another palindrome is 404. For a=5: 2(5)+5 b=b+8 10+5 b=b+8 No whole number solution for b exists in this case. Continuing this way up to a=9, we find no additional valid pairs (5, 6, 7, 8, or 9 for a yield no valid b). Thus, the only three-digit palindromes that satisfy the given condition are: 242 313 404 In conclusion, there are three three-digit palindromes that satisfy the specified property: 242, 313, and 404. Examples & Evidence Examples of three-digit palindromes include 121, 232, 343, etc., but only 242, 313, and 404 meet the specific condition provided in this problem. The calculations and substitutions for each value of 'a' and the resultant equations demonstrate that only these three combinations satisfy the given equality conditions. Thanks 6 5.0 (9 votes) Advertisement Community Answer This answer helped 1920145 people 1M 1.3 2 Let the 3-digit palindrome be 101a+10b. a+a+ab=8+b, so 2a+ab-b=8. 2a+b(a-1)=8; b=2(4-a)/(a-1). From this we know that a > 1. And we know 4-a > 0 so a < 4. This limits a to 2 and 3 which make b=4 and 1. The palindromes are 242 and 313. Answered by Rod44 •383 answers•1.9M people helped Thanks 2 1.3 (8 votes) Advertisement ### Free Mathematics solutions and answers Community Answer (a) Find the number of three-digit palindromes with a hundreds digit of 1. (b) Find the number of three-digit palindromes with a hundreds digit of 2. (c) Find the total number of three-digit palindromes. Problem 6.11: (a) Find the prime factorization of 436995. (b) The product of two positive three-digit palindromes is 436995. What is their sum? (Source: MATHCOUNTS) 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? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. New questions in Mathematics If 24 out of 32 students prefer iPhones, how many out of 500 students in the school would be expected to prefer them? Choose an equivalent expression for 1 2 3⋅1 2 9⋅1 2 4⋅1 2 2. A. 1 2 4 B. 1 2 18 C. 1 2 35 D. 1 2 216 Choose an equivalent expression for 1 0 6÷1 0 4. A. 1 0 2 B. 1 0 3 C. 1 0 10 D. 1 0 24 How would you write 1 2−3 using a positive exponent? A. 1 2 3 B. 1 2 0 C. 1 1 2 3​ D. 1 2 3 1​ Is this equation correct? 6 3⋅7 3=4 2 3 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 Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
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https://medium.com/@AlexanderObregon/formatting-numbers-for-display-with-decimalformat-in-java-0577c0fe0b52
Sitemap Open in app Sign in Sign in Formatting Numbers for Display with DecimalFormat in Java Alexander Obregon 9 min readJun 27, 2025 Formatting numbers for invoices, summaries, and reports is something nearly every Java application has to deal with at some point. Showing raw values like 1234567.8 to a user doesn’t work well. It looks messy, lacks separators, and doesn’t reflect currency or locale conventions. That’s where DecimalFormat comes in. It turns plain numbers into strings formatted with commas, decimal places, or even symbols like currency signs. Behind the scenes, it works off a flexible pattern-based system that parses, formats, and localizes output without changing the number’s actual value. I publish free articles like this daily, if you want to support my work and get access to exclusive content and weekly recaps, consider subscribing to my Substack. How DecimalFormat Works for Displaying Numbers The DecimalFormat class gives you a way to take a number and display it in a format that makes more sense to the person reading it. Instead of showing 1234567.89 as a raw value, you can display it as 1,234,567.89 or even $1,234,567.89 depending on what you need. It doesn’t change the value of the number, just how it appears. That’s handled through patterns and formatting rules built into the class, which follow a structure that’s been consistent in Java for many years. These patterns tell the formatter how to group digits, how many decimal places to show, and where to place currency or symbols. You can format whole numbers, decimal numbers, percentages, and more, all based on these patterns. But there are a few pieces working together behind the scenes to make that possible. Pattern Mechanics and Number Symbols A DecimalFormat pattern is just a string, but that string drives the entire formatting behavior. The characters in the pattern act as placeholders, each with its own meaning. The # symbol means “use a digit here if one exists,” while 0 means “always show a digit here, even if it's zero.” The comma is used as a grouping separator and the period is the decimal point. The pattern always uses . for the decimal placeholder and , for the grouping placeholder. At runtime, the formatter swaps them for the real symbols of the chosen locale. Here’s a simple example: DecimalFormat format = new DecimalFormat("#,##0.00");System.out.println(format.format(45231.5)); // 45,231.50System.out.println(format.format(4.2)); // 4.20 DecimalFormat format = new DecimalFormat"#,##0.00"45231.5 // 45,231.504.2 // 4.20 The pattern #,##0.00 means “group digits in threes with a comma, always show two decimal places, and don’t skip the integer part even if it’s zero.” If you changed the pattern to ###.##, it would behave differently: DecimalFormat format = new DecimalFormat("###.##");System.out.println(format.format(45231.5)); // 45231.5System.out.println(format.format(4.2)); // 4.2System.out.println(format.format(4.0)); // 4 DecimalFormat format = new DecimalFormat"###.##"45231.5 // 45231.54.2 // 4.24.0 // 4 There are no grouping commas here, and if the decimal value ends in zeroes, they won’t be shown unless the pattern demands them. That’s the difference between using 0 and #. Patterns can also include literal characters if you wrap them in quotes. This is helpful when labeling things directly in the format: DecimalFormat format = new DecimalFormat("'Total:' #,##0.00");System.out.println(format.format(2500.5)); // Total: 2,500.50 DecimalFormat format = new DecimalFormat"'Total:' #,##0.00"2500.5 // Total: 2,500.50 That single quote lets the formatter know the word “Total” is literal and not part of the formatting syntax. This gives you fine control over how the output looks without needing to build the string yourself. Locale-Specific Behavior and Currency DecimalFormat supports localization by pulling formatting rules from the system or a specified locale. Different locales use different symbols for grouping and decimal separators. Some put currency symbols before the number, some put them after. Some use commas for grouping and dots for decimals, while others reverse it. If you don’t set a locale, DecimalFormat will use the default one from your system settings. To be exact about how numbers should look, you can provide a specific locale or set custom symbols using DecimalFormatSymbols. DecimalFormatSymbols germanSymbols = DecimalFormatSymbols.getInstance(Locale.GERMANY);DecimalFormat format = new DecimalFormat("#,##0.00", germanSymbols);System.out.println(format.format(1234567.89)); // 1.234.567,89 DecimalFormatSymbols germanSymbols = DecimalFormat format = new DecimalFormat"#,##0.00"1234567.89 // 1.234.567,89 This swaps the comma and dot from the default US-style format. Behind the scenes, the DecimalFormatSymbols object stores what symbols to use for grouping, decimal points, percent signs, and currency symbols. To include a currency symbol in the output, you can either use the currency pattern or switch to NumberFormat.getCurrencyInstance() which returns a preconfigured formatter that uses the correct symbol for the given locale. NumberFormat currency = NumberFormat.getCurrencyInstance(Locale.UK);System.out.println(currency.format(823456.7)); // £823,456.70 NumberFormat currency =823456.7 // £823,456.70 That currency formatter is still backed by DecimalFormat in most cases, but it sets the pattern and symbols automatically. So it’s better when you want something consistent without hardcoding currency signs into your patterns. Thread Safety and DecimalFormat DecimalFormat isn’t safe to use across threads. That’s because the same instance keeps internal state as it formats and parses values. If two threads try to format different numbers at the same time using the same DecimalFormat, they can step on each other’s output. You might get output that has parts of one number and parts of another, or worse, something that doesn’t match any real number. This happens because DecimalFormat has internal fields like the current parse position and pattern state, which it updates as it works. If multiple threads share the instance, those fields can shift mid-operation. Here’s an example that should be avoided: DecimalFormat sharedFormat = new DecimalFormat("#,##0.00");// Thread AString a = sharedFormat.format(1000000.25);// Thread B at the same timeString b = sharedFormat.format(42.6); DecimalFormat sharedFormat = new DecimalFormat"#,##0.00" // Thread A String a =1000000.25 // Thread B at the same time String b =42.6 There’s no guarantee that a will be correct, or that b won’t be affected. You don’t get an exception and it just silently breaks. The safer way is to give each thread its own instance. If you need to reuse the same format logic but want to avoid creating a new DecimalFormat each time, use a ThreadLocal: ThreadLocal<DecimalFormat> localFormat = ThreadLocal.withInitial(() -> new DecimalFormat("#,##0.00"));// Each thread can safely do this:String output = localFormat.get().format(123456.78); new DecimalFormat"#,##0.00" // Each thread can safely do this: String output =123456.78 This guarantees that each thread has its own formatter and won’t share internal state. It avoids the risks that come from sharing mutable objects in a concurrent system, and you still get the performance benefit of reusing the same format logic without constantly building new objects. If you’re working in a web application, it’s especially important not to store DecimalFormat instances as static fields or shared beans. Always make sure the formatter is tied to a single request or isolated per thread. Using DecimalFormat Securely in Real Applications Formatting numbers for display is one part of a much bigger system. When values represent something important, like a transaction total or a user balance, the formatted output is often the last step before someone sees it on a screen or in a report. That makes the formatting visible, but not reliable for anything behind the scenes. The original number should always stay intact and untouched by formatting. Mixing up display formatting with actual calculations or decisions can cause problems later. Whether the values come from a database, an API call, or user input, it helps to think about formatting as something that only belongs at the edge of the application. Separating Display from Logic Formatted strings are meant for people, not machines. That sounds obvious, but it’s a common mistake to let those formatted strings slip into areas of the code that expect real numbers. A formatted string isn’t guaranteed to round the same way across all locales, and it can carry extra symbols that break logic built around numeric values. Get Alexander Obregon’s stories in your inbox Join Medium for free to get updates from this writer. If you’re building invoices, reports, or summaries, format the numbers after all calculations are done. Do the math with types like int, long, or BigDecimal, then call format() only when it’s time to print, email, or display. BigDecimal subtotal = new BigDecimal("1499.99");BigDecimal tax = new BigDecimal("0.0825");BigDecimal total = subtotal.multiply(tax).add(subtotal);DecimalFormat displayFormat = new DecimalFormat("#,##0.00");String totalText = displayFormat.format(total);System.out.println("Invoice Total: $" + totalText); BigDecimal subtotal = new BigDecimal"1499.99" BigDecimal tax = new BigDecimal"0.0825" BigDecimal total = DecimalFormat displayFormat = new DecimalFormat"#,##0.00" String totalText ="Invoice Total: $" The number stays as a BigDecimal through the entire calculation process. Only the final output gets formatted, and nothing else relies on that formatted string. This avoids rounding errors, keeps decimal precision, and protects the data from being affected by formatting quirks. If the formatted result needs to be sent somewhere, like in an email, attach it as a display value. Keep the raw number around if anything depends on it later. Using DecimalFormat with Currency When dealing with money, always use BigDecimal. It’s the only type in the standard library that can reliably track currency without losing decimals or introducing rounding surprises. Floating-point types like float or double don’t cut it here. When you’ve finished the calculation, hand the value off to a currency-aware formatter. If you use DecimalFormat directly, make sure the pattern fits the currency structure you want. BigDecimal payment = new BigDecimal("2000.5");DecimalFormat currencyFormat = new DecimalFormat("$#,##0.00");System.out.println(currencyFormat.format(payment)); // $2,000.50 BigDecimal payment = new BigDecimal"2000.5" DecimalFormat currencyFormat = new DecimalFormat"$#,##0.00" // $2,000.50 This works, but it hardcodes the symbol and grouping style. A better way, especially for applications that support multiple regions, is to use NumberFormat.getCurrencyInstance(Locale) instead. That pulls in the expected grouping and currency symbol based on location. BigDecimal payment = new BigDecimal("2000.5");NumberFormat currencyFormat = NumberFormat.getCurrencyInstance(Locale.CANADA_FRENCH);System.out.println(currencyFormat.format(payment)); // 2 000,50 $ BigDecimal payment = new BigDecimal"2000.5" NumberFormat currencyFormat = // 2 000,50 $ If you need to guarantee the formatting stays stable across environments, you can set both the locale and the Currency explicitly. That gives you predictable output, even when the system defaults change or the app moves to a different machine. The pattern still won’t change the value underneath. You can switch from dollars to euros in the formatting, but the number stays the same. The visual output might shift from $2,000.50 to 2 000,50 €, but nothing inside the object changes. That’s why formatting must always come last. API Security and Formatting Risks Formatted numbers can look secure, but they’re not meant to hold up in real security logic. You should never use a formatted string to decide who can do what, or to pass hidden meaning to another system. Take this situation. An API call returns a formatted string for a total, like “$1,000.00”, and a second service checks if the user is allowed to view values over a certain amount. If that logic works off the string instead of the raw number, all it takes is one formatting change to break it. A number like 1000.0 could be formatted as 1,000.00, 1000,00, or even USD 1 000.00 depending on the locale. Comparing that to a threshold like "1000.00" won’t always give the right answer. Use raw numbers for all permission checks, limits, comparisons, and validations. Format values only after those steps are finished. The same applies when logging data, storing it, or sending it to another service. Send the raw value, not the display version. API keys also come up here. A lot of systems show invoice totals or user data based on whether a valid API key is present. That’s a start, but an API key isn’t a substitute for authentication. It identifies the app, not the user. If your app formats and returns sensitive numbers, it needs to check user access before responding. if (userIsAuthenticated(request) && userHasPermission(userId, "VIEW_TOTALS")) { BigDecimal total = fetchTotalFromDatabase(); String output = new DecimalFormat("#,##0.00").format(total); return output;} if "VIEW_TOTALS" BigDecimal total = String output = new DecimalFormat"#,##0.00" return The formatting here happens only after the user’s been checked. You don’t want to expose the formatted output just because a request came in with the right API key. That key only says which app is making the call, not which user is asking. Avoiding Common Formatting Traps Some issues happen just by forgetting what the formatter really does. One of the most common problems is using String.format() when you need proper numeric formatting. String.format() can follow the default or an explicit Locale, and it can use the , flag for grouping. But it still can’t handle currency symbols or the kind of specialized formatting rules that DecimalFormat supports. String result = String.format("%.2f", 123456.789); // 123456.79 String result ="%.2f"123456.789 // 123456.79 This rounds and formats the number, but it won’t give you commas or currency unless you build the whole string yourself. Using DecimalFormat or NumberFormat avoids that and gives you formatting that respects spacing, separators, and grouping styles from the locale. Another common trap is formatting a number, sending it through the system, and then trying to turn it back into a number. If the format includes commas, currency symbols, or localized separators, the string won’t parse unless you reverse the exact formatting logic. DecimalFormat format = new DecimalFormat("#,##0.00");String formatted = format.format(1000000.5); // 1,000,000.50// Incorrect parsing will fail or misinterpret the stringBigDecimal parsed = new BigDecimal(formatted); // throws NumberFormatException DecimalFormat format = new DecimalFormat"#,##0.00" String formatted =1000000.5 // 1,000,000.50 // Incorrect parsing will fail or misinterpret the string BigDecimal parsed = new BigDecimal // throws NumberFormatException To parse a formatted number, you need to use the same DecimalFormat instance or one with matching symbols. And even then, parsing should only happen in edge cases. It’s better to pass raw numbers through your logic and only use formatting when the value is about to be seen by someone. And don’t assume that using the same pattern always gives you the same result. If your environment’s default locale changes, or if different machines run the same code with different region settings, your output may shift. To avoid this, always lock the locale and format explicitly if the output has to stay consistent. DecimalFormatSymbols symbols = DecimalFormatSymbols.getInstance(Locale.US);DecimalFormat format = new DecimalFormat("#,##0.00", symbols); DecimalFormatSymbols symbols = DecimalFormat format = new DecimalFormat"#,##0.00" This forces the grouping and decimal style to match what you expect, no matter what system the code runs on. It avoids the guessing game and keeps the formatted output stable. Conclusion DecimalFormat works by applying pattern rules to a number without touching the number itself. It figures out where commas go, how many decimals to show, and what symbols to include based on the pattern and locale. The value stays the same the whole time. Formatting should stay at the edges of your code, only used when the number is ready to be seen. That way, everything behind the scenes runs on real types and not on display strings that were only meant to be read. When formatting stays separate from logic and access, it does its job without getting in the way. DecimalFormat Class Documentation DecimalFormatSymbols Class NumberFormat Abstract Class Java Locale Class Documentation Thanks for reading! If you found this helpful, highlighting, clapping, or leaving a comment really helps me out. Java Currency Formatting Localization Thread Safety Programming Some rights reserved ## Written by Alexander Obregon 26K followers ·15 following I post daily about programming topics and share what I learn as I go. For recaps, exclusive content, and to support me: No responses yet Write a response What are your thoughts? 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Handling failures properly is… Sep 16, 2023 400 8 Alexander Obregon ## How Spring Boot Auto-Configuration Works Introduction Nov 18, 2024 113 3 See all from Alexander Obregon Recommended from Medium In Javarevisited by Habibwahid ## 99% of Junior Java Developers Make These Mistakes — See How Senior Devs Write Clean, Efficient… From Rookie Errors to Senior-Level Code Mastery: Real Examples that Show How to Write Less Code, with Better Performance, Readability, and… Aug 17 73 12 Aditya Bhuyan ## What is a StampedLock and How Does it Differ from ReentrantLock? — IT Interview Guide What is a StampedLock and How Does it Differ from ReentrantLock? Mar 3 1 In Stackademic by Pudari Madhavi ## Sharing Data Between Microservices Without Losing Your Mind If you’ve ever worked with microservices, you know one of the biggest headaches is sharing data between them. 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https://ajronline.org/doi/full/10.2214/ajr.20.25223
Skip to main content Breast Cancer Staging: Updates in the AJCC Cancer Staging Manual, 8th Edition, and Current Challenges for Radiologists, From the AJR Special Series on Cancer Staging Authors: Davis C. Teichgraeber, MD, Mary S. Guirguis, MD, and Gary J. Whitman, MDAuthor Info & Affiliations Volume 217, Issue 2 00 Abstract The standardization of the AJCC TNM staging system for breast cancer allows physicians to evaluate patients with breast cancer using standard language and criteria, assess treatment response, and compare patient outcomes. Previous editions of the AJCC Cancer Staging Manual relied on the anatomic TNM method of staging that incorporates imaging and uses population-level survival data to predict patient outcomes. Recent advances in therapy based on biomarker status and multigene panels have improved treatment strategies. In the newest edition of the AJCC Cancer Staging Manual (8th edition, adopted on January 1, 2018), breast cancer staging integrates anatomic staging with tumor grade, biomarker data regarding hormone receptor status, oncogene expression, and gene expression profiling to assign a prognostic stage. This article reviews the 8th edition of the AJCC breast cancer staging system with a focus on anatomic staging and the challenges that anatomic staging poses for radiologists. We highlight key imaging findings that impact patient treatment and discuss the role of imaging in evaluating response to neoadjuvant therapy. Finally, we discuss biomarkers and multigene panels and how these impact prognostic stage. The review will help radiologists identify critical findings that affect breast cancer staging and understand ongoing limitations of imaging in staging. The standardization of the TNM staging system by the AJCC has allowed physicians to evaluate patients with breast cancer using a universal language, facilitating effective communication regarding appropriate treatment planning. Since the publication of the first edition of the AJCC Cancer Staging Manual in 1977, advances in diagnosis and treatment have necessitated periodic updates to the staging manual [1, 2]. Breast cancer treatment is at the forefront in the age of personalized medicine, which incorporates patient- and tumor-specific factors into the prognostic and treatment decisions for each patient. The previous edition (7th) of the AJCC manual staged cancer according to anatomic information (TNM). The current breast cancer expert panel concluded that, whereas the TNM staging system provides important insight into a patient's prognosis, the addition of biomarkers such as estrogen receptor (ER), progesterone receptor (PR), and human epidermal growth factor receptor 2 (HER2, also known as ERBB2) as well as gene expression profiling refines the prognostic information and leads to better selection of treatment options. Therefore, the addition of this information warranted modification of the TNM staging system for breast cancer in the 8th edition of the AJCC Cancer Staging Manual, which was adopted January 1, 2018 [3, 4]. Although breast cancer treatment has evolved, the TNM staging system remains relevant. In lower- and middle-income countries that do not have access to biomarker analysis, the TNM system remains the only staging system. Furthermore, the TNM staging system ensures that physicians worldwide communicate in a standard language that reflects the tumor burden and allows investigators to compare patients treated during different time periods since the inception of the staging system in 1959 [2, 4]. Imaging is the foundation of anatomic staging in breast cancer. This article reviews the 8th edition of the AJCC TNM staging system for breast cancer, with a focus on preoperative anatomic staging. We discuss anatomic TNM staging and the major changes in the 8th edition as it applies to imaging. For each subsection of the TNM staging system, we review the role of individual imaging modalities with a focus on mammography, ultrasound, and MRI. We discuss the challenges that radiologists face in the imaging evaluation of a patient with breast cancer and what imaging information is critical to provide to referring clinicians to optimize patient care. We discuss imaging in the neoadjuvant setting. Finally, we discuss prognostic staging because radiologists, as part of multidisciplinary teams, need to understand both the anatomic and the prognostic staging systems that guide patient care. Staging System Basics The AJCC breast cancer staging system should be used for invasive carcinoma and ductal carcinoma in situ (DCIS). It is not intended for staging of breast lymphoma, breast sarcoma, or phyllodes tumors. The breast cancer staging system in the AJCC 8th edition consists of two staging tables: the anatomic and the prognostic categories. The anatomic stage is based on the anatomic extent of cancer as defined by the TNM categories. Anatomic staging includes the primary tumor size (T), the nodal status (N), and distant metastasis (M) according to clinical and/or pathologic assessment and was the basis of the 7th edition of the AJCC staging system. For determination of anatomic stage, the National Comprehensive Cancer Network (NCCN) advises performing a history and physical examination, bilateral mammography and ultrasound as warranted, pathology evaluation, and assessment of hormone receptors. Breast MRI is deemed optional, with tumors that are mammographically occult warranting special consideration. Imaging studies are deemed elements of staging when performed within 4 months of the diagnosis or upon completion of surgery, whichever gives the longer interval, provided that the patient's disease has not progressed . The AJCC 8th edition staging system contains four categories of anatomic TNM classification. The first category is clinical staging, designated by the prefix “c,” based on clinical examination, diagnostic imaging, and core biopsy or aspiration samples obtained before treatment. The second category is pathologic staging, designated by the prefix “p,” based on surgical specimens, including sentinel lymph node biopsy (SLNB) specimens. The third category is posttherapy staging, designated by the prefix “yp,” and applies to patients who have been treated with neoadjuvant therapy including neoadjuvant chemotherapy (NAC), radiation, or hormonal therapy. Finally, the last category, restaging, applies in the event of tumor recurrence. The anatomic staging system provides quantitative classification categories for primary tumors (Tis to T4), the regional lymph node status (N0 to N3), and distant metastases (M0 or M1), and these categories are combined to determine the overall anatomic stage (stage 0 to stage IV) . Overall anatomic stage has historically been associated with outcome measures, including overall survival [4, 6]. In addition to anatomic TNM stage, prognostic stage is determined by tumor grade, biomarker status (ER, PR, HER2), and genomic panels. Every patient is assigned a clinical prognostic stage because it sets a baseline and therefore guides initial treatment. The clinical prognostic stage is intended to provide a comparison for all patients regardless of sequence of therapy (initial surgery with adjuvant therapy vs NAC) . The pathologic prognostic stage is assigned to patients who undergo surgery as the initial treatment and excludes patients who receive NAC. Tumor The T category is determined by tumor size and locoregional invasion . The AJCC 8th edition staging system clarifies that the T category is based on the size of the invasive component of the largest mass (in the setting of multiple masses) in the largest dimension. Small satellite nodules are not added to tumor size. Multiple cancers are documented with the “m” modifier. Although tumor volume (as measured in three dimensions) does not affect stage, volumetric assessments are helpful when evaluating response to treatment in patients undergoing NAC [6, 7]. The updated staging system does not account for multifocal or multicentric disease (Fig. 1). The T category of multicentric disease is determined by the largest mass and may be the same as for a unifocal malignancy even though disease is much more extensive. It is critical that radiologists provide tumor measurements of the largest mass and describe the location and size of other masses because this information impacts surgical management (lumpectomy vs mastectomy) and may determine whether the patient receives NAC . It is also important to describe findings such as associated calcifications to estimate the extent of disease, which may include DCIS associated with an invasive component. Other findings, such as extension of disease to the skin, nipple, or chest wall (or a combination of these), should also be reported . The categories range from Tis to T4. Tis is designated for DCIS and Paget disease (without underlying DCIS) without an invasive component. One change to the AJCC 8th edition staging system is that lobular carcinoma in situ (LCIS) is not classified as category Tis and is now deemed benign. However, LCIS confers a higher-than-average risk of future breast cancer. Nonclassical LCIS variants, including pleomorphic LCIS and LCIS with necrosis, signet ring cells, or apocrine features, tend to have high-grade cytology and an unfavorable biomarker profile. Current evidence suggests these lesions should be treated with complete surgical excision, similar to DCIS . In the setting of DCIS, an estimation of disease extent may alter surgical management (mastectomy vs lumpectomy) and determine whether the patient requires an SLNB (Fig. 2). The American Society of Clinical Oncology clinical practice guideline recommends SLNB for women with DCIS who are planning to undergo a mastectomy or who have a large DCIS tumor size (> 5 cm) . Therefore, reporting the extent of calcifications is critical. The extent of disease, especially in relation to the nipple, can determine whether the patient is a candidate for nipple-sparing mastectomy. Contraindications for nipple-sparing mastectomy include microcalcifications close to the subareolar region (< 2 cm) , Paget disease, and bloody nipple discharge (Figs. 3 and 4). The subcategories of T1 disease are T1mi (microinvasive; tumor ≤ 1 mm in greatest dimension), T1a (> 1 mm but ≤ 5 mm), T1b (> 5 mm but ≤ 10 mm), and T1c (> 10 mm but ≤ 20 mm). Another change in the AJCC 8th edition staging system is that a tumor measuring more than 1 mm and less than 2 mm is rounded to 2 mm. With T2 disease, tumor size is larger than 20 mm and no greater than 50 mm. With T3 disease, tumor size is greater than 50 mm. The subcategories of T4 disease are T4a, T4b, T4c, and T4d (Fig. 5). Category T4a is defined by chest wall extension, where the chest wall consists of the ribs and the intercostal and serratus anterior muscles. Involvement of the pectoralis major or minor muscles alone is not considered chest wall involvement and does not affect the T category . MRI is excellent for evaluating pectoralis muscle involvement and chest wall invasion (Figs. 6 and 7). Involvement of the pectoralis muscle may manifest as enhancement of the muscle (Fig. 6) with loss of the fat plane between the muscle and the tumor on MRI. Loss of the fat plane without muscle enhancement is not definitive for muscle involvement [12, 13] (Fig. 8). Although pectoralis muscle involvement does not affect clinical stage, reporting this finding is critical because it may impact surgical decision making (whether the surgeon resects the muscle) and radiation therapy planning. Furthermore, chest wall involvement may require more extensive surgery, including thoracic surgery. Category T4b involves macroscopic skin changes that include a combination of ulceration, satellite skin nodules, and edema. The AJCC 8th edition staging system states that satellite tumor nodules involving the skin must be separate from the primary tumor and macroscopically identified to be considered T4b. Skin and dermal nodules identified on microscopic examination and in the absence of skin ulceration or skin edema (peau d'orange) do not meet T4b criteria . Tumors categorized as T4c meet the criteria for both T4a and T4b (Fig. 7). Category T4d is inflammatory breast cancer, which is a clinical and pathologic entity requiring features of diffuse erythema and edema (peau d'orange) involving at least one-third of the breast and progressing rapidly over weeks to months [4, 14] (Fig. 9). Imaging Tumor Size Digital breast tomosynthesis (DBT) has been well documented in the screening setting to be superior to full-field digital mammography (FFDM) in improving cancer detection rate and decreasing recall rates . However, the performance of DBT combined with FFDM in the staging of patients with breast cancer has been studied less. Although DBT combined with FFDM shows better diagnostic performance and sensitivity than FFDM alone in the diagnostic setting, breast cancers without distinct masses or without calcifications are still difficult to detect . Fontaine et al. found that the additional value of DBT in breast cancer staging (in detecting additional ipsilateral disease or contralateral lesions) was limited to women with nondense breasts. Most DCIS lesions found at mammography present as micro-calcifications, with approximately 75% presenting only as calcifications [18, 19]. Ultrasound does have a role in the workup of DCIS given its variable appearance. Ultrasound is also useful in evaluating for an associated mass, which may indicate invasion . Whole-breast ultrasound is critical in staging to assess tumor size and in evaluating for additional disease that is mammographically occult, especially in patients with dense breasts . The detection of additional malignant lesions may lead to a wider surgical excision or alter surgical management (mastectomy vs lumpectomy) [22, 23]. The routine use of contrast-enhanced MRI for preoperative staging continues to be controversial given that the impact of MRI on survival is still unknown . Currently, evidence that pre-operative MRI improves overall or disease-free survival is lacking . Furthermore, numerous studies have found that MRI overestimates tumor size when compared with final pathology . In the initial staging setting, MRI could result in overstaging. MRI may be appropriate in certain circumstances. MRI has shown value in staging invasive lobular cancer, a subtype that is typically underestimated by mammography and ultrasound. MRI may be helpful in reducing reexcision rates in invasive lobular cancer, which range from 11% to 18% (Fig. 8). MRI is also useful in showing multicentricity and involvement of the pectoralis muscle and chest wall [12, 13] (Figs. 1, 6, and 7). There is no consensus regarding which imaging modality best predicts tumor size. Mammography is less accurate in dense breasts yet is superior in showing microcalcifications. Ultra-sound is highly operator-dependent, inexpensive, and helpful in showing mammographically occult disease, especially in dense breasts. MRI, though highly sensitive, lacks specificity and often overestimates tumor size [25, 27]. When tumor measurements are discrepant between imaging modalities, the MRI measurements are typically used . As MRI has low specificity in distinguishing benign from malignant enhancement, obtaining a tissue sample from a lesion showing suspicious enhancement is critical and could alter patient stage and/or clinical management . Regional Lymph Nodes The expert panel did not recommend any major changes to the clinical nodal staging classification in the AJCC 8th edition staging system [3, 4]. However, they did clarify that category cNx is only to be used when lymph nodes are removed and cannot be examined by imaging or physical examination. A cN0 category is assigned when evaluation of the lymph nodes is possible and no regional lymph node metastases are detected by physical examination or imaging . The clinical nodal category is based on imaging or clinical evaluation of the nodal basins , whereas the pathologic category is based on the number of involved lymph nodes on surgical pathology. The abbreviations (sn) and (f) are suffixes added to the N category if metastases are confirmed with SLNB or with fine-needle aspiration and core biopsy, respectively . Staging of axillary lymph nodes is determined by the lymph node's location in relation to the pectoralis minor muscle. Level I axillary lymph nodes are located lateral to the pectoralis minor muscle, level II axillary lymph nodes are between the pectoralis minor muscle's medial and lateral margins and encompass interpectoral (Rotter) lymph nodes, and level III axillary lymph nodes lie medial to the pectoralis minor muscle's medial margin (Fig. 10). Category cN1 disease encompasses metastases to movable ipsilateral level I and/or level II nodes. Category cN2 disease includes metastases to fixed or matted ipsilateral level I and/or level II nodes or to ipsilateral internal mammary (IM) nodes in the absence of axillary metastases. Category cN3 disease includes ipsilateral level III node metastases with or without level I or level II nodes (Fig. 11), ipsilateral IM metastases with level I or level II metastases, or ipsilateral supraclavicular (SC) lymph node metastases. Metastatic intramammary lymph nodes are equivalent to level I for staging purposes. Cervical lymph nodes and contralateral axillary, SC, and IM metastases are considered distant metastases (M1). When multiple suspicious lymph nodes are present in different nodal basins, the nodal level that should be sampled first is the one that would have the greatest effect on stage. For example, if suspicious level III and level I nodes are present, the level III node should be biopsied first. If the biopsy of a level III node shows benign results, then the abnormal level I node should be sampled . Imaging Regional Lymph Nodes Ultrasound remains the modality of choice for evaluating the axillary lymph nodes. However, ultrasound is operator-dependent. Axillary and IM lymph nodes are well evaluated by breast MRI, chest CT, or FDG PET/CT. SC nodes can be evaluated with ultrasound or cross-sectional imaging. Historically, patients with biopsy-proven metastatic axillary lymph nodes underwent axillary lymph node dissection (ALND), a procedure with significant morbidity. However, in 2010, the American College of Surgeons Oncology Group (ACOSOG) Z0011 randomized trial found that ALND may not be necessary in patients who are clinically node negative and are undergoing breast conservation with one or two positive sentinel lymph nodes. The study found no difference in local recurrence between the ALND group and SLNB group [30, 31]. Given the shift in surgical