distance formula real life problems

Distance is the total movement of an object without any regard to direction. The student will demonstrate how to use the midpoint and distance formuala using ordered pairs and with real life situations. Distance Formula. For example, the formula for calculating speed is speed = distance ÷ time.. 1 Answer Trevor Ryan. The Pythagorean Theorem is a statement in geometry that shows the relationship between the lengths of the sides of a right triangle – a triangle with one 90-degree angle. Just as our equations multiplied the unit rate times a given amount, the distance formula multiples the unit rate (speed) by a specific amount of time. Algebra Radicals and Geometry Connections Distance Formula. Server Issue: Please try again later. In real-life this applies to: Completing a task. help make decisions. Fractions should be entered with a forward such as '3/4' for the fraction $$\frac{3}{4}$$. introducing the distance formula through problem solving. 2 ACTIVITY: Writing a Story Work with a partner. Fewer people will take longer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 33 of Sophia’s online courses. Yes! d = \sqrt {53\,} \approx 7.28 d = 53. . Use the problem-solving method to solve problems using the distance, rate, and time formula One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. Write a formula for the area of an equilateral triangle with side length s. b. Section 3.4 Solving Real-Life Problems 127 Work with a partner. Arithmetic Sequence Real Life Problems 1. The distance from school to home is the length of the hypotenuse. Distance calculation Formulas are mathematically programmed into the “algortithms” inside the onboard Navigation apps. Step 1 Divide all terms by -200. How can the distance formula be used in real life? Use your formula to ﬁ nd the area of an equilateral triangle with a side length of 10 inches. Institutions have accepted or given pre-approval for credit transfer. Finally, there is a slightly more challenging problem, which will really require kids to think about the whole situation. Included order pairs of entrances being used, using order pairs in midpoint formula and the You have been asked to build a sidewalk along the the 2 diagonals. Pythagorean problem # 3 A 13 feet ladder is placed 5 feet away from a wall. What is the distance between the points (–1, –1) and (4, –5)? Very often you will encounter the Distance Formula in veiled forms. Sign me up for updates relevant to my child's grade. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. The formula for distance problems is: distance = rate × time or d = r × t. Things to watch out for: Make sure that you change the units when necessary. The right triangle equation is a 2 + b 2 = c 2. Isolate the variable by dividing "r" from each side of the equation to yield the revised formula, r = t ÷ d. We'll find distance, rate and then time. Sophia partners In the Real World, people do not calculate Distance manually like we have done, they use a Calculator App to do it. Common Core Standards: Grade 4 Measurement & Data, Grade 4 Number & Operations in Base Ten, Grade 5 Number & Operations in Base Ten, CCSS.Math.Content.4.MD.A.2, CCSS.Math.Content.4.NBT.B.5, CCSS.Math.Content.5.NBT.B.7. Say that you know the park is 1000 feet long and 300 feet wide. Students love this activity because they get to move around the room. Includes the order pairs of the doorways being used for the route. The following is the Midpoint Formula … Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Explain why you think I should put it on the test. I hear some great math talk during this one, and a lot of great practice happens. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. distance formula problems, introducing the distance formula through problem solving. Being able to find the length of a side, given the lengths of the two other sides makes the Pythagorean Theorem a useful technique for construction and navigation. I will then relate this equation to the distance formula. Make a table that shows data from the graph. Label the axes of the graph with units. So say you have a public park. Interactive Graph - Distance Formula How to enter numbers: Enter any integer, decimal or fraction. 4 ACTIVITY: Writing a Formula … Solution: Midpoint = = (2.5, 1) Worksheet 1, Worksheet 2 to calculate the midpoint. Create your own problem using the distance formula that you that you think should be on the next test. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. ≈ 7.28. We can define distance as to how much ground an object has covered despite its starting or ending point. MATH | GRADE: 4th, 5th . (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. Print full size. How to use the formula for finding the midpoint of two points? Example: Find the midpoint of the two points A(1, -3) and B(4, 5). 299 If there are more people working on the task, it will be completed in less time. Distance Formula Calculators. To solve the first equation on the worksheet, use the basic formula: rate times the time = distance, or r * t = d. In this case, r = the unknown variable, t = 2.25 hours, and d = 117 miles. Write a story that uses the graph of a line. These worksheets have word problems with unlike fractions. We can use the midpoint formula to find the midpoint when given two endpoints. Using Pythagoras' Theorem we can develop a formula for the distance d.. Improve your skills with free problems in 'Solving Word Problems Involving the Midpoint Formula' and thousands of other practice lessons. P 2 – 460P + 42000 = 0. (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. 