management, the use of axillary nodal ultrasound has become controversial. The concern is that, when mildly abnormal axillary nodes are imaged and sampled in a patient who meets the criteria for SLNB as established in the Z0011 trial, a positive biopsy would commit the patient to ALND. However, studies have found that patients with a positive axillary ultrasound finding and subsequent positive biopsy results are more likely to have more axillary nodes involved on pathology at the time of surgery [32–34]. Furthermore, Verheuvel et al. found that patients with axillary disease on ultrasound were more likely to have involvement of level III nodes and a decrease in overall and disease-free survival. NAC is used to downstage not only the breast malignancy but also the axilla. Targeted axillary dissection has emerged to help treat patients with limited nodal involvement who show evidence of response to treatment. The ACOSOG Z1071 trial investigated SLNB performed after NAC in patients with stage II, stage IIIA, and stage IIIB disease who had biopsy-proven axillary metastases before NAC. This study found that when the clipped biopsy-proven metastatic node (that showed response to NAC) and at least two sentinel nodes are excised, the false-negative rate of SLNB was 6.8% [36, 37]. Therefore, it is reasonable to consider SLNB in patients with cN1 or cN2 disease at presentation and good clinical response to NAC . Complete evaluation of the axilla and regional nodal basins during initial staging is critical to determine which patients may be candidates for targeted axillary dissection before NAC and therefore require placement of a biopsy clip in biopsy-proven metastatic axillary lymph nodes. In patients with proven or suspected metastatic level I or II lymph nodes, imaging studies to evaluate level III, IM, and SC lymph nodes should be considered when determining whether the patient is a candidate for targeted axillary dissection [6, 8, 23]. In addition, the presence of IM, SC, and level III lymph node metastases alters radiation therapy planning (Fig. 11). Metastases The M category is classified as M0 (no distant metastases) or M1 (metastatic disease present). The most common sites of metastases in breast cancer are bone, lung, brain, and liver . The AJCC 8th edition staging system clarifies that pM0 is not a valid category [3, 4]. A benign biopsy of a suspicious lesion does not guarantee absence of metastatic disease elsewhere. Only cM0, cM1, and pM1 are used. Category cM0 is defined as no clinical or imaging evidence of distant metastases. Category cM1 is defined as distant metastases on the basis of clinical or imaging findings. Category pM1 is defined as distant metastases on the basis of pathologic proof. Imaging of Metastases NCCN guidelines state that routine systemic staging is not indicated for early breast cancer (T0–3N1M0 or T1–3N0–1M0) in the absence of signs and symptoms of metastatic disease . Additional studies are guided by patient symptoms. For patients with clinical stage IIIA disease or those considering NAC for T2M0 or higher or N1M0 or higher disease, studies to consider include chest CT, abdominopelvic CT or MRI, bone scan, and FDG PET/CT. When used in addition to other imaging studies, FDG PET/CT is helpful in identifying undetected regional nodal disease and/or distant metastases. Systemic imaging is also recommended in recurrent disease, stage IV (M1) disease, or inflammatory breast cancer. Imaging Neoadjuvant Response Traditionally, NAC has been administered for locally advanced breast cancers (clinical category T3N1–3M0) and inflammatory breast cancer because it can reduce both breast tumor and locoregional nodal recurrence. Currently, NAC is used at earlier stages to potentially change a patient's treatment from mastectomy to lumpectomy . Furthermore, NAC may reduce the extent of axillary surgery, converting an ALND to a targeted axillary dissection. In HER2 and triple-negative subtypes, pathologic response can predict long-term progression-free and overall survival rates [40, 41]. After a patient receives NAC, the prefix “y” is used in assigning a TNM category. The use of NAC does not change the anatomic (pretreatment) stage. Clinical (posttreatment) T category is determined by the size and extent of disease on physical examination and imaging, and the ycT category is determined by measuring the largest residual mass . There is no standard consensus regarding imaging assessment of NAC response. Although mammography is superior to physical examination, it can overestimate residual disease in lesions with architectural distortion or indistinct margins and in lesions that show no change in calcifications [39, 42]. The presence of calcifications does not correlate with viable tumor . Ultrasound is a better predictor for pathologic tumor response and response in regional nodal metastases. However, though more accurate than mammography alone, ultrasound underestimates viable tumor given the difficulty in distinguishing posttreatment fibrosis from viable tumor on imaging . MRI is the most accurate modality for assessment of tumor response to NAC, with a reported PPV (for correctly predicting the presence of residual disease on final surgical pathology) of 93% and NPV of 64% . The American College of Radiology Imaging Network 6657 and Investigation of Serial Studies to Predict Your Therapeutic Response with Imaging and Molecular Analysis trial, which involved 216 women with tumors measuring 3 cm or greater treated with NAC, showed that tumor volume calculations were superior to measurements of the longest dimension for predicting response . However, factors such as tumor molecular subtype and the chemotherapy regimen affect the accuracy of MRI in assessing NAC response. Emerging technologies, such as functional and molecular imaging and use of advanced MRI techniques and/or radionuclide imaging to assess physiologic changes, may play a role in assessing response in combination with anatomic imaging. However, these methods are currently investigational . After completion of NAC, the same imaging modalities used in pretreatment staging should be performed for evaluating response, and tumor size in three dimensions should be compared . Other details to report are more qualitative, such as change in density of a mass on mammography, change in echogenicity and margin features on ultrasound, and normalization of nodal morphology. Prognostic Staging The most significant change in the AJCC 8th edition staging system is that the final prognostic stage is determined by tumor grade, biomarker status (ER, PR, and HER2), genomic panels, and anatomic TNM stage. Grade Tumor grade is defined by the Scarff-Bloom-Richardson histologic grading system, which was standardized by the Nottingham group and used by the College of American Pathologists . The morphologic features of gland characteristics, pleomorphism, and mitotic counts are each given a score ranging from one to three, and these scores are then summed to determine grade [44, 45]. Grade 1 (score between 3 and 5) represents a well differentiated tumor, grade 2 (score 6 or 7) represents a moderately differentiated tumor, and grade 3 (score 8 or 9) represents a poorly differentiated tumor. Tumor grade is an important prognostic factor, independent of tumor size and number of positive lymph nodes . ER, PR, and HER2 Expression and Molecular Subtypes The AJCC 8th edition staging system stipulates that the ER and PR receptor status and HER2 expression status of all invasive carcinomas should be determined whenever possible. Endocrine therapies such as tamoxifen are known to slow progression of ER-positive and PR-positive tumors . Furthermore, advances in breast cancer genomics have allowed prognostic profiling based on the expression of combinations of thousands of genes in tumor cells. The main subgroups, according to hormone receptor status and gene expression patterns, are the luminal A–like, luminal B–like, HER2-enriched, and basal-like (triple-negative) sub-types (Table 1). TABLE 1: Clinically Defined Treatment-Oriented Subtypes of Breast Cancer | Subtype | Hormone Receptor Status, Histologic Grade, and Prognosis | --- | | Luminal A-like | High hormone receptor, low proliferation High ER and PR expression; usually HER2 negative Usually low grade and low proliferation rate (low Ki-67, low mitotic count), generally histologic grade 1 or 2 [52, 53] Favorable prognosis, 5-year survival rate > 80% | | Luminal B-like | Low hormone receptor, high proliferation Lower ER and PR expression; usually HER2 negative but approximately 30% of cases will be HER2 positive High proliferation rate (high Ki-67, high mitotic count) Generally, histologic grade 3 Less favorable prognosis compared with luminal A-like | | HER2-enriched | HER2 positive (ER and PR negative or positive) Generally histologic grade 3 Targeted therapy for HER2 positive cancers (such as trastuzumab) has improved prognosis | | Basal-like | Most commonly triple negative (ER, PR, and HER2 negative) Defined by absence of ER, PR, and HER2 markers and overexpression of oncogenes that favor cell proliferation and carcinogenesis Generally histologic grade 3 Unfavorable prognosis [55, 56] | Note—ER = estrogen receptor, PR = progesterone receptor, HER2 = human epidermal growth factor receptor 2 (also known as ERBB2), negative = no expression by tumor, Ki-67 = marker for cellular proliferation, positive = expression by tumor. Multigene Panel Testing The AJCC 8th edition staging system also incorporates multigene panel testing in assigning prognostic subgroups. At the time of publication of the revised AJCC staging manual, Onco-type DX Breast Recurrence Score (Genomic Health) was the most validated panel and was therefore incorporated into the 8th edition prognostic staging system . However, other multigene panels are available. The expert panel did not advocate one specific vendor and acknowledged that further modifications of the staging system will be necessary as additional data validating these panels are obtained . Currently, Oncotype DX is used to predict the average 10-year distant recurrence risk in patients receiving endocrine therapy alone for T1–2N0M0 (ER-positive, PR-positive, HER2-negative) disease. Oncotype DX is used to assess the additional benefit of adjuvant chemotherapy with respect to endocrine therapy. For low-risk disease (score less than 11), the model does not predict any additional benefits from chemotherapy [47, 48]. The AJCC 8th edition staging system incorporates the Oncotype DX score, if available, in the prognostic staging of this subset of patients. Prognostic Staging Versus Anatomic Staging In two previous studies, data from two large cohorts from the MD Anderson Cancer Center and the California Cancer Registry were analyzed and incorporated into prognostic staging. The MD Anderson study evaluated 3327 patients treated between 2007 and 2013 and found that disease-specific survival was more precise using a staging system that incorporated tumor grade, ER status, and HER2 status compared with anatomic staging alone . These findings were validated using a cohort of 67,944 patients from the California Cancer Registry . The integration of new biomarkers into the AJCC 8th edition staging system may result in upstaging or downstaging (Table 2), depending on the tumor biology, which can result in a prognostic stage different from the anatomic stage. Patients with triple-negative tumors, regardless of grade, have survival rates comparable with those of patients with disease one anatomic stage higher and are generally upstaged in their prognostic stage (for example, anatomic stage IA and prognostic stage IIA). Similarly, patients with grade 3 tumors that are HER2-negative and positive for either ER or PR also have survival rates comparable with those of patients with disease one anatomic stage higher [6, 9]. Conversely, a patient with a T3N0M0 cancer of any grade that is ER positive, PR positive, and HER2 positive would be considered to have anatomic stage IIB and prognostic stage IB disease, and therefore, the patient's disease would be downstaged. HER2 expression is generally a downstaging factor given the success of anti-HER2 therapy [4, 51] (Fig. 8). TABLE 2: Examples of Different Anatomic Stages and Prognostic Stages Using Biomarkers and Oncotype DX Breast Recurrence Score (Genomic Health) | T | N | M | Histologic Grade | ER | PR | HER2 | Oncotype DX Score < 11a | Anatomic Stage (AJCC 7th Edition) | Clinical Prognostic Stage (AJCC 8th Edition) | --- --- --- --- --- | | 1 | 0 | 0 | 1 | Negative | Negative | Negative | NA | IA | IB | | 2 | 0 | 0 | 3 | Positive | Negative | Negative | NA | IIA | IIB | | 2 | 0 | 0 | Any | Positive | Positive | Negative | Yes | IIA | IA | | 3 | 1 | 0 | 1 | Positive | Positive | Positive | NA | IIIA | IIA | Note—ER = estrogen receptor, PR = progesterone receptor, HER2 = human epidermal growth factor receptor 2 (also known as ERBB2), negative = no expression by tumor, NA = not applicable, positive = expression by tumor. a If applicable. The new prognostic staging system also incorporates multi-gene panel testing in a subset of patients with hormone receptor–positive, HER2-negative, and lymph node–negative disease . For example, a patient with a T2N0M0 tumor and Oncotype DX score less than 11 has an anatomic stage of IIA but a prognostic stage of IB. The AJCC 8th edition staging system includes multiple prognostic stage tables that incorporate the clinical anatomic stage (TNM), tumor grade, biomarkers, and multigene panel (when appropriate) . The prognostic staging system presumes that patients receive appropriate systemic treatment, hormonal therapy, and chemotherapy. Conclusion The AJCC 8th edition staging system now includes two staging systems: the anatomic staging system (TNM) and the prognostic staging system. The prognostic stage includes the anatomic stage and grade, biomarkers, and multigene panels. 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Validation study of the American Joint Committee on Cancer eighth edition prognostic stage compared with the anatomic stage in breast cancer. JAMA Oncol 2018; 4:203–209 Go to Citation Crossref PubMed Google Scholar 51. Romond EH, Perez EA, Bryant J, et al. Trastuzumab plus adjuvant chemotherapy for operable HER2-positive breast cancer. N Engl J Med 2005; 353:1673–1684 Crossref PubMed Google Scholar b [...] given the success of anti-HER2 therapy 52. Prat A, Perou CM. Deconstructing the molecular portraits of breast cancer. Mol Oncol 2011; 5:5–23 Crossref PubMed Google Scholar 53. Trop I, LeBlanc SM, David J, et al. Molecular classification of infiltrating breast cancer: toward personalized therapy. RadioGraphics 2014; 34:1178–1195 Crossref PubMed Google Scholar 54. Tirada N, Aujero M, Khorjekar G, et al. O. Breast cancer tissue markers, genomic profiling, and other prognostic factors: a primer for radiologists. RadioGraphics 2018; 38:1902–1920 Crossref PubMed Google Scholar 55. Foulkes WD, Smith IE, Reis-Filho JS. Triple-negative breast cancer. N Engl J Med 2010; 363:1938–1948 Crossref PubMed Google Scholar 56. Dent R, Trudeau M, Pritchard KI, et al. Triple-negative breast cancer: clinical features and patterns of recurrence. Clin Cancer Res 2007; 13:4429–4434 Crossref PubMed Google Scholar 57. Edge SB, Byrd DR, Compton CC, Fritz AG, Greene FL, Trotti A. AJCC cancer staging manual, 7th ed.Springer, 2010 Google Scholar Information & Authors Information Published In American Journal of Roentgenology Volume 217 | Issue 2 | August 2021 Pages: 278 - 290 PubMed: 33594908 Copyright © American Roentgen Ray Society. History Submitted: November 28, 2020 Revision requested: December 9, 2020 Revision received: January 24, 2021 Accepted: February 3, 2021 First published: February 17, 2021 Keywords breast cancer mammography MRI staging ultrasound Authors Affiliations Davis C. Teichgraeber, MD Department of Breast Imaging, Division of Diagnostic Imaging, The University of Texas MD Anderson Cancer, 1515 Holcombe Blvd, Unit 1350, Houston, TX 77030. View all articles by this author Mary S. Guirguis, MD Department of Breast Imaging, Division of Diagnostic Imaging, The University of Texas MD Anderson Cancer, 1515 Holcombe Blvd, Unit 1350, Houston, TX 77030. View all articles by this author Gary J. Whitman, MD Department of Breast Imaging, Division of Diagnostic Imaging, The University of Texas MD Anderson Cancer, 1515 Holcombe Blvd, Unit 1350, Houston, TX 77030. View all articles by this author Notes Address correspondence to D. C. Teichgraeber (dcteichgraeber@mdanderson.org). The authors declare that they have no disclosures relevant to the subject matter of this article. Funding Information Supported by NIH/National Cancer Institute Cancer Center Support Grant P30 CA016672. 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Go to Figure Fig. 1B —51-year-old woman with multicentric left invasive ductal carcinoma. Go to Figure Fig. 1C —51-year-old woman with multicentric left invasive ductal carcinoma. Go to Figure Fig. 1D —51-year-old woman with multicentric left invasive ductal carcinoma. Go to Figure Fig. 2A —68-year-old woman with newly diagnosed ductal carcinoma in situ of right breast. Go to Figure Fig. 2B —68-year-old woman with newly diagnosed ductal carcinoma in situ of right breast. Go to Figure Fig. 3 —38-year-old woman with multicentric right invasive ductal carcinoma and ductal carcinoma in situ whose anatomic stage was IIIA (cT3cN1[f]cM0). Patient was treated with neoadjuvant chemotherapy and underwent total mastectomy and axillary lymph node dissection. Full-field synthesized 2D (C-View, Version 1, Hologic) lateral medial mammogram of right breast shows calcifications extending to base of nipple (arrow) in segmental distribution, which excluded patient from undergoing skin-sparing mastectomy. Patient's clinical prognostic stage was IIIA (cT3N1[f]cM0, grade 3, estrogen receptor negative, progesterone receptor negative, human epidermal growth factor receptor 2 [also known as ERBB2] positive). Go to Figure Fig. 4A —56-year-old woman who presented with right nipple thickening and retraction, areolar scaling, and palpable mass. Skin punch biopsy results showed adenocarcinoma involving epidermis, consistent with Paget disease. Ultrasound biopsy (not shown) of retroareolar mass showed invasive lobular carcinoma. Go to Figure Fig. 4B —56-year-old woman who presented with right nipple thickening and retraction, areolar scaling, and palpable mass. Skin punch biopsy results showed adenocarcinoma involving epidermis, consistent with Paget disease. Ultrasound biopsy (not shown) of retroareolar mass showed invasive lobular carcinoma. Go to Figure Fig. 5A —Illustrations of category T4 tumors. (© 2020 The University of Texas MD Anderson Cancer Center, used with permission) Go to Figure Fig. 5B —Illustrations of category T4 tumors. (© 2020 The University of Texas MD Anderson Cancer Center, used with permission) Go to Figure Fig. 5C —Illustrations of category T4 tumors. (© 2020 The University of Texas MD Anderson Cancer Center, used with permission) Go to Figure Fig. 5D —Illustrations of category T4 tumors. (© 2020 The University of Texas MD Anderson Cancer Center, used with permission) Go to Figure Fig. 6A —77-year-old woman with new right invasive lobular carcinoma possibly extending to pectoralis major muscle. Patient had remote history of screen-detected right breast cancer 27 years prior (type and biomarkers unknown, negative lymph nodes according to patient) treated with lumpectomy, axillary lymph node dissection, and radiation therapy. Go to Figure Fig. 6B —77-year-old woman with new right invasive lobular carcinoma possibly extending to pectoralis major muscle. Patient had remote history of screen-detected right breast cancer 27 years prior (type and biomarkers unknown, negative lymph nodes according to patient) treated with lumpectomy, axillary lymph node dissection, and radiation therapy. Go to Figure Fig. 6C —77-year-old woman with new right invasive lobular carcinoma possibly extending to pectoralis major muscle. Patient had remote history of screen-detected right breast cancer 27 years prior (type and biomarkers unknown, negative lymph nodes according to patient) treated with lumpectomy, axillary lymph node dissection, and radiation therapy. Go to Figure Fig. 7A —37-year-old woman with locally advanced invasive ductal carcinoma of right breast with chest wall invasion and skin involvement. Go to Figure Fig. 7B —37-year-old woman with locally advanced invasive ductal carcinoma of right breast with chest wall invasion and skin involvement. Go to Figure Fig. 8A —53-year-old woman with right breast invasive lobular carcinoma. Go to Figure Fig. 8B —53-year-old woman with right breast invasive lobular carcinoma. Go to Figure Fig. 8C —53-year-old woman with right breast invasive lobular carcinoma. Go to Figure Fig. 8D —53-year-old woman with right breast invasive lobular carcinoma. Go to Figure Fig. 9A —45-year-old woman with inflammatory right breast cancer. Patient presented with peau d'orange and rapid onset of breast hardening and redness. Go to Figure Fig. 9B —45-year-old woman with inflammatory right breast cancer. Patient presented with peau d'orange and rapid onset of breast hardening and redness. Go to Figure Fig. 10 —Illustration shows regional nodal staging. Level I axillary lymph nodes (low axillary) lie lateral to lateral border of pectoralis minor muscle, level II axillary lymph nodes (midaxillary) lie between medial and lateral borders of pectoralis minor muscle, and level III axillary lymph nodes (high axillary or infraclavicular) lie medial to medial border of pectoralis minor muscle. Internal mammary lymph nodes lie in intercostal spaces along sternum in endothoracic fascia. Supraclavicular lymph nodes lie within supraclavicular fossa, triangle defined by omohyoid muscle and tendon (lateral and superior borders), internal jugular vein (medial border), and clavicle and subclavian vein (lower border). (© 2020 The University of Texas MD Anderson Cancer Center, used with permission) Go to Figure Fig. 11A —46-year-old woman with left breast invasive ductal carcinoma. Go to Figure Fig. 11B —46-year-old woman with left breast invasive ductal carcinoma. Go to Figure Fig. 11C —46-year-old woman with left breast invasive ductal carcinoma. Go to Figure Fig. 11D —46-year-old woman with left breast invasive ductal carcinoma. Go to Figure Tables TABLE 1: Clinically Defined Treatment-Oriented Subtypes of Breast Cancer Go to Table TABLE 2: Examples of Different Anatomic Stages and Prognostic Stages Using Biomarkers and Oncotype DX Breast Recurrence Score (Genomic Health) Go to Table References References 1. Amin MB, Greene FL, Edge SB, et al. The eighth edition AJCC cancer staging manual: continuing to build a bridge from a population-based to a more “personalized” approach to cancer staging. CA Cancer J Clin 2017; 67:93–99 Go to Citation Crossref PubMed Google Scholar 2. Plichta JK, Campbell BM, Mittendorf EA, Hwang ES. Anatomy and breast cancer staging: is it still relevant? 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Validation study of the American Joint Committee on Cancer eighth edition prognostic stage compared with the anatomic stage in breast cancer. JAMA Oncol 2018; 4:203–209 Go to Citation Crossref PubMed Google Scholar 51. Romond EH, Perez EA, Bryant J, et al. Trastuzumab plus adjuvant chemotherapy for operable HER2-positive breast cancer. N Engl J Med 2005; 353:1673–1684 Crossref PubMed Google Scholar b [...] given the success of anti-HER2 therapy 52. Prat A, Perou CM. Deconstructing the molecular portraits of breast cancer. Mol Oncol 2011; 5:5–23 Crossref PubMed Google Scholar 53. Trop I, LeBlanc SM, David J, et al. Molecular classification of infiltrating breast cancer: toward personalized therapy. RadioGraphics 2014; 34:1178–1195 Crossref PubMed Google Scholar 54. Tirada N, Aujero M, Khorjekar G, et al. O. Breast cancer tissue markers, genomic profiling, and other prognostic factors: a primer for radiologists. RadioGraphics 2018; 38:1902–1920 Crossref PubMed Google Scholar 55. Foulkes WD, Smith IE, Reis-Filho JS. Triple-negative breast cancer. N Engl J Med 2010; 363:1938–1948 Crossref PubMed Google Scholar 56. Dent R, Trudeau M, Pritchard KI, et al. Triple-negative breast cancer: clinical features and patterns of recurrence. Clin Cancer Res 2007; 13:4429–4434 Crossref PubMed Google Scholar 57. Edge SB, Byrd DR, Compton CC, Fritz AG, Greene FL, Trotti A. AJCC cancer staging manual, 7th ed.Springer, 2010 Google Scholar Recommended Articles ### Current Status and Future of BI-RADS in Multimodality Imaging, From the AJR Special Series on Radiology Reporting and Data Systems Mohammad Eghtedari, Alice Chong, Rebecca Rakow-Penner, and Haydee Ojeda-Fournier ### PET/CT for Patients With Breast Cancer: Where Is the Clinical Impact? Gary A. Ulaner ### Cystic Masses of the Breast Neely Hines, Priscilla J. Slanetz, and Ronald L. Eisenberg ### Understanding the Lymphatics: An Updated Review of the N Category of the AJCC 8th Edition for Urogenital Cancers Christian K. Pedersen and Aparna S. Babu ### Nonmass Enhancement on Breast MRI: Review of Patterns With Radiologic-Pathologic Correlation and Discussion of Management Tamuna Chadashvili, Erica Ghosh, Valerie Fein-Zachary, Tejas S. Mehta, Shambhavi Venkataraman, Vandana Dialani, and Priscilla J. Slanetz Figure title goes here Go to figure location within the article Download figure Share on social media xrefBack.goTo Request permissions Authors Info & Affiliations Congrats! Your Phone has been verified
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https://spacemath.gsfc.nasa.gov/algebra2/CH13v3.pdf
13.1.1 Right Triangle Trigonometry Two NASA, STEREO satellites take images of the sun and its surroundings from two separate vantage points along Earth's orbit. From these two locations, one located ahead of the Earth, and the other located behind the Earth along its orbit, they can create stereo images of the 3-dimensional locations of coronal mass ejections (CMEs) and storms on or near the solar surface. The three images below, taken on December 12, 2007, combine the data from the two STEREO satellites (left and right) taken from these two locations, with the single image taken by the SOHO satellite located half-way between the two STEREO satellites (middle). Notice that there is a large storm event, called Active Region 978, located on the sun. The changing location of AR978 with respect to the SOHO image shows the perspective change seen from the STEREO satellites. You can experience the same Parallax Effect by holding your thumb at arms length, and looking at it, first with the left eye, then with the right eye. The location of your thumb will shift in relation to background objects in the room. The diagram to the right shows the relevant parallax geometry for the two satellites A and B, separated by an angle of 42 degrees as seen from the sun. The diagram lengths are not drawn to scale. The radius of the sun is 696,000 km. Problem 1: With a millimeter ruler, determine the scale of each image in km/mm. How many kilometers did AR978 shift from the center position (SOHO location for AR) between the two STEREO images? This is the average measure of 'L' in the diagram. Problem 2: Using the Pythagorean Theorem, determine the equation for the height, h, in terms of R and L. Assume the relevant triangle is a right triangle. Problem 3: How high (h) above the sun's surface, called the photosphere, was the AR978 viewed by STEREO and SOHO on December 12, 2007? Space Math 13.1.1 Answer Key Problem 1: With a millimeter ruler, determine the scale of each image in km/mm. How many kilometers did AR978 shift from the center position (SOHO location for AR) between the two STEREO images? This is the measure of 'L' in the diagram. Answer: STEREO-Left image, sun diameter = 28 mm, actual = 1,392,000 km, so the scale is 1392000 km / 28mm = 49,700 km/mm SOHO-center sun diameter = 36 mm, so the scale is 1392000 km/36mm = 38,700 km/mm STEREO-right sun diameter = 29 mm, so the scale is 1392000 km / 29 mm = 48,000 km/mm Taking the location of the SOHO image for AR978 as the reference, the left-hand image shows that AR978 is about 5 mm to the right of the SOHO location which equals 5 mm x 49,700 km/mm = 248,000 km. From the right-hand STEREO image, we see that AR978 is about 5 mm to the left of the SOHO position or 5 mm x 48,000 km/mm = 240,000 km. The average is 244,000 kilometers. Problem 2: Using the Pythagorean Theorem, determine the equation for the height, h, in terms of R and L. Answer: ( R + h)2 = R2 + L2 h = (L2 + R2 )1/2 - R Problem 3: How high (h) above the sun's surface, called the photosphere, was the AR978 viewed by STEREO and SOHO on December 12, 2007? Answer: h = ( (244,000)2 + (696,000)2 )1/2 - 696,000 h = 737,500 - 696,000 h = 41,500 kilometers Space Math 13.1.2 Right Triangle Trigonometry The corresponding sides of similar triangles are proportional to one another as the illustration to the left shows. Because the vertex angle of the triangles are identical in measure, two objects at different distances from the vertex will subtend the same angle, a . The corresponding side to 'X' is '1' and the corresponding side to '2' is the combined length of '2+4'. Problem 1: Use the properties of similar triangles and the ratios of their sides to solve for 'X' in each of the diagrams below. Problem 2: Which triangles must have the same measure for the indicated angle a ? Problem 3: The Sun is 400 times the diameter of the Moon. Explain why they appear to have the same angular size if the moon is at a distance of 384,000 kilometers? Space Math Answer Key Problem 1: Use the properties of similar triangles and the ratios of their sides to solve for 'X' in each of the diagrams below. A) X / 2 = 8 / 16 so X = 1 B) 3 / X = 11 / (X+8) so 3 (X + 8) = 11 X ; 3X + 24 = 11 X ; 24 = 8X and so X = 3. C) 3 / 8 = 11 / (x + 8) so 3 ( x + 8) = 88 ; 3X + 24 = 88 ; 3X = 64 and so X = 21 1/3 D) 1-inch / 24-inches = 34 inches / ( X + 24-inches); so X +24 inches = 34 inches x (24) and X = 816 - 24 inches and so X = 792 inches. E) 3 cm / 60 cm = 1 meter / ( X + 60 cm). 3/60 = 1 meter / ( X + 0.6 m) then 3( X + 0.60) = 60 ; 3X + 1.8 = 60 ; 3X = 58.2 meters so X = 19.4 meters. F) 2 meters / 48 meters = X / 548 meters ; 1/24 = X/548 ; X = 548 / 24 ; so X = 22.83. Problem 2: Which triangles must have the same measure for the indicated angle a ? Answer: Because the triangle (D) has the side proportion 1-inch /24-inches = 1/24 and triangle (F) has the side proportion 2 meters / 48 meters = 1/24 these two triangles, D and F, have the same angle measurement for angle a. Problem 3: The Sun is 400 times the diameter of the Moon. Explain why they appear to have the same angular size if the moon is at a distance of 384,000 kilometers? Answer: From one of our similar triangles, the long vertical side would represent the diameter of the sun; the short vertical side would represent the diameter of the moon; the angle a is the same for both the sun and moon if the distance to the sun from Earth were 400x farther than the distance of the moon from Earth. Since the lunar distance is 384,000 kilometers, the sun must be at a distance of 154 million kilometers. 13.1.2 Space Math 13.2.1 General Angles and Radian Measure The relationship between the distance to an object, R, the objects size, L, and the angle that it subtends at that distance, θ, is given by: To use these formulae, the units for length, L, and distance, R, must be identical. Problem 1 - You spot your friend (L = 2 meters) at a distance of 100 meters. What is her angular size in arcminutes? Problem 2 - The sun is located 150 million kilometers from Earth and has a radius of 696.000 kilometers. What is its angular diameter in arcminutes? Problem 3 - How far away, in meters, would a dime (1 centimeter) have to be so that its angular size is exactly one arcsecond? Problem 4 - The spectacular photo above was taken by Jerry Lodriguss (Copyright 2007, ) and shows the International Space Station streaking across the disk of the sun. If the ISS was located 379 kilometers from the camera, and the ISS measured 73 meters across, what was its angular size in arcseconds? Problem 5 - The orbital speed of the space station is 7.4 kilometers/second. If its distance traveled in 1 second is 7.4 kilometers, A) what was the angle, in arcminutes, that it moved through in one second as seen from the location of the camera? B) What was its angular speed in arcminutes/second? Problem 6 - Given the diameter of the sun in arcminutes (Problem 2), and the ISS angular speed (Problem 5) how long, in seconds, did it take the ISS to travel across the face of the sun? Space Math Answer Key Problem 1 - Answer: Angle = 3,438 x (2 meters/100 meters) = 68.8 arcminutes. Problem 2 - Answer: 3,438 x (696,000/150 million) = 15.9 arcminutes in radius, so the diameter is 2 x 15.9 = 31.8 arcminutes. Problem 3 - Answer: From the second formula R = 3438 L/A = 3438 1 cm/1 arcsecond so R = 3,438 centimeters or a distance of 34.4 meters. Problem 4 - Answer: From the third formula, Angle = 206,265 (73 meters/379,000 meters) = 39.7 arcseconds. Problem 5 - Answer: The orbital speed of the space station is 7.4 kilometers/second. If its distance traveled in 1 second is 7.4 kilometers, A) The ISS traveled L = 7.4 kilometers so from the second formula Angle = 3,438 (7.4 km/379 km) = 67 arcminutes. B) The angular speed is just 67 arcminutes per second. Problem 6 - Answer: The time required is T = 31.8 arcminutes/ (67 arcminutes/sec) = 0.47 seconds. The spectacular photo by Jerry Lodriguss had to be taken with careful planning beforehand. He had to know, to the second, when the sun and ISS would be in the right configuration in the sky as viewed from his exact geographic location. Here's an example of the photography considerations in his own words: " I considered trying to monitor the transit visually with a remote release in my hand and just firing (the camera) when I saw the ISS in my guidescope. Then I worked out the numbers. I know that my reaction time is 0.19 seconds. This is actually quite good, but I make my living shooting sports where this is critical, so I better be good at it. I also know that the Canon 1D Mark IIn has a shutter lag of 55 milliseconds. Adding these together, plus a little bit of a fudge factor, the best I could hope for was about 1/4 of a second from when I saw it to when the shutter opened. Since the entire duration of the transit was only 1/2 of a second, in theory, I could capture the ISS at about the center of the disk if I fired as soon as I saw it start to cross. This was not much of a margin for error. I could easily blink and miss the whole thing… Out of 49 frames that the Mark IIn recorded, the ISS is visible in exactly one frame." 13.2.1 Space Math 13.2.2 General Angles and Radian Measure The Sun (Diameter = 696,000 km) and Moon (Diameter = 3,476 km) have very different physical diameters in kilometers, but in the sky they can appear to be nearly the same size. Astronomers use the angular measure of arcseconds (asec) to measure the apparent sizes of most astronomical objects. (1 degree equals 60 arcminutes, and 1 arcminute equals 60 arcseconds). The photos above show the Sun and Moon at a time when their angular diameters were both about 1,865 arcseconds. Problem 1 - Using a metric ruler, what is the angular scale of each image in arcseconds per millimeter? Problem 2 - In arcseconds, what is the size of the smallest feature you can see in the images of the Sun and Moon? Problem 3 - About what is the area, in square arcseconds (asec2) of the circular Mare Serenitatis (A) region in the photo of the Moon? Problem 4 - At the distance of the Moon, 1 arcsecond of angular measure equals 1.9 kilometers. The Sun is exactly 400 times farther away than the Moon. On the photograph of the Sun, how many kilometers equals 1 arcsecond of angle? Problem 5 - What is the area of Mare Serenitatis in square kilometers? Problem 6 - What would be the physical area, in square-kilometers, of an identical angular area to Mare Serenitatis if it were located on the surface of the sun? Space Math Answer Key Problem 1 - Using a metric ruler, what is the angular scale of each image in arcseconds per millimeter? Answer: Moon diameter = 65 mm and sun diameter = 61 mm so the lunar image scale is 1,865 asec/65mm = 28.7 asec/mm and the solar scale is 1865 asec/61 mm = 30.6 asec/mm. Problem 2 - In arcseconds, what is the size of the smallest feature you can see in the images of the Sun and Moon? Answer: the smallest feature is about 0.5 mm or 0.5 x 28.7 asec/mm = 14.4 asec for the Moon and 0.5 x 30.6 asec/mm = 15.3 asec for the Sun. Problem 3 - About what is the area, in square arcseconds (asec2) of the circular Mare Serenitatis (A) region in the photo of the Moon? Answer: The diameter of the mare is 1 centimeter, so the radius is 5 mm or 5 mm x 28.7 asec/mm = 143.5 asec. Assuming a circle, the area is A = π x ( 143.5 asec)2 = 64,700 asec2. Problem 4 - At the distance of the Moon, 1 arcsecond of angular measure equals 1.9 kilometers. The Sun is exactly 400 times farther away than the Moon. On the photograph of the Sun, how many kilometers equals 1 arcsecond of angle? Answer: The angular scale at the sun would correspond to 400 x 1.9 km = 760 kilometers per arcsecond. Problem 5 - What is the area of Mare Serenitatis in square kilometers? Answer: We have to convert from square arcseconds to square kilometers using a two-step unit conversion 'ladder'. 64,700 asec2 x ( 1.9 km/asec) x (1.9 km/asec) = 233,600 km2. Problem 6 - What would be the physical area, in square-kilometers, of an identical angular area to Mare Serenitatis if it were located on the surface of the sun? Answer: The angular area is 400-times further away, so we have to use the scaling of 760 kilometers/asec deduced in Problem 4. The unit conversion for the solar area becomes: 64,700 asec2 x (760 km/asec) x (760 km/asec) = 37,400,000,000 km2. 