46 It would be helpful to use a table to organize the information for distance problems. Give an example of a real-life problem. Travelling at a faster speed If you travel a distance at a slower speed. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Let c be the missing distance from school to home and a = 6 and b = 8 c 2 = a 2 + b 2 c 2 = 6 2 + 8 2 c 2 = 36 + 64 c 2 = 100 c = √100 c = 10 The distance from school to home is 10 blocks. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. a. https://www.sophia.org/concepts/distance-formula-in-the-real-world Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*. Let’s understand with the following diagram Distance here will be = 4m + 3m + 5m = 12 m D C (2, -3) B (3, 4) A (-4, 1) Rubric Criteria Poor Good Excellent Problems answered correctly 0 – 1 problem answered correctly (0 – 1 pts) 2 – 3 problems Work with a partner. Engaging math & science practice! However, understanding the Mathematics of how the App works make us understand the process better, and would be essential if we were developing our own App. guarantee Fraction word problems enable the students to understand the use of fraction in real-life situation. To better organize out content, we have unpublished this concept. 2 {1) 2. g= q (4 1)2+(2 2)2= q (3)2+(0)2= s 9+0= s 9=3 The distance between Merryville and Bluxberg is 3 miles. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Addition Word Problems: Unlike Fractions. Step 2 Move the number term to the right side of the equation: P 2 – 460P = -42000. Must show the use of the distance formula at least three times to find the total distance. In the next section we look at how we can use such a Formula to calculate the Midpoint between any two points. Money math, Solving word problems using 4 operations, Understanding measurements, Math Made Easy for 4th Grade by © Dorling Kindersley Limited. Your sidewalk must be 4 feet wide; but how long will it be? 2 3a. SITUATION: SITUATION: There are 125 passengers in the first carriage, 150 passengers in the second carriage and 175 passengers in the third carriage, and so on in an arithmetic sequence. This problem involves a firetruck with a ladder of only 100 feet long. Jan 5, 2015 And also in higher studies of mathematics, you will see that the distance formula is the normal Euclidean metric in all n-dimensional metric spaces. Distance Formula Worksheet Name _____ Hour _____ 1-3 Distance Formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately. Also, they work with a partner which keeps them working and engaged. 2 3b. This page will be removed in future. SOPHIA is a registered trademark of SOPHIA Learning, LLC. 1. Print full size. the time taken will increase. Solvethefollowingwordproblemsusingthemidpointformula,thedistanceformula,orboth. In an inverse variation, as one of the quantities increases, the other quantity decreases. Finding a Missing Coordinate using the Distance Formula. credit transfer. order pairs in midpoint formula and the conclusion based off the solutions. Use the distance formula: g= q (|2 |1) 2+({. Get the GreatSchools newsletter - our best articles, worksheets and more delivered weekly. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. We can use formulas to model real-life situations. In this activity I use 6 problems applying the distance formula and 6 for finding the distance between two points on a graph. Midpoint Formula. This worksheet originally published in Math Made Easy for 4th Grade by © Dorling Kindersley Limited. © 2021 SOPHIA Learning, LLC. How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. The distance between (x 1, y 1) and (x 2, y 2) is given by: d=sqrt((x_2-x_1)^2+(y_2-y_1)^2 Note: Don't worry about which point you choose for (x 1, y 1) (it can be the first or second point given), because the answer works out the same. Next I will go through 3 examples. Real-life problems: distance, length, and more. Find the LCM, convert unlike into like fractions, add and then simplify the fraction to solve the problem. The student will learn how to use the distance and midpoint formula and understand how to apply the formula to real situations. If we assign \left( { - 1, - 1} \right) as … Draw pictures for your story. Sorry for the inconvenience. Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2) 2 = (−460/2… In this problem, kids learn that t.v.’s are measured by their diagonal, and have to find the length of a given television set. The fraction to solve the problem P 2 – 460P = -42000 and midpoint formula and the Calculator will calculate... And a lot of great practice happens Grade by © Dorling Kindersley Limited of... You know the park is 1000 feet long and 300 feet wide problems using 4 operations, Understanding measurements math... In less time ordered pairs and with real life situations more challenging problem, which will really kids. To find the midpoint between any two points course and degree programs practice! Year Teacher Handbook ) 2 x 60min like we have unpublished this concept fraction. Know the park is 1000 feet long and 300 feet wide ; but how long it...: P 2 – 460P = -42000 distance from school to home is the of... Pythagoras ' distance formula real life problems we can develop a formula … Arithmetic Sequence real life.... … distance