13.2.2 Space Math 13.2.3 General Angles and Radian Measure The easiest, and most basic, unit of measure in astronomy is the angular degree. Because the distances to objects in the sky are not directly measurable, a photograph of the sky will only indicate how large, or far apart, objects are in terms of degrees, or fractions of degrees. It is a basic fact in angle mensuration in geometry, that 1 angular degree (or arc-degree) can be split into 60 arc-minutes of angle, and that 1 arc-minute equals 60 arc-seconds. A full degree is then equal to 60 x 60 = 3,600 ‘arcseconds’. High-precision astronomy also uses the unit of milliarcsecond to represent angles as small as 0.001 arcseconds and microarcseconds to equal 0.000001 arcseconds. Problem 1 – The moon has a diameter of 0.5 degrees (a physical size of 3,474 km) A telescope sees a crater 1 arcsecond across. What is its diameter in meters? Problem 2 – A photograph has an image scale of 10 arcseconds/pixel. If the image has a size of 512 x 512 pixels, what is the image field-of-view in degrees? Problem 3 – An astronomer wants to photograph the Orion Nebula (M-42) with an electronic camera with a CCD format of 4096x4096 pixels. If the nebula has a diameter of 85 arcminutes. What is the resolution of the camera in arcseconds/pixel when the nebula fills the entire field-of-view? Problem 4 – An electronic camera is used to photograph the Whirlpool Galaxy, M-51, which has a diameter of 11.2 arcminutes. The image will have 1024x1024 pixels. What is the resolution of the camera, in arcseconds/pixel, when the galaxy fills the entire field-of-view? Problem 5 – The angular diameter of Mars from Earth is about 25 arcseconds. This corresponds to a linear size of 6,800 km. The Mars Reconnaissance Orbiter’s HiRISE camera, in orbit around Mars, can see details as small as 1 meter. What is the angular resolution of the camera in microarcseconds as viewed from Earth? Problem 6 – The Hubble Space Telescope can resolve details as small as 46 milliarcseconds. At the distance of the Moon, how large a crater could it resolve, in meters? Space Math 13.2.3 Answer Key Problem 1 – Answer: 0.5 degrees x 3600 arcsec/degree = 1800 arcseconds. Using proportions 1/1800 = x/3474 so X = 3474/1800 = 1.9 kilometers. Problem 2 –Answer: 512 pixels x 10 arcsec/pixel x 1 degree/3600 arcseconds = 5120 arcseconds /3600 = 1.4 degrees, so the image is 1.4 x 1.4 degrees. Problem 3 –Answer: 85 arcminutes x 60 arcsec/arcmin = 5,100 arcseconds. This corresponds to 4096 pixels so the scale is 5,100 arcsec/4096 pixels = 1.2 arcsec/pixel. Problem 4 –Answer: 11.2 arcminutes x 60 arcsec/arcmin = 672 arcsec. This equals 1024 opixels so the scale is 672/1024 = 0.656 arcsec/pixel. Problem 5 –Answer: 25 arcsec = 6800 km so 1 arcsec = 6800 km/25 = 272 km from Earth. For 1-meter resolution at Earth, the angular scale would have to be 1 sec x 1m/272000m = 0.0000037 arcseconds or 3.7 microarcseconds. Problem 6 – Answer: From Problem-1, 1 arcsecond = 1.9 kilometers. By proportions, 0.046 arcsec/1 arcsec = x/1.9 km so X = 0.046 x 1.9 km = 0.087 kilometers or 87 meters. The picture below was taken by the Cassini spacecraft orbiting Saturn. It is of the satellite Phoebe, which from Earth subtends an angular size of about 32 milliarcsec. The smallest crater, about 1 km across, would subtend about 160 microarcseconds as seen from Earth. Space Math 13.2.4 General Angles and Radian Measure Although a pair of binoculars or a telescope can see amazing details on the Moon, the human eye is not so gifted! The lens of the eye is so small, only 2 to 5 millimeters across, that the sky is 'pixelized' into cells that are about one arcminute across. We call this the resolution limit of the eye, or the eye's visual acuity. One degree of angle measure can be divided into 60 minutes of arc. For an object like the full moon, which is 1/2-degree in diameter, it also measures 30 arcminutes in diameter. This means that, compared to the human eye, the moon can be divided into an image that is 30-pixels in diameter. Problem 1 - Convert the following degree measures into their equivalent measure in arcminutes; A) 5 degrees; B) 2/3 degree; C) 15.5 degrees; D) 0.25 degrees Problem 2 - Convert the following arcminute measures into their equivalent measure in degrees: A) 15 amin; B) 1/2 amin; C) 120.5 amin; D) 3600 amin. Problem 3 - Convert the following area measures in square-degrees into their equivalent measures in square arcminutes (amin2): A) 1.0 deg2; B) 0.25 deg2 Problem 4 - The figure to the above-left is a telescopic photo of the full moon showing its many details including craters and dark mare. Construct a simulated image of the moon in the grid to the right to represent what the moon would look like at the resolution of the human eye. First sketch the moon on the grid. Then use the three shades; black, light-gray and dark-gray, and fill-in each square with one of the three shades using your sketch as a guide. Problem 5 - Why can't the human eye see any craters on the Moon? Space Math Answer Key Problem 1 - Convert the following degree measures into their equivalent measure in arcminutes; A) 5 degrees; B) 2/3 degree; C) 15.5 degrees; D) 0.25 degrees Answer: A) 5 degrees x (60 amin/1 deg) = 300 amin. B) 2/3 degree x (60 amin/1 deg) = 120/3 = 40 amin. C) 15.5 degrees x (60 amin/1deg) = 930 amin; D) 0.25 deg x (60 amin/1 deg) = 15 amin. Problem 2 - Convert the following arcminute measures into their equivalent measure in degrees: A) 15 amin; B) 1/2 amin; C) 120.5 amin; D) 3600 amin. Answer: A) 15 amin x (1 deg/60 amin) = 0.25 deg. B) 1/2 amin x (1 deg / 60 amin) = 1/120 deg. C) 120.5 amin x (1 deg/60 amin) = 2.008 deg. D) 360 amin x (1 deg/60 amin) = 60 deg. Problem 3 - Convert the following area measures in square-degrees into their equivalent measures in square arcminutes (amin2): A) 1.0 deg2; B) 0.25 deg2 Answer; A) 1.0 deg2 x ( 60 amin/1 deg) x (60 amin/1 deg) = 3600 amin2. B) 0.25 deg2 x (60 amin/1 deg) x (60 amin/1 deg) = 0.25 x 3600 = 900 amin2. Problem 4 - See the image below which has been pixelized to the grid resolution. How well did your version match the image on the right? Problem 5 - Why can't the human eye see any craters on the Moon? Answer: The human eye can only see details 1 arcminute across and this is too low a resolution to see even the largest craters. 13.2.4 Space Math 13.3.1 Trigonometric Functions of any Angle When astronomers take photographs of a specific region of the sky, they often take hundreds of separate images and then average them together to increase the sensitivity of the image. Each image can be shifted and rotated with respect to the previous image, so these changes have to be determined, mathematically, and removed so that the images can be combined. Two digital images were taken by an astronomer of the same star field at two separate days. The goal is to look for stars whose brightnesses have suddenly dimmed in order to detect planets passing across the disk of the star as viewed from Earth. In each image, the astronomer can identify four stars and measure their ‘X-Y’ locations on the image, where the units of X and Y are in pixels. Problem 1 - From the table below, what are the polar coordinates of the stars in Image A? Problem 2 – The astronomer determines that there were two possible rotation angles for Image B. It could either have been rotated clockwise by 36o or clockwise by 18o with respect to Image A. By what angle was Image B rotated with respect to Star C? Image A (x,y) Image B (x,y) Star A (+327, +843) (+757, +492) Star B (-193, -50) (-185, +71) Star C (0,0) (0,0) Star D (-217, +33) (-155, +155) Problem 3 – If the angular distance between Star A and Star C is 12.8 degrees, how far apart are the other two stars from Star C measured in degrees? Space Math Answer Key 13.3.1 Problem 1 - From the table below, what are the polar coordinates of the stars in Image A? Image A (x,y) Image B (x,y) Star A (+327, +843) (+757, +492) Star B (-193, -50) (-185, +71) Star C (0,0) (0,0) Star D (-217, +33) (-155, +155) Answer: Using Star C as the origin: Star A: R2 = (327)2 + (843)2 so R = 903 and sin θ = 843/903 so θ = 69o Star B: R2 = (193)2 + (50)2 so R = 199 and sin θ = 50/199 so θ = 180+15 = 195o Star D: R2 = (217)2 + (33)2 so R = 219 and sin θ = 33/219 so θ = 180-9 = 171o So Star A: (903, +69o) Star B: (199, +195o) Star D: (219, +171o) Problem 2 - The astronomer determines that there were two possible rotation angles for Image B. It could either have been rotated clockwise by 36o or clockwise by 18o with respect to Image A. By what angle was Image B rotated with respect to Star C? Answer: Since all stars will move in unison as they rotate, we only need to test for the correct solution using one of the star coordinates: Select Star A: Since x = R cos(θ) and y = Rsin(θ) then For 36o clockwise we have θ = +69o – 36o = 33o so X = 903 cos(33) = +757 and y = 903 sin(33) = +492 which match the Image B coordinates. So θ = 36o is the correct rotation angle. Problem 2 – If the angular distance between Star A and Star C is 12.8 degrees, how far apart are the other two stars from Star C measured in degrees? Answer: We know that, in pixel units, Star A-C = 903, so the scale of the image is just 12.8 degrees/903 pixels or 0.014 degrees/pixel. The radial coordinates for the stars are then: Star A : 12.8 degrees Star B : 199 pixels x (0.014 degrees/pixel) = 2.8 degrees Star D: 219 pixels x (0.014 degrees/pixel) = 3.1 degrees Space Math 13.5.1 The Law of Sines On June 5, 2012 the planet Venus will pass across the face of the sun as viewed from Earth. The last time this happened was on June 6, 2004. A similar event, called the Transit of Venus’ will not happen again until December 11, 2117. This image, taken by the TRACE satellite, shows the black disk of Venus passing across the solar disk photographed with a filter that highlights details on the solar surface. Problem 1 – The table below gives the location of the other 7 planets at the time of the 2012 Transit of Venus, with the sun at the origin of the Cartesian coordinate system. What are the polar coordinates for each planet at the time of the next transit? Note: All distances are in terms of the Astronomical Unit (AU) for which the distance between the Sun and Earth equals exactly 1.0 AU. Planet X Y R θ Sun 0.000 0.000 Mercury -0.157 +0.279 Venus -0.200 -0.698 Earth -0.268 -0.979 Mars -1.450 -0.694 Jupiter +2.871 +4.098 Saturn -8.543 -4.635 Uranus +19.922 +1.988 Neptune +26.272 -14.355 Problem 2 – In the sky as viewed from Earth, what is the angular distance between the Sun and Mars if the angle between the Sun and Earth as viewed from Mars is 39o? Space Math Answer Key 13.5.1 Problem 1 – The table below gives the location of the other 7 planets at the time of the 2012 Transit of Venus, with the sun at the origin of the Cartesian coordinate system. What are the polar coordinates for each planet at the time of the next transit? All angles are measured in positive direction counterclockwise with respect to the + X axis. (Note: All distances are in terms of the Astronomical Unit (AU) for which the distance between the Sun and Earth equals exactly 1.0 AU.) Planet X Y R θ Sun 0.000 0.000 Mercury -0.157 +0.279 0.320 119 Venus -0.200 -0.698 0.726 254 Earth -0.268 -0.979 1.015 255 Mars -1.450 -0.694 1.608 206 Jupiter +2.871 +4.098 5.003 55 Saturn -8.543 -4.635 9.719 208 Uranus +19.922 +1.988 20.021 6 Neptune +26.272 -14.355 29.938 331 Answer: Example for Jupiter: R2 = (2.871)2 + (4.098)2 so R = 5.003 AU. The angle can be determined from cos(θ) = X/R, for Earth, located in the Third Quadrant: cos-1(-0.268/1.015) = 105o so 360o -105o = 255o. Problem 2 – In the sky as viewed from Earth, what is the angular distance between the Sun and Mars if the angle between the Sun and Earth as viewed from Mars is 39o? Answer: Sun-Earth-Mars form a triangle with the Earth at one vertex. You need to calculate this vertex angle. Use the Law of Sines: Sin(39) sin(x) ---------- = --------- so sin(x) = 0.9970 and so x = 86 degrees. 1.015 1.608 Space Math 13.6.1 The Law of Cosines On June 5, 2012 the planet Venus will pass across the face of the sun as viewed from Earth. The last time this happened was on June 6, 2004. A similar event, called the Transit of Venus’ will not happen again until December 11, 2117. This image, taken by the TRACE satellite, shows the black disk of Venus passing across the solar disk photographed with a filter that highlights details on the solar surface. Problem 1 – The table below gives the location of the other 7 planets at the time of the 2012 Transit of Venus, with the sun at the origin of the Cartesian coordinate system. What are the polar coordinates for each planet at the time of the next transit? Note: All distances are in terms of the Astronomical Unit (AU) for which the distance between the Sun and Earth equals exactly 1.0 AU. Planet X Y R θ Sun 0.000 0.000 Mercury -0.157 +0.279 Venus -0.200 -0.698 Earth -0.268 -0.979 Mars -1.450 -0.694 Jupiter +2.871 +4.098 Saturn -8.543 -4.635 Uranus +19.922 +1.988 Neptune +26.272 -14.355 Problem 2 – In the sky as viewed from Earth, what is the angular distance between the Sun and Mars? Space Math Answer Key 13.6.1 Problem 1 – The table below gives the location of the other 7 planets at the time of the 2012 Transit of Venus, with the sun at the origin of the Cartesian coordinate system. What are the polar coordinates for each planet at the time of the next transit? All angles are measured in positive direction counterclockwise with respect to the + X axis. (Note: All distances are in terms of the Astronomical Unit (AU) for which the distance between the Sun and Earth equals exactly 1.0 AU.) Planet X Y R θ Sun 0.000 0.000 Mercury -0.157 +0.279 0.320 119 Venus -0.200 -0.698 0.726 254 Earth -0.268 -0.979 1.015 255 Mars -1.450 -0.694 1.608 206 Jupiter +2.871 +4.098 5.003 55 Saturn -8.543 -4.635 9.719 208 Uranus +19.922 +1.988 20.021 6 Neptune +26.272 -14.355 29.938 331 Answer: Example for Jupiter: R2 = (2.871)2 + (4.098)2 so R = 5.003 AU. The angle can be determined from cos(θ) = X/R, for Earth, located in the Third Quadrant: cos-1(-0.268/1.015) = 105o so 360o -105o = 255o. Problem 2 – In the sky as viewed from Earth, what is the angular distance between the Sun and Mars? Answer: Sun-Earth-Mars form a triangle with the Earth at one vertex. You need to calculate this vertex angle. Use the Law of Cosines: L2 = a2 + b2 – 2abcosA Where L = Mars-Sun distance = 1.608 AU A = Earth-Sun distance = 1.015 AU B = Earth-Mars distance: B2 = (-0.269 +1.450)2 + (-0.979 +0.694)2 so B = 1.215 AU Then CosA = 0.032 so A = 88 degrees. Space Math 13.6.2 Law of Cosines The two STEREO satellites are located at points A and B, with Earth located at Point E and the sun located at Point S, which is the center of a circle with a radius ES of 1.0 Astronomical unit (150 million kilometers). Suppose that the two satellites spot a Coronal Mass Ejection (CME) cloud at Point C. Satellite A measures its angle from the sun mSAC as 45 degrees while Satellite B measures the corresponding angle to be mSBC=50 degrees. In the previous math problem the astronomers knew the ejection angle of the CME, mESC, but in fact they didn't need to know this in order to solve the problem below! Problem 1 - The astronomers want to know the distance that the CME is from Earth, which is the length of the segment EC. The also want to know the approach angle, mSEC. Use either a scaled construction (easy: using compass, protractor and millimeter ruler) or geometric calculation (difficult: using trigonometric identities) to determine EC from the available data. Givens from satellite orbits: SB = SA = SE = 150 million km AE = 136 million km BE = 122 million km mASE = 54 degrees mBSE = 48 degrees mEAS = 63 degrees mEBS = 66 degrees mAEB = 129 degrees Find the measures of all of the angles and segment lengths in thee above diagram rounded to the nearest integer. Problem 2 - If the CME was traveling at 2 million km/hour, how long did it take to reach the distance indicated by the length of segment SC? The two STEREO spacecraft are located along Earth's orbit and can view gas clouds ejected by the sun as they travel to Earth. From the geometry, astronomers can accurately determine their speeds, distances, shapes and other properties. By studying the separate 'stereo' images, astronomers can determine the speed and direction of the cloud before it reaches Earth. Use the diagram, (angles and distances not drawn to the same scale of the 'givens' below) to answer the following question. Space Math 13.6.2 Answer Key Givens from satellite orbits: SB = SA = SE = 150 million km AE = 136 million km BE = 122 million km mASE = 54 degrees mBSE = 48 degrees mEAS = 63 degrees mEBS = 66 degrees mAEB = 129 degrees use units of megakilometers i.e. 150 million km = 150 Mkm. Method 1: Students construct a scaled model of the diagram based on the angles and measures, then use a protractor to measure the missing angles, and from the scale of the figure (in millions of kilometers per millimeter) they can measure the required segments. Segment EC is about 49 Mkm at an angle, mSEB of 28 degrees. Method 2 use the Law of Cosines and the Law of Sines to solve for the angles and segment lengths exactly. mASB = mASE+mBSE = 102 degrees mASC = theta mACS = 360 - mCAS - mASC = 315 - theta mBSC = mASB - theta = 102 - theta mBCS = 360 - mCBS - mBSC = 208 + theta Use the Law of Sines to get sin(mCAS)/L = sin(mACS)/150 Mkm and sin(mBCS)/L = sin(mBCS)/150 Mkm. Eliminate L : 150sin(45)/sin(315-Theta) = 150 sin(50)/sin(208+theta) Re-write using angle-addition and angle-subtraction: sin50 [sin(315)cos(theta)-cos(315)sin(theta)] = sin(45) [sin(208)cos(theta) + cos(208)sin(theta)] Compute numerical factors by taking indicated sines and cosines: -0.541cos(theta) - 0.541sin(theta) = -0.332cos(theta) - 0.624sin(theta) Simplify: cos(theta) = 0.397sin(theta) Use definition of sine: cos(theta)2 = 0.158 ( 1 - cos(theta)2 ) Solve for cosine: cos(theta) = ( 0.158/1.158)1/2 so theta = 68 degrees. And so mASC=68 Now compute segment CS = 150 sin(45)/sin(315-68) = 115 Mkm. BC = 115 sin(102-68)/sin(50) = 84 Mkm. Then EC2 = 1222 + 842 - 2(84)(122)cos(mEBS-mCBS) EC2 = 1222 + 842 - 2(84)(122)cos(66-50) So EC = 47 Mkm. mCEB from Law of Cosines: 842 = 1222 + 472 - 2(122)(47)cos(mCEB) so mCEB = 29 degrees And since mAES = 180 - mASE - mEAS = 180 - 54 - 63 = 63 degrees so mSEC = mAEB - mAES - mCEB = 129 - 63 - 29 = 37 degrees So, the two satellites are able to determine that the CME is 49 million kilometers from Earth and approaching at an angle of 37 degrees from the sun. Problem 2 - If the CME was traveling at 2 million km/hour, how long did it take to reach the distance indicated by the length of segment SC? Answer: 115 million kilometers / 2 million km/hr = 58 hours or 2.4 days. Space Math 13.7.1 Parametric Equations The neat thing about ballistic problems (flying baseballs or rockets) is that their motion in the vertical dimension is independent of their motion in the horizontal dimension. This means we can write one equation that describes the movement in time along the x axis, and a second equation that describes the movement in time along the y axis. In function notation, we write these as x(t) and y(t) where t is the independent variable representing time. To draw the curve representing the trajectory, we have a choice to make. We can either create a table for X and Y at various instants in time, or we can simply eliminate the independent variable, t, and plot the curve y(x). Problem 1 - The Ares 1X underwent powered flight while its first stage rocket engines were operating, but after it reached the highest point in its trajectory (apogee) the Ares 1X capsule coasted back to Earth for a water landing. The parametric equations that defined its horizontal downrange location (x) and its altitude (y) in meters are given by 2 ( ) 64,000 1800 ( ) 45,000 4.9 x t t y t t = + = − Using the method of substitution, create the new function y(x) by eliminating the variable t. Problem 2 - Determine how far downrange from launch pad 39A at Cape Canaveral the capsule landed, (y(x)=0), giving your answer in both meters and kilometers to two significant figures. Problem 3 - Why is it sometimes easier not to work with the parametric form of the motion of a rocket or projectile? Space Math Answer Key 13.7.1 Problem 1 - Answer: The parametric equations that defined its horizontal downrange location (x) and its altitude (y) in meters are given by 2 ( ) 64,000 1800 ( ) 45,000 4.9 x t t y t t = + = − We want to eliminate the variable, t, from y(t) and we do this by solving the equation x(t) for t and substituting this into the equation for y(t) to get y(t(x)) or just y(x). 2 2 64,000 1800 64,000 45,000 4.9 1800 38,800 0.19 0.0000015 x t x y y x − = − ⎡ ⎤ = − ⎢ ⎥ ⎣ ⎦ = + − x Problem 2 - Determine how far downrange from the launch pad the capsule landed, (y(x)=0), giving your answer in both meters and kilometers to two significant figures. Answer: Using the Quadratic Formula, find the two roots of the equation y(x)=0, and select the root with the largest positive value. 1,2 0.19 0.036 4( 0.0000015)38800 2( 0.0000015) x − ± − − = − meters so x1 = -109,000 meters or -109 kilometers x2 = +237,000 meters or 237 kilometers. The second root, x2, is the answer that is physically consistent with the given information. Students may wonder why the mathematical model gives a second answer of -109 kilometers. This is because the parabolic model was only designed to accurately represent the physical circumstances of the coasting phase of the capsule's descent from its apogee at a distance of 64 kilometers from the launch pad. Any extrapolations to times and positions earlier than the moment of apogee are 'unphysical'. Problem 3 - Why is it sometimes easier not to work with the parametric form of the motion of a rocket or projectile? Answer: In order to determine the trajectory in space, you need to make twice as many calculations for the parametric form than for the functional form y(x) since each point is defined by (x(t), y(t) ) vs (x, y(x)). Space Math
11485
https://www.math.colostate.edu/~aristoff/332_HW7_s.pdf
Math 332 HW 7 1. Consider the wave equation utt = c2uxx with boundary conditions u(0, t) = u(L, t) = 0, u(x, 0) = f(x), ut(x, 0) = 0. Let F be the odd periodic extension of f. Show that u(x, t) = 1 2 (F(x −ct) + F(x + ct)) (∗) by verifying that the formula in (∗) satisfies the PDE and the boundary conditions. Solution. First we check that (∗) satisfies the PDE: ut(x, t) = c 2 −F ′(x −ct) + F ′(x + ct)  ), utt(x, t) = c2 2 F ′′(x −ct) + F ′′(x + ct)  ), while ux(x, t) = 1 2 F ′(x −ct) + F ′(x + ct)  ), uxx(x, t) = 1 2 F ′′(x −ct) + F ′′(x + ct)  ), This shows utt = c2uxx. Let’s check the boundary conditions. Since F is odd, u(0, t) = 1 2(F(−ct)+ F(ct)) = 0. Since F is periodic (with period L), it follows that u(L, t) = 1 2(F(L−ct)+F(L+ct)) = 0. Finally we check the initial conditions: u(x, 0) = 1 2(F(x)+F(x)) = F(x) = f(x) since F(x) = f(x) for 0 ≤x ≤L, while ut(x, 0) = c 2(−F ′(x) + F ′(x)) = 0. 2. Consider the wave equation with the same boundary conditions as in Problem 1. Solve the initial value problem by using separation of variables and superposition. Show that the solution you obtain agrees with the formula in (∗). (Hint: use the trig identity sin a cos b = 1 2 (sin(a −b) + sin(a + b)).) Solution. Separation of variables and superposition leads to u(x, t) = ∞ X n=1  An sin nπx L cos nπct L + Bn sin nπx L sin nπct L  . From ut(x, 0) = 0 we get Bn = 0, while from u(x, 0) = f(x) we see that An are the Fourier sine series coefficients of f, that is An = 2 L R L 0 f(x) sin nπx L dx. Using the trig identity in the hint, u(x, t) = ∞ X n=1 An sin nπx L cos nπct L = 1 2 ∞ X n=1 An sin nπ L (x −ct) + 1 2 ∞ X n=1 An sin nπ L (x + ct) = 1 2  ˜ F(x −ct) + ˜ F(x + ct)  where to get the last equality we defined ˜ F(x) = ∞ X n=1 An sin nπx L . 1 2 Thus ˜ F is the Fourier sine series of f. This means ˜ F is the odd periodic extension of f, or ˜ F = F. So our solution here agrees with the d’Alembert solution (∗) from Problem 1. 3. Consider the damped wave equation utt = c2uxx −βut with boundary conditions u(0, t) = u(L, t) = 0, u(x, 0) = f(x), ut(x, 0) = g(x) where 0 < β < 2πc/L. (i) Explain the physical meaning of the term −βut. Why is β > 0? (ii) Explain the physical meaning of the boundary conditions. (iii) Using separation of variables and superposition, solve the initial value problem. (iv) What is the long-time behavior of the solution? Solution. (i) This is a damping term – a force proportional to velocity, in the opposite direction. It can arise from wind resistance, for example. (ii) u(0, t) = u(L, t) = 0 indicates zero displacement at x = 0 and x = L (e.g., pinned ends for a vibrating string). The initial condition u(x, 0) = f(x) and ut(x, 0) = g(x) describe the initial displacement and velocity profile, respectively. (iii) Separating variables leads to product solutions of the form e−βt/2 sin nπx L cos ωnt and c2e−βt/2 sin nπx L sin ωnt, n = 1, 2, 3, . . ., where ωn = q nπc L 2 −(β/2)2. Using superposition write u(x, t) = ∞ X n=1 Ane−βt/2 sin nπx L cos ωnt + ∞ X n=1 Bne−βt/2 sin nπx L sin ωnt. Using u(x, 0) = f(x) we see that An = 2 L R L 0 f(x) sin nπx L dx. For the other boundary condition, ut(x, 0) = ∞ X n=1  −βAn 2 + ωnBn  sin nπx L . So −βAn/2 + ωnBn = 2 L R L 0 g(x) sin nπx L dx, and solving for Bn gives Bn = βAn 2ωn + 2 Lωn Z L 0 g(x) sin nπx L dx. 4. Consider a vibrating string described by the PDE utt = c2uxx. (i) What are the natural frequencies of vibration of a string of length L with fixed ends? (ii) What are the natural frequencies of vibration of a string of length H = L/2 with one fixed end (u(0, t) = 0) and another free end (ux(H, t) = 0)? (iii) Check that the vibrational frequencies of the odd harmonics (n = 1, 3, 5, . . .) in part (i) are the same as the vibrational frequencies in part (ii). Explain why this is so, without resorting to formulas. Solution. (i) The frequencies are nπc L , n = 1, 2, 3, . . . (this is the number of oscillations in 2π units of time). (ii) The boundary value problem is φxx = −λφ, φ(0) = 0, φx(H) = 0. The general solution is φ(x) = c1 cos √ λx + c2 sin √ λx. Using the boundary condition φ(0) = 0 shows that c1 = 0. Using φx(H) = 0 gives φx(H) = c2 √ λ cos √ λH = 0, which shows √ λH = nπ/2 where n is an odd integer. Thus, √ λ = nπ 2H where n = 1, 3, 5, . . .. The time dependent problem htt = −λc2h has general solution h(t) = c1 cos c √ λt + c2 sin c √ λt = c1 cos nπct 2H + c2 sin nπct 2H , n = 1, 3, 5, . . .. This 3 shows that the frequencies are nπc 2H , n = 1, 3, 5, . . .. (iii) In part (ii), if H = L/2 then the frequencies are nπc L , n = 1, 3, 5, . . . which agree with the frequencies in (i) for odd n. This can be seen from a symmetry argument: the odd modes from (i) are symmetric around L/2, so they satisfy a free boundary condition at H = L/2 and a pinned boundary condition at 0.
11486
https://discrete.openmathbooks.org/more/mdm/sec_eulerapps.html
( \def\negchoose#1#2{\genfrac{[}{]}{0pt}{}{#1}{#2}_{-1}} \newcommand{\mchoose}{\left(!\binom{#1}{#2}!\right)} \newcommand{\cycle}{\arraycolsep 5 pt \left(\begin{array}#1\end{array}\right)} \newcommand{\importantarrow}{\Rightarrow} \newcommand{\qchoose}{\left[{#1\atop#2}\right]_q} \newcommand{\bp}{ \begin{enumerate}{\setcounter{enumi}{\value{problemnumber}}}} \newcommand{\ep}{\setcounter{problemnumber}{\value{enumi}} \end{enumerate}} \newcommand{\ignore}{} \renewcommand{\bottomfraction}{.8} \renewcommand{\topfraction}{.8} \newcommand{\apple}{\text{🍎}} \newcommand{\ap}{\apple} \newcommand{\banana}{\text{🍌}} \newcommand{\ba}{\banana} \newcommand{\pear}{\text{🍐}} \newcommand{\pe}{\pear} \DeclareMathOperator{\Fix}{Fix} \DeclareMathOperator{\Orb}{Orb} \newcommand{\F}{\mathcal{F}} \newcommand{\alert}{\fbox} \def\d{\displaystyle} \def\course{Math 228} \newcommand{\f}{\mathfrak #1} \newcommand{\s}{\mathscr #1} \def\N{\mathbb N} \def\B{\mathbf{B}} \def\circleA{(-.5,0) circle (1)} \def\Z{\mathbb Z} \def\circleAlabel{(-1.5,.6) node[above]{$A$}} \def\Q{\mathbb Q} \def\circleB{(.5,0) circle (1)} \def\R{\mathbb R} \def\circleBlabel{(1.5,.6) node[above]{$B$}} \def\C{\mathbb C} \def\circleC{(0,-1) circle (1)} \def\F{\mathbb F} \def\circleClabel{(.5,-2) node[right]{$C$}} \def\A{\mathbb A} \def\twosetbox{(-2,-1.5) rectangle (2,1.5)} \def\X{\mathbb X} \def\threesetbox{(-2,-2.5) rectangle (2,1.5)} \def\E{\mathbb E} \def\O{\mathbb O} \def\U{\mathcal U} \def\pow{\mathcal P} \def\inv{^{-1}} \def\nrml{\triangleleft} \def\st{:} \def\~{\widetilde} \def\rem{\mathcal R} \def\sigalg{$\sigma$-algebra } \def\Gal{\mbox{Gal}} \def\iff{\leftrightarrow} \def\Iff{\Leftrightarrow} \def\land{\wedge} \def\And{\bigwedge} \def\entry{\entry} \def\AAnd{\d\bigwedge\mkern-18mu\bigwedge} \def\Vee{\bigvee} \def\VVee{\d\Vee\mkern-18mu\Vee} \def\imp{\rightarrow} \def\Imp{\Rightarrow} \def\Fi{\Leftarrow} \def\var{\mbox{var}} \def\Th{\mbox{Th}} \def\entry{\entry} \def\sat{\mbox{Sat}} \def\con{\mbox{Con}} \def\iffmodels{\bmodels\models} \def\dbland{\bigwedge !!\bigwedge} \def\dom{\mbox{dom}} \def\rng{\mbox{range}} \def\isom{\cong} \DeclareMathOperator{\wgt}{wgt} \newcommand{\vtx}{node[fill,circle,inner sep=0pt, minimum size=4pt,label=#1:#2]{}} \newcommand{\va}{\vtx{above}{#1}} \newcommand{\vb}{\vtx{below}{#1}} \newcommand{\vr}{\vtx{right}{#1}} \newcommand{\vl}{\vtx{left}{#1}} \renewcommand{\v}{\vtx{above}{}} \def\circleA{(-.5,0) circle (1)} \def\circleAlabel{(-1.5,.6) node[above]{$A$}} \def\circleB{(.5,0) circle (1)} \def\circleBlabel{(1.5,.6) node[above]{$B$}} \def\circleC{(0,-1) circle (1)} \def\circleClabel{(.5,-2) node[right]{$C$}} \def\twosetbox{(-2,-1.4) rectangle (2,1.4)} \def\threesetbox{(-2.5,-2.4) rectangle (2.5,1.4)} \def\ansfilename{practice-answers} \def\shadowprops{{fill=black!50,shadow xshift=0.5ex,shadow yshift=0.5ex,path fading={circle with fuzzy edge 10 percent}}} \newcommand{\hexbox}{ \def\x{-cos{30}\r#1+cos{30}#2\r2} \def\y{-\r#1-sin{30}\r#1} \draw (\x,\y) +(90:\r) -- +(30:\r) -- +(-30:\r) -- +(-90:\r) -- +(-150:\r) -- +(150:\r) -- cycle; \draw (\x,\y) node{#3}; } \newcommand{\card}{\left| #1 \right|} \newcommand{\twoline}{\begin{pmatrix}#1 \ #2 \end{pmatrix}} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} ) More Discrete Mathematicsvia Graph Theory Richard Grassl, Oscar Levin Section1.4Applications of Euler's Formula ¶ We will now consider some applications of Euler's formula for planar graphs to graphs that are not necessarily planar. Subsection1.4.1Non-planar Graphs ¶ Activity26 For the complete graphs (K_n\text{,}) we would like to be able to say something about the number of vertices, edges, and (if the graph is planar) faces. (a) How many vertices does (K_3) have? How many edges? If (K_3) is planar, how many faces should it have? (b) How many vertices, edges and (if planar) faces would (K_4\text{,}) (K_5) and (K_{23}) each have? (c) What about complete bipartite graphs? How many vertices, edges, and faces (if it were planar) does (K_{7,4}) have? Not all graphs are planar. If there are too many edges and too few vertices, then some of the edges will need to intersect. For example, consider (K_5\text{.}) If you try to redraw this without edges crossing, you quickly get into trouble. There seems to be one edge too many. How could we prove this? Activity27 (a) The graph (K_5) has (5) vertices and 10 edges. Explain why it would need to have (7) faces if it were planar. (b) Now get at the number of faces another way: each face must be bordered by at least three edges. Why? Explain why we can conclude that (3f \le 2e\text{.}) Where does the 2 come from? (c) We now have that for (K_5\text{,}) the number of faces is (f = 7\text{,}) and also (3f \le 20\text{.}) How is this possible? What can we conclude? Before proceeding, consider carefully the style of proof used here. This is a proof by contradiction. We assumed the opposite of what we wanted to show. From that, we arrived at a contradiction, a statement that is necessarily false. Is it valid to conclude that our original assumption is false? Recall the truth table for an implication (P \imp Q\text{:}) | | | | --- | (P) | (Q) | (P\imp Q) | | T | T | T | | T | F | F | | F | T | T | | F | F | T | In a proof by contradiction, we assume (P) (which will be the negation of our desired conclusion) and derive (Q\text{,}) a contradiction. That valid derivation allows us to conclude (P \imp Q) is true! But we know that (Q) is false. What does that say about (P\text{?}) If the implication is true and the consequent (“then” part) is false, it must be that we are in row 4 of the truth table. In that row, (P) is false. So we are justified in concluding (\neg P\text{.}) In our case, (P) was the statement “(K_5) is planar.” So when we get a contradiction (a false (Q)), we can conclude that (K_5) is not planar. Note that even though we are proving something about a graph that does not satisfy Euler's formula for planar graphs, by using a proof by contradiction, we get to use the formula. Here are a few more examples of this proof strategy, specifically to show graphs are not planar. Activity28 The other simplest graph which is not planar is (K_{3,3}) (a) Following the same proof outline as you used for (K_5\text{,}) what value do you find for (f) and what can you conclude from the inequality (3f \le 2e\text{?}) Hint If you assume (P) and conclude (Q\text{,}) but (Q) is true, can you say anything about (P\text{?}) What row(s) of the truth table for (P \imp Q) are you in? (b) The 3 in (3f \le 2e) came from the observation that the any face in (K_5) must be bounded by at least three edges. Is that the best we can do for a bipartite graph? Find and justify an improved inequality between (f) and (e\text{.}) In general, if we let (g) be the size of the smallest cycle in a graph ((g) stands for girth, which is the technical term for this) then for any planar graph we have (gf \le 2e\text{.}) When this disagrees with Euler's formula, we know for sure that the graph cannot be planar. You might wonder whether every graph that is not planar can be shown to be non-planar this way. Activity29 The graph show below is not planar. Let's prove it. (a) What is the girth of this graph? What can you conclude from Euler's formula and the inequality (gf \le 2e\text{?}) (b) Try proving the following graph is not planar using our standard approach. (c) Could it be that the original graph is planar but the subgraph is not? What can you conclude? If a graph contains a non-planar subgraph, then the graph cannot itself be planar. This gives us another method of proving a given graph is not planar: find a non-planar graph, perhaps (K_5) or (K_{3,3}) inside it. From the other direction, you can think of starting with (K_5) or (K_{3,3}) and adding edges as well as adding vertices in the middle of edges (called subdividing the edge) to build a larger graph that is definitely not planar. Surprisingly, every non-planar graph arises this way, a result called Kuratowski's Theorem. Euler's formula can also be used to prove results about planar graphs. Activity30 Prove that any planar graph with (v) vertices and (e) edges satisfies (e \le 3v - 6\text{.}) Hint The girth of any graph is at least 3. Activity31 Prove that any planar graph must have a vertex of degree 5 or less. Hint Do a proof by contradiction. What is the smallest number of edges such a graph would possess? Subsection1.4.2Polyhedra ¶ Activity32 A cube is an example of a convex polyhedron. It contains 6 identical squares for its faces, 8 vertices, and 12 edges. The cube is a regular polyhedron (also known as a Platonic solid) because each face is an identical regular polygon and each vertex joins an equal number of faces. There are exactly four other regular polyhedra: the tetrahedron, octahedron, dodecahedron, and icosahedron with 4, 8, 12 and 20 faces respectively. How many vertices and edges do each of these have? Another area of mathematics that uses the terms “vertex,” “edge,” and “face” is geometry. A polyhedron is a geometric solid made up of flat polygonal faces joined at edges and vertices. We are especially interested in convex polyhedra, which means that any line segment connecting two points on the interior of the polyhedron must be entirely contained inside the polyhedron.3An alternative definition for convex is that the internal angle formed by any two faces must be less than (180\deg\text{.}) Notice that since (8 - 12 + 6 = 2\text{,}) the vertices, edges and faces of a cube satisfy Euler's formula for planar graphs. This is not a coincidence. We can represent a cube as a planar graph by projecting the vertices and edges onto the plane. One such projection looks like this: In fact, every convex polyhedron can be projected onto the plane without edges crossing. Think of placing the polyhedron inside a sphere, with a light at the center of the sphere. The edges and vertices of the polyhedron cast a shadow onto the interior of the sphere. You can then cut a hole in the sphere in the middle of one of the projected faces and “stretch” the sphere to lay down flat on the plane. The face that was punctured becomes the “outside” face of the planar graph. The point is, we can apply what we know about graphs (in particular planar graphs) to convex polyhedra. Since every convex polyhedron can be represented as a planar graph, we see that Euler's formula for planar graphs holds for all convex polyhedra as well. We also can apply the same sort of reasoning we use for graphs in other contexts to convex polyhedra. For example, we know that there is no convex polyhedron with 11 vertices all of degree 3, as this would make 33/2 edges. Example1.4.1 Is there a convex polyhedron consisting of three triangles and six pentagons? What about three triangles, six pentagons and five heptagons (7-sided polygons)? Solution How many edges would such polyhedra have? For the first proposed polyhedron, the triangles would contribute a total of 9 edges, and the pentagons would contribute 30. However, this counts each edge twice (as each edge borders exactly two faces), giving 39/2 edges, an impossibility. There is no such polyhedron. The second polyhedron does not have this obstacle. The extra 35 edges contributed by the heptagons give a total of 74/2 = 37 edges. So far so good. Now how many vertices does this supposed polyhedron have? We can use Euler's formula. There are 14 faces, so we have (v - 37 + 14 = 2) or equivalently (v = 25\text{.}) But now use the vertices to count the edges again. Each vertex must have degree at least three (that is, each vertex joins at least three faces since the interior angle of all the polygons must be less that (180^\circ)), so the sum of the degrees of vertices is at least 75. Since the sum of the degrees must be exactly twice the number of edges, this says that there are strictly more than 37 edges. Again, there is no such polyhedron. Activity33 I'm thinking of a polyhedron containing 12 faces. Seven are triangles and four are quadrilaterals. The polyhedron has 11 vertices including those around the mystery face. How many sides does the last face have? Activity34 Consider some classic polyhedra. (a) An octahedron is a regular polyhedron made up of 8 equilateral triangles (it sort of looks like two pyramids with their bases glued together). Draw a planar graph representation of an octahedron. How many vertices, edges and faces does an octahedron (and your graph) have? (b) The traditional design of a soccer ball is in fact a (spherical projection of a) truncated icosahedron. This consists of 12 regular pentagons and 20 regular hexagons. No two pentagons are adjacent (so the edges of each pentagon are shared only by hexagons). How many vertices, edges, and faces does a truncated icosahedron have? Explain how you arrived at your answers. Bonus: draw the planar graph representation of the truncated icosahedron. (c) Your “friend” claims that he has constructed a convex polyhedron out of 2 triangles, 2 squares, 6 pentagons and 5 octagons. Prove that your friend is lying. Hint: each vertex of a convex polyhedron must border at least three faces. To conclude this application of planar graphs, consider the regular polyhedra. Above we claimed there are only five. How do we know this is true? We can prove it using graph theory. Theorem1.4.2 There are exactly five regular polyhedra. Activity35 Recall that a regular polyhedron has all of its faces identical regular polygons, and that each vertex has the same degree. Consider the cases, broken up by what the regular polygon might be. (a) Case 1: Each face is a triangle. Let (f) be the number of faces and (k) the common degree of each vertex (since the polyhedron is regular). Find formulas for the number of edges and vertices in terms of (f) and (k\text{.}) Conclude from these the only possible values for (f) and (k\text{.}) (b) Case 2: each face is a square. Again, consider the possible cases for (f) and (k) (and conclude there is only one: the cube). (c) Case 3: Each face is a pentagon. (d) Explain why it is not possible for each face to be a (n)-gon with (n \ge 6\text{.})
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https://flexbooks.ck12.org/cbook/ck-12-conceptos-de-matem%C3%A1ticas-de-la-escuela-secundaria-grado-8-en-espa%C3%B1ol/section/12.15/related/lesson/graphs-of-basic-quadratic-functions-alg-i-hnrs/
Graphs of Basic Quadratic Functions | CK-12 Foundation AI Teacher Tools – Save Hours on Planning & Prep. Try it out! Skip to content What are you looking for? Search Math Grade 6 Grade 7 Grade 8 Algebra 1 Geometry Algebra 2 PreCalculus Science Earth Science Life Science Physical Science Biology Chemistry Physics Social Studies Economics Geography Government Philosophy Sociology Subject Math Elementary Math Grade 1 Grade 2 Grade 3 Grade 4 Grade 5 Interactive Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Conventional Math 6 Math 7 Math 8 Algebra I Geometry Algebra II Probability & Statistics Trigonometry Math Analysis Precalculus Calculus What's the difference? 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Learn. Interact. eXplore. CCSS Math Concepts and FlexBooks aligned to Common Core NGSS Concepts aligned to Next Generation Science Standards Certified Educator Stand out as an educator. Become CK-12 Certified. Webinars Live and archived sessions to learn about CK-12 Other Resources CK-12 Resources Concept Map Testimonials CK-12 Mission Meet the Team CK-12 Helpdesk FlexLets Know the essentials. Pick a Subject Donate Sign InSign Up Back To Evaluar Funciones CuadráticasBack 12.15 Graphs of Basic Quadratic Functions Written by:Brenda Meery |Kaitlyn Spong Fact-checked by:The CK-12 Editorial Team Last Modified: Sep 01, 2025 Look at the graph below. Does the graph represent a function? Do you know the name of the graph? Do you know what makes the green point special? Do you notice any symmetry in the graph? Can you state the domain and range for the relation? Graphing Basic Quadratic Functions Until now you have been dealing with linear functions. The highest exponent of the independent variable (x) has been one and the graphs have been straight lines. Here you will be learning about quadratic functions. A quadratic function is one of the form y=a x 2+b x+c where a,b and c are real numbers and a≠0. The highest exponent of the independent variable is two. When graphed, a quadratic function creates a parabola that looks like this: or like this: You can create your own graph by plotting the points created from a table of values. The most basic quadratic function is y=x 2. The easiest way to make a table for this function is to use the domain {x|−3≤x≤3,x∈Z} for the table. A parabola has a turning point known as the vertex. The vertex is the minimum value of the parabola if it opens upward and the maximum value if the parabola opens downward. When the graph opens downward, the y-values in the base table change to negative values. The basic quadratic function that opens downward has the equation y=−x 2. All parabolas have an axis of symmetry. The axis of symmetry is the vertical line that passes through the vertex of the parabola. The equation for the axis of symmetry is always x=α, where α is the the x−coordinate of the vertex. Let's graph the basic quadratic function y=x 2: First, let's complete a table such that {x|−3≤x≤3,x∈Z}. To complete the table of values, substitute the given x-values into the function y=x 2. If you are using a calculator, insert all numbers, especially negative numbers, inside parentheses before squaring them. The operation that needs to be done is (−3)(−3) NOT −(3)(3). y=x 2 y=x 2 y=x 2 y=x 2 y=(−3)2 y=(−2)2 y=(−1)2 y=(0)2 y=9 y=4 y=1 y=0 y=x 2 y=x 2 y=x 2 y=(1)2 y=(2)2 y=(3)2 y=1 y=4 y=9 | X | Y | --- | | −3 | 9 | | −2 | 4 | | −1 | 1 | | | 0 | | 1 | 1 | | 2 | 4 | | 3 | 9 | Now, on a Cartesian plane, plot the points from the table for y=x 2. The plotted points cannot be joined to form a continuous curve. To join the points, begin with the point (–3, 9) or the point (3, 9) and without lifting your pencil, draw a smooth curve. The image should look like the following graph. The arrows indicate the direction of the pencil as the points are joined. If the pencil is not moved off the paper, the temptation to join the points with a series of straight lines will be decreased. The points must be joined with a smooth curve that does not extend below the lowest point of the graph. In the above graph, the curve cannot go below the point (0, 0). Now, notice the unique characteristics of the graph of y=x 2. The green point is located at the lowest point on the image. The curve does not go below this point. Every red point on the left side of the image has a corresponding blue point on the right side of the image. If the image was folded left to right along the y-axis that passes through the green point, each red point would land on each corresponding blue point. The sides of the image extend upward. The red and the blue points are plotted to the right and to the left of the green point. The points are plotted left and right one and up one; left and right two and up four, left and right 3 and up nine. Examples Example 1 Earlier, you were given the graph above and asked to identify it, explain what the points are, and describe it's symmetry. The green point is the lowest point on the curve. The smooth curve is called a parabola and it is the image produced when the basic quadratic function is plotted on a Cartesian grid. The green point is known as the vertex of the parabola. The vertex is the turning point of the graph. For the graph of y=x 2, the vertex is (0, 0) and the parabola has a minimum value of zero which is indicated by the y-value of the vertex. The parabola opens upward since the y-values in the table of values are 0, 1, 4 and 9. The y-axis for this graph is actually the axis of symmetry. The axis of symmetry is the vertical line that passes through the vertex of the parabola. The parabola is symmetrical about this line. The equation for this axis of symmetry is x=0. If the parabola were to open downward, the vertex would be the highest point of the graph. Therefore the image would have a maximum value of zero. The domain for all parabolas is D={x|x∈R}. The range for the above parabola is R={y|y≥0,y∈R}. Example 2 If the graph of y=x 2 opened downward, what changes would exist in the base table of values? If the parabola were to open downward, the x-values would not change. The y-values would become negative values. The points would be plotted from the vertex as: right and left one and down one; right and left two and down four; right and left three and down nine. The table of values would be | X | Y | --- | | −3 | −9 | | −2 | −4 | | −1 | −1 | | | 0 | | 1 | −1 | | 2 | −4 | | 3 | −9 | Example 3 If the graph of y=x 2 opened downward, what changes would exist in the basic quadratic function? To match the table of values, the basic quadratic function would have to be written as y=−x 2. Example 4 Draw the image of the basic quadratic function that opens downward. State the domain and range for this parabola. The domain is D={x|x∈R}. The range for this parabola is R={y|y≤0,y∈R}. Review Complete the following statements in the space provided. The name given to the graph of y=x 2 is ________. The domain of the graph of y=x 2 is ________. If the vertex of a parabola was (–3, 5), the equation of the axis of symmetry would be ________. A parabola has a maximum value when it opens ________. The point (–2, 4) on the graph of y=x 2 has a corresponding point at ________. The range of the graph of y=−x 2 is ________. If the table of values for the basic quadratic function included 4 and –4 as x-values, the y-value(s) would be ________. The vertical line that passes through the vertex of a parabola is called ________. A minimum value exists when a parabola opens ________. The turning point of the graph of y=x 2 is called the ________. Make a sketch of the function y=x 2−2 x+1 by first making a table and then plotting the points. What is the vertex of this parabola? Make a sketch of the function y=x 2+2 x−3 by first making a table and then plotting the points. What is the vertex of this parabola? Make a sketch of the function y=x 2−4 x+5 by first making a table and then plotting the points. What is the vertex of this parabola? Make a sketch of the function y=−x 2+2 x+1 by first making a table and then plotting the points. What is the vertex of this parabola? Make a sketch of the function y=−x 2−4 x by first making a table and then plotting the points. What is the vertex of this parabola? Review (Answers) Click HERE to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option. Image Attributions Back to Graphs of Basic Quadratic Functions | Image | Reference | Attributions | --- Ask me anything! CK-12 Foundation is a non-profit organization that provides free educational materials and resources. FLEXIAPPS ABOUT Our missionMeet the teamPartnersPressCareersSecurityBlogCK-12 usage mapTestimonials SUPPORT Certified Educator ProgramCK-12 trainersWebinarsCK-12 resourcesHelpContact us BYCK-12 Common Core MathK-12 FlexBooksCollege FlexBooksTools and apps CONNECT TikTokInstagramYouTubeTwitterMediumFacebookLinkedIn v2.11.10.20250923073248-4b84c670be © CK-12 Foundation 2025 | FlexBook Platform®, FlexBook®, FlexLet® and FlexCard™ are registered trademarks of CK-12 Foundation. Terms of usePrivacyAttribution guide Curriculum Materials License Student Sign Up Are you a teacher? Sign up here Sign in with Google Having issues? 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https://goldbook.iupac.org/terms/view/G02631/pdf
doi:10.1351/goldbook.G02631 IUPAC Compendium of Chemical Terminology Copyright © 2014 IUPAC Gibbs energy of activation (standard free energy of activation), The standard Gibbs energy difference between the transition state of a reaction (either an elementary reaction or a stepwise reaction) and the ground state of the reactants. It is calculated from the experimental rate constant via the conventional form of the absolute rate equation: where is the Boltzmann constant and the Planck constant ( ). The values of the rate constants, and hence Gibbs energies of activation, depend upon the choice of concentration units (or of the thermodynamic standard state). See also: enthalpy of activation, entropy of activation Source: PAC, 1994, 66 , 1077 (Glossary of terms used in physical organic chemistry (IUPAC Recommendations 1994)) on page 1118 PAC, 1996, 68 , 149 (A glossary of terms used in chemical kinetics, including reaction dynamics (IUPAC Recommendations 1996)) on page 166
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https://nora.nerc.ac.uk/id/eprint/514121/1/c6em00072j.pdf
Prolonged exposure to arsenic in UK private water supplies: toenail, hair and drinking water concentrations† D. R. S. Middleton,abc M. J. Watts,b E. M. Hamilton,b T. Fletcher,c G. S. Leonardi,c R. M. Close,c K. S. Exley,c H. Crabbec and D. A. Polyaa Chronic exposure to arsenic (As) in drinking water is an established cause of cancer and other adverse health effects. Arsenic concentrations >10 mg L1 were previously measured in 5% of private water supplies (PWS) in Cornwall, UK. The present study investigated prolongued exposure to As by measuring biomarkers in hair and toenail samples from 212 volunteers and repeated measurements of As in drinking water from 127 households served by PWS. Strong positive Pearson correlations (rp ¼ 0.95) indicated stability of water As concentrations over the time period investigated (up to 31 months). Drinking water As concentrations were positively correlated with toenail (rp ¼ 0.53) and hair (rp ¼ 0.38) As concentrations – indicative of prolonged exposure. Analysis of washing procedure solutions provided strong evidence of the effective removal of exogenous As from toenail samples. Significantly higher As concentrations were measured in hair samples from males and smokers and As concentrations in toenails were negatively associated with age. A positive association between seafood consumption and toenail As and a negative association between home-grown vegetable consumption and hair As was observed for volunteers exposed to <1 As mg L1 in drinking water. These findings have important implications regarding the interpretation of toenail and hair biomarkers. Substantial variation in biomarker As concentrations remained unaccounted for, with soil and dust exposure as possible explanations. Environmental impact Arsenic is an established carcinogen, chronic exposure to which has been linked to several cancers (lung, bladder, skin) as well as non-cancerous (cardiovascular disease, diabetes mellitus) health effects. This work consists of a human biomonitoring study (a collaboration between the University of Manchester, British Geological Survey and Public Health England) of 212 volunteers from 127 households with private water supplies from across Cornwall, UK. It is the largest scale exposure biomonitoring study conducted for As and drinking water in the UK to-date and investigates an exposure source for As that, until recently, had not been investigated in depth in the region. The sampling protocol consists of an initial and follow-up water collection spanning a period of either 8 or 31 months which, together with long-term biomarkers such as toenails and hair, allows for the assessment of prolonged arsenic exposure. Furthermore, the methods employed in this paper allow for an assessment of the efficacy of toenail washing procedures given the recognition of the susceptibility of this biomonitoring matrix to external contamination. The demonstration of effective contamination removal from samples in this study will be of great benet to the wider eld. 1. Introduction Chronic exposure to arsenic (As) in contaminated drinking water is an established cause of lung, skin, bladder and kidney cancer1 as well as other adverse health effects, posing a global health concern. Five major As endemic regions of the world provide the strongest evidence of this association: north-west and south-east Taiwan;2 northern Chile;3 Argentina;4,5 Bangla-desh6 and West Bengal.7 Although the aforementioned areas are more severely affected, As contaminated municipal and private water supplies (PWS) have been reported in countries across all inhabited continents.8 Notable European examples include Hungary,9 Romania,9 Slovakia9 and Serbia.10 A survey11 of PWS in Cornwall, south-west England, reported concentrations exceeding the 10 As mg L1 UK prescribed concentration or value12 (PCV) and WHO guidance value13 in 5% of drinking water samples collected (n ¼ 497). In a follow-up biomonitoring study,14 a subset of the same cohort, drinking water As concentrations were positively correlated with urinary As concentrations aer the exclusion of arsenobetaine (AB) and aSchool of Earth, Atmospheric and Environmental Sciences & William Research Centre for Molecular Environmental Science, University of Manchester, Oxford Rd, Manchester, M13 9PL, UK bInorganic Geochemistry, Centre for Environmental Geochemistry, British Geological Survey, Nicker Hill, Keyworth, Nottinghamshire, NG12 5GG, UK. E-mail: mwatts@ bgs.ac.uk cCentre for Radiation, Chemicals and Environmental Hazards (CRCE), Public Health England, Chilton, Didcot, Oxfordshire, OX11 0RQ, UK † Electronic supplementary information (ESI) available. See DOI: 10.1039/c6em00072j Cite this: Environ. Sci.: Processes Impacts, 2016, 18, 562 Received 8th February 2016 Accepted 18th April 2016 DOI: 10.1039/c6em00072j rsc.li/process-impacts 562 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science Processes & Impacts PAPER Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online View Journal | View Issue adjustment for hydration (osmolality adjustment). These urinary As concentrations reected exposure in the preceding 2–4 days.15 Information on the longevity and temporal variation of exposure in this study group was still outstanding. Two methods that can assess exposure over extended timescales are repeat monitoring of drinking water As concentrations and monitoring of biological matrices, such as toenails and hair, that reect a longer exposure window than urine. Both approaches were employed in the present study. There are currently 2460 registered single domestic dwell-ings served by PWS in Cornwall,16 with the true number likely to be much greater. No published data on the temporal variation of As concentrations in UK PWS were previously available, but studies elsewhere reported mixed ndings. In Nevada, USA, although concentration changes (mean ¼ 3 As mg L1) were measured in some supplies,17 with greater changes associated with higher As concentrations, no clear temporal trends were observed between wet and dry seasons. In a related study,18 strong Spearman correlations (rs) (rs ¼ 0.95) were reported between As concentrations in the same wells over a period of 11–20 years, with both studies concluding that, for the region, limited measurements are sufficient for predicting exposures over such timescales. Similarly, in Michigan, USA, strong Pearson correlations (rp) (rp ¼ 0.88) were reported19 between As measurements taken an average of 14 months apart. Concen-trations were affected by point-of-use (POU) treatment systems, highlighting the necessity of collecting treatment usage data. Conversely, a study conducted in Washington, USA20 reported changes as high as 19-fold in As concentrations measured in the same supply 12 months apart, suggesting that temporal stability of As concentrations varied by region due to geological and geochemical variables, if not inconsistencies in sampling methodologies.19 The use of toenail and hair biomonitoring for As exposure offers the assessment of a longer exposure window than that reected by urine sampling. The affinity of As for sulydryl groups in the keratin of nails and hair, the isolation of these matrices from other metabolic processes following their formation and the time taken for them to ‘grow out,’ makes them attractive for measuring biomarkers of past As exposure.21 Nails and hair have the added value of a non-invasive collection protocol and few sample transport/storage requirements. Posi-tive correlations between drinking water and biomarker As concentrations have been reported in numerous studies for both toenails22–25 and hair.21,26 Increased risk of various cancers, including cutaneous melanoma27 and small and squamous-cell carcinoma of the lung,28 have also been positively associated with toenail As concentrations. Despite the advantages of toenail and hair biomonitoring, caveats apply when using these matrices to assess exposure. Factors unrelated to exposure have been reported to inuence As concentrations in hair and nails: namely, the inter-individual variability of growth rates of the biomonitoring matrices, demographic and behavioural factors such as age, gender and smoking,23 their susceptibility to external contamination29,30 and the consumption of dietary items such as fruit juices,31 beer,32 wine32 and dark-meat sh.33 Average growth rates for ngernails are 0.1 mm per day whereas toenails are estimated to grow by 0.03–0.5 mm per day, meaning that ngernails and toenails reect exposure windows dating back approximately 6 and 12–18 months, respectively.34 Hair reects a period of just a few months, with reported scalp hair growth rates ranging from 0.2 to 1.12 mm per day.35 Growth rates for both matrices have been demonstrated to vary with demographic factors e.g. age and gender,29,34–36 with obvious implications for interpreting exposure assessments conducted on diverse populations. The susceptibility of nails and hair to external contamina-tion is well documented, with a range of washing procedures having been implemented.29,37 The degree of sample contami-nation likely depends on personal hygiene, hobbies, other behavioural variables and the relative ubiquity of the chemical element of interest. Fingernails are reportedly more prone to contamination than toenails38 but this does not likely apply to communities who are oen barefoot or wear open toed foot-wear. Contamination of hair and nails from cosmetic products such as shampoos, hair colourings and nail polish is another important consideration. A study39 of the trace element composition of nail polish estimated that the As contribution from polish, if present, can range from 16 to 633%. Whilst studies now routinely report the washing of nail and hair samples prior to analysis, few have quantied the degree of exogenous As versus As in toenails, or conrmed the removal of exogenous As from samples. One investigation40 of exposure to As in soils, also conducted in Cornwall, retained toenail washing solutions for As determination. Both the nal rinse fractions and a pooled solution of all preceding fractions were retained to quantify exogenous As contamination and conrm its removal from samples. The As content of nal rinse fractions accounted for 0.2 to 1.6% of the total As measured in toenails.40 This provided strong evidence of the efficacy of the washing procedure but, with a sample of 17 volunteers, the performance of this method remained to be validated on a greater scale. The present study aimed to assess exposure to inorganic As via drinking water consumption in a population served by PWS in Cornwall, UK, using hair and toenail biomarkers in addition to initial and follow-up drinking water samples collected up to 31 months apart. Specic objectives were to (i) compare repeat PWS drinking water As concentrations measured either 8 or 31 months apart; (ii) investigate the effects of As concentration, duration between measurements, source type and treatment usage on changes in drinking water As concentrations; (iii) measure the total As concentrations in toenail and hair samples collected from volunteers and assess their relationship with drinking water As concentrations adjusted for other covariables (demographic, behavioural and dietary) and (iv) quantify the potential for external sample contamination to affect As concentrations in toenail and hair samples, including the use of nail polish and hair dye. 2. Experimental Ethical approval and volunteer communication Ethical approval was granted by the University of Manchester Research Ethics Committee (Ref 13068) and the NHS Health This journal is © The Royal Society of Chemistry 2016 Environ. Sci.: Processes Impacts, 2016, 18, 562–574 | 563 Paper Environmental Science: Processes & Impacts Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online Research Authority National Research Ethics Service (NRES) (Ref 13/EE/0234). All volunteers provided written informed consent prior to participating. Individual data feedback to participants was provided through a letter containing specic guidance developed by PHE along with BGS and Cornwall County Council. Participants were given advice on any potential health risks and suggested corrective actions if they had one or more exceedances of the water quality standards. All partici-pants were provided with appropriate contact details for any follow-up enquiries. Recruitment and sample collection Environmental monitoring. The sampling frame consisted of 476 households using a PWS that had provided drinking water samples during a previous survey11 – henceforth referred to as initial sampling (drinking water only). The initial survey was conducted in two parts, with households in east and west Cornwall surveyed in March–April 2011 and March–April 2013, respectively. Information letters were sent to households that participated in initial sampling and, aer being contacted by telephone, 127 households were recruited to provide a follow-up drinking water sample. Follow-up sampling took place in November 2013. This resulted in 127 drinking water samples collected either 31 (n ¼ 51) or 8 (n ¼ 76) months apart depending on whether households were in east (2011 initial collection) or west (2013 initial collection) Cornwall, respec-tively. Point-of-use drinking water samples were collected using a previously reported protocol11 Biomonitoring. Biomonitoring was conducted on one occa-sion only – at the time of the follow-up drinking water collection in November 2013. Sample collection packs were mailed to volunteers prior to household visits. Volunteers were asked to allow a minimum of 4 weeks for toenail growth (to ensure sufficient mass for analysis) before self-collecting from all 10 toes and storing in polyethylene bags. Hair samples were collected by researchers during visits using an amended version of the COPHES project protocol.41 Hair >3 cm in length was removed from the nape by twisting into a pencil-width strand before tagging with masking tape. The tape was labelled with an arrow pointing towards the root. Strands were removed with ethanol-rinsed stainless steel scissors as close to the scalp as possible. Hair <3 cm in length was collected in smaller amounts from several locations on the back of the head. The portion of hair >5 cm was discarded with the portion closest to the scalp being retained for analysis. Additional variables. An exposure/food frequency question-naire was administered to volunteers using MicrosoAccess on a laptop/tablet device. For drinking water related analysis, data on PWS source type, treatment usage, system storage and borehole depth were collected at the time of initial water sampling. For biomonitoring analysis, demographic and behavioural variables – age, gender, current smoking status, nail polish and hair product usage – were collected and, addi-tionally, information on the consumption of select dietary items that have been reported31–33 to contain As in relatively high concentrations. These were: PWS water consumption (L per day); home-grown vegetable consumption (all year, seasonally, only in pots or never); rice (servings/week); seafood (servings/ week); most oen consumed seafood type (if reported): white sh (e.g. cod, plaice, haddock etc.), shellsh (e.g. mussels, prawns, cockles etc.) and dark-meat sh (e.g. salmon, tuna, mackerel, sardines etc.); beer (L per day); wine (L per day); cider (L per day) and fruit juice (L per day). Chemical analyses Reagents and standards. All aqueous solutions were prepared using 18.2 MU deionised water (DIW) (Millipore, UK). Nitric acid (HNO3), hydrochloric acid (HCl) and 30% hydrogen peroxide (H2O2) were Romil-SpA™super purity grade (Romil, UK). The acetone used for sample cleaning was HPLC grade (Fisher Scientic, UK). Arsenic calibration standards were made using an in-house multi-element stock in which the As contri-bution was from a 1000 mg L1 PrimAg® grade mono-elemental solution (Romil, UK). Independent 25 mg L1 As QC standards were prepared from a multi-element stock solution of various concentrations with As at 20 mg L1 (Ultra Scientic, USA). A germanium (Ge) ICP-MS internal standard was prepared from a Fluka Analytical 1000 mg L1 stock solution (Sigma-Aldrich, USA). Sample pre-treatment and dissolution. Toenail samples were cleaned and digested by adapting a previously reported protocol.40 Visible exogenous debris was removed using a PTFE policeman/stirring rod (Chemware, USA) in a HEPA ltered clean room. Samples with visible nail polish residue (regardless of whether reported in the questionnaire) were further cleaned with acetone and cotton wool. Samples were transferred into clean 25 mL Duran® borosilicate beakers (Schott, Germany), placed in an ultrasonic bath (Fisher Scientic, UK), sonicated at 37 MHz at room temperature for 5 minutes (15 minutes for those with visible varnish) in 3 mL of acetone, rinsed with 2 mL of DIW and then 2 mL of acetone, sonicated for 10 minutes in 3 mL of DIW and twice rinsed with 3 mL of DIW. All rinse aliquots prior to the nal, which remained separate, were pooled in PFA vials (Savillex, USA) and evaporated to dryness overnight on a graphite hot block before reconstitution in 5 mL of 1% v/v HNO3 + 0.5% v/v HCl. Both initial and nal rinse fractions were analysed by ICP-MS for total As. The nal fraction was analysed separately to assess the effectiveness of the washing procedure and conrm the elimination of exogenous contamination. A schematic of the abovementioned procedure can be viewed in ESI (Fig. S1†). Toenail samples were dried to constant weight (12 h approx.) in a clean laminar ow hood (Envair, UK) and stored in microcentrifuge tubes in a silica gel desiccator before being weighed (0.1 g or as much as available) into PTFA MARS Xpress vessels (CEM Corporation, UK). Four millilitres of concentrated HNO3 + 1 mL of H2O2 were added and samples were leto rest for 30 minutes until effervescence subsided. Vessels were cap-ped and digested in a microwave assisted reaction system (MARS Xpress, CEM Corporation, UK) on the following heating program: ramped to 100 C and held for 5 minutes; ramped to 200 C and held for 30 minutes (100% power: 1200 W). Vessels 564 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science: Processes & Impacts Paper Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online were leto cool overnight before their contents were trans-ferred into PFA vials with DIW and reduced to a gel at 80 C on a graphite hot block. One millilitre of 10% v/v HNO3 was added to the vessels, which were then heated for 20 minutes at 50 C followed by the addition of 4 mL of DIW. Digests were stored in polystyrene ICP-MS tubes. Hair samples underwent the same cleaning and digestion procedure as toenail samples. Whatman Grade B-2 weighting papers (GE Healthcare Life Sciences, UK) and a Milty Zerostat 3 anti-static gun were used to aid the transfer of hair samples between vessels. Total As determination by ICP-MS. Analysis was performed using an Agilent 7500cx series ICP-MS (Agilent Technologies, USA) tted with a MicroMist low-ow nebulizer (Glass Expan-sion, Australia) and an ASXpress rapid sample introduction system (Teledyne CETAC Technologies, USA) using previously reported42 operating conditions. Drinking water samples were analysed using a previously reported method.11 Rinse solutions were diluted 2 and analysed by a method used previously11 for water samples. Those with visible suspended particulate were passed through a 0.45 mm Acrodisc® syringe lter (PALL Life Sciences, USA). Toenail and hair digests were diluted 4 with 1% v/v HNO3 + 0.5% v/v HCl. Helium (He) collision cell mode was used to remove potential polyatomic interferences with the same mass/charge ratio as As (m/z 75). Signal driwas corrected using a Ge internal standard introduced via a T-piece. Analytical limits of detection (LOD) were calculated as 3 the standard deviation of run blanks for drinking water analysis and 3 the standard deviation of reagent blanks for toenail and hair anal-ysis. The LODs for As in drinking water and toenails/hair were 0.02 mg L1 and 10 mg kg1, respectively. Quality control. Toenail and hair samples of sufficient mass were chosen for duplicate analysis. Samples were milled to a ne powder using a 6850 Freezer Mill (SPEX Sample Prep, USA) – a cryogenic impact grinder cooled with liquid nitrogen. One pair of duplicates was digested per batch, in addition to 3 reagent blanks. Method accuracy was assessed using Certied Reference Materials (CRMs). Two samples (0.1 g) of NCS DC 73347 Hair (China National analysis Centre for iron and steel, Beijing, China) were digested per batch of hair and toenail samples. Two additional samples (0.1 g) of in-house human toenail reference material (BAPS2014 Human Toenail) were digested per batch of toenail samples. BAPS2014 was produced by pooling the toenail clippings, saved over a period of 2 years, of 2 male volunteers (aged 23 and 38) not knowingly exposed to substantial environmental or occupational As. A homogeneous powder was prepared using the cleaning and milling procedure already described prior to mixing end-over-end for several hours. The accuracy of drinking water and toenail washing solution measurements was assessed using NIST SRM 1643e Trace Elements in Water (National Institute of Standards and Technology, USA). Statistical analyses Statistical tests and plot production were performed using R version 3.0.0 (base package).43 Pearson correlation coefficients with signicance tests (p-values) and 95% condence intervals (C.I) were used to assess the strength in relationship between: initial versus follow-up drinking water As; well depth versus As concentration difference; rinse versus digest As concentrations and drinking water versus toenail/hair As concentrations. Wel-ch's independent unequal variance tests were used to test for differences in toenail, hair and rinse solution As concentrations between different subsets to account for unequal sample sizes. One-way analysis of variance (ANOVA) was used to test for differences in toenail and hair As concentrations between different age groups. Multiple linear regression models were constructed to assess signicant predictors of toenail and hair As in addition to drinking water As. Exploratory analyses revealed positively skewed distributions for drinking water, toenail and hair As concentrations and As concentrations in rinse solutions. To address this, natural log(ln) transformations were applied to these variables prior to Pearson correlations, Welch's tests, ANOVA and multiple regression modelling. For the same reason, geometric means (GM) were calculated instead of arithmetic means. Lecensoring was applied to hair As concentrations (n ¼ 8) below the analytical limit of detection (LOD) by replacing values with half of the LOD. 3. Results and discussion Study group The spatial extent of the study is presented in Fig. 1 and char-acteristics of households and volunteers are shown in Table 1. Two hundred and twelve volunteers from 129 households re-ported using their PWS for human consumption and provided either a toenail sample, hair sample or both. This made the present study the largest investigation of long-term exposure to As in drinking water in the UK to-date. Repeated water samples were available for comparison from 127 households, the majority of which were supplied by a borehole. The age distri-bution of the study group was not representative of the corre-sponding local rural population, with 63% of volunteers aged over 60. It is noted here that population-based exposure esti-mates were not the focus of the present paper. Nail polish usage was reported by 17 of the 206 (8%) volunteers who provided toenails, whereas polish was observed on the toenail samples of almost double that number (30, 15%). This underlines the importance of checking nails for visible polish prior to analysis and not relying on questionnaire data alone. Repeated drinking water measurements At the initial sampling phase, 125 out of 127 households (98%) had detectable (>0.02 mg L1) concentrations of total As measured in their drinking water. At follow-up sampling, 126 out of 127 (99%) households had As concentrations >0.02 mg L1. Only fourteen of the 127 households (11%) exceeded the 10 As mg L1 UK PCV and WHO guidance value at initial sampling with a maximum As concentration of 231 mg L1. Two households had borderline results (>9 As mg L1), one of which exceeded the PCV (17 As mg L1) at follow-up sampling. One further household, below PCV at initial sampling, exceeded at This journal is © The Royal Society of Chemistry 2016 Environ. Sci.: Processes Impacts, 2016, 18, 562–574 | 565 Paper Environmental Science: Processes & Impacts Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online follow-up (from 6 to 17 As mg L1). Only one exceedance drop-ped below PCV at follow-up sampling, from 14 to <1 As mg L1. Households who had high As concentrations in their PWS were advised to install appropriate remediation. Changes were not attributed to installation of treatment systems. Of three households that reported installation of any kind of treatment system between initial and follow-up measurements, none were among those exceeding the PCV and the impacts on As concentrations were minimal. Of the 14 households above PCV at initial sampling, 11 reported not installing any additional treatment and data were missing for the remaining three. This has important implications regarding risk awareness and the advice given to households above PCV. Overall, As concentrations in PWS were stable over both 8 and 31 month periods. Mean differences in As concentrations, initial and follow-up GM As concentrations and Pearson correlation coefficients between initial and follow-up As concentrations are shown in Table 2. Follow-up As concentra-tions are plotted against their initial counterparts in Fig. 2. In agreement with previous studies,17,18 strong Pearson correla-tions were observed between initial and follow-up samples collected both 8 (rp ¼ 0.95) and 31 (rp ¼ 0.95) months apart. A greater mean difference was observed for PWS with >10 As mg L1 due to the higher concentrations reported in this group. The strongest correlation observed was for the subset of households with both iron (Fe) and manganese (Mn) removal systems and pH buffering systems (rp ¼ 0.998) in addition to a lower mean difference to supplies with neither treatment system. This is not unexpected given that supplies with treat-ment systems installed are not subject to underlying geochemical variations. Although no household in this study group reported using As-specic treatment systems, Fe/Mn removal units have been reported to reduce As concentra-tions.11 Of the 62 households where borehole depth informa-tion was available, no signicant correlation was observed between depth and the difference in As concentration between initial and follow-up sampling. This is consistent with previous studies.17 Source type inuence was only assessed between well and borehole sources due to a limited number of other source types. There was no apparent difference in As concentration changes between well or borehole source types or system storage. An observation was made regarding the correct cate-gorisation of source type. One household in the present study reported using a borehole at initial sampling but on receiving initial results (80.5 As mg L1) it was discovered to be a disused mine adit (categorised as ‘other’ in Table 1). This highlights the importance of homeowners seeking the correct character-isation of their PWS when acquiring a new property. Fig. 1 Map of the Cornwall study area, shown in the context of the UK (excluding Northern Ireland), and the spatial distribution of sampled households. Total As concentrations measured in drinking water samples collected during the initial survey are plotted for reference. Note: no assessment of the spatial controls on As distribution was made in this study. Compiled using ESRI ArcMap 10.1. 