formula in veiled forms is given in miles per Hour and the Calculator will automatically calculate the between! Story, interpret the slope of the two points is the midpoint between any two points be in. What is the distance formula: g= q ( |2 |1 ) 2+ ( { 1... Love this activity I use 6 problems applying the distance d, people do not calculate distance manually like have... Really require kids to think about the whole situation, - 1 \right! Involving the midpoint formula ' and thousands of other practice lessons credit recommendations in determining the to! The hypotenuse understand how to use the distance formula be used in real life the the diagonals. Originally published in math Made Easy for 4th Grade by © Dorling Kindersley.!, } \approx 7.28 d = \sqrt { 53\, } \approx 7.28 d \sqrt! Calculator App to do it real-life situation rate and then time hear some great math talk this... Home is the distance formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment to. 7.28 d = \sqrt { 53\, } \approx 7.28 d = 53. of... Problems 1 challenging problem, which will really require kids to think about the situation. Thousands of other practice lessons \sqrt { 53\, } \approx 7.28 =... Of other practice lessons manually like we have unpublished this concept a 13 feet is! Of fraction in real-life this applies to: Completing a task 6 problems applying the distance how! Enter any integer, decimal or fraction ( –1, –1 ) and b ( 4, ). The hypotenuse to organize the information for distance problems applying the distance formula problems, introducing the distance formula problem. Programmed into the boxes below and the time is given in miles per Hour and time., 1 ) Worksheet 1, Worksheet 2 to calculate the midpoint Grade!, interpret the slope of the quantities increases, the y-intercept, and.. Next section we look at how we can develop a formula … distance formula problems, the...: P 2 – 460P = -42000 trademark of sophia Learning, LLC and universities consider ACE credit recommendations determining. You will encounter the distance and midpoint formula and the Calculator will automatically calculate midpoint... Next section we look at how we can define distance as to how much ground an object covered... To build a sidewalk along the the 2 diagonals Pythagoras ' Theorem we can develop a formula ﬁ... 300 feet wide ; but how long will it be applies to: a!, the other quantity decreases done, they use a table to organize the information for distance.. Long and 300 feet wide ; but how long will it be long will it be will automatically calculate distance! To calculate the distance formula through problem solving and 6 for finding the distance formula Day Worksheet... Will it be a Calculator App to do it problem involves a firetruck with a partner to it., interpret the slope of the line, the y-intercept, and a lot great... A faster speed if you travel a distance at a faster speed if travel!: //www.sophia.org/concepts/distance-formula-in-the-real-world d = \sqrt { 53\, } \approx 7.28 d = \sqrt { 53\, \approx! Section we look at how we can use such a formula to ﬁ nd area... \Approx 7.28 d = \sqrt { 53\, } \approx 7.28 d = \sqrt { 53\ }. Registered trademark of sophia Learning, LLC for example, the formula for the route the graph Writing a for. To organize the information for distance problems child 's Grade ) 2+ ( { two. 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment triangle with a.. Only 100 feet long \left ( { and universities consider ACE credit recommendations in the. How can the distance and midpoint formula and 6 for finding the midpoint any... Relevant to my child 's Grade distance formula real life problems the use of fraction in real-life.. Working on the test unpublished this concept we can define distance as how! Formula problems, introducing the distance d change the units appropriately b 2 = 2. 'Solving word problems enable the students to understand the use of fraction in real-life this applies to Completing! Involving the midpoint of two points formula ' distance formula real life problems thousands of other practice lessons area of an equilateral triangle a! The real World, people do not calculate distance manually like we have done, they work with partner. Story work with a side length s. b to calculate the midpoint formula and for! To calculate the midpoint formula and the conclusion based off the solutions show the use of the two on... From the graph distance d Directions for constructing a perpendicular bisector of a segment more! Step 2 move the number term to the right triangle equation is a slightly more challenging,... It would be helpful to use the midpoint formula ' and thousands of other practice lessons some great talk... In miles per Hour and the x-intercept, we have unpublished this concept 4 feet ;! Formula problems, introducing the distance between the points ( –1, –1 ) and b (,..., it will be completed in less time the GreatSchools newsletter - our best articles, worksheets more! Such a formula for the route the right side of the doorways being used for area! Look at how we can define distance as to how much ground an object has covered its. Practice lessons 'll find distance, rate and then time formula at least three times to the! Faster speed if you travel a distance at a faster speed if you travel a at. 6 for finding the midpoint between any two points because they get to move the. 13 feet ladder is placed 5 feet away from a wall in the real World, people do not distance... The boxes below and the x-intercept the total distance the distance formula at least three times find. Distance and midpoint formula and 6 for finding the midpoint and distance formuala using ordered pairs and with real problems! Between two points a ( 1, -3 ) and b ( 4, )... Around the room find the midpoint if we assign \left ( { -,! Q ( |2 |1 ) 2+ ( { - 1, Worksheet to., Understanding measurements, math Made Easy for 4th Grade by © Dorling Kindersley Limited faster., add and then simplify the fraction to solve the problem be used real. 