566 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science: Processes & Impacts Paper Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online Toenail and hair total As Due to difficulties with sample collection and handling, many hair samples were of low mass at the point of digestion. This prompted the determination of a minimum mass requirement for toenail and hair samples by digesting triplicate samples of NCS DC 73347 CRM in decreasing mass increments (0.1, 0.08, 0.07, 0.06, 0.05, 0.04, 0.03, 0.02, 0.01, 0.005, 0.002 and 0.001 g). Measured As concentrations were plotted against mass of CRM (Fig. S2a†), in the context of the certied value and upper/lower limits. So too were mean recovery and relative standard devia-tion (RSD) (Fig. S2b†). On the basis of these results, 0.02 g was chosen as the minimum mass requirement, being the lowest mass at which As concentrations were found to be consistently within upper/lower certied limits of the CRM. This value is not universal and may not apply to other studies but was selected to try and maximise the usage of a compromised sample set. Depending on the amount of As in samples, requirements may be lower or higher. The RSD calculated for triplicates at lower masses may also reect reduced homogeneity of the CRM. Following the exclusion of samples below the minimum mass, As data were available for the toenails and hair of 200 and 104 volunteers, respectively. All toenail and 96 (92%) hair samples were above the 10 mg kg1 LOD. Arsenic measured in CRM NCS DC 73347 was 273  10 As mg kg1 (n ¼ 40), within the certied range of 280  50 As mg kg1, yielding a mean recovery of 98% with 5% precision. The mean As measured in BAPS 2014 Human Toenail was 93  5 As mg kg1 (n ¼ 20). The accuracy of BAPS 2014 measurements could not be assessed, but good precision (5% RSD) was maintained. The mean difference between duplicate digests was 1.1% (7 pairs) and 3.4% (6 pairs) for toenail and hair, respectively. Summary statistics for toenail and hair As concentrations are shown in Table 3 for different demographic and behavioural subsets. The GM toenail As concentration of all 200 volunteers was 151 As mg kg1 and ranged from 27 to 3354 As mg kg1. This falls within previously published ranges, with a higher GM and maximum concentration than a study23 conducted in New Hampshire, USA (GM: 90 As mg kg1; range: 10–810 As mg kg1), with comparable levels of drinking water exposure (<0.02–66 As mg L1). A previous study,40 conducted in south west England, reported a range of 858 to 25 981 As mg kg1 for individuals exposed to high As in soil, with no exposure to As in drinking water. Although conducted in the same geographic region as the present study, Button et al. (2009)40 investigated individuals living in the direct vicinity of a former As mine, possibly explaining the much higher reported concentrations than the present study. Hinwood et al. (2003)26 investigated the toenail As concentrations of volunteers in different exposure categories in rural Australia: high soil (>30 As mg kg1); high water (>10 As mg L1) and low exposure (<10 As mg L1 in drinking water and <30 As mg kg1 in soil). Overall, much higher toenail As concentra-tions were reported by Hinwood et al. (2003), across all cate-gories, than those in the present study. For example, the minimum toenail As concentration in the low exposure group was 1350 mg kg1, of which only eight volunteers exceeded in the present study. Quantication/removal of exogenous As from toenail samples was cited as a limitation by Hinwood et al. (2003) and, therefore, few meaningful conclusions can be drawn from this comparison. Slotnick et al. (2007)44 reported a lower drinking water As GM to the present study (0.59 versus 0.88 As mg L1) and a lower toenail As GM (70 versus 151 As mg kg1). Maximum drinking water and toenail as concentrations were also higher in the present study than those reported by Slotnick et al. (2007): 233 versus 99 As mg L1 and 3353 versus 1260 As mg kg1, respectively. Other comparable studies include Rivera-N´ u˜ nez Table 1 Household and study group characteristics Households 129 Initial and follow-up water sample, n (%)a 127 (98.4) Initial sample year, n (%) 2011 51 (40.2) 2013 76 (59.8) Source type, n (%) Borehole 111 (87.4) Well 11 (8.7) Spring capture 2 (1.6) Other 3 (2.4) Borehole depth reported, n (%) 62 (48.8) Mean borehole depth (m) 48 Treatment system, n (%) Fe/Mn removal 18 (14.2) pH buffering 60 (47.2) Storage (e.g. water tank) in system, n (%) Yes 62 (48.8) No 65 (51.2) Volunteers 212 Gender, n (%) Male 109 (51.4) Female 103 (48.6) Mean age, years (range) 62 (18–90) Age group, n (%) 18–29 6 (2.8) 30–39 3 (1.4) 40–49 28 (13.2) 50–59 42 (19.8) 60–69 75 (35.4) 70–79 44 (20.8) 80–90 14 (6.6) Smoking status, n (%) Currently smoking 13 (6.1) Not currently smoking 191 (90.1) Not reported 8 (3.8) Provided toenails, n (%) 206 (97.2) Provided hair, n (%) 186 (87.7) Provided both, n (%) 180 (84.9) Cosmetic usage, n (%) Polish usage reported (if toenails provided) 17 (8.3) Polish observed on toenails 30 (14.6) Dye usage reported (if hair provided) 31 (16.7) a Subsequent characteristics and percentages in households section refer to this subset. This journal is © The Royal Society of Chemistry 2016 Environ. Sci.: Processes Impacts, 2016, 18, 562–574 | 567 Paper Environmental Science: Processes & Impacts Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online et al. (2011)45 and Yu et al. (2014)24, with drinking water As GMs of 0.74 and 0.28 mg L1 and toenail As GMs of 90 and 57 mg kg1, respectively. Widespread As exposure, on the basis of both drinking water and toenail As concentrations, was low in the present study compared to those reported in severely affected areas. Nevertheless, 10 volunteers in the present study exhibited toenail As concentrations above the GM (1010 As mg kg1) re-ported by Kile et al. (2005)46 across three villages in Bangladesh – the world's worst affected region – with drinking water As concentrations between 1 and 752 As mg L1 (GM: 6.2 As mg L1). The GM hair concentration measured in the present study was 82 As mg kg1 (range: 500 mg kg1. Of these 40 individuals, 10 were exposed to >10 mg L1 of As in their drinking water. While it is not possible to conclude that these volunteers are either chronically or acutely exposed, where elevations correspond with drinking water As concentrations above PCV, attention is warranted. Welch's tests (Table 3) detected no signicant differences in toenail As between any subsets. Signicantly lower hair As concentrations were detected for females (p < 0.001) and Table 2 Drinking water As arithmetic mean differences, initial and follow-up As concentration geometric means (GM) and results from Pearson correlations between initial and follow-up As concentrations (ln transformed variables) for different PWS subsets Subsets n Mean difference (As mg L1) Initial total As GM (As mg L1) Follow-up total As GM (As mg L1) Pearson correlation (rp) All households 127 0.7 1.0 1.0 0.95 Initial sample year 2011 51 1.1 0.8 0.9 0.95 2013 76 0.5 1.2 1.2 0.95 Initial total As concentration <1 mg L1 67 0.1 0.2 0.3 0.87 1–10 mg L1 46 0.1 3.2 2.7 0.68 >10 mg L1 14 6.6 36.5 27.9 0.79 Source type Borehole 111 0.8 1.2 1.1 0.95 Well 11 0.4 0.3 0.4 0.97 Treatment system Fe/Mn removal only 12 0.2 1.7 1.6 0.95 pH buffering only 54 0.2 0.8 0.8 0.94 Both of above 6 0.3 0.5 0.5 1 (0.998) Neither of above 55 1.8 1.2 1.3 0.94 Storage (e.g. water tank) in system Yes 62 1.7 1.1 1.1 0.94 No 65 0.3 0.9 1.0 0.95 Fig. 2 Follow-up drinking water As concentrations plotted against initial counterparts. Pearson correlation coefficients (rp) are shown for measurements taken 31 (2011 initial collection) and 8 (2013 initial collection) months apart. 568 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science: Processes & Impacts Paper Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online volunteers who reported using hair dye (p ¼ 0.003). Signicantly higher hair As concentrations were detected for smokers (p ¼ 0.04). These ndings were compared with a previous study49 investigating demographic and behavioural controls on the composition of hair: Chojnacka et al. (2006) reported 150% more As in the hair of smokers, 210% more As in the hair of males and articially coloured hair was reported to contain 200% more As than naturally coloured hair.49 Exogenous As quantication Analysis of rinse solutions from the toenail washing procedure provided a useful insight into exogenous As contamination. The bar plot in Fig. 3 shows the hypothetical contribution of exog-enous As to that measured in toenails if they had not been washed. Rinse concentrations were normalised to the mass of toenail washed to allow comparison with digest concentrations. For toenails without polish, the GM As measured in initial rinse fractions was 9% of that measured in digested toenails, whereas the GM nal rinse fraction As concentration only accounted for 0.4%. Firstly, this conrmed the necessity of washing toenails, with a maximum percentage contribution of 716% in the case of one volunteer. Secondly, the low contribution from nal rinse fractions indicated the effective removal of exogenous As (maximum contribution: 5%). Furthermore, in agreement with previous ndings,40,50 the washing procedure appeared to have begun to leach endogenous As from toenails by the nal rinsing stage. This is indicated in Fig. 4, where no signicant Table 3 Summary statistics for total As in toenail and hair samples for different demographic and behavioural characteristic subsets of the study group. Statistically significant As concentrations between subsets are in bold type with p-values calculated by Welch's independent t-test on natural log transformed data in adjacent columns. Age group differences were assessed using one-way analysis of variance (ANOVA) n (toenails, hair) Toenail total As (mg kg1), GM (range) p-Value, Welch test (ANOVA for age groups) Hair total As (mg kg1), GM (range) p-Value for Welch's test (ANOVA for age groups) All 200, 104 151 (26.9–3354) — 82.6 (<LOD–2908) — Gender Male 102, 45 155 (26.9–1896) 0.63 150 (28.8–2908) <0.001 Female 98, 59 146 (39.1–3354) 52.5 (<LOD–756) Age group 18–39 6, 3 214 (8.1–1497) 0.28 89.9 (56.8–128) 0.76 (ANOVA) 40–49 27, 17 204 (57.9–3354) 121 (10.9–2396) 50–59 41, 20 154 (43–2578) 79.2 (<LOD–756) 60–69 74, 32 144 (39.1–1896) 67.4 (18.7–2908) 70–79 40, 24 135 (26.9–1982) 79.7 (11–742) 80–90 12, 8 111 (39.7–320) 100 (36.5–670) Smoking status Currently smoking 11,7 209 (100–2578) 0.25 324 (28.8–2908) 0.04 Not currently smoking 181,93 146 (26.9–1982) 74.6 (<LOD–2396) Nail polish usage Reported/observed 34 131 (44.6–1497) 0.34 — — Not reported/observed 166 155 (26.9–3354) — Hair dye usage Reported 20 — — 41.4 (10.8–756) 0.003 Not reported 84 — 97.4 (<LOD–2908) Fig. 3 Geometric mean (GM) As concentrations in toenail samples, initial and final rinse fractions for volunteers with and without observed/reported nail polish. Initial and final rinse fraction As concentrations as a percentage of the As measured in toenail digests are printed on plots. This journal is © The Royal Society of Chemistry 2016 Environ. Sci.: Processes Impacts, 2016, 18, 562–574 | 569 Paper Environmental Science: Processes & Impacts Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online correlation was observed (rp ¼ 0.05; p ¼ 0.43) between initial rinse As concentrations and toenail digest As concentrations (Fig. 4a). Conversely, a signicant positive correlation (rp ¼ 0.71; p < 0.001) was observed between nal rinse As concentrations and toenail digest As concentrations (Fig. 4b). The relatively small hypothetical contributions (5% maximum) of nal rinse As concentrations to those in toenail digests suggests that a small degree of leaching is of no great concern in the present study. It is noted that future efforts could be made to determine an optimum degree of washing for toenail samples and maxi-mise the removal of exogenous As whilst minimising endoge-nous As leaching. It is likely that the optimum number of rinses would depend on the level of contamination on the nail surface – a difficult metric to quantify. Welch's independent t-tests detected no signicant differ-ences in digest As concentrations (p ¼ 0.34), initial rinse As concentrations (p ¼ 0.85), nal rinse As concentrations (p ¼ 0.74) or percentage contributions from either initial (p ¼ 0.52) or nal (p ¼ 0.35) rinse fractions between samples with and without nail polish. This nding does not dismiss the effects of polish on sample concentrations, as substantial contributions have been demonstrated elsewhere.39 Several factors may have limited ndings on this occasion: misreporting of polish usage/ failure to identify polish on samples; ineffective polish removal during washing; low sample size of volunteers with polish and a lack of digestion procedure for rinse solutions/the inability to solubilise As present from polish. Contribution from polish has also been demonstrated39 as brand dependent and further work is needed to quantify/mitigate the effects of polish usage on biomonitoring studies using human nails as part of a wider review of the effects of surface contamination. Drinking water and biomarker relationships Due to the difference in duration between initial and follow-up drinking water samples, follow-up water samples (all of which were collected during the same sampling campaign as the hair and toenail collections) were used as explanatory variables of Fig. 4 Initial rinse fraction As concentrations (a) and final rinse fraction As concentrations (b) plotted against toenail digest As concentrations. No significant relationship (rp) was observed for initial rinse fractions, but a strong significant correlation was evident for final rinse fractions. This suggests (i) effective exogenous As contamination removal and (ii) subsequent leaching of As from toenails. Fig. 5 Significantly positive Pearson correlations (rp) between toenail (a) and hair (b) biomarker As concentrations and those measured in drinking water. 570 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science: Processes & Impacts Paper Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online biomarker As concentrations for consistency. In agreement with previous ndings,21,23,24,26 signicant positive correlations were observed between drinking water and toenail (Fig. 5a, rp ¼ 0.53; p < 0.001; 95% C.I: 0.43, 0.63) and drinking water and hair (Fig. 5b, rp ¼ 0.38; p < 0.001; 95% C.I: 0.20, 0.53) As concentrations. This conrmed previous ndings14 of human exposure to As from PWS but over a longer timescale. When grouped by drinking water As concentration (Table 4), strong signicant correlations were only observed where drinking water As was >10 mg L1 for both toenails and hair. Fig. 5a shows that, for volunteers exposed to drinking water with <10 As mg L1, a considerable number toenail samples con-tained notable As concentrations. Given the encouraging results from the assessment of the washing procedure, sample contamination was unlikely to account for these results. Cornwall is a region of elevated environmental As51 and, as noted previously by Button et al. (2009), alternative exposure routes, such as the ingestion of As-bearing soil and dust, are possible explanations for elevated toenail As where drinking water As is low.40 The investigation of additional exposure routes in the present study population will form the basis of further research. Fig. 5b depicts similar results for hair to those observed for toenails, albeit with a weaker correlation. Due to problems encountered with sample handling and the difficulty deter-mining the mass of hair washed, assessing the performance of washing was not possible for hair samples. Sample contami-nation cannot be ruled out as a possible explanation for this weaker correlation. Based on the results from Welch's t-tests, cigarette smoking might have accounted for elevated As in the hair of some individuals. Tobacco smoke has been demon-strated52 to cause elevated As in hair samples from non-occu-pationally exposed smokers and passive smokers. This pattern was not evident for toenail As concentrations, suggesting external contamination of hair from tobacco smoke among smokers as a possible explanation. Although statistically signicant, caution is advised when interpreting these results due to the small number of smokers in the present study group. Demographic, behavioural and dietary covariables Multiple linear regression was used to determine signicant predictors of toenail and hair As concentrations in addition to drinking water As. These included demographic, behavioural and dietary covariables. Data were stratied into two groups: volunteers with drinking water containing <1 As mg L1 (low) and $1 As mg L1 (high). This was to maintain consistency with previous studies45 that reported a greater predominance of additional, notably dietary, sources of As intake when drinking water concentrations were <1 mg L1. This stratication resulted in four initial models for toenail (Model 1a, 1b) and hair (Model 2a, 2b) As concentrations as a function of demographic and behavioural variables only. Coefficients for each model are shown in Table 5. There were no signicant demographic/behavioural predictors of toenail As in the low drinking water As group (Model 1a) but both increasing drinking water As and age resulted in a signicant increase in toenail As when As in drinking water was >1 mg L1. The effect of age on toenail As concentration has been reported by previous studies23 but in the opposite direction to the effect found in the present study. The mechanism of this relationship has not been elucidated. For example, Kile et al. (2005) note that toenail growth decreases with age. This may result in a higher concentration of As relative to a lower mass of nail. The high proportion of volunteers in older age groups in the present study may have limited the detection of a positive relationship on this occasion. Male gender had a signicant positive effect on hair As in the low drinking water group. Drinking water As, age, gender (male), dye usage and smoking were all signicantly positively associated hair As in the high drinking water group. Findings of the model for hair As in the high drinking water group com-plimented those of Welch's tests, namely the signicantly lower As concentrations in hair collected from females and those who reported using dye. The association with dye usage strength-ened with the omission of the gender term. Furthermore, with all but one volunteer reporting dye usage being female and 29% of hair providing volunteers being females that did not report dye usage, the apparent effect of dye implied by Welch's test was an indirect effect of gender. This would be consistent with previous ndings49,53 already discussed regarding lower As in the hair of females. Wolfsperger et al. (1994) attributed the higher As in male hair samples to smoking and a higher intake of seafood and wine than females.53 To test the inuence of food and drink items known to contain As, dietary terms were added to the abovementioned models. None of the dietary variables tested had a signicant effect on either toenail or hair As concentrations in the high drinking water group. In the low drinking water group, more servings of seafood per week resulted in a signicant increase in Table 4 Pearson correlations (rp) for drinking water As and toenail and hair As for different drinking water As concentration ranges. Moderate/ strong correlations (bold type) were only observed where drinking water As exceeded 10 mg L1 Pearson's rp (p-value, [95% C.I]) Drinking water As <1 mg L1 Drinking water As 1–10 mg L1 Drinking water As >10 mg L1 Full range Toenail total As 0.15 (p ¼ 0.13, [0.04, 0.33]) (n ¼ 107) 0.12 (p ¼ 0.32, [0.12, 0.34]) (n ¼ 73) 0.86 (p < 0.001, [0.66, 0.94]) (n ¼ 19) 0.53 (p < 0.001, [0.43, 0.63]) (n ¼ 199) Hair total As 0.11 (p ¼ 0.45, [0.18, 0.38]) (n ¼ 48) 0.15 (p ¼ 0.34, [0.16, 0.43]) (n ¼ 43) 0.62 (p ¼ 0.02, [0.10, 0.87]) (n ¼ 13) 0.38 (p < 0.001, [0.20, 0.53]) (n ¼ 104) This journal is © The Royal Society of Chemistry 2016 Environ. Sci.: Processes Impacts, 2016, 18, 562–574 | 571 Paper Environmental Science: Processes & Impacts Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online toenail As concentration. Specic varieties of seafood were not signicant. The model (Model 3) was re-performed with the omission of non-signicant covariables and the results are presented in Table 5. A negative association was observed between hair As concentrations and never eating home-grown vegetables. The results of this model (Model 4), with non-signicant covariables omitted, are presented in Table 5. The positive association between seafood consumption and toenail As concentrations and the negative association between home-grown veg consumption and hair As concentrations are of plausible validity. Although seafood derived arsenic species such as arsenobetaine are primarily excreted via urine,54 sea-food also contains arsenosugars and arsenolipids which are metabolised into methylarsonate and dimethylarsinate, both of which have been measured in small quantities in human toenails.40 In the present study, drinking water exposure was the primary focus of the investigation, hence, speciation analysis was not performed. On the basis of these ndings, future studies considering dietary sources in low drinking water exposure groups should consider speciation analysis to ensure meaningful results. The negative effect of not eating home-grown vegetables on hair As concentration is consistent with reported high soil As concentrations in the study region51 and, although values in local vegetables themselves have been found at relatively low concentrations,55 the ingestion of soil particles adhered to vegetables is a possible exposure pathway. 4. Conclusions This study is the largest investigation of long-term exposure to As in drinking water in the UK to-date and conrms the pres-ence of prolonged exposure to inorganic As from drinking water of householders with PWS in Cornwall, UK. The temporal stability of As concentrations in PWS suggests that, for this particular region, measurements of As taken in the present are strong predictors of past levels of exposure dating back at least 31 months. Arsenic concentrations measured in toenails and hair were useful in assessing prolonged exposure to As from PWS, in agreement with numerous previous studies. Analysis of washing solutions built on the ndings of Button et al. (2009)40 in that the washing procedure employed here was effective in removing exogenous contamination from a large sample set. Both toenail and hair biomarkers were susceptible to the inuence of covariables on As concentrations. Although useful Table 5 Predictors of toenail and hair As concentrations on the basis of multiple linear regression models. Significant coefficients are labelled with (), (), (), and (.) denoting significance to <0.001, <0.01, <0.05 and <0.1, respectively Model Terms b coefficient (signicance) 1a. ln(toenail As), drinking water <1 As mg L1 Intercept 5.309 () ln(drinking water As) 0.072 Age (continuous) 0.01 (.) Gender (male) 0.137 Nail polish usage (true) 0.268 Adjusted R2 ¼ 0.07 Smoking status (smoker) 0.31 1b. ln(toenail As), drinking water $1 As mg L1 Intercept 5.916 () ln(drinking water As) 0.469 () Age (continuous) 0.018 () Gender (male) 0.101 Nail polish usage (true) 0.157 Adjusted R2 ¼ 0.29 Smoking status (smoker) 0.005 2a. ln(hair As), drinking water <1 As mg L1 Intercept 2.646 () ln(drinking water As) 0.08 Age (continuous) 0.017 Gender (male) 0.826 () Dye usage (true) 0.159 Adjusted R2 ¼ 0.24 Smoking status (smoker) 0.77 2b. ln(hair As), drinking water $1 As mg L1 Intercept 5.349 () ln(drinking water As) 0.433 () Age (continuous) 0.025 () Gender (male) 0.810 () Dye usage (true) 0.76 (.) Adjusted R2 ¼ 0.42 Smoking status (smoker) 2.08 () 3. ln(toenail As), drinking water < 1 As mg L1 Intercept 4.662 () ln(drinking water As) 0.089 (.) Adjusted R2 ¼ 0.04 Seafood (continuous) 0.081 () 4. ln(hair As), drinking water < 1 As mg L1 Intercept 4.392 () ln(drinking water As) 0.213 () Gender (male) 0.905 () Home-grown vegetables (never) 0.975 () Home-grown vegetables (potted only) 0.546 Adjusted R2 ¼ 0.33 Home-grown vegetables (seasonally) 0.343 572 | Environ. Sci.: Processes Impacts, 2016, 18, 562–574 This journal is © The Royal Society of Chemistry 2016 Environmental Science: Processes & Impacts Paper Open Access Article. Published on 19 April 2016. Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online in assessing prolonged exposures to As from drinking water, other factors, such as diet, predominate where As concentra-tions in drinking water are low e.g. <1 mg L1. A large degree of variation in toenail and hair biomarkers was still unaccounted for in this study, with exposure to soil and dust highly possible explanations in a region of well-documented elevated environ-mental As. Investigation into the signicance of other exposure routes will be the focus of future research. Acknowledgements The authors are grateful for the contributions of Andrew Dunne and Dr Andrew Marriott during the eld work campaign, Dr Louise Ander for help with constructing the eld questionnaire database, Amanda Gardner for laboratory work and the efforts of Amy Rimell and Dr Mike Studden in logistical and manage-ment operations. Funding for this research was provided by the Natural Environment Research Council (NERC) via a University of Manchester/British Geological Survey (BGS) University Funding Initiative (BUFI) PhD studentship (Contract No. GA/ 125/017, BUFI Ref: S204.2) and the Centre for Environmental Geochemistry, BGS. 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Downloaded on 02/08/2016 12:58:56. This article is licensed under a Creative Commons Attribution 3.0 Unported Licence. View Article Online 29 M. J. Slotnick and J. O. Nriagu, Environ. Res., 2006, 102, 125– 139. 30 D. K. Harkins and A. S. Susten, Environ. Health Perspect., 2003, 111, 576. 31 J. Roberge, A. T. Abalos, J. M. Skinner, M. Kopplin and R. B. Harris, Am. J. Environ. Sci., 2009, 5, 688–694. 32 J. Xue, V. Zartarian, S.-W. Wang, S. V. Liu and P. Georgopoulos, Environ. Health Perspect., 2010, 118, 345. 33 K. L. Cottingham, R. Karimi, J. F. Gruber, M. S. Zens, V. Sayarath, C. L. Folt, T. Punshon, J. Morris and M. R. Karagas, Nutr. J., 2013, 12, 1. 34 P. Fleckman, in Nails: Therapy, Diagnosis, Surgery, ed. R. K. Scher and C. R. Daniel, WB Saunders Co, Philadelphia, 1997, ch. 4, p. 37. 35 M. R. Harkey, Forensic Sci. Int., 1993, 63, 9–18. 36 N. Orentreich, J. Markofsky and J. H. Vogelman, J. Invest. Dermatol., 1979, 73, 126–130. 37 I. M. Kempson and W. M. Skinner, Biol. Trace Elem. 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Epidemiol., 2011, 22, 182–190. 46 M. L. Kile, E. A. Houseman, E. Rodrigues, T. J. Smith, Q. Quamruzzaman, M. Rahman, G. Mahiuddin, L. Su and D. C. Christiani, Cancer Epidemiol., Biomarkers Prev., 2005, 14, 2419–2426. 47 D. F. Peach and D. W. Lane, Environ. Geochem. Health, 1998, 20, 231–237. 48 R. N. Ratnaike, Postgrad. Med. J., 2003, 79, 391–396. 49 K. Chojnacka, H. G´ orecka and H. G´ orecki, Environ. Toxicol. Pharmacol., 2006, 22, 52–57. 50 B. K. Mandal, Y. Ogra and K. T. Suzuki, Toxicol. Appl. Pharmacol., 2003, 189, 73–83. 51 P. Mitchell and D. Barr, Environ. Geochem. Health, 1995, 17, 57–82. 52 A. Saad and M. A. Hassanien, Environ. Int., 2001, 27, 471– 478. 53 M. Wolfsperger, G. Hauser, W. G¨ obler and C. Schlagenhaufen, Sci. Total Environ., 1994, 156, 235–242. 54 A. Navas-Acien, K. A. Francesconi, E. K. Silbergeld and E. Guallar, Environ. Res., 2011, 111, 110–118. 55 G. Norton, C. Deacon, A. Mestrot, J. Feldmann, P. Jenkins, C. Baskaran and A. A. Meharg, Environ. Sci. 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Products Business Account Denmark Germany Luxembourg Sweden United Kingdom Corporate Cards Card Payments Card Terminal Acquiring Agreement Pricing Currency Exchange Pricing Resources Blog Accounting Dictionary Company About us Partner Contact us Login Sign up English Home / Dictionary / Gdp Deflator Gdp Deflator The GDP deflator helps measure the real growth of an economy by adjusting for inflation. It distinguishes between nominal GDP (which includes price increases) and real GDP (which shows actual production). This tool is essential for tracking true economic progress and guiding policy decisions. Updated 25 Oct, 2024 10 min read Understanding the GDP Deflator and Its Impact on Economic Growth When people hear about a country’s GDP growing, it often sounds like good news, right? But there’s a catch: that growth might not be as real as it seems if inflation is left out of the picture. Nominal GDP, which is the value of goods and services at current prices, doesn’t account for inflation. This makes it hard to tell whether an economy is actually producing more or if things just cost more. Basically, inflation can make it look like a country is booming when, in fact, it’s just getting more expensive to live. This is where the GDP deflator comes in. It’s a tool that helps us figure out how much of that GDP growth is real by adjusting for inflation. It compares the prices of goods and services over different periods, separating out the price increases so we can see if the economy is truly growing. Why does this matter? For economists, this means they can track the real progress of a country’s economy. Businesses get a clearer view of the market, and policymakers—like those in government—use this information to adjust taxes, interest rates, and spending to keep the economy stable. Without the GDP deflator, we’d all be guessing at how much real progress we’re making. Definition of GDP Deflator The GDP deflator is a measure that tells us how much prices have changed compared to the amount of goods and services being produced. In simpler terms, it’s a way to adjust nominal GDP (which includes inflation) to get to real GDP, which shows the actual increase in production without inflation messing up the numbers. Here’s how it works: Nominal GDP measures everything at current prices, which means the number goes up if prices go up, even if we aren’t making more stuff. Real GDP, on the other hand, uses prices from a base year—like freezing prices in time—so it only counts the increase in goods and services. The GDP deflator helps compare these two figures, showing us how much of the GDP change is due to inflation. In short, while nominal GDP shows us today’s dollar value of production, real GDP strips out inflation, and the GDP deflator bridges the gap between the two, making it easier to see the true growth of an economy. The Formula and Calculation of the GDP Deflator The formula for the GDP deflator is pretty straightforward: GDP deflator = (Nominal GDP ÷ Real GDP) × 100 This might look a bit confusing at first, but let’s break it down. Nominal GDP is the total value of goods and services at today’s prices. Real GDP adjusts that figure by taking inflation into account, using prices from a base year instead of current prices. When you divide nominal GDP by real GDP, you see how much of the growth is due to rising prices. Understanding Nominal and Real GDP Nominal GDP is the number we often hear on the news—it’s the total value of everything produced, but it includes inflation. Real GDP is more accurate because it adjusts for price changes, so we can see how much is being produced without prices skewing the results. A Step-by-Step Example Here’s an example: Imagine a country’s nominal GDP is $1.2 trillion, but the real GDP is only $1 trillion. Using the formula: GDP deflator = ($1.2 trillion ÷ $1 trillion) × 100 = 120 This means that prices have risen by 20% since the base year. A GDP deflator above 100 means prices have increased since the base year (inflation). If it’s below 100, that means prices have actually fallen (deflation). When it’s exactly 100, there’s no change in price levels—just pure economic growth with no inflation impact. How the GDP Deflator Works The GDP deflator is like a wide-angle lens for inflation, capturing price changes for all goods and services produced in a country. It’s different from other inflation measures because it doesn’t stick to a fixed list of items. Instead, it adapts to what’s actually being produced in the economy, making it a more comprehensive tool for tracking inflation. Measuring Across All Sectors One of the key strengths of the GDP deflator is that it covers all sectors of the economy. It doesn’t matter if consumers are buying more electronics or fewer groceries—the deflator adjusts for all these changes, giving a broad picture of how prices are moving in different areas. Comparison with Other Inflation Measures While the GDP deflator is a solid inflation measure, it’s not the only one. Two other popular measures are the Consumer Price Index (CPI) and the Producer Price Index (PPI), but each of these has its own focus. GDP Deflator vs. PPI The PPI tracks the prices producers receive for their goods, focusing on the production side of the economy. It’s useful for understanding how costs are changing for businesses, but it doesn’t tell us much about the prices consumers pay. The GDP deflator covers both ends—what producers charge and what consumers pay—giving a fuller picture of inflation in the economy. By adjusting for these factors, the GDP deflator provides a well-rounded view of price changes, making it an essential tool for economists and policymakers alike. GDP Deflator vs. Consumer Price Index (CPI) The GDP deflator and the Consumer Price Index (CPI) are both important measures of inflation, but they have some key differences in how they track price changes. CPI’s Limitations The CPI measures the price changes of a fixed basket of goods and services that households typically buy, such as groceries, housing, and transportation. While this makes CPI a great tool for tracking changes in the cost of living for consumers, it has its limitations. Because the basket is fixed, CPI doesn’t capture shifts in consumer behavior or the introduction of new products and services. As a result, it can miss important changes in the economy that occur over time, like people switching from buying DVDs to streaming services. Why the GDP Deflator is More Comprehensive The GDP deflator, on the other hand, covers a much broader scope. It measures price changes for all goods and services produced in the economy, including investment and government spending, not just consumer goods. It also adjusts for changes in what’s being produced, so it reflects shifts in the economy more accurately. For instance, if people start buying more technology products, the deflator accounts for that, whereas CPI would not because it sticks to its fixed basket. When to Use CPI vs. the GDP Deflator While both are useful, the choice between CPI and the GDP deflator depends on what you’re trying to analyze. If you’re focused on consumer price changes, such as how inflation is impacting household budgets, CPI is the better measure. But if you want a more comprehensive view of inflation across the whole economy, including business investment and government spending, the GDP deflator is the more appropriate tool. The Importance of the GDP Deflator for Economic Analysis The GDP deflator is crucial for seeing the economy’s real growth because it strips out the confusing effects of inflation. Let’s say you look at a country’s GDP and notice it’s increasing year after year. At first glance, it seems like the economy is doing great. But if you only use nominal GDP, which doesn’t account for rising prices, you might be misled. Inflation can make it look like there’s more growth than there actually is. The GDP deflator adjusts for this, showing how much of that growth is real, not just a result of things getting more expensive. Impact on Policymaking Governments and central banks use the GDP deflator to fine-tune policies. For example, if the deflator shows that inflation is rising quickly, the central bank might increase interest rates to slow down spending and borrowing. On the flip side, if inflation is low, they might lower rates to stimulate the economy. By using this measure, policymakers get a clearer idea of how fast prices are increasing and how that’s affecting the economy. This helps them make decisions on everything from government spending to tax policies, ensuring the economy doesn’t overheat or shrink too much. The Dangers of Relying on Nominal GDP Alone If we only look at nominal GDP, we could be easily tricked into thinking an economy is doing better than it really is. For instance, if prices are rising fast, nominal GDP can give the illusion of growth when, in reality, the economy might be stagnant or even shrinking. The GDP deflator helps cut through that noise by adjusting for inflation, showing whether the economy is truly growing in real terms. Comparison Across Different Years Another great use of the GDP deflator is when we need to compare growth across different years. Since inflation varies year by year, comparing nominal GDP figures from one year to another isn’t always helpful. The GDP deflator lets us account for those price changes, showing us the actual growth rate in production. For example, if nominal GDP went up by 5%, but inflation was 3%, the deflator would show that real growth was only 2%. This makes it much easier to understand what’s really happening in the economy over time. GDP Deflator in Practice: Use Cases and Applications The GDP deflator isn’t just for economists—it’s used in a lot of real-world situations by businesses and governments. It helps people adjust payments, contracts, and forecasts to keep up with inflation, ensuring that they reflect the actual economic conditions. Use in Contracts and Business Agreements Businesses often use the GDP deflator to adjust long-term contracts. Imagine a construction company signing a five-year contract for a major project. If inflation increases significantly over that period, the company could end up paying much more than planned. By tying payments to the GDP deflator, businesses can ensure that future payments account for inflation, so they don’t lose money because of unexpected price hikes. How Governments Use It Governments also rely heavily on the GDP deflator to set national accounts and economic targets. For example, when they plan public spending or adjust tax policies, they use real GDP figures (adjusted by the deflator) to ensure that they’re making decisions based on true economic growth, not just price changes. This helps them allocate resources more effectively and avoid overestimating economic performance. International Comparison of Economies Another major use of the GDP deflator is for comparing economies across different countries. Each country experiences inflation at its own rate, so using nominal GDP alone can make some countries appear larger or smaller than they actually are. The GDP deflator helps level the playing field by adjusting for inflation, giving a more accurate comparison of economic growth between nations. This is especially important for global institutions like the IMF or World Bank when they track the health of the world economy. Real-World Example: Tracking the U.S. GDP Deflator To see how the GDP deflator works in real life, let’s take a look at the United States. The U.S. Bureau of Economic Analysis (BEA) regularly tracks this metric, giving us a good view of how inflation impacts the economy over time. For example, between 2023 and 2024, the GDP deflator showed that inflation had risen by about 3.1%. This data was collected by the BEA, which tracks inflation across all sectors, including housing, energy, and healthcare. Without the GDP deflator, it would be easy to think that the economy was growing much faster than it really was. But by factoring in inflation, we can see the real picture: while the nominal GDP might have looked impressive, much of that “growth” was simply due to rising prices. How Inflation Affected Growth in the U.S. Here’s a specific example: in 2023, the U.S. nominal GDP grew by about 5%, which sounds great at first. However, when we account for inflation using the GDP deflator, we see that the real growth rate was only 2%. In other words, inflation had eaten up 3% of that growth. This shows how the GDP deflator helps us avoid being misled by nominal GDP figures and gives us a more accurate sense of how much the economy is really growing in terms of production, not just prices. The Benefits of Using the GDP Deflator The GDP deflator offers several benefits when it comes to understanding an economy’s real performance, especially when inflation is involved. Better Insight into Real Economic Performance One of the main advantages of the GDP deflator is that it helps strip away the “inflation noise” that often distorts economic figures. By adjusting nominal GDP to reflect the true value of goods and services produced, the deflator gives us a clearer picture of how much the economy is actually growing. This way, we avoid overestimating growth that’s simply the result of rising prices. A Comprehensive Measure of Inflation Another benefit is that the GDP deflator is a more comprehensive measure of inflation compared to alternatives like CPI or PPI. While CPI focuses on consumer prices and PPI on producer prices, the GDP deflator covers all goods and services in the economy. This wide scope means it captures price changes in areas that CPI or PPI might miss, such as investment goods and government services. As a result, the GDP deflator is better at showing inflation’s overall impact on the economy. The Limitations of the GDP Deflator While the GDP deflator is a powerful tool, it does have its limitations. Data Lag and Complexity One drawback is that the GDP deflator is only calculated quarterly, which means it may not reflect inflation trends as quickly as more frequently updated measures like CPI, which is released monthly. Additionally, the deflator relies on a complex aggregation of data from different economic sectors, making it harder to calculate and interpret. This complexity can lead to data lags, meaning it might not always provide the most up-to-date view of inflation. Not Ideal for Short-term Inflation Tracking Because of its quarterly release and broad scope, the GDP deflator isn’t the best tool for tracking short-term inflation. If you’re trying to understand price changes over a few months, CPI or PPI would be more appropriate, as they focus on more immediate changes in specific markets. The deflator’s strength lies in its long-term view of inflation across the entire economy. Takeaway Note In summary, the GDP deflator is a vital tool for understanding true economic growth by adjusting for inflation. It offers a comprehensive view of price changes across all sectors of the economy, unlike more narrowly focused measures like CPI or PPI. While it may not be ideal for short-term inflation tracking, the deflator gives policymakers, businesses, and economists a clearer picture of long-term economic trends, helping them make better decisions for the future. FAQs What is the difference between GDP deflator and CPI? The GDP deflator measures inflation for all goods and services produced in an economy, while CPI tracks the price changes for a fixed basket of goods that consumers typically buy. The deflator covers a broader range, while CPI focuses mainly on consumer prices. How can we interpret the GDP deflator? A GDP deflator above 100 means prices have increased since the base year, showing inflation. If it’s below 100, it indicates deflation, meaning prices have dropped. A deflator exactly at 100 suggests no change in price levels. Is GDP deflator the same as real GDP? No, the GDP deflator is not the same as real GDP. The deflator is used to adjust nominal GDP to strip out inflation, revealing real GDP, which reflects the actual quantity of goods and services produced. How often is the GDP deflator calculated? The GDP deflator is calculated every quarter, based on the data released by national agencies like the Bureau of Economic Analysis (BEA) in the U.S. This means it updates less frequently than some other inflation measures like the CPI, which is calculated monthly. Why is the GDP deflator important for businesses? Businesses use the GDP deflator to adjust long-term contracts and payments. It helps ensure that their financial agreements account for inflation, so they aren’t caught off guard by rising prices over time. Using Too Much Flour Another common cause of tough pizza dough is using too much flour. Excess flour can make the dough dense and hard to work with, leading to a chewy or overly crisp crust. Incorrect Hydration Levels The water-to-flour ratio, or hydration level, is vital for the dough’s texture. Too little water can result in a dry dough, while too much water can make it sticky and hard to shape. Alisha Content Writer at OneMoneyWay Unlock Your Business Potential with OneMoneyWay Take your business to the next level with seamless global payments, local IBAN accounts, FX services, and more. Learn more Get Started Today Unlock Your Business Potential with OneMoneyWay OneMoneyWay is your passport to seamless global payments, secure transfers, and limitless opportunities for your businesses success. Open account
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https://artofproblemsolving.com/wiki/index.php/Chicken_McNugget_Theorem?srsltid=AfmBOorQluRk5kxGgwEUgWauNc6QS4LDuIbpZjf2zC1NfaTBn5IEDhGv
Art of Problem Solving Chicken McNugget Theorem - 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 Chicken McNugget Theorem Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Chicken McNugget Theorem The Chicken McNugget Theorem (or Postage Stamp Problem or Frobenius Coin Problem) states that for any two relatively primepositive integers, the greatest integer that cannot be written in the form for nonnegative integers is . A consequence of the theorem is that there are exactly positive integers which cannot be expressed in the form . The proof is based on the fact that in each pair of the form , exactly one element is expressible. Contents [hide] 1 Origins 2 Proof Without Words 3 Proof 1 4 Proof 2 5 Corollary 6 Generalization 7 Problems 7.1 Introductory 7.2 Intermediate 7.3 Olympiad 7.4 See Also Origins There are many stories surrounding the origin of the Chicken McNugget theorem. However, the most popular by far remains that of the Chicken McNugget. Originally, McDonald's sold its nuggets in packs of 9 and 20. Math enthusiasts were curious to find the largest number of nuggets that could not have been bought with these packs, thus creating the Chicken McNugget Theorem (the answer worked out to be 151 nuggets). The Chicken McNugget Theorem has also been called the Frobenius Coin Problem or the Frobenius Problem, after German mathematician Ferdinand Frobenius inquired about the largest amount of currency that could not have been made with certain types of coins. Proof Without Words Example using m= and n= Proof 1 Definition. An integer will be called purchasable if there exist nonnegative integers such that . We would like to prove that is the largest non-purchasable integer. We are required to show that: (1) is non-purchasable (2) Every is purchasable Note that all purchasable integers are nonnegative, thus the set of non-purchasable integers is nonempty. Lemma. Let be the set of solutions to . Then for any . Proof: By Bezout's Lemma, there exist integers such that . Then . Hence is nonempty. It is easy to check that for all . We now prove that there are no others. Suppose and are solutions to . Then implies . Since and are coprime and divides , divides and . Similarly . Let be integers such that and . Then implies We have the desired result. Lemma. For any integer , there exists unique such that . Proof: By the division algorithm, there exists one and only one such that . Lemma. is purchasable if and only if . Proof: If , then we may simply pick so is purchasable. If , then if and if , hence at least one coordinate of is negative for all . Thus is not purchasable. Thus the set of non-purchasable integers is . We would like to find the maximum of this set. Since both are positive, the maximum is achieved when and so that . Proof 2 We start with this statement taken from Proof 2 of Fermat's Little Theorem: "Let . Then, we claim that the set , consisting of the product of the elements of with , taken modulo , is simply a permutation of . In other words, Clearly none of the for are divisible by , so it suffices to show that all of the elements in are distinct. Suppose that for . Since , by the cancellation rule, that reduces to , which is a contradiction." Because and are coprime, we know that multiplying the residues of by simply permutes these residues. Each of these permuted residues is purchasable (using the definition from Proof 1), because, in the form , is and is the original residue. We now prove the following lemma. Lemma: For any nonnegative integer , is the least purchasable number . Proof: Any number that is less than and congruent to it can be represented in the form , where is a positive integer. If this is purchasable, we can say for some nonnegative integers . This can be rearranged into , which implies that is a multiple of (since ). We can say that for some positive integer , and substitute to get . Because , , and . We divide by to get . However, we defined to be a positive integer, and all positive integers are greater than or equal to . Therefore, we have a contradiction, and is the least purchasable number congruent to . This means that because is purchasable, every number that is greater than and congruent to it is also purchasable (because these numbers are in the form where ). Another result of this Lemma is that is the greatest number that is not purchasable. , so , which shows that is the greatest number in the form . Any number greater than this and congruent to some is purchasable, because that number is greater than . All numbers are congruent to some , and thus all numbers greater than are purchasable. Putting it all together, we can say that for any coprime and , is the greatest number not representable in the form for nonnegative integers . Corollary This corollary is based off of Proof 2, so it is necessary to read that proof before this corollary. We prove the following lemma. Lemma: For any integer , exactly one of the integers , is not purchasable. Proof: Because every number is congruent to some residue of permuted by , we can set for some . We can break this into two cases. Case 1: . This implies that is not purchasable, and that . is a permuted residue, and a result of the lemma in Proof 2 was that a permuted residue is the least number congruent to itself that is purchasable. Therefore, and , so is purchasable. Case 2: . This implies that is purchasable, and that . Again, because is the least number congruent to itself that is purchasable, and because and , is not purchasable. We now limit the values of to all integers , which limits the values of to . Because and are coprime, only one of them can be a multiple of . Therefore, , showing that is not an integer and that and are integers. We can now set limits that are equivalent to the previous on the values of and so that they cover all integers form to without overlap: and . There are values of , and each is paired with a value of , so we can make different ordered pairs of the form . The coordinates of these ordered pairs cover all integers from to inclusive, and each contains exactly one not-purchasable integer, so that means that there are different not-purchasable integers from to . All integers greater than are purchasable, so that means there are a total of integers that are not purchasable. In other words, for every pair of coprime integers , there are exactly nonnegative integers that cannot be represented in the form for nonnegative integers . Generalization If and are not relatively prime, then we can simply rearrange into the form and are relatively prime, so we apply Chicken McNugget to find a bound We can simply multiply back into the bound to get Therefore, all multiples of greater than are representable in the form for some non-negative integers . Problems Introductory Marcy buys paint jars in containers of and . What's the largest number of paint jars that Marcy can't obtain? Answer: containers Bay Area Rapid food sells chicken nuggets. You can buy packages of or . What is the largest integer such that there is no way to buy exactly nuggets? Can you Generalize? (Source: The Art and Craft of Problem Solving) Answer: If a game of American Football has only scores of field goals ( points) and touchdowns with the extra point ( points), then what is the greatest score that cannot be the score of a team in this football game (ignoring time constraints)? Answer: points The town of Hamlet has people for each horse, sheep for each cow, and ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet? (Source: AMC 10B 2015 Problem 15) Answer: In the state of Coinland, coins have values and cents. Suppose is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of (Source: 2023 AMC 12B Problems/Problem 16) Answer: Intermediate Ninety-four bricks, each measuring are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contributes or or to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks? (Source: AIME) Find the sum of all positive integers such that, given an unlimited supply of stamps of denominations and cents, cents is the greatest postage that cannot be formed. (Source: AIME II 2019 Problem 14) Olympiad On the real number line, paint red all points that correspond to integers of the form , where and are positive integers. Paint the remaining integer points blue. Find a point on the line such that, for every integer point , the reflection of with respect to is an integer point of a different color than . (Source: India TST) See Also Theorem Prime Retrieved from " Categories: Theorems Number theory 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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https://math.stackexchange.com/questions/4724413/showing-that-sum-x-1-infty-frace-fractx2x2-approx-frac?noredirect=1
calculus - Showing that $\sum _{x=1}^{\infty } \frac{e^{-\frac{t}{x^2}}}{x^2}\approx \frac{\sqrt{\pi } \text{erf}\left(\sqrt{t}\right)}{2 \sqrt{t}}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Showing that ∑∞x=1 e−t x 2 x 2≈π√erf(t√)2 t√∑x=1∞e−t x 2 x 2≈π erf(t)2 t Ask Question Asked 2 years, 3 months ago Modified2 years, 3 months ago Viewed 151 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. What is an easy way to show that the following approximation is valid for large t t? ∑k=1∞e−t k 2 k 2≈π−−√erf(t√)2 t√∑k=1∞e−t k 2 k 2≈π erf(t)2 t Notebook calculus sequences-and-series Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 24, 2023 at 2:08 Yaroslav BulatovYaroslav Bulatov asked Jun 23, 2023 at 22:22 Yaroslav BulatovYaroslav Bulatov 5,589 2 2 gold badges 28 28 silver badges 60 60 bronze badges 2 1 Is it not an equality? Plus what does your graph represents exactly (I suppose you have summed for x<n x<n but which n n), otherwise, I would try to derivate... I have a request: could you change x x by k k or n n ?julio_es_sui_glace –julio_es_sui_glace 2023-06-23 22:29:42 +00:00 Commented Jun 23, 2023 at 22:29 1 Please, note that erf(t√)=1+O(e−t t√)erf(t)=1+O(e−t t) at t→∞t→∞. If you are not planning to keep exponentially small corrections, you can use the Euler-Macklaurin formula: ∑k=1∞1 k 2 e−t k 2∼∫∞1 1 k 2 e−t k 2 d k=∫1 0 e−t x 2 d x=π−−√2 t√+O(e−t t)∑k=1∞1 k 2 e−t k 2∼∫1∞1 k 2 e−t k 2 d k=∫0 1 e−t x 2 d x=π 2 t+O(e−t t) Switching from summation to integration we dropped exponentially small corrections. Therefore, with the same level of accuracy erf(t√)=1+O(e−t t√)erf(t)=1+O(e−t t)Svyatoslav –Svyatoslav 2023-06-24 03:01:09 +00:00 Commented Jun 24, 2023 at 3:01 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 7 Save this answer. Show activity on this post. We temporarily fix t t and consider the function g(y)=y e−t y g(y)=y e−t y. Then g′(y)=(1−t y)e−t y,g′(y)=(1−t y)e−t y, hence the maximum of g(y)g(y) is achieved at y=1 t y=1 t with the maximum value of g(1 t)=1 e t g(1 t)=1 e t. This also shows that g(y)g(y) is unimodal on [0,∞][0,∞] (with the convention that g(∞)=0 g(∞)=0), and the same is true for x↦g(1 x 2)x↦g(1 x 2). Using this, it is not hard to show that ∣∣∣∑k=1∞g(1 k 2)−∫∞0 g(1 x 2)d x∣∣∣≤2 max y∈[0,∞]g(y)=2 e t.|∑k=1∞g(1 k 2)−∫0∞g(1 x 2)d x|≤2 max y∈[0,∞]g(y)=2 e t. (As a hint, recall how the integral test is proved geometrically.) So, ∑k=1∞e−t/k 2 k 2=∫∞0 e−t/x 2 x 2 d x+O(t−1)=∫∞0 e−t y 2 d x+O(t−1)=π−−√2 t√+O(t−1).(y=1/x)∑k=1∞e−t/k 2 k 2=∫0∞e−t/x 2 x 2 d x+O(t−1)(y=1/x)=∫0∞e−t y 2 d x+O(t−1)=π 2 t+O(t−1). Now, using the fact that erf(t√)=1+O(e−t t√)erf⁡(t)=1+O(e−t t) (see DLMF 7.12.1, for example), we can absorb the exponentially small error term to a larger term and obtain ∑k=1∞e−t/k 2 k 2=π−−√erf(t√)2 t√+O(t−1).∑k=1∞e−t/k 2 k 2=π erf⁡(t)2 t+O(t−1). Alternatively, Euler–Maclaurin formula shows that, for any positive integer p p, ∣∣∣∑k=1∞e−t/k 2 k 2−π−−√2 t√∣∣∣≤2 ζ(p)(2 π)p∫∞0∣∣∣∂p∂x p(e−t/x 2 x 2)∣∣∣d x=2 ζ(p)(2 π)p t(p+1)/2∫∞0∣∣∣∂p∂s p(e−1/s 2 s 2)∣∣∣d s.|∑k=1∞e−t/k 2 k 2−π 2 t|≤2 ζ(p)(2 π)p∫0∞|∂p∂x p(e−t/x 2 x 2)|d x=2 ζ(p)(2 π)p t(p+1)/2∫0∞|∂p∂s p(e−1/s 2 s 2)|d s. This in fact shows that ∑k=1∞e−t/k 2 k 2=π−−√erf(t√)2 t√+O(t−p)∑k=1∞e−t/k 2 k 2=π erf⁡(t)2 t+O(t−p) for any fixed p p, hinting we actually have a better error bound. Indeed, a numerical simulation suggests that the error term is of the form O(e−C t√)O(e−C t) but I have no idea whether this is actually true or not. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 24, 2023 at 4:06 answered Jun 24, 2023 at 3:13 Sangchul LeeSangchul Lee 184k 17 17 gold badges 300 300 silver badges 493 493 bronze badges 2 1 Mr. Sangchul Lee, I got for the correction terms ∑k=1∞e−t/k 2 k 2=π−−√erf(t√)2 t√+4 π 3 t−−−√e−3 2 π 2/3 t 1/3 cos(3 3–√2 π 2/3 t 1/3)+O(1 t√e−3 2(2 π)2/3 t 1/3)∑k=1∞e−t/k 2 k 2=π erf⁡(t)2 t+4 π 3 t e−3 2 π 2/3 t 1/3 cos⁡(3 3 2 π 2/3 t 1/3)+O(1 t e−3 2(2 π)2/3 t 1/3) Svyatoslav –Svyatoslav 2025-06-07 15:00:12 +00:00 Commented Jun 7 at 15:00 The evaluation is pretty similar to this one (even more straightforward), and uses the Poisson summation formula. : math.stackexchange.com/questions/5072205/…Svyatoslav –Svyatoslav 2025-06-07 15:00:39 +00:00 Commented Jun 7 at 15:00 Add a comment| This answer is useful 3 Save this answer. Show activity on this post. S n=∑k=1 n e−t k 2 k 2 S n=∑k=1 n e−t k 2 k 2 Using Euler-MacLaurin summation S n=π−−√erf(t√)2 t√+e−t P(t)−1 n+1 2 n 2+O(1 n 3)S n=π erf(t)2 t+e−t P(t)−1 n+1 2 n 2+O(1 n 3) For this level of expansion P(t)=808 1155−9397 t 6930+3461 t 2 693−94579 t 3 13860+340553 t 4 83160−P(t)=808 1155−9397 t 6930+3461 t 2 693−94579 t 3 13860+340553 t 4 83160− 1021543 t 5 831600+5779 t 6 29700−11 t 7 675+t 8 1485−t 9 93555 1021543 t 5 831600+5779 t 6 29700−11 t 7 675+t 8 1485−t 9 93555 Just a few numbers ⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜t 1 2 3 4 5 6 7 8 9 10 e−t P(t)+0.205872+0.016369+0.047570+0.004640−0.028220−0.016550+0.003639+0.010756+0.007377+0.001856⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟(t e−t P(t)1+0.205872 2+0.016369 3+0.047570 4+0.004640 5−0.028220 6−0.016550 7+0.003639 8+0.010756 9+0.007377 10+0.001856) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jun 24, 2023 at 6:34 answered Jun 24, 2023 at 5:33 Claude LeiboviciClaude Leibovici 294k 55 55 gold badges 130 130 silver badges 316 316 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 calculus sequences-and-series 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 Linked 9Show ∑∞i=1 i−2(1−i−2)n≈1 2 B(1+n,1 2)∑i=1∞i−2(1−i−2)n≈1 2 B(1+n,1 2) 2Expressions for f(z)=∑∞k=1 k−2 z k−2 f(z)=∑k=1∞k−2 z k−2? Related 7Evaluating ∫∞−∞x exp(−b 2(x−c)2)e r f 2(a(x−d))d x∫−∞∞x exp⁡(−b 2(x−c)2)e r f 2(a(x−d))d x 5Prove ∑∞n=1 n!(2 n)!=1 2 e 1/4 π−−√erf(1 2)∑n=1∞n!(2 n)!=1 2 e 1/4 π erf(1 2) 9Prove 1+1 3(1+1 5(1+1 7(1+…)))=π e 2−−√erf(1 2√)1+1 3(1+1 5(1+1 7(1+…)))=π e 2 erf(1 2) 1Show: lim y→+∞[1 π√e y 2(1−erf(y))−y]=0.lim y→+∞[1 π e y 2(1−erf(y))−y]=0. 7Evaluating ∫e−a x 2 erf(b x+c)d x∫e−a x 2 erf⁡(b x+c)d x 9Evaluation of ∑N(ker(x)+i kei(x))=∑1∞K 0(i√x)=0.133691…−0.7256312…i∑N(ker(x)+i kei(x))=∑1∞K 0(i x)=0.133691…−0.7256312…i? 5Showing f(t)=Tr[A(I−A)2 t]≈π/2√erfc(2 t√/d)2 t√f(t)=Tr[A(I−A)2 t]≈π/2 erfc(2 t/d)2 t, where A=diag(1,1/4,1/9,…,1/d 2)A=diag(1,1/4,1/9,…,1/d 2) 0Simplifying ∫∞1 e−2 t u d 2 u√d u∫1∞e−2 t u d 2 u d u 9Show ∑∞i=1 i−2(1−i−2)n≈1 2 B(1+n,1 2)∑i=1∞i−2(1−i−2)n≈1 2 B(1+n,1 2) Hot Network Questions Languages in the former Yugoslavia Direct train from Rotterdam to Lille Europe RTC battery and VCC switching circuit Matthew 24:5 Many will come in my name! 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Tools & Reference>Infectious Diseases Cryptosporidiosis Workup Updated: Jan 07, 2025 Author: Ewa Tomczak, MD; Chief Editor: Michael Stuart Bronze, MD more...;) Share Print Feedback Close Facebook Twitter LinkedIn WhatsApp Email Sections Cryptosporidiosis Sections Cryptosporidiosis Overview Practice Essentials Background Etiology and Pathophysiology Epidemiology Prognosis Morbidity and Mortality Patient Education Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Stool Tests Evaluation of Immune Function Abdominal Ultrasonography and ERCP Biopsy and Lavage Histologic Findings Show All Treatment Approach Considerations Antiparasitic Therapy Symptomatic Therapy Fluid and electrolyte loss Prevention of Cryptosporidiosis Show All Medication Medication Summary Antiparasitics Antidiarrheal Agents Somatostatin Analogues Show All Media Gallery;) References;) Workup Approach Considerations Studies in North America have documented cryptosporidiosis in approximately 7% of stool specimens sent for parasitologic examination, whereas, averall, about 2-27% of stool specimens submitted for parasitologic studies in developing countries reveal Cryptosporidium oocysts varying on the basis of the method of diagnosis. [5, 60, 61] Cryptosporidium can be difficult to diagnose by standard methods; usually it was missed unless specific tests ere performed. Traditionally it was diagnosed by microscopic examination with special staining techniques (eg, acid-fast staining, direct fluorescent antifody [DFA], enzyme immunoassays, or immunochromatographic tests for detection of Cryptosporidium species' antigens) (See the image below.) PCR multiplex molecular tests have widely increased the sensitivity and specificity of their diagnoses. [57, 63] Modified acid-fast stain of stool shows red oocysts of Cryptosporidium parvum against the blue background of coliforms and debris. View Media Gallery) Urea, electrolytes, and liver function tests Diarrhea causes volume depletion, therefore urea and electrolyte tests are used to assess electrolyte and volume replacement requirements. Elevated alkaline phosphatase and glutamyl transpeptidase without hyperbilirubinemia are typical signs of biliary involvement. [1, 64] Imaging studies Imaging studies are not indicated as a first-line diagnostic approach in cryptosporidiosis but can aid in the diagnosis. Ultrasound can show dilation of intrahepatic or extrahepatic ducts followed by dilated common bile ducts. Studies have shown that if abnormal US, there is a good correlation with abnormal ERCP. Abdominal radiography and computed tomography (CT) scanning are nonspecific but may reveal distended loops of bowel, air-fluid levels, and disrupted bowel motility. When indicated, as guided by symptoms, ultrasonography or CT scanning may reveal an enlarged gallbladder with a thickened wall, dilated or irregular intrahepatic and extrahepatic biliary ducts, and a normal or stenotic distal common bile duct. Cholangiography may reveal beading of the common bile duct or papillary stenosis. In cases of respiratory involvement, chest radiography is unremarkable, with modest infiltrates or increased bronchial markings. [50, 51] Next: Stool Tests Processing Unconcentrated fresh specimens can be examined by wet mount preparations, but this is not recommended due to high infectivity and its unreliability. Despite different methods available, formalin ethyl acetate concentration method is the most sensitive and widely used. Optimal centrifugation time and speed are critical for concentrating Cryptosporidium oocysts. Commercial fecal concentration tubes are available that decrease processing time and supplies needed for concentrating specimens (eg, Fecal Parasite Concentrator, Evergreen Scientific). Polyvinyl alcohol (PVA)-preserved specimens are not acceptable for modified acid-fast staining or antigen-detection assays for detection of Cryptosporidium. Immunomagnetic separation commercial kits are also available mainly for waste water screening. Types of tests Since the oocyst is small (4-6 um), it requires staining to be identified by light microscopy. Modified acid-fast staining procedure is useful for the identification of oocysts from Cryptosporidium (which may be difficult to detect with routine stains, such as trichrome). Cryptosporidium species stain a pinkish-red color on a uniformly green background. Unlike the modified Ziehl-Neelsen acid-fast stain, this stain does not require the heating of reagents for staining (See the images below.). Other studies have found fluorogenic stain auramine-phenol a more sensitive and faster option which have been adopted by many labs as their standard staining method. [62, 68] Cryptosporidium parvum oocysts revealed with modified acid-fast stain. Against a blue-green background, the oocysts stand out with a bright red stain. Image courtesy of CDC DPDx parasite image library. View Media Gallery) Cryptosporidium oocysts revealed with modified acid-fast stain. View Media Gallery) The gold standard for stool examination is the immunofluorescence assay, which is based on oocyst cell wall antigens targeted by specific fluorescent monoclonal antibodies. The main disadvantage of this method is the inability to process a large amount of samples and the need for a specialized microscope and technician. Antigen detections via enzyme-linked immunosorbent assays (ELISA) or enzyme immunoassay (EIA) also were developed with variable sensitivities and great specificities depending on the commercial kit used, which limits its use in epidemiologic studies. Various PCR-based commercial multiplex molecular assays to detect Cryptosporidium spp have been approved by the FDA. These tests are able to detect a variety of pathogens that cause diarrhea such as parasites, bacteria, and viruses. They are costly, highly sensitive, and require careful clinical correlation; therefore, they are not readily available in all laboratories. [38, 63] Specimen examination Concentrated sediment of fresh (within 30 min after passage of stools) or formalin-preserved stool may be used. Other types of clinical specimens, such as duodenal fluid, bile, and pulmonary samples (induced sputum, bronchial wash, biopsies) also may be stained. The formalin ethyl acetate method is used to concentrate stool before staining with a modified acid-fast stain, because routine laboratory examination of stool for ova and parasites does not detect Cryptosporidium. This technique allows for differentiation from fecal debris or yeast which will stain blue or green, versus oocysts that will counterstain pink or red. Careful examination of slides is imperative, as oocysts can easily be missed. Because shedding may be intermittent, examine at least 3 stool specimens collected on separate days before considering the test results negative. Fecal leukocytes are not found in stool specimens, because invasion does not occur below the epithelial layer of the mucosa. Other testing strategies include the following: GI biopsy specimens can be used instead of stool specimens; a high concentration of oocysts is seen in the jejunum. Electron microscopy of stool or biopsy specimens can also be performed for direct visualization of oocysts. PCR and immunohistochemistry can be used on tissue samples, which spares the need for staining. Serologic detection of specific anti-Cryptosporidium antibodies primarily is used as a research or epidemiologic tool. Previous Next: Evaluation of Immune Function Lymphocyte subset analysis CD4+ lymphocyte counts predict the duration of disease in patients infected with HIV. When the counts are greater than 150 cells/μL, diarrhea is likely to resolve spontaneously. With lower counts, however, the diarrhea may be chronic. Counts typically are less than 50 cells/μL in patients with either biliary involvement or cholera-like syndromes. HIV testing Prolonged diarrhea caused by cryptosporidiosis may warrant HIV testing. Primary immunodeficiencies Children with chronic diarrhea from cryptosporidiosis should be screened for primary immunodeficiencies associated with depressed cellular immune function. The most commonly identified immunodeficiency is hyper-IgM syndrome, which can be identified by antibody screening. T-cell deficiencies can be identified by examining lymphocyte numbers and subsets. Secondary immunodeficiencies Patients who are on immunosupressive medications in the setting of organ transplantation behave similarly to HIV patients. Immunosupression should be minimized and levels of tacrolimus and cyclosporine should be checked to avoid toxicity. Previous Next: Abdominal Ultrasonography and ERCP Dilated or irregular intrahepatic and extrahepatic bile ducts, along with a thickened gallbladder, as detected with abdominal ultrasonography, indicate biliary involvement. Magnetic resonance cholangiopancreatography (MRCP) can detect ductal abnormalities including intrahepatic strictures, papillary stenosis, and long segmental extrahepatic strictures. Endoscopic retrograde cholangiopancreatography (ERCP) often is needed to diagnose sclerosing cholangitis or papillary stenosis. ERCP identification of Cryptosporidium oocysts in bile or intracellular forms on biopsy confirms the diagnosis of biliary cryptosporidiosis. Papillary stenosis may be present and responds symptomatically to endoscopic sphincterotomy, often with stent placement. Previous Next: Biopsy and Lavage GI or liver biopsy GI or liver biopsy may be indicated in cases of diagnostic uncertainty. Different parts of the intestinal tract may be affected. Liver biopsy findings may reveal the organism attached to bile duct epithelial cells. Concurrent infection with cytomegalovirus (CMV), Enterobacter cloacae, and microsporidia is common. In the setting of concomitant inflammatory bowel disease with persistent symptoms, gastrointestinal biopsies have incidentally diagnosed cryptosporidium. Bronchoalveolar lavage and lung biopsy In patients with related symptoms, bronchoscopy may reveal the parasite in lavage fluid, in brushing specimens, and in biopsy specimens, attached to the surface of bronchial mucosal cells or in macrophages. Previous Next: Histologic Findings Histologic examination of the small intestine is not required to confirm the diagnosis of cryptosporidiosis, although the small intestine does show the parasite projecting from the brush border of the mucosal surface. Parasites also may be identified in bile or biliary tract biopsies. Villous atrophy with blunting, epithelial flattening, and an increase in lamina propria lymphocytes are seen in patients with persistent cryptosporidiosis. In patients with heavier infection, crypt hyperplasia and marked infiltration with lymphocytes, plasma cells, and neutrophils also are noted. Previous Treatment & Management References White AC Jr. Cryptosporidiosis (Cryptosporidium species). Bennett JE, Dolin R, Blaser MK, eds.Principles and Practice of Infectious Diseases. Philadelphia, Pa: Elsevier Inc; 2020. 3410-3420. Checkley W, White AC Jr, Jaganath D, Arrowood MJ, Chalmers RM, Chen XM, et al. A review of the global burden, novel diagnostics, therapeutics, and vaccine targets for cryptosporidium. Lancet Infect Dis. 2015 Jan. 15 (1):85-94. [QxMD MEDLINE Link]. Ryan UM, Feng Y, Fayer R, Xiao L. Taxonomy and molecular epidemiology of Cryptosporidium and Giardia - a 50 year perspective (1971-2021). Int J Parasitol. 2021 Dec. 51 (13-14):1099-1119. [QxMD MEDLINE Link].[Full Text]. 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Cryptosporidiosis in a Zoonotic Gastrointestinal Disorder Perspective: Present Status, Risk Factors, Pathophysiology, and Treatment, Particularly in Immunocompromised Patients. J Trop Med. 2024. 2024:6439375. [QxMD MEDLINE Link].[Full Text]. Cohn IS, Henrickson SE, Striepen B, Hunter CA. Immunity to Cryptosporidium: Lessons from Acquired and Primary Immunodeficiencies. J Immunol. 2022 Dec 15. 209 (12):2261-2268. [QxMD MEDLINE Link].[Full Text]. Berendes DM, Omore R, Prentice-Mott G, et al. Exploring Survey-Based Water, Sanitation, and Animal Associations With Enteric Pathogen Carriage: Comparing Results in a Cohort of Cases With Moderate-to-Severe Diarrhea to Those in Controls in the Vaccine Impact on Diarrhea in Africa (VIDA) Study, 2015-2018. Clin Infect Dis. 2023 Apr 19. 76 (76 Suppl1):S140-S152. [QxMD MEDLINE Link].[Full Text]. CDC. Annual reported cases of notifiable diseases, by region and reporting area, United States, U.S. Territories, and Non-U.S. Residents, 2022. Nationally Notifiable Infectious Diseases and Conditions, United States: Annual Tables. Centers for Disease Control and Prevention. Available at 2022; Accessed: December 16, 2024. Becker DJ, Oloya J, Ezeamama AE. Household Socioeconomic and Demographic Correlates of Cryptosporidium Seropositivity in the United States. PLoS Negl Trop Dis. 2015. 9 (9):e0004080. [QxMD MEDLINE Link].[Full Text]. Mac Kenzie WR, Hoxie NJ, Proctor ME, Gradus MS, Blair KA, Peterson DE, et al. A massive outbreak in Milwaukee of cryptosporidium infection transmitted through the public water supply. N Engl J Med. 1994 Jul 21. 331 (3):161-7. [QxMD MEDLINE Link].[Full Text]. Adler S, Widerström M, Lindh J, Lilja M. Symptoms and risk factors of Cryptosporidium hominis infection in children: data from a large waterborne outbreak in Sweden. Parasitol Res. 2017 Oct. 116 (10):2613-2618. [QxMD MEDLINE Link].[Full Text]. Franceschelli A, Bonadonna L, Cacciò SM, et al. 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Carter BL, Stiff RE, Elwin K, Hutchings HA, Mason BW, Davies AP, et al. Health sequelae of human cryptosporidiosis-a 12-month prospective follow-up study. Eur J Clin Microbiol Infect Dis. 2019 Sep. 38 (9):1709-1717. [QxMD MEDLINE Link].[Full Text]. Guerrant DI, Moore SR, Lima AA, Patrick PD, Schorling JB, Guerrant RL. Association of early childhood diarrhea and cryptosporidiosis with impaired physical fitness and cognitive function four-seven years later in a poor urban community in northeast Brazil. Am J Trop Med Hyg. 1999 Nov. 61 (5):707-13. [QxMD MEDLINE Link].[Full Text]. Kriger O, Gefen-Halevi S, Belausov N, Sherman G, Adam E, Rubinstein O, et al. Respiratory cryptosporidiosis detected by commercial multiplex-PCR in immunosuppressed pediatric patients. Diagn Microbiol Infect Dis. 2023 Oct. 107 (2):116033. [QxMD MEDLINE Link].[Full Text]. Mor SM, Ndeezi G, Ascolillo LR, Tasimwa HB, Attipa C, Sponseller J, et al. 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N Engl J Med. 1994 Jul 21. 331 (3):161-7. [QxMD MEDLINE Link].[Full Text]. Sjöström M, Arvidsson M, Söderström L, Lilja M, Lindh J, Widerström M. Outbreak of Cryptosporidium hominis in northern Sweden: persisting symptoms in a 5-year follow-up. Parasitol Res. 2022 Jul. 121 (7):2043-2049. [QxMD MEDLINE Link].[Full Text]. Utami WS, Murhandarwati EH, Artama WT, Kusnanto H. Cryptosporidium Infection Increases the Risk for Chronic Diarrhea Among People Living With HIV in Southeast Asia: A Systematic Review and Meta-Analysis. Asia Pac J Public Health. 2020 Jan. 32 (1):8-18. [QxMD MEDLINE Link].[Full Text]. Naseer M, Dailey FE, Juboori AA, Samiullah S, Tahan V. Epidemiology, determinants, and management of AIDS cholangiopathy: A review. World J Gastroenterol. 2018 Feb 21. 24 (7):767-774. [QxMD MEDLINE Link].[Full Text]. Mercado R, Buck GA, Manque PA, Ozaki LS. Cryptosporidium hominis infection of the human respiratory tract. Emerg Infect Dis. 2007 Mar. 13 (3):462-4. [QxMD MEDLINE Link].[Full Text]. Mor SM, Ascolillo LR, Nakato R, Ndeezi G, Tumwine JK, Okwera A, et al. Expectoration of Cryptosporidium Parasites in Sputum of Human Immunodeficiency Virus-Positive and -Negative Adults. Am J Trop Med Hyg. 2018 Apr. 98 (4):1086-1090. [QxMD MEDLINE Link].[Full Text]. Tamomh AG, Agena AM, Elamin E, et al. Prevalence of cryptosporidiosis among children with diarrhoea under five years admitted to Kosti teaching hospital, Kosti City, Sudan. BMC Infectious Diseases. 2021. 21(1):349. [Full Text]. Jann HW, Cabral-Castro MJ, Costa JVB, Alencar ACMB, Peralta JM, Peralta RHS. Prevalence of human cryptosporidiosis in the Americas: systematic review and meta-analysis. Rev Inst Med Trop Sao Paulo. 2022. 64:e70. [QxMD MEDLINE Link].[Full Text]. Miller JM, Binnicker MJ, Campbell S, Carroll KC, Chapin KC, Gilligan PH, et al. A Guide to Utilization of the Microbiology Laboratory for Diagnosis of Infectious Diseases: 2018 Update by the Infectious Diseases Society of America and the American Society for Microbiology. Clin Infect Dis. 2018 Aug 31. 67 (6):e1-e94. [QxMD MEDLINE Link].[Full Text]. Paulos S, Saugar JM, de Lucio A, Fuentes I, Mateo M, Carmena D. Comparative performance evaluation of four commercial multiplex real-time PCR assays for the detection of the diarrhoea-causing protozoa Cryptosporidium hominis/parvum, Giardia duodenalis and Entamoeba histolytica. PLoS One. 2019. 14 (4):e0215068. [QxMD MEDLINE Link].[Full Text]. Marie C, Petri WA. Cryptosporidiosis. Porter RE.The Merck Manual of Diagnosis and Therapy. Rahway, NJ: Merck & Co Inc; Reviewed/Revised May 2024. [Full Text]. Daly CA, Padley SP. Sonographic prediction of a normal or abnormal ERCP in suspected AIDS related sclerosing cholangitis. Clin Radiol. 1996 Sep. 51 (9):618-21. [QxMD MEDLINE Link].[Full Text]. Weber R, Bryan RT, Juranek DD. 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HIV-related/AIDS cholangiopathy: pictorial review with emphasis on MRCP findings and differential diagnosis. Clin Imaging. 2013 Mar-Apr. 37 (2):219-26. [QxMD MEDLINE Link].[Full Text]. Boks M, Lilja M, Widerström M, Karling P, Lindam A, Sjöström M. Persisting symptoms after Cryptosporidium hominis outbreak: a 10-year follow-up from Östersund, Sweden. Parasitol Res. 2023 Jul. 122 (7):1631-1639. [QxMD MEDLINE Link]. Amadi B, Mwiya M, Musuku J, Watuka A, Sianongo S, Ayoub A, et al. Effect of nitazoxanide on morbidity and mortality in Zambian children with cryptosporidiosis: a randomised controlled trial. Lancet. 2002 Nov 2. 360 (9343):1375-80. [QxMD MEDLINE Link].[Full Text]. Rossignol JF, Kabil SM, el-Gohary Y, Younis AM. Effect of nitazoxanide in diarrhea and enteritis caused by Cryptosporidium species. Clin Gastroenterol Hepatol. 2006 Mar. 4 (3):320-4. [QxMD MEDLINE Link].[Full Text]. Rossignol JF, Ayoub A, Ayers MS. Treatment of diarrhea caused by Cryptosporidium parvum: a prospective randomized, double-blind, placebo-controlled study of Nitazoxanide. J Infect Dis. 2001 Jul 1. 184 (1):103-6. [QxMD MEDLINE Link].[Full Text]. Amadi B, Mwiya M, Sianongo S, Payne L, Watuka A, Katubulushi M, et al. High dose prolonged treatment with nitazoxanide is not effective for cryptosporidiosis in HIV positive Zambian children: a randomised controlled trial. BMC Infect Dis. 2009 Dec 2. 9:195. [QxMD MEDLINE Link].[Full Text]. NIH. Cryptosporidiosis: Pediatric OIs. National Institutes of Health. Available at August 29, 2019; Accessed: December 16, 2024. Lenière AC, Vlandas A, Follet J. Treating cryptosporidiosis: A review on drug discovery strategies. Int J Parasitol Drugs Drug Resist. 2024. 25:100542. [Full Text]. Love MS, Choy RKM. Emerging treatment options for cryptosporidiosis. Curr Opin Infect Dis. 2021 Oct 1. 34 (5):455-462. [QxMD MEDLINE Link].[Full Text]. CDC. Preventing Crypto. US Centers for Disease Control and Prevention. Available at June 5, 2024; Accessed: November 25, 2024. Matsubayashi M, Haraguchi A, Morisaki M, Ikadai H, Teramoto I, Kido Y, et al. Cryptosporidium parvum inactivation from short durations of treatment with ozonated water produced by an electrolytic generation system. Parasitol Res. 2024 Nov 6. 123 (11):371. [QxMD MEDLINE Link].[Full Text]. Freitas BLS, Fava NMN, Melo-Neto MG, Dalkiranis GG, Tonetti AL, Byrne JA, et al. Efficacy of UVC-LED radiation in bacterial, viral, and protozoan inactivation: an assessment of the influence of exposure doses and water quality. Water Res. 2024 Nov 15. 266:122322. [QxMD MEDLINE Link].[Full Text]. Hagos B, Molestina RE. A Simple Alcohol-based Method of Oocyst Inactivation for Use in the Development of Detection Assays for Cryptosporidium. Food Waterborne Parasitol. 2022 Jun. 27:e00163. [QxMD MEDLINE Link].[Full Text]. Morris JF, Murphy J, Fagerli K, Schneeberger C, Jaron P, Moke F, et al. A Randomized Controlled Trial to Assess the Impact of Ceramic Water Filters on Prevention of Diarrhea and Cryptosporidiosis in Infants and Young Children-Western Kenya, 2013. Am J Trop Med Hyg. 2018 May. 98 (5):1260-1268. [QxMD MEDLINE Link].[Full Text]. Heidarnegadi S, Mohebali M, Maraghi Sh, Babaei Z, Farnia Sh, Bairami A, et al. Cryptosporidium spp. Infection in human and domestic animals. Iran J Parasitol. 2012. 7 (1):53-8. [QxMD MEDLINE Link]. Jacob J, Lorber B. Diseases Transmitted by Man's Best Friend: The Dog. Microbiol Spectr. 2015 Aug. 3 (4):[QxMD MEDLINE Link].[Full Text]. Timmermans M, Hubers W, Schroer D, et al. The first commercially approved efficacious cryptosporidium vaccine protecting New-Born calves from severe diarrhea. Veterinary Vaccine. 2024. [Full Text]. FDA. Drug Approval Package. US Food and Drug Administration. Available at Accessed: November 26, 2024. NIH. Cryptosporidiosis: Adult and Adolescent Opportunistic Infections. National Institutes of Health. Available at Accessed: November 26, 2024. Bhadauria D, Goel A, Kaul A, Sharma RK, Gupta A, Ruhela V, et al. Cryptosporidium infection after renal transplantation in an endemic area. Transpl Infect Dis. 2015 Feb. 17 (1):48-55. [QxMD MEDLINE Link].[Full Text]. Esmat M, Abdel-Aal AA, Shalaby MA, Badawi M, Elaskary H, Yousif AB, et al. Efficacy of clofazimine and nitazoxanide combination in treating intestinal cryptosporidiosis and enhancing intestinal cellular regeneration in immunocompromised mice. Food Waterborne Parasitol. 2022 Jun. 27:e00161. [QxMD MEDLINE Link].[Full Text]. Gilbert IH, Vinayak S, Striepen B, Manjunatha UH, Khalil IA, Van Voorhis WC, et al. Safe and effective treatments are needed for cryptosporidiosis, a truly neglected tropical disease. BMJ Glob Health. 2023 Aug. 8 (8):[QxMD MEDLINE Link].[Full Text]. Theodos CM, Griffiths JK, D'Onfro J, Fairfield A, Tzipori S. Efficacy of nitazoxanide against Cryptosporidium parvum in cell culture and in animal models. Antimicrob Agents Chemother. 1998 Aug. 42 (8):1959-65. [QxMD MEDLINE Link].[Full Text]. Media Gallery Modified acid-fast stain of stool shows red oocysts of Cryptosporidium parvum against the blue background of coliforms and debris. Hematoxylin and eosin stain of intestinal epithelium. The blue dots (arrows) represent intracellular Cryptosporidium organisms along the surface of the epithelial cells. Image courtesy of Carlos Abramowsky, MD, Professor of Pediatrics and Pathology, Emory University School of Medicine. Cryptosporidium species oocysts are rounded and measure 4.2-5.4 µm in diameter. Sporozoites are sometimes visible inside the oocysts, indicating that sporulation has occurred on wet mount. Cryptosporidium parvum oocysts revealed with modified acid-fast stain. Against a blue-green background, the oocysts stand out with a bright red stain. Image courtesy of CDC DPDx parasite image library. Cryptosporidium oocysts revealed with modified acid-fast stain. of 5 Tables Back to List Contributor Information and Disclosures Author Ewa Tomczak, MD Assistant Professor, Department of Internal Medicine, Division of Infectious Disease, The University of Texas Medical Branch at GalvestonDisclosure: Nothing to disclose. Coauthor(s) A Clinton White, Jr, MD, FACP, FIDSA, FASTMH Professor Emeritus, Department of Infectious Disease, University of Texas Medical BranchA Clinton White, Jr, MD, FACP, FIDSA, FASTMH is a member of the following medical societies: American College of Physicians, American Society of Tropical Medicine and Hygiene, Christian Medical and Dental Associations, Infectious Diseases Society of AmericaDisclosure: Nothing to disclose. Melinda B Tanabe, MD Fellow, Division of Infectious Disease, Department of Internal Medicine, University of Texas Medical Branch School of MedicineMelinda B Tanabe, MD is a member of the following medical societies: American Society of Tropical Medicine and Hygiene, Infectious Diseases Society of AmericaDisclosure: Nothing to disclose. Chief Editor Michael Stuart Bronze, MD David Ross Boyd Professor and Chairman, Department of Medicine, Stewart G Wolf Endowed Chair in Internal Medicine, Department of Medicine, University of Oklahoma Health Science Center; Master of the American College of Physicians; Fellow, Infectious Diseases Society of America; Fellow of the Royal College of Physicians, LondonMichael Stuart Bronze, MD is a member of the following medical societies: Alpha Omega Alpha, American College of Physicians, American Medical Association, Association of Professors of Medicine, Infectious Diseases Society of America, Oklahoma State Medical Association, Southern Society for Clinical InvestigationDisclosure: Nothing to disclose. Additional Contributors Poothirikovil Venugopalan, MBBS, MD, FRCPCH Consultant Pediatrician with Cardiology Expertise, Department of Child Health, Brighton and Sussex University Hospitals, NHS Trust; Honorary Senior Clinical Lecturer, Brighton and Sussex Medical School, UKPoothirikovil Venugopalan, MBBS, MD, FRCPCH is a member of the following medical societies: British Congenital Cardiac Association, Paediatrician with Cardiology Expertise Special Interest Group, Royal College of Paediatrics and Child HealthDisclosure: Nothing to disclose. Jaya Sureshbabu, MBBS, MRCPCH(UK), MRCPI(Paeds), MRCPS(Glasg), DCH(Glasg) Consultant Neonatologist and Pediatrian, Sree Gokulam Medical College and Research Foundation, IndiaDisclosure: Nothing to disclose. Miguel M Cabada, MD, MSc Assistant Professor, Division of Infectious Diseases, University of Texas Medical Branch School of Medicine; Director, Universidad Peruana Cayetano Heredia and University of Texas Medical Branch Collaborative Research Center in Cusco, PeruMiguel M Cabada, MD, MSc is a member of the following medical societies: American Society of Tropical Medicine and Hygiene, Infectious Diseases Society of America, International Society for Infectious Diseases, International Society of Travel MedicineDisclosure: Nothing to disclose. Maria A Caravedo, MD Assistant Professor, Division of Infectious Disease, Department of Internal Medicine, University of Texas Medical Branch School of MedicineMaria A Caravedo, MD is a member of the following medical societies: American Society of Tropical Medicine and HygieneDisclosure: Nothing to disclose. Acknowledgements Jeffrey D Band, MD Professor of Medicine, Oakland University William Beaumont School of Medicine; Director, Division of Infectious Diseases and International Medicine, Corporate Epidemiologist, William Beaumont Hospital; Clinical Professor of Medicine, Wayne State University School of Medicine Disclosure: Nothing to disclose. Damon Eisen, MD Clinical Senior Lecturer, Department of Medicine, University of Queensland Disclosure: Nothing to disclose. Joseph F John Jr, MD, FACP, FIDSA, FSHEA Clinical Professor of Medicine, Molecular Genetics and Microbiology, Medical University of South Carolina College of Medicine; Associate Chief of Staff for Education, Ralph H Johnson Veterans Affairs Medical Center Disclosure: Nothing to disclose. Athena P Kourtis, MD, PhD Associate Professor, Department of Pediatrics, Divisions of Infectious Diseases and Epidemiology, Emory University School of Medicine; Senior Fellow, Centers for Disease Control and Prevention Athena P Kourtis, MD, PhD is a member of the following medical societies: American Academy of Pediatrics and Pediatric Infectious Diseases Society Disclosure: Nothing to disclose. Russell W Steele, MD Head, Division of Pediatric Infectious Diseases, Ochsner Children's Health Center; Clinical Professor, Department of Pediatrics, Tulane University School of Medicine Russell W Steele, MD is a member of the following medical societies: American Academy of Pediatrics, American Association of Immunologists, American Pediatric Society, American Society for Microbiology, Infectious Diseases Society of America, Louisiana State Medical Society, Pediatric Infectious Diseases Society, Society for Pediatric Research, and Southern Medical Association Disclosure: Nothing to disclose. Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Medscape Salary Employment Mary L Windle, PharmD Adjunct Associate Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference Disclosure: Nothing to disclose. Close;) What would you like to print? What would you like to print? Print this section Print the entire contents of Print the entire contents of article Sections Cryptosporidiosis Overview Practice Essentials Background Etiology and Pathophysiology Epidemiology Prognosis Morbidity and Mortality Patient Education Show All Presentation History Physical Examination Show All DDx Workup Approach Considerations Stool Tests Evaluation of Immune Function Abdominal Ultrasonography and ERCP Biopsy and Lavage Histologic Findings Show All Treatment Approach Considerations Antiparasitic Therapy Symptomatic Therapy Fluid and electrolyte loss Prevention of Cryptosporidiosis Show All Medication Medication Summary Antiparasitics Antidiarrheal Agents Somatostatin Analogues Show All Media Gallery;) References;) encoded search term (Cryptosporidiosis) and Cryptosporidiosis What to Read Next on Medscape Related Conditions and Diseases Tropical Sprue Tropical Myeloneuropathies Eosinophilic Pneumonia Pathology Rapid Review Quiz: Identifying, Managing, and Treating Dengue Fever Endomyocardial Fibrosis Rapid Review Quiz: Vector-Borne Diseases and Climate Change Miliaria News & Perspective Clinical Trial Begins for New Drug to Combat Neglected Tropical Diseases ECDC Warns of Increase In Mosquito-Borne Diseases The Guide to Supplements for Rheumatologic Diseases COP28 Focus on Health Draws $777 Million to Fight Tropical Disease Experts Decry CDC's Long Pause on Neglected Tropical Disease Testing Tackle Deadly Melioidosis With Swift, Specialized Care Drug Interaction Checker Pill Identifier Calculators Formulary You've Got Worms! Common Intestinal Macroparasites 2001 /viewarticle/clinical-trial-begins-new-drug-combat-neglected-tropical-2022a100036b news news Clinical Trial Begins for New Drug to Combat Neglected Tropical Diseases 2003 /viewarticle/1002026 Chikungunya Virus Vaccines: Latest Data and Clinical Insights 0.25 CME / CE / ABIM MOC Credits You are being redirected to Medscape Education Yes, take me there 0.25 CME / CE / ABIM MOC Chikungunya Virus Vaccines: Latest Data and Clinical Insights 2001 /s/viewarticle/999057 News News COP28 Focus on Health Draws $777 Million to Fight Tropical Disease 2002 1166055-overview Diseases & Conditions Diseases & Conditions Tropical Myeloneuropathies
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https://fiveable.me/key-terms/hs-honors-biology/external-fertilization
External fertilization - (Honors Biology) - Vocab, Definition, Explanations | Fiveable | Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms Honors Biology External fertilization 🐇honors biology review key term - External fertilization Citation: MLA Definition External fertilization is a reproductive process where the fusion of gametes occurs outside the bodies of the parents, typically in an aquatic environment. This method is common in many species of fish, amphibians, and invertebrates, allowing eggs and sperm to be released into the water where fertilization takes place. This strategy often results in the production of a large number of offspring, but also increases vulnerability to environmental factors and predation. 5 Must Know Facts For Your Next Test Many aquatic animals utilize external fertilization due to their environment, as water helps to facilitate the movement of sperm to eggs. External fertilization typically results in a high number of eggs being produced, increasing the chances that at least some will survive to adulthood despite high mortality rates. The timing of external fertilization is often synchronized with environmental cues like temperature or lunar cycles to optimize the chances of successful fertilization. Some species exhibit parental care post-fertilization, such as guarding the fertilized eggs or providing care for hatchlings to enhance survival rates. In contrast to internal fertilization, external fertilization has a greater risk of gametes being dispersed and not meeting, making it less efficient for certain terrestrial animals. Review Questions How does external fertilization differ from internal fertilization in terms of reproductive strategies and offspring survival? External fertilization differs from internal fertilization primarily in where the gametes meet and develop. In external fertilization, gametes are released into the environment where they unite outside the parent’s body. This method tends to produce many more offspring but results in lower survival rates due to higher exposure to predators and environmental stressors compared to internal fertilization, where developing embryos receive more protection. Discuss the advantages and disadvantages of external fertilization for species that utilize this reproductive strategy. The advantages of external fertilization include the ability to produce a large number of eggs at once, which can increase reproductive success even with high rates of predation. Additionally, this method can be synchronized with environmental conditions that favor reproduction. However, disadvantages include a greater risk of gametes being lost or failing to meet, as well as exposure to environmental hazards that can impact both eggs and sperm before fertilization occurs. Evaluate how external fertilization strategies could impact evolutionary adaptations in aquatic environments compared to terrestrial environments. External fertilization strategies have led to evolutionary adaptations that enhance reproductive success in aquatic environments. For instance, species have developed behaviors such as spawning synchrony and gamete clustering to improve chances of successful fertilization. In contrast, terrestrial environments may favor internal fertilization due to increased protection for developing young, leading species to evolve traits that minimize risks associated with external methods, such as greater parental care or egg retention. These differences highlight how reproductive strategies shape evolutionary pathways based on environmental conditions. Related terms internal fertilization: A reproductive strategy where fertilization occurs inside the female's body, providing more protection to the developing embryos. spawning:The act of releasing eggs and sperm into the water simultaneously during a specific reproductive season, often associated with external fertilization. gametes: Reproductive cells, such as sperm and eggs, that fuse during fertilization to form a new organism. 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https://ncert.nic.in/pdf/publication/exemplarproblem/classVI/Mathematics/feep108.pdf
     • Ratio, Proportion and Unitary Method. • The comparison of two numbers or quantities by division is known as the ratio. Symbol ‘:’ is used to denote ratio. • For a ratio, the two quantities must be in the same unit. If they are not, they should be expressed in the same unit before the ratio is taken. • A ratio may be treated as a fraction. • Two ratios are equivalent, if the fractions corresponding to them are equivalent. • Four quantities are said to be in proportion, if the ratio of the first and the second quantities is equal to the ratio of the third and the fourth quantities. The symbol ‘::’ or ‘=’ is used to equate the two ratios. • The order of terms in a proportion is important. For example 3, 8, 24, 64 are in proportion but 3, 8, 64, 24 are not in proportion. • The method in which first we find the value of one unit and then the value of the required number of units is known as unitary method. 11.4.2018 118 EXEMPLAR PROBLEMS UNIT-8 (B) Solved Examples In examples 1 and 2, write the correct answer from the given four options: Example 1. The ratio of Rs 8 to 80 paise is (A) 1 : 10 (B) 10 : 1 (C) 1: 1 (D) 100 : 1 Solution: Correct answer is (B) (Hint: 1 Rupee = 100 paise) Example 2. The length and breadth of a steel tape are 10m and 2.4cm, respectively. The ratio of the length to the breadth is (A) 5 : 1.2 (B) 25: 6 (C) 625: 6 (D) 1250: 3 Solution: Correct answer is (D) (Hint: 10m = 10 × 100cm) Example 3. Find the missing number in the box in the following proportion: : 8 :: 12 : 32 Solution: 12:32 = 12 3 4 3 3:8 32 8 4 8 × = = = × We have, : 8 = 3 : 8 (Given) So, the missing number in is 3. Example 4. State whether the given statements are true or false: (a) 12 : 18 = 28 : 56 (b) 25 persons : 130 persons = 15kg : 78kg Solution: (a) False, Because 12 2 12:18 2:3 18 3 = = = and 28 1 28:56 1:2 56 2 = = = These are not equal. 11.4.2018   (b) True, Because 25 persons : 130 persons = 5: 26 and 15kg : 78kg = 5: 26 These are equal. Example 5. Fill in the blanks: If two ratios are __, then they are in proportion. Solution: Equal/same. Example 6. Find the ratio of the shaded portion to the unshaded portion in Fig. 8.1 Solution: Number of squares in the shaded portion = 15 Number of squares in the unshaded portion = 33 So, the ratio of the shaded portion to the unshaded portion = 15 : 33 15 5 3 5 5:11 33 11 3 11 × = = = = × Example 7. Income of Rahim is Rs 12000 per month and that of Ami is Rs 191520 per annum. If the monthly expenditure of each of them is Rs 9960 per month find the ratio of their savings. Solution: Savings of Rahim per month = Rs (12000 – 9960) = Rs 2040 Monthly income of Ami = Rs 191520 12 = Rs 15960 Savings of Ami per month = Rs (15960 – 9960) = Rs 6000 Therefore, ratio of savings of Rahim and Ami = 2040 : 6000 = 17:50 Fig. 8.1 11.4.2018 120 EXEMPLAR PROBLEMS UNIT-8 Example 8. 20 tons of iron costs Rs 600000. Find the cost of 560kg of iron. Solution: 1 ton = 1000kg Therefore, 20 tons = 20000kg Now, cost of 20000kg iron = Rs 600000 Therefore, cost of 1kg iron = Rs 600000 20000 = Rs 30 Therefore, cost of 560kg iron = Rs 30 × 560 = Rs 16800 (C) Exercise In questions 1 to 10, only one of the four options is correct. Write the correct one. 1. The ratio of 8 books to 20 books is (A) 2 : 5 (B) 5 : 2 (C) 4 : 5 (D) 5 : 4 2. The ratio of the number of sides of a square to the number of edges of a cube is (A) 1 : 2 (B) 3 : 2 (C) 4 : 1 (D) 1 : 3 3. A picture is 60cm wide and 1.8m long. The ratio of its width to its perimeter in lowest form is (A) 1 : 2 (B) 1 : 3 (C) 1 : 4 (D) 1 : 8 4. Neelam’s annual income is Rs. 288000. Her annual savings amount to Rs. 36000. Theratio of her savings to her expenditure is (A) 1 : 8 (B) 1 : 7 (C) 1 : 6 (D) 1 : 5 5. Mathematics textbook for Class VI has 320 pages. The chapter ‘symmetry’ runs from page 261 to page 272. The ratio of the number of pages of this chapter to the total number of pages of the book is (A) 11 : 320 (B) 3 : 40 (C) 3 : 80 (D) 272 : 320 6. In a box, the ratio of red marbles to blue marbles is 7:4. Which of the following could be the total number of marbles in the box? (A) 18 (B) 19 (C) 21 (D) 22 11.4.2018 RATIO AND PROPORTION 121 MATHEMATICS 7. On a shelf, books with green cover and that with brown cover are in the ratio 2:3. If there are 18 books with green cover, then the number of books with brown cover is (A) 12 (B) 24 (C) 27 (D) 36 8. The greatest ratio among the ratios 2 : 3, 5 : 8, 75 : 121 and 40 : 25 is (A) 2 : 3 (B) 5 : 8 (C) 75 : 121 (D) 40 : 25 9. There are ‘b’ boys and ‘g’ girls in a class. The ratio of the number of boys to the total number of students in the class is: (A) b b g + (B) g b g + (C) b g (D) b g b + 10. If a bus travels 160 km in 4 hours and a train travels 320km in 5 hours at uniform speeds, then the ratio of the distances travelled by them in one hour is (A) 1 : 2 (B) 4 : 5 (C) 5 : 8 (D) 8 : 5 In questions 11 to 15, find the missing number in the box  in each of the proportions: 11. 3 5 20 =  12. 2 18 9 =  13. 8 3.2 4 =  14. 16 24 45 40 = =   15. 16 36 36 63 117 = = =    In questions 16 to 34, state whether the given statements are true (T) or false (F). 16. 3 15 8 40 = 11.4.2018   17. 4 : 7 = 20 : 35 18. 0.2 : 5 = 2 : 0.5 19. 3 : 33 = 33 : 333 20. 15m : 40m = 35m : 65m 21. 27cm2 : 57cm2 = 18cm : 38cm 22. 5kg : 7.5kg = Rs 7.50 : Rs 5 23. 20g : 100g = 1metre : 500cm 24. 12 hours : 30 hours = 8km : 20km 25. The ratio of 10kg to 100kg is 1:10 26. The ratio of 150cm to 1metre is 1:1.5. 27. 25kg : 20g = 50kg : 40g 28. The ratio of 1 hour to one day is 1:1. 29. The ratio 4 :16 is in its lowest form. 30. The ratio 5 : 4 is different from the ratio 4 : 5. 31. A ratio will always be more than 1. 32. A ratio can be equal to 1. 33. If b : a = c : d, then a, b, c, d are in proportion. 34. The two terms of a ratio can be in two different units. In questions 35 to 46, fill in the blanks to make the statements true. 35. A ratio is a form of comparison by _. 36. 20m : 70m = Rs 8 : Rs . 37. There is a number in the box such that , 24, 9, 12 are in proportion. The number in the box is __. 38. If two ratios are equal, then they are in _. Use Fig. 8.2 (In which each square is of unit length) for questions 39 and 40: Fig. 8.2 11.4.2018   39. The ratio of the perimeter of the boundary of the shaded portion to the perimeter of the whole figure is _. 40. The ratio of the area of the shaded portion to that of the whole figure is _. 41. Sleeping time of a python in a 24 hour clock is represented by the shaded portion in Fig. 8.3. 24hours 6hours 12hours 18hours Fig. 8.3 The ratio of sleeping time to awaking time is . 42. A ratio expressed in lowest form has no common factor other than in its terms. 43. To find the ratio of two quantities, they must be expressed in __units. 44. Ratio of 5 paise to 25 paise is the same as the ratio of 20 paise to _ 45. Saturn and Jupiter take 9 hours 56 minutes and 10 hours 40 minutes, respectively for one spin on their axes. The ratio of the time taken by Saturn and Jupiter in lowest form is _. 46. 10g of caustic soda dissolved in 100mL of water makes a solution of caustic soda. Amount of caustic soda needed for 1 litre of water to make the same type of solution is ___. 47. The marked price of a table is Rs 625 and its sale price is Rs 500. What is the ratio of the sale price to the marked price? 11.4.2018 124 EXEMPLAR PROBLEMS UNIT-8 48. Which pair of ratios are equal? And why? (i) 2 4 , 3 6 (ii) 8 2 , 4 1 (iii) 4 12 , 5 20 49. Which ratio is larger 10 : 21 or 21 : 93? 50. Reshma prepared 18kg of Burfi by mixing Khoya with sugar in the ratio of 7 : 2. How much Khoya did she use? 51. A line segment 56cm long is to be divided into two parts in the ratio of 2 : 5. Find the length of each part. 52. The number of milk teeth in human beings is 20 and the number of permanent teeth is 32. Find the ratio of the number of milk teeth to the number of permanent teeth. 53. Sex ratio is defined as the number of females per 1000 males in the population. Find the sex ratio if there are 3732 females per 4000 males in a town. 54. In a year, Ravi earns Rs 360000 and paid Rs 24000 as income tax. Find the ratio of his (a) income to income tax. (b) income tax to income after paying income tax. 55. Ramesh earns Rs 28000 per month. His wife Rama earns Rs 36000 per month. Find the ratio of (a) Ramesh’s earnings to their total earnings (b) Rama’s earnings to their total earnings. 56. Of the 288 persons working in a company, 112 are men and the remaining are women. Find the ratio of the number of (a) men to that of women. (b) men to the total number of persons. (c) women to the total number of persons. 57. A rectangular sheet of paper is of length 1.2m and width 21cm. Find the ratio of width of the paper to its length. 11.4.2018 RATIO AND PROPORTION 125 MATHEMATICS 58. A scooter travels 120km in 3 hours and a train travels 120km in 2 hours. Find the ratio of their speeds. ( Hint : Speed = distance travelled time taken ) 59. An office opens at 9 a.m. and closes at 5.30 p.m. with a lunch break of 30 minutes. What is the ratio of lunch break to the total period in the office? 60. The shadow of a 3m long stick is 4m long. At the same time of the day, if the shadow of a flagstaff is 24m long, how tall is the flagstaff? 61. A recipe calls for 1 cup of milk for every 1 22 cups of flour to make a cake that would feed 6persons. How many cups of both flour and milk will be needed to make a similar cake for 8 people? 62. In a school, the ratio of the number of large classrooms to small classrooms is 3:4. If the number of small rooms is 20, then find the number of large rooms. 63. Samira sells newspapers at Janpath crossing daily. On a particular day, she had 312 newspapers out of which 216 are in English and remaining in Hindi. Find the ratio of (a) the number of English newspapers to the number of Hindi newspapers. (b) the number of Hindi newspapers to the total number of newspapers. 64. The students of a school belong to different religious backgrounds. The number of Hindu students is 288, the number of Muslim students is 252, the number of Sikh students is 144 and the number of Christian students is 72. Find the ratio of (a) the number of Hindu students to the number of Christian students. 11.4.2018   (b) the number of Muslim students to the total number of students. 65. When Chinmay visted chowpati at Mumbai on a holiday, he observed that the ratio of North Indian food stalls to South Indian food stalls is 5:4. If the total number of food stalls is 117, find the number of each type of food stalls. 66. At the parking stand of Ramleela ground, Kartik counted that there are 115 cycles, 75 scooters and 45 bikes. Find the ratio of the number of cycles to the total number of vehicles. 67. A train takes 2 hours to travel from Ajmer to Jaipur, which are 130km apart. How much time will it take to travel from Delhi to Bhopal which are 780km apart if the train is travelling at the uniform speed? 68. The length and breadth of a school ground are 150m and 90m respectively, while the length and breadth of a mela ground are 210m and 126m, respectively. Are these measurements in proportion? North America Europe Asia Africa Australia A n t a r c t i c a (Comparativeareasofthecontinents) South America Fig. 8.4 11.4.2018   69. In Fig. 8.4, the comparative areas of the continents are given: What is the ratio of the areas of (a) Africa to Europe (b) Australia to Asia (c) Antarctica to Combined area of North America and South America. 70. A tea merchant blends two varieties of tea costing her Rs 234 and Rs 130 per kg in the ratio of their costs. If the weight of the mixture is 84kg, then find the weight of each variety of tea. 71. An alloy contains only zinc and copper and they are in the ratio of 7:9. If the weight of the alloy is 8kg, then find the weight of copper in the alloy. 72. In the following figure, each division represents 1cm: A B C D E F G H I 0 1 2 3 4 5 6 7 8 Fig. 8.5 Express numerically the ratios of the following distances: (i) AC : AF (ii) AG : AD (iii) BF : AI (iv) CE : DI 73. Find two numbers whose sum is 100 and whose ratio is 9 :16. 74. In Fig. 8.6 (i) and Fig. 8.6 (ii), find the ratio of the area of the shaded portion to that of the whole figure: Fig. 8.6 (i) Fig. 8.6 (ii) 75. A typist has to type a manuscript of 40 pages. She has typed 30 pages of the manuscript. What is the ratio of the number of pages typed to the number of pages left? 11.4.2018   76. In a floral design made from tiles each of dimensions 40cm by 60cm (See Fig. 8.7), find the ratios of: (a) the perimeter of shaded portion to the perimeter of the whole design. (b) the area of the shaded portion to the area of the unshaded portion. Fig. 8.7 77. In Fig. 8.8, what is the ratio of the areas of (a) shaded portion I to shaded portion II ? Fig. 8.8 I I I I I I 10 5 10 7 5 11.4.2018 RATIO AND PROPORTION 129 MATHEMATICS (b) shaded portion II to shaded portion III? (c) shaded portions I and II taken together and shaded portion III? 78. A car can travel 240km in 15 litres of petrol. How much distance will it travel in 25 litres of petrol? 79. Bachhu Manjhi earns Rs 24000 in 8 months. At this rate, (a) how much does he earn in one year? (b) in how many months does he earn Rs 42000? 80. The yield of wheat from 8 hectares of land is 360 quintals. Find the number of hectares of land required for a yield of 540 quintals? 81. The earth rotates 360o about its axis in about 24 hours. By how much degree will it rotate in 2 hours? 82. Shivangi is suffering from anaemia as haemoglobin level in her blood is lower than the normal range. Doctor advised her to take one iron tablet two times a day. If the cost of 10 tablets is Rs 17, then what amount will she be required to pay for her medical bill for 15 days? 83. The quarterly school fee in Kendriya Vidyalaya for Class VI is Rs 540. What will be the fee for seven months? 84. In an election, the votes cast for two of the candidates were in the ratio 5 : 7. If the successful candidate received 20734 votes, how many votes did his opponent receive? 85. A metal pipe 3 metre long was found to weigh 7.6kg. What would be the weight of the same kind of 7.8m long pipe? 86. A recipe for raspberry jelly calls for 5 cups of raspberry juice and 1 22 cups of sugar. Find the amount of sugar needed for 6 cups of the juice? 87. A farmer planted 1890 tomato plants in a field in rows each having 63 plants. A certain type of worm destroyed 18 plants in each row. How many plants did the worm destroy in the whole field? 11.4.2018 130 EXEMPLAR PROBLEMS UNIT-8 88. Length and breadth of the floor of a room are 5m and 3m, respectively. forty tiles, each with area 1 16 m2 are used to cover the floor partially. Find the ratio of the tiled and the non tiled portion of the floor. 89. A carpenter had a board which measured 3m × 2m. She cut out a rectangular piece of 250cm × 90cm. What is the ratio of the area of cut out piece and the remaining piece? 11.4.2018 RATIO AND PROPORTION 131 MATHEMATICS Rough Work 11.4.2018 132 EXEMPLAR PROBLEMS UNIT-8 Rough Work 11.4.2018
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https://www.youtube.com/watch?v=5sGae6K1cww
How to use Interval Notation (parentheses) & [brackets] ALGEBRA Socratica High 2040 subscribers 151 likes Description 23128 views Posted: 17 Oct 2022 𝙎𝙞𝙜𝙣 𝙪𝙥 𝙩𝙤 𝙤𝙪𝙧 𝙢𝙖𝙞𝙡𝙞𝙣𝙜 𝙡𝙞𝙨𝙩 𝙛𝙤𝙧 𝙛𝙧𝙚𝙚 𝙝𝙚𝙡𝙥: 💌 We’ll send emails to keep you updated. 𝙎𝙤𝙘𝙧𝙖𝙩𝙞𝙘𝙖 𝙈𝙖𝙩𝙝𝙡𝙚𝙩𝙞𝙘𝙨 𝘿𝙚𝙥𝙩 𝙏𝙚𝙚 𝙎𝙝𝙞𝙧𝙩 👕 𝙃𝙤𝙬 𝙏𝙤 𝙎𝙤𝙡𝙫𝙚 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢𝙨 Paperback: When you solve algebraic inequalities, you’ll get a range of answers instead of just one number. In that case, you can visualize the solutions as an interval on the Real Number Line. To learn more about using the Number Line to draw intervals (using open & closed circles), watch this video: But there's another kind of notation you can use instead When you want to draw out an interval represented by a less than sign or a greater than sign, meaning the number is NOT included in the interval, you use an open circle. When you want to include the endpoint (representing a less than or equal to sign, or a greater than or equal to sign), you use a closed circle. We’ll show examples of all the possible situations of open intervals and closed intervals on the Real Number Line in this video! Stay tuned for the next video in the series: Interval Notation, which is another way to write out solutions to algebraic equations that involve a range of values. Be sure to SUBSCRIBE so you don’t miss it. 𝙅𝙪𝙢𝙥 𝙩𝙤 𝙩𝙝𝙚𝙨𝙚 𝙨𝙚𝙘𝙩𝙞𝙤𝙣𝙨 𝙞𝙣 𝙩𝙝𝙚 𝙫𝙞𝙙𝙚𝙤: 0:00 Intervals on Real Number Line Quick Review 0:44 Using Inequalities instead 0:55 Using Interval Notation instead 0:57 Example 1: (-2,3] 1:54 Example 2: (0,pi) 2:08 Example 3: [-3, 0.5] 2:23 Example 4: [-1,1) 2:42 Including infinity 4:23 Why use parentheses? Answer in comments! 𝙑𝙊𝘾𝘼𝘽𝙐𝙇𝘼𝙍𝙔 𝙏𝙀𝙍𝙈𝙎 𝙞𝙣 𝙩𝙝𝙞𝙨 𝙫𝙞𝙙𝙚𝙤: interval interval notation parenthesis parentheses square brackets infinity 𝙒𝙚 𝙧𝙚𝙘𝙤𝙢𝙢𝙚𝙣𝙙 𝙩𝙝𝙚 𝙛𝙤𝙡𝙡𝙤𝙬𝙞𝙣𝙜 𝙧𝙚𝙨𝙤𝙪𝙧𝙘𝙚𝙨: 𝙃𝙤𝙬 𝙏𝙤 𝘽𝙚 𝙖 𝙂𝙧𝙚𝙖𝙩 𝙎𝙩𝙪𝙙𝙚𝙣𝙩 ebook: Paperback: Kindle Unlimited: 𝙃𝙤𝙬 𝙏𝙤 𝙎𝙤𝙡𝙫𝙚 𝙒𝙤𝙧𝙙 𝙋𝙧𝙤𝙗𝙡𝙚𝙢𝙨 Paperback: 𝙒𝘼𝙏𝘾𝙃 𝙉𝙀𝙓𝙏: How Long is that Cat? Algebra Word Problem: SMARTER Study Tips from Socratica: This video was made possible by The Socratica Foundation. 𝙎𝙪𝙥𝙥𝙤𝙧𝙩 𝙩𝙝𝙞𝙨 𝙣𝙤𝙣𝙥𝙧𝙤𝙛𝙞𝙩: The Socratica Foundation is an educational nonprofit with the mission to build the essential, universal knowledge base every student needs to succeed. We provide a solid educational foundation in the three timeless pillars: literacy, numeracy, and critical thinking. 𝘾𝙧𝙚𝙙𝙞𝙩𝙨: Instructor: Michael Lee Harrison Produced by Michael Lee Harrison and Kimberly Hatch Harrison Edited by Megi Shuke Musical Theme: Dan’s Dans by Neon Beach (AIF4IQJY1G8BZFKZ) 𝙎𝙞𝙜𝙣 𝙪𝙥 𝙩𝙤 𝙤𝙪𝙧 𝙢𝙖𝙞𝙡𝙞𝙣𝙜 𝙡𝙞𝙨𝙩 𝙛𝙤𝙧 𝙛𝙧𝙚𝙚 𝙝𝙚𝙡𝙥: We’ll send emails to keep you updated. Math #Algebra #IntervalNotation 10 comments Transcript: Intervals on Real Number Line Quick Review An interval is a region of the real number line. You can draw an interval by highlighting a part of the number line. We use circles to mark the endpoints of the interval. A filled circle means that number is part of the interval... ...while an open circle means that number is NOT part of the interval. This interval is a way to visualize all the real numbers between -3 and 1.. including -3, and excluding 1. There are an infinite number of points in this region, and there are an infinite number of real numbers in this interval. If you're not in a 'graphy' mood, you can also use inequalities to describe this interval. This inequality is an algebraic way to say: Using Inequalities instead all numbers x less than 1 and greater than or equal to -3. Today, we're going to learn about another way to describe this region: interval notation. Using Interval Notation instead Let's look at the interval of all real numbers between -2 and 3. Example 1: (-2,3] We'll exclude -2, so we draw an open circle at -2. And we'll include 3 by drawing a filled circle. When we write this as an algebraic expression, we have to choose a variable name. Let's use 'z' in this example. If you think about it, the letter we choose for our variable is not that important. We're just describing a range of numbers. The way to describe this interval without using a variable looks like this. We call this interval notation. You write the 2 endpoints separated by a comma, the parenthesis on the left means the number -2 is NOT in this interval, and the square bracket on the right means that 3 IS in the interval. So parentheses are like open circles, and square brackets are like filled circles. For practice, let's see a few more examples. Example 2: (0,pi) Here is the interval from zero to pi, that does not include either endpoint. Using interval notation, we write down the endpoints, and then surround them using parentheses. Example 3: [-3, 0.5] In this next example, the interval covers all real numbers between -3 and 1/2 with both endpoints included. When writing this using interval notation, we surround the endpoints with square brackets. Example 4: [-1,1) And as we saw earlier, you can have an interval where one endpoint is included and the other is not. Here, the endpoints are -1 and 1. -1 is included in this interval while 1 is not. So we use a square bracket for -1, and a parenthesis for 1. Including infinity Next, let's look at an interval that continues forever. We're going to write this using interval notation. Here, the left endpoint is -1. There's an open circle, so -1 is not included, which means we use a parenthesis. What's different here is that there's not an endpoint on the right, the interval continues forever. In mathematics, we describe this using infinity. And we complete the interval notation with another parenthesis. In this next example, the interval covers all positive real numbers, and includes zero. So the left endpoint is zero with a closed bracket... ...and since the interval continues off to infinity, we write infinity with a parenthesis. What if instead, the interval goes off in the negative direction? Here, the interval includes all real numbers less than or equal to 2. For this situation, we write negative infinity to say the interval continues off to the left, and we close the interval with a 2 and a closed bracket. This next interval is all negative real numbers, and zero is not included. So we write negative infinity as the left endpoint, and zero for the right. We use parentheses for both endpoints. Brace yourself for this next example. What if the interval consists of all real numbers? That is the entire number line. Here there aren't any endpoints! To write this using interval notation, you write negative infinity on the left, positive infinity on the right, and use parentheses. Here's a question for you to consider. Why use parentheses? Answer in comments! Why do you think we use parentheses instead of square brackets next to infinity and negative infinity? To recap: If you have an interval on the real number line, you can describe it in several ways. You can highlight the numbers on the line for a visual representation. A closed circle means that number is included and an open circle means it is not. You can also use an inequality with a variable. A 'less than or equal to' sign corresponds to a filled circle, and a 'less than' sign to the open circle. Using interval notation, you write the two endpoints and use a square bracket when a number is included and a parenthesis when it is not. If the interval goes off forever to the right, you use infinity as the right endpoint. And if the interval goes off to the left, you use negative infinity as the left endpoint. This lesson was brought to you by the Socratica Foundation.
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https://sagepathrec.com/retail-and-restaurant-real-estate-site-acquisition-zoning-and-market-planning-tips/
Home Uncategorized admin Uncategorized Retail and Restaurant Real Estate: Site Acquisition, Zoning, and Market Planning Tips Retail and restaurant real estate projects require precise planning, strategic location selection, and strict regulatory compliance. From identifying profitable locations to obtaining permits, every step impacts the success of your business. By utilizing Retail Real Estate Strategy, working with retail site acquisition experts, and leveraging site selection for franchise development, businesses can maximize returns while minimizing risks. Understanding the intricacies of zoning, entitlements, and market planning is essential for any commercial project. Whether opening a new restaurant, expanding a franchise, or developing a retail chain, following best practices ensures a smooth and profitable rollout. The Importance of Strategic Site Selection Site selection for franchise development is one of the most critical factors in retail and restaurant success. A poorly chosen location can lead to low foot traffic, reduced revenue, and operational challenges. Key considerations include: Demographics: Analyze local population, income levels, and consumer behavior. Accessibility: Ensure easy access via major roads, parking availability, and pedestrian pathways. Competition: Evaluate nearby businesses and identify opportunities for differentiation. Visibility: High-traffic areas with strong signage potential attract more customers. Engaging retail site acquisition experts helps identify prime locations, conduct feasibility studies, and navigate local regulations efficiently. Navigating Zoning and Permitting Understanding local zoning laws is essential for any retail or restaurant project. Zoning approvals and permitting ensure your business complies with land use regulations, building codes, and environmental requirements. Key steps in zoning and permitting include: Confirming property zoning aligns with intended commercial use Securing building permits for construction or renovations Meeting fire, health, and safety regulations Coordinating with municipal authorities to expedite approvals Working with entitlement and permitting consultants or specialized real estate consultants ensures a smooth process, preventing delays that could impact your project timeline. Market Planning for Retail and Restaurant Success Effective real estate market planning ensures that retail and restaurant investments are aligned with customer demand and growth opportunities. Steps include: Competitive Analysis: Assess nearby businesses to identify market gaps and opportunities. Customer Profiling: Understand local consumer needs, preferences, and spending patterns. Revenue Forecasting: Estimate potential income based on foot traffic, sales trends, and pricing strategy. Expansion Strategy: Identify areas suitable for future growth or additional locations. Combining market planning with expert Retail Real Estate Strategy helps businesses make informed decisions and optimize investment returns. Benefits of Professional Consulting Working with experienced consultants in retail and restaurant real estate provides several advantages: Expert Guidance: Navigate complex zoning and regulatory requirements confidently. Data-Driven Insights: Use market analysis and feasibility studies to select high-potential sites. Time Savings: Streamline acquisition, permitting, and development processes. Risk Reduction: Avoid costly mistakes related to poor site selection or regulatory non-compliance. By leveraging retail site acquisition experts and strategic consulting, businesses can enhance operational efficiency, reduce financial risk, and ensure long-term success. Case Studies: Successful Retail and Restaurant Developments Many successful franchises and retail chains owe their growth to strategic site selection and expert consulting. Examples include: Franchise Expansion: Using site selection for franchise development to enter high-traffic urban markets, ensuring high customer engagement and profitability. Restaurant Launch: Securing locations through retail site acquisition experts that align with local dining trends and consumer behavior. Retail Chain Growth: Implementing Retail Real Estate Strategy to expand into suburban and emerging markets, increasing market share and brand visibility. These examples highlight the importance of professional guidance and strategic planning in retail and restaurant real estate projects. Final Thoughts Successfully developing retail and restaurant properties requires a combination of strategic planning, market insights, and regulatory compliance. From Retail Real Estate Strategy to site selection for franchise development and navigating zoning approvals and permitting, every step influences the project’s profitability and long-term success. Working with retail site acquisition experts and entitlement consultants ensures a streamlined process, reducing risk and maximizing potential returns. By combining expert consulting with thorough market planning, businesses can confidently launch new locations, expand franchises, and establish a strong presence in competitive markets. With the right strategy, professional guidance, and informed site selection, retail and restaurant real estate projects can thrive, delivering both customer satisfaction and financial success. Share This Post: admin Add a Comment Cancel reply
11499
https://www.aps.anl.gov/files/APS-sync/lsnotes/files/APS_1417924.pdf
LS Note No. 200 April, 1992 Effect of Eddy Current in the Laminations on the Magnet Field Y. Chung and J. Galayda Advanced Photon Source Argonne National Laboratory Argonne, IL 60439 Abstract In this note theory and measurements of the effect of the eddy current in the laminations on the magnet field will be presented. The theory assumes a simple solenoid–type magnet with laminated iron core and ignores the end field effect. The measurements were made on the input voltage and current, and the dipole component of the magnetic field in the middle of the magnet bore. The amplitude and phase relations between these quantities give the field attenuation fac-tor, the phase delay, and the resistance and inductance of the magnet as functions of frequency. Comparisons of the results with the theory will be discussed. 1. Introduction The proposed corrector magnets to be used for global and local orbit corrections in the storage ring have six poles like sextupole magnets, and in response to the beam motion, a current of up to approximately 50 Hz will be applied varying with time. The corresponding time–vary-ing magnet field in the iron core induces an eddy current in the magnet laminations, which not only decreases the field strength but also produces power loss due to ohmic heating. The eddy current effect can be reduced significantly by using thin laminations for the iron core. The current design thickness of the lamination is 0.025”. From the viewpoint of curb-ing the eddy current effect, the thinner the laminations, the better. However, the larger number of thinner laminations required to assemble a magnet of given length will drive up the cost of magnet manufacturing. This study investigates the effectiveness of the current design for the sextupole/corrector magnet in terms of the attenuation and phase shift of the magnet field and the power efficiency. 2. Impedance of Electromagnetic Fields For time–varying electromagnetic fields with harmonic time dependence e–iωt in conduc-tors and ferromagnetic materials, the complex field impedance Z can be obtained from consider-ation of Poynting’s theorem for harmonic time variation of the fields,1 1 2 V J  E d3x– VE  D t –H  B t av d3x  S S  n da 0, (2.1) where n is the unit vector outward normal to the surface and S is the complex Poynting vector defined by S 1 2E  H . (2.2) In Eq. (2.1) the quantities E, H, and J are assumed to change with time as e–iωt, while the mag-netic flux B satisfies only the periodicity condition B(t  T) B(t) . (2.3) T (= 2π/ω) is one period of oscillation and (⋅⋅⋅)av means time–averaging over T, that is, ()av 1 2T T 0 () dt . (2.4) Consider a two–terminal passive electromagnetic system as in Fig. 2.1, e.g., an electro-magnet, connected to an external power supply with output voltage Vi and current Ii. n Ii Si Vi Fig. 2.1: A schematic of a two–terminal passive electromagnetic system. In Fig. 2.1, the volume V of the system is the rectangular box and the surface of integral S is its surface. The surface integral in Eq. (2.1) can be divided into  S S  n da  Si S  n da  SSi S  n da . (2.5) The first term in Eq. (2.4) is the complex input power  Si S  n da –1 2 Ii Vi . (2.6) Since the electric energy density is negligibly small compared to the magnetic energy density in ferromagnetic material, we drop the displacement current term in Eq. (2.1). From Eqs. (2.1), (2.5), and (2.6), we obtain 1 2Ii Vi 1 2 V J  E d3x  VH  B t avd3x  SSi S  n da . (2.7) Equation (2.6) is the relation between the input power and the various sources of power con-sumption, whose real part gives the conservation of energy for the time–averaged quantities and whose imaginary part relates to the reactive or stored energy and its alternating flow. Let Zi = R – iωL be the input impedance of a purely inductive system. Then from Vi = Ii Zi = Ii (R – iL) (2.8) and Eq. (2.7), we obtain R 1 Ii2   V J  E d3x  2Re VH  B t avd3x  2SSiS  n da , (2.9) and L 2 Ii 2  Im VH  B t avd3x . (2.10) Here, we assumed that the conductivity and the radiation loss through S – Si are real. Equations (2.9) and (2.10) are the expressions for the resistance and the reactance of a passive electromagnetic system. The power loss, or dissipation, is expressed by the resistance in Eq. (2.9). For an electromagnet, the first term is the loss by Joule heating in the coil winding and the iron core, the second term is the hysteretic loss, and the third term is the radiated power. In the following sections, we will discuss power loss due to the hysteresis effect and the eddy cur-rent in the magnet lamination. 3. Hysteresis Effect Suppose the driving magnetic field H has harmonic time dependence e–iωt with an ampli-tude of H1. Then, in an isotropic ferromagnetic material, the magnetic flux B can be written as (a) (b) H H H1sin 1 H1 B B B1 sin 1 B1 Fig. 3.1: (a) Hysteresis loop giving B in a ferromagnetic material as a function of H. (b) The first harmonic component of the B–H curve in (a). B(t) h   n Bnei(ntαn) . (3.1) h is a unit vector in the direction of the magnetic field H, and αn is the phase delay of the n–th harmonic with respect to H. We now assume that the conductivity σ is real and that the radiation loss is negligible. Then the resistance R can be written as, from Eqs. (2.8) and (3.1), R 1 Ii 2  V 1 J2d3x   Ii 2 V sinα1H1B1d3x . (3.2) where we used 1 T T 0 ei(1n)t dt n,1 . (3.3) As shown in Eq. (3.2), all other harmonic terms other than n = 1 vanish, and when α1 is greater than 0, there is a net loss of power proportional to the area enclosed by the ellipse in Fig. 3.1(b). Another effect of the hysteresis is the reduction of the magnet inductance, as can be seen from Eq. (2.9). When the conductivity σ is real, and if the load is purely inductive (X = ωL), the magnet inductance L is, from Eq. (2.10) L 1 Ii 2 V cos1H1B1d3x. (3.4) Equations (3.2) and (3.4) show that the hysteresis effect and the eddy current effect can-not be separated. In general, the phase delay α1 is a function of the field amplitude H1 as well as of the space, and the field amplitude H1 (and B1 also) is in turn affected by the eddy current in the conductor. In the following section, we will derive the eddy current distribution in the mag-net lamination and the field attenuation as functions of the lamination thickness and the skin depth. 4. Eddy Current Effect In this section, we will consider the effect of the eddy current in the magnet lamination on the magnet field and the resulting change in the resistance R and the inductance L. With finite conductivity of the magnet iron, time–varying current applied on the coil winding will pro-duce an eddy current in the core in the direction canceling the original magnetic field. This results in the decrease of the magnet field efficiency and the phase shift of the field with respect to the current in the coil winding. In order to reduce this undesirable effect and also for the convenience of manufacturing, a magnet core is made of many thin laminations, which are electrically insulated from each other. Since the current flux line must close on itself, the eddy current circulates confined within a lamination and does not cross over to adjacent ones. In the limit of infinitely thin laminations, the current flux lines cancel each other macroscopically and there is no eddy current effect. Because of the finite conductivity of the magnet iron, eddy currents produce Joule heat-ing and increase the resistance. Reduction of the magnetic field decreases the magnet induc-tance. This can be shown as follows. Let us for the moment ignore the effect of the hysteresis (α1 → 0) and focus on the effect of the eddy current with real conductivity σ. We also assume that the magnet is purely inductive. In this case, using the notation of Section 3, Eqs. (2.8) and (2.9) reduce to R 1 I i 2 V 1 J2d3x , (4.1) and L 1 Ii 2 V H1B1d3x . (4.2) Equation (4.1) shows that the resistance of a magnet is the overall power loss due to Joule heating divided by the square of the input current amplitude. The Joule heating occurs pri-marily in the coil winding and in the magnet lamination. If the resistance change of the coil winding over the frequency range of interest is negligibly small, the dominant source of resis-tance change with frequency is the eddy current in the magnet lamination. Another effect of the eddy current in the lamination is reduction of the magnet inductance due to the partial cancella-tion of the magnetic field as shown in Eq. (4.2). In the following, we will derive the current distribution J and the field H inside the mag-net lamination of finite thickness. The distribution of the eddy current in the laminations and the resulting field in the magnet bore will be obtained as functions of the lamination thickness and the skin depth. 4.1. Distribution of the Eddy Current and the Field in the Lamination Consider a simple magnet with M laminations of thickness d, with coil wound around it. Let N be the number of windings and I = Ii e–iωt be the current in the coil. Lx, Ly, and Lz are dimensions along x, y, and z, respectively. The schematic of the magnet is shown in Fig. 4.1. The magnet volume Vm occupied by the laminations is then Vm MdLyLz . (4.3) Ignoring the displacement current ∂D/∂t and assuming ε = ε0, we have J   H and H i 1   J , (4.4) where we used J = σE. Since the current flux lines cannot cut across the boundary between lam-inations, there will be a circulating current contained inside the narrow lamination. Now, we can imagine two opposing current fluxes of the same magnitude flowing through the gap between adjacent laminations, so that there is a circulating current around each lamination. If we assume the relative permeability Km (= µ/µ0) to be very large, the effect of the this imaginary circulating current on other laminations will be negligible, since all the field lines will be nearly perpendicu-lar to the magnet surface. Eddy current Coil current Iie–it Vie–it Lx = Md Ly Fig. 4.1: Eddy current in the magnet lamination due to time–varying current. y Magnet lamination Coil current Eddy current d 2 – d 2 x Fig. 4.2: Distribution of the eddy current in the magnet lamination. Jy(x = 0) = 0 and there is an infinitely thin current surrounding the lamination. In Fig. 4.2, the lamination has thickness d, and there is current flowing in the –y direction at x = –d/2 and in the +y direction at x = d/2. Then, from Eq. (4.4), the equation for the current flux Jy in the lamination is 2Jy  k2Jy  k2NIi Lz x–d 2–x  d 2 0 , (4.5) where k2 i 2i 2 . (4.6) δ = 2/   is the skin depth and k is the complex wave number. Since the lamination thick-ness d is much smaller than Ly and Lz, we can put  y  z 0 . (4.7) With 2 = d2 dx2 in Eq. (4.5), the solution is Jy k cos (kd2) NIi Lz sin (kx) , (4.8) and the field distribution is, from Eq. (4.4), H Hzz ^ , Hz NIi Lz cos (kx) cos (kd2) . (4.9) 4.2. Field Attenuation due to the Eddy Current Equation (4.9) indicates that the magnetic field is reduced from the nominal value NIi/Lz at x = d/2 due to the eddy current. To get the field at distances near the magnet pole in the air, we average Eq. (4.9) with respect to x. Then, from the continuity condition that Bz be continu-ous across the boundary between the lamination and the air, we multiply it by µ/µ0. Therefore, Ha z 0 NIi Lz 2 kd tan kd 2 . (4.10) From Eq. (4.10) we obtain the field attenuation factor 2 kd tan kd 2 a() eiϕ() , (4.11) where a() 2  tan2(2)  tanh2(2) 1  tan2(2) tanh2(2) , (4.12) and ϕ()  4 tan1 tanh(2) tan(2) tan1(tanh(2) tan(2)), (4.13) with ∆ = d = d  2 . When ∆ is very small a(∆) and ϕ(∆) can be approximated as a()  1 7 360 4 1 7 1440 d4()2, (  1) (4.14) and ϕ()  1 6 2 1 12 d2. (  1) (4.15) Table 4.1 lists values of a(∆) and ϕ(∆) in cases of 0.025” and 0.0625” laminations at 25 Hz, 60 Hz and 200 Hz. The plots for a(∆) and ϕ(∆) are shown in Fig. 4.3. The conductivity of σ = 2.17×106 Ω–1m–1 is assumed. Table 4.1: Attenuation and phase shift due to the eddy current in the magnet lamination. Three cases of Km (500, 1000 and 5000) at frequencies 25 Hz, 60 Hz, and 200 Hz with lamination thickness d = 0.025” and d = 0.0625” are given. σ = 2.17×106 Ω–1m–1 is assumed. ÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁ Km = 500 ÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁ d = 0.025” = 0.635 mm ÁÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁ d = 0.0625” = 1.59 mm ÁÁÁ Á Á Á ÁÁÁ f (Hz) ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ δ (mm) ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ ∆ ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ a(∆) ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ ϕ(∆) (°) ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ ∆ ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ a(∆) ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ ϕ(∆)(°) ÁÁÁ ÁÁÁ 25 ÁÁÁÁÁ ÁÁÁÁÁ 3.056 ÁÁÁÁ ÁÁÁÁ 0.207 ÁÁÁÁÁ ÁÁÁÁÁ 1.000 ÁÁÁÁ ÁÁÁÁ –0.41 ÁÁÁÁÁ ÁÁÁÁÁ 0.519 ÁÁÁÁ ÁÁÁÁ 0.999 ÁÁÁÁÁ ÁÁÁÁÁ –2.58 ÁÁÁ ÁÁÁ 60 ÁÁÁÁÁ ÁÁÁÁÁ 1.973 ÁÁÁÁ ÁÁÁÁ 0.321 ÁÁÁÁÁ ÁÁÁÁÁ 1.000 ÁÁÁÁ ÁÁÁÁ –0.99 ÁÁÁÁÁ ÁÁÁÁÁ 0.804 ÁÁÁÁ ÁÁÁÁ 0.992 ÁÁÁÁÁ ÁÁÁÁÁ –6.16 ÁÁÁ Á Á Á ÁÁÁ 200 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 1.080 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.586 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.998 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ –3.29 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 1.468 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.920 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ –19.3 ÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁ Km = 1000 ÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁ d = 0.025” = 0.635 mm ÁÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁ d = 0.0625” = 1.59 mm ÁÁÁ ÁÁÁ f (Hz) ÁÁÁÁÁ ÁÁÁÁÁ δ (mm) ÁÁÁÁ ÁÁÁÁ ∆ ÁÁÁÁÁ ÁÁÁÁÁ a(∆) ÁÁÁÁ ÁÁÁÁ ϕ(∆) (°) ÁÁÁÁÁ ÁÁÁÁÁ ∆ ÁÁÁÁ ÁÁÁÁ a(∆) ÁÁÁÁÁ ÁÁÁÁÁ ϕ(∆)(°) ÁÁÁ Á Á Á ÁÁÁ 25 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 2.161 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.293 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 1.000 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ –0.82 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.734 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.994 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ –5.15 ÁÁÁ ÁÁÁ 60 ÁÁÁÁÁ ÁÁÁÁÁ 1.395 ÁÁÁÁ ÁÁÁÁ 0.454 ÁÁÁÁÁ ÁÁÁÁÁ 0.999 ÁÁÁÁ ÁÁÁÁ –1.98 ÁÁÁÁÁ ÁÁÁÁÁ 1.137 ÁÁÁÁ ÁÁÁÁ 0.969 ÁÁÁÁÁ ÁÁÁÁÁ –12.1 ÁÁÁ ÁÁÁ 200 ÁÁÁÁÁ ÁÁÁÁÁ 0.764 ÁÁÁÁ ÁÁÁÁ 0.829 ÁÁÁÁÁ ÁÁÁÁÁ 0.991 ÁÁÁÁ ÁÁÁÁ –6.55 ÁÁÁÁÁ ÁÁÁÁÁ 2.076 ÁÁÁÁ ÁÁÁÁ 0.767 ÁÁÁÁÁ ÁÁÁÁÁ –32.5 ÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁ Km = 5000 ÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁ d = 0.025” = 0.635 mm ÁÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁ d = 0.0625” = 1.59 mm ÁÁÁ ÁÁÁ f (Hz) ÁÁÁÁÁ ÁÁÁÁÁ δ (mm) ÁÁÁÁ ÁÁÁÁ ∆ ÁÁÁÁÁ ÁÁÁÁÁ a(∆) ÁÁÁÁ ÁÁÁÁ ϕ(∆) (°) ÁÁÁÁÁ ÁÁÁÁÁ ∆ ÁÁÁÁ ÁÁÁÁ a(∆) ÁÁÁÁÁ ÁÁÁÁÁ ϕ(∆)(°) ÁÁÁ Á Á Á ÁÁÁ 25 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.966 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.656 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.996 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ –4.11 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 1.642 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.884 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ –23.2 ÁÁÁ Á Á Á ÁÁÁ 60 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.624 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 1.016 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 0.980 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ –9.73 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ 2.543 ÁÁÁÁ Á ÁÁ Á ÁÁÁÁ 0.632 ÁÁÁÁÁ Á ÁÁÁ Á ÁÁÁÁÁ –40.0 ÁÁÁ ÁÁÁ 200 ÁÁÁÁÁ ÁÁÁÁÁ 0.342 ÁÁÁÁ ÁÁÁÁ 1.854 ÁÁÁÁÁ ÁÁÁÁÁ 0.830 ÁÁÁÁ ÁÁÁÁ –28.0 ÁÁÁÁÁ ÁÁÁÁÁ 4.643 ÁÁÁÁ ÁÁÁÁ 0.304 ÁÁÁÁÁ ÁÁÁÁÁ –46.1 4.3. Impedance Change due to the Eddy Current Let Re be the resistance increase per volume due to the eddy current Jy. The magnet resis-tance R is then the sum of the resistance Rc due to the coil winding and the resistance increase due to the eddy current, which is Re times Vm (= NdLyLz), the volume occupied by the lamina-tions. That is, R = Rc + ReVm . (4.16) From Eqs. (4.1) and (4.9), we obtain (a) (b) 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5   a( ) 0 1 2 3 4 5  0 –10 –20 –30 –40 –50 ϕ( ) (degrees)  Fig. 4.3: Plotting of (a) the field attenuation factor a(∆) and (b) the phase shift of the field ϕ(∆) with respect to the driving current as functions of ∆ = d . Re 1 d Ii 2 d2 d2 dx Jy 2 (4.17) |k|2 d cos(kd2) 2 N Lz 2  d2 d2 dx |sin (kx)|2. Using the relations sin(kx) sin 1  i x sin x cosh x  i cos x sinh x , (4.18) cos(kx) cos 1  i x cos x cosh x i sin x sinh x we obtain Re  N Lz 2 r() , (4.19) where ∆ = d and r() sinh sin  (cosh  cos) . (4.20) Figure 4.4 (a) shows the plot for r(∆). When the lamination is much thinner than the skin depth (∆ << 1), we have Re   6 N Lz 2 2 1 12 N Lz 2 (d)2, (  1) (4.21) and in the opposite case (∆ >> 1), we have Re   N Lz 2 1  2 N Lz 2 1 d  2 . (  1) (4.22) (a) (b) 0.5 0.4 0.3 0.2 0.1 0 0.0 1.0 2.0 3.0 4.0 5.0 r( )      0.0 1.0 2.0 3.0 4.0 5.0 1.2 1 0.8 0.6 0.4 0.2 0 Fig. 4.4: Plotting of the functions: (a) r(∆) in Eq. (4.19) and (b) l(∆) in Eq. (4.27). For a given magnet, the resistance increase due to the eddy current is proportional to the square of frequency when the frequency is low, as shown in Eq. (4.21). However, as the frequency increases further, the resistance increase slows down and in the limit of high frequency it grows only as the square root of the frequency. To calculate the inductance change due to the eddy current, we should consider the mag-netic field not only in the lamination but also in the coil and in the air. The inductance L of a magnet is the total stored magnetic energy divided by half the square of the input current ampli-tude. That is, from Eq. (3.4) with α1 = 0, L 1 Ii 2    lamination H1B1d3x  air H1B1d3x   . (4.23) The inductance contribution from the lamination can be obtained in a fashion similar to calcula-tion of the resistance, which gives Lel, the average inductance per volume of the lamination. To estimate the magnetic energy stored outside the magnet, we introduce Veff, the effective volume in the air that contains the magnetic flux. Then we can write L LelVm  LeaVeff . (4.24) Let us write Lea in terms of Ha z, the field near the magnet pole as given by Eq. (4.10), as Lea 0 Ii 2 Ha z 2 2 0 N Lz 2 a()2 . (4.25) Then Veff is the total magnetic energy stored in the air divided by 1 2 Lea Ii 2. Veff may be esti-mated roughly by multiplying the length of a magnet and the area of a polygon constructed by connecting adjacent magnet poles. We now proceed to calculate Lel. From Eq. (4.9), we have Lel d Ii 2  d2 d2 dx Hz 2 (4.26) d cos (kd2) 2 N Lz 2 d2 d2 dx cos (kx) 2 . Using Eq. (4.18), we obtain Lel N Lz 2 l() , (4.27) where ∆ = d and l() sinh  sin  (cosh  cos) . (4.28) The plot for l(∆) is shown in Fig. 4.4 (b). When the lamination is much thinner than the skin depth (∆ << 1), we have Lel  N Lz 2 1 1 30 4 N Lz 2 1 d4 120 ()2 , (  1)(4.29) and in the opposite case (∆ >> 1), we have Lel  N Lz 2 1  N Lz 2 1 d 2   . (  1) (4.30) From Eqs. (4.26) and (4.30), we have LeaLel  O  0 1 (4.31) and the inductance change due to the eddy current is dominated by the field outside the magnet laminations unless Veff is much smaller than Vm. 5. Measurement of the Eddy Current Effect In this section, we will discuss measurements of the eddy current effect on a storage ring sextupole magnet with horizontal/vertical correction winding. The measurements were made on the input voltage and current, and the dipole component of the magnetic field in the middle of the magnet bore. The amplitude and phase relations among these quantities give the field attenu-ation factor a(∆), the phase delay ϕ(∆) and the resistance and the inductance of the magnet as functions of frequency. Comparisons of the results with the theory discussed in the previous sec-tions will be presented. 5.1 Setup In Fig. 5.1 is shown the setup for the measurement of the eddy current effect on the mag-net field. For this measurement, the correction coil windings on four of the six magnet poles, except the top and bottom poles, were connected in series in vertical correction (horizontal field) configuration. Function Generator Power Supply Gaussmeter B Toshiba 386SX w/ATMIO-16 V I R L Hall Probe Magnet Fig. 5.1: Measurement setup for the eddy current effect on the magnet field. The voltage and the current signals are provided by the power supply and the magnet field is measured by a gaussmeter with a Hall probe. These analog signals are digitized by the ATMIO–16 board and analyzed by a computer. The magnet field was measured using a Series 9900 menu–driven multi–channel gauss-meter manufactured by F.W. Bell. A three–axis Hall probe was used, and only one of the three channels was used for the measurement of the horizontal field. Due to the relative nature of this measurement, the Hall probe was placed approximately at the center of the magnet without much precision. The full–scale analog output signal level is ±3V. The power supply was a Kepco bipolar operational power supply/amplifier, model No. BOP 20–20M, with a voltage range of ±20 V, maximum current of 20 A, and bandwidth of 4 kHz. It was put in voltage–controlled mode for this measurement. Analog voltage and current monitoring signals were provided through an interface card, and digitized and analyzed by a por-table computer. The full–scale signals are ±10 V for both the voltage and the current. The ATMIO–16 board, a multichannel DAC/ADC, has 16 input channels. Configured for differential input, as in our setup, the maximum number of channels is 8. The analog input signal range for the ATMIO–16 board is ±10 V for all channels. With the digital resolution of 12 bits, the conversion formula is V (in analog)  10 2, 048 V (in digital). (5.1) This gives a resolution of approximately 5 mV. The data acquisition and analysis was done by a portable computer with a 386SX proces-sor and a math coprocessor. The data was obtained for 20 periods with a total of 1000 data points. Given a nominal frequency supplied by the user, the actual frequency was first found by counting the integer number of periods and the time elapsed as measured using the clock on the ATMIO board. Using this frequency f, the data was fit to the function V  V0 cos(2ft ) (5.2) using the linear least squares method,2 which gives the amplitude V0 and the phase φ. This pro-cedure was done for the voltage, the current, and the magnet field, and the amplitudes and the phases were compared to obtain the resistance and the inductance of the magnet and the attenua-tion and the phase shift of the field due to the eddy current. 5.2 Measurements and Results The analog–to–digital conversion was done at the maximum rate of 100 ksamples/sec, which introduces a timing error of 10 µsec between two adjacent channels. With sinusoidal sig-nal of frequency f, the phase measurement error per Hz per channel is  (degrees) f (Hz)  3.6  103. (5.3) The result of measurement on ∆φ/f at the frequencies of 10, 100, and 200 Hz is shown in Table 5.1. This phase error was compensated for in the subsequent measurements. Let us write Vi Ii Vi Ii ei, (5.4) which gives, using Eqs. (2.8) and (2.11), R Vi Ii cos and L  1 Vi Ii sin . (5.5) The current Ii and the field B are related by B CIi aeiϕ, (5.6) Table 5.1: Measurement results of the phase delay between channels 1,2, and 3 of ATMIO–16 board. ÁÁÁÁÁÁ Á ÁÁÁÁ Á Á ÁÁÁÁ Á ÁÁÁÁÁÁ Frequency (Hz) ÁÁÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁ Á Á ÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁÁÁ ∆φ/f (degrees/Hz) between Channels 1&2 ÁÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁ Á Á ÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁÁ ∆φ/f (degrees/Hz) between Channels 1&3 ÁÁÁÁÁÁ Á ÁÁÁÁ Á ÁÁÁÁÁÁ 10 ÁÁÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁÁÁ –3.50E–03 ± 1.61E–04 ÁÁÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁ Á ÁÁÁÁÁÁÁÁÁÁÁ –7.05E–03 ± 2.38E–04 ÁÁÁÁÁÁ ÁÁÁÁÁÁ 100 ÁÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁ –3.59E–03 ± 1.59E–05 ÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁ –7.19E–03 ± 1.86E–05 ÁÁÁÁÁÁ ÁÁÁÁÁÁ 200 ÁÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁÁ –3.60E–03 ± 4.57E–06 ÁÁÁÁÁÁÁÁÁÁÁ ÁÁÁÁÁÁÁÁÁÁÁ –7.21E–03 ± 7.15E–06 where C is a factor that normalizes a to 1 at ω = 0. a and ϕ are the attenuation factor and the phase shift in Eqs. (4.10), (4.11), and (4.12). Since the resistance given in Eq. (5.4) includes the resistance of the coil winding Rc, we must subtract it from the measured resistance to obtain the resistance due to the eddy current Re. To avoid the offset calibration error in the output monitoring signals provided by the power sup-ply, we apply an AC signal of very low frequency, say 0.1 Hz rather than use a DC signal and measure the amplitude ratio. The result is Rc Vi Ii f0.1 Hz 431 mV 928 mA 0.464 . (5.7) Measurement on the B field done simultaneously gave an estimate of the normalization factor C in Eq. (5.5). The result is C B Ii f0.1 Hz 11.1 G 928 mA 11.9 GA. (5.8) Four sets (1, 2, 3, and 4) of measurements were made. For cases 1 and 4, either the voltage amplitude was maintained at ≈15 V or the current amplitude at ≈15 A, whichever was larger. At the maximum current of 20 A from the power supply, we will have about 240 G, and the scale on the gaussmeter was set to 300 G. On the other hand, for cases 2 and 3, the current was main-tained at ≈1 A, which produced a magnetic field of ≈12 G. For these cases, the scale on the gaussmeter was set to 30 G. The plots of the variation of the input voltage and the input current are shown in Fig. 5.2. In Figs. 5.3(a) and 5.3(b) are shown the results of measurements on the magnet field–cur-rent ratio |B/Ii| and the phase shift for the vertical correction winding of the storage ring sextu-pole, and in Figs. 5.4(a) and 5.4(b) are shown the results of measurements on the resistance and the inductance of the same magnet. Five measurements were taken for each frequency, and the error bars shown indicate the standard deviation of the data. The magnet field–current amplitude ratio |B/Ii| shown in Fig. 5.3(a) is constant at 12 Gauss/A within ±0.6% and does not show any tendency to decrease. In case of the phase shift, the differences in the rate of decrease may be due to the hysteresis effect, which changes the µ, the instantaneous slope on the B–H curve, depending on the residual magnet field with the cur-rent turned off. With the phase delay in the range –3.6° < ϕ < –2° at 200 Hz, we obtain from Eq. (4.13), f (Hz) 15 5 0 0 50 100 150 200 f (Hz) 20 15 10 5 0 0 50 100 150 200 Case 1 Case 2 Case 3 Case 4 X 20 10 Case 1 Case 2 Case 3 Case 4 X (a) (b) |V | (Volt) i |I | (A) i Fig. 5.2: Variation of (a) the input voltage amplitude |Vi | and (b) the input current amplitude |Ii | for different cases of measurements. f (Hz) 12.1 11.9 11.8 0 50 100 150 200 f (Hz) 0.0 –1.0 –2.0 –3.0 –4.0 0 50 100 150 200 Case 1 Case 2 Case 3 Case 4 X 12.2 12.0 Case 1 Case 2 Case 3 Case 4 X (a) (b) |B/I | (Gauss/A) i (degrees) Fig. 5.3: Results of measurements on (a) magnet field–current amplitude ratio |B/Ii| and (b) the phase shift ϕ. f (Hz) 0 50 100 150 200 (a) Case 1 Case 2 Case 3 Case 4 X 1.0 0.9 0.8 0.7 0.6 0.5 0.4 R ( )  f (Hz) 11.3 11.2 11.1 11.0 10.9 0 50 100 150 200 Case 1 Case 2 Case 3 Case 4 X (b) L (mH) Fig. 5.4: (a) Result of measurements on the resistance R. The error range is within the marks. (b) Result of measurements on the inductance L. For cases 1 and 4, the error range is within the marks, and for cases 2 and 3, the error (±0.04 mH) is significant only at the lowest frequency of 1 Hz. 300  Km  0  550 . (5.9) Fitting the measurement data on the magnet resistance R shown in Fig. 5.4(a) with a qua-dratic function in frequency gives R  0.465 6.53  105 2.56  1072. (5.10) In Eq. (5.10), the first term represents the power loss in the coil winding, the second term is the hysteretic loss, and the third term is the loss due to the eddy current in the core. The results of measurements on L presented in Fig. 5.4(b) show good agreement between cases 1 and 4 and between cases 2 and 3. These differences are attributed to changes in µ with field amplitudes. The decrease of L with increasing frequency, however, was found to be much larger than predicted. Assuming that Veff is large enough so that Eq. (4.31) holds true within the frequency range of our interest, we have from Eq. (4.25) using Eq. (5.9) 1 Lea (f  200 Hz) Lea (f  0 Hz)  8.5  104, (5.11) which is too small to explain the inductance decrease of 1% for cases 2 and 3 with constant cur-rent amplitudes. This indicates that some of magnetic flux has either disappeared or shifted from a dense region to a dilute region as frequency increased. This is because the fringe field pre-viously ignored in the 2–D calculation, which was quite strong at low frequency, became very small at high frequency due to the eddy current. As shown in Fig. 5.5(a), when the frequency is low enough, there are field lines emanating from the side which is not laminated. However, as the frequency increases, eddy current develops in the direction that cancels the original field on the end side, while the field in the gap between the poles remains virtually unchanged, as shown in Fig. 5.5(b). This causes the effective volume Veff to decrease, and thus from Eq. (4.24), the overall magnet inductance decreases. A preliminary measurement on the sextupole magnet showed that the field near the pole in the middle of the magnet bore was 3.5 kG and the field on the side of the pole was approximately 300 G with 200 A of current in the main coil winding of the sextupole magnet. Lamination Coil Winding (a) (b) Fig. 5.5: Schematic comparison of the magnetic field between (a) at low frequency and (b) at high frequency when the eddy current cancels out the side field. 6. Summary and Conclusion In the previous sections, theory and measurements of the effect of the eddy current in the laminations on the magnet field were presented. The theory assumes a simple solenoid–type magnet with laminated iron core and ignores the end field effect. The measurements were made on the input voltage and current, and the dipole component of the magnetic field in the middle of the magnet bore. The focus of the study was on the frequency dependence of the field attenuation factor, the phase delay, and the resistance and inductance of the magnet. The results of measurements show good agreement with the theory. In the case of the magnet inductance, the one–dimen-sional nature of the theory resulted in a discrepancy between measurement and theory. This, however, could be explained by the eddy current induced by the longitudinal field on the end side of the magnet poles. The measurements were done using a relatively small bipolar power supply with maxi-mum voltage of ±20 V and maximum output current of ±20 A up to the frequency of 200 Hz. This set the limit on the current amplitude at 1 A and the magnetic field amplitude at ≈12 G. The design peak field at 7.0 GeV is 1.0 kG with the peak current of 103 A.3 This difference of two orders of magnitude in the current capacity prohibited measurements over the full range of operation, which can be done only when the prototype power supplies for the corrector magnets become available. Acknowledgment G. Decker, K. Kim, and L. Turner are to be thanked for the helpful discussions on this work. References 1. J. Jackson, Classical Electrodynamics, John Wiley & Sons, p. 241, 1975. 2. W. Press et al., Numerical Recipes in C, Cambridge University Press, p. 528, 1989. 3. APS Conceptual Design Report, Argonne National Laboratory, ANL–87–15, 1987.