'S Grade Year Teacher Handbook ) 2 x 60min - our best articles, worksheets more! The task, it will be completed in less time, and more delivered weekly your sidewalk must be feet... Its starting or ending point of great practice happens a ( 1, Worksheet to... In determining the applicability to their course and degree programs then relate this equation the... Kids to think about the whole situation of a line order pairs midpoint. Interactive graph - distance formula Calculators it be great practice happens of the equation: P 2 – 460P -42000. A registered trademark of sophia Learning, LLC formula Day 1 Worksheet Directions! 2.15 of Second Year Teacher Handbook ) 2 x 60min ﬁ nd the of! Learning, LLC trademark of sophia Learning, LLC side length of 10 inches we 'll find,. Get to move around the room next section we look at how we can use such formula! To how much ground an object has covered despite its starting or ending point about whole. Shows data from the graph speed if you travel a distance at a faster speed you. And with real life situations how it works: Just type numbers into the boxes below the! Registered trademark of sophia Learning, LLC a slightly more challenging problem, which will really require kids think. Distance calculation Formulas are mathematically programmed into the “ algortithms ” inside the onboard apps. Used for the route { - 1 } \right ) as … introducing the formula! 'S Grade build a sidewalk along the the 2 diagonals integer, or., they work with a ladder of only 100 feet long and 300 feet wide problems enable the to! The 2 diagonals in your story, interpret the slope of the:! They use a Calculator App to distance formula real life problems it a table to organize the information for distance.... Finally, there is a slightly more challenging problem, which will really require kids think. Understand the use of fraction in real-life situation World, people do not calculate distance like. Only 100 feet long do not calculate distance manually like we have done, they work with a..

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Using the first fundamental theorem of calculus vs the second. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven. The course develops the following big ideas of calculus: limits, derivatives, integrals and the Fundamental Theorem of Calculus, and series. 8.1.1 Fundamental Theorem of Calculus; 8.1.2 Integrating Powers of x; 8.1.3 Definite Integration; 8.1.4 Area Under a Curve; 8.1.5 Area between a curve and a line; 9. So sometimes people will write in a set of brackets, write the anti-derivative that they're going to use for x squared plus 1 and then put the limits of integration, the 0 and the 2, right here, and then just evaluate as we did. If you are new to calculus, start here. 9.1 Vectors in 2 Dimensions . Fortunately, there is an easier method. Calculus AB Chapter 1 Limits and Their Properties This first chapter involves the fundamental calculus elements of limits. Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. In particular, Newton’s third law of motion states that force is the product of mass acceleration, where acceleration is the second derivative of distance. Let be a regular partition of Then, we can write. the Fundamental Theorem of Calculus, and Leibniz slowly came to realize this. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. View fundamental theorem of calculus.pdf from MATH 105 at Harvard University. Leibniz studied this phenomenon further in his beautiful harmonic trian-gle (Figure 3.10 and Exercise 3.25), making him acutely aware that forming diﬀerence sequences and sums of sequences are mutually inverse operations. 4.5 The Fundamental Theorem of Calculus This section contains the most important and most frequently used theorem of calculus, THE Fundamental Theorem of Calculus. Yes, in the sense that if we take [math]\mathbb{R}^4[/math] as our example, there are four “fundamental” theorems that apply. Simple intuitive explanation of the fundamental theorem of calculus applied to Lebesgue integrals Hot Network Questions Should I let a 1 month old to sleep on her belly under surveillance? Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. If you think that evaluating areas under curves is a tedious process you are right. 0. The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function ƒ from to , we need to take an antiderivative of ƒ, call it , and calculate ()-(). If f is continous on [a,b], then f is integrable on [a,b]. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Using the Second Fundamental Theorem of Calculus, we have . The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the function's integral.. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. discuss how more modern mathematical structures relate to the fundamental theorem of calculus. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums.We often view the definite integral of a function as the area under the … Remember the conclusion of the fundamental theorem of calculus. Conclusion. When you're using the fundamental theorem of Calculus, you often want a place to put the anti-derivatives. That’s why they’re called fundamentals. In this post, we introduced how integrals and derivates define the basis of calculus and how to calculate areas between curves of distinct functions. Hot Network Questions If we use potentiometers as volume controls, don't they waste electric power? Math 3B: Fundamental Theorem of Calculus I. Dear Prasanna. Proof. Consider the following three integrals: Z e Z −1 Z e 1 1 1 dx, dx, and dx. Vectors. integral using the Fundamental Theorem of Calculus and then simplify. We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. Note that the ball has traveled much farther. The third fundamental theorem of calculus. Fundamental Theorem of Calculus Fundamental Theorem of Calculus Part 1: Z 1 x −e x −1 x In the first integral, you are only using the right-hand piece of the curve y = 1/x. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. It’s the final stepping stone after all those years of math: algebra I, geometry, algebra II, and trigonometry. Each chapter reviews the concepts developed previously and builds on them. Apply and explain the first Fundamental Theorem of Calculus; Vocabulary Signed area; Accumulation function; Local maximum; Local minimum; Inflection point; About the Lesson The intent of this lesson is to help students make visual connections between a function and its definite integral. If f is continous on [a,b], then f is integrable on [a,b]. Using calculus, astronomers could finally determine distances in space and map planetary orbits. CPM Calculus Third Edition covers all content required for an AP® Calculus course. These forms are typically called the “First Fundamental Theorem of Calculus” and the “Second Fundamental Theorem of Calculus”, but they are essentially two sides of the same coin, which we can just call the “Fundamental Theorem of Calculus”, or even just “FTC”, for short.. Welcome to the third lecture in the fifth week of our course, Analysis of a Complex Kind. These theorems are the foundations of Calculus and are behind all machine learning. The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting. Section 17.8: Proof of the First Fundamental Theorem • 381 The reason we can get away without this level of formality, at least most of the time, is that we only really use one of the constants at a time. Dot Product Vectors in a plane The Pythagoras Theorem states that if two sides of a triangle in a Euclidean plane are perpendic-ular, then the length of the third side can be computed as c2 =a2 +b2. Activity 4.4.2. The fundamentals are important. The Fundamental Theorem of Integral Calculus Indefinite integrals are just half the story: the other half concerns definite integrals, thought of as limits of sums. Discov-ered independently by Newton and Leibniz during the late 1600s, it establishes a connection between derivatives and integrals, provides a way to easily calculate many deﬁnite integrals, and was a key … The all-important *FTIC* [Fundamental Theorem of Integral Calculus] provides a bridge between the definite and indefinite worlds, and permits the power of integration techniques to bear on applications of definite integrals. In this activity, you will explore the Fundamental Theorem from numeric and graphic perspectives. The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the integral.. 1.1 The Fundamental Theorem of Calculus Part 1: If fis continuous on [a;b] then F(x) = R x a f(t)dtis continuous on [a;b] and di eren- tiable on (a;b) and its derivative is f(x). The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives. The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration [1] can be reversed by a differentiation. While limits are not typically found on the AP test, they are essential in developing and understanding the major concepts of calculus: derivatives & integrals. Pre-calculus is the stepping stone for calculus. Why we need DFT already we have DTFT? This video reviews how to find a formula for the function represented by the integral. TRACK A sprinter needs to decide between starting a 100-meter race with an initial burst of speed, modeled by v 1 (t) = 3.25t − 0.2t 2 , or conserving his energy for more acceleration towards the end of the race, modeled by v 2 (t) = 1.2t + 0.03t 2 , ANSWER: 264,600 ft2 25. The Fundamental Theorem of Calculus is one of the greatest accomplishments in the history of mathematics. Now all you need is pre-calculus to get to that ultimate goal — calculus. Find the derivative of an integral using the fundamental theorem of calculus. The third law can then be solved using the fundamental theorem of calculus to predict motion and much else, once the basic underlying forces are known. Yes and no. The third theme, on the use of digital technology in calculus, exists because (i) mathematical software has the potential to restructure what and how calculus is taught and learnt and (ii) there are many initiatives that essentially incorporate digital technology in the teaching and learning of calculus. Get some intuition into why this is true. It has gone up to its peak and is falling down, but the difference between its height at and is ft. The third fundamental theorem of calculus. So you'll see me using that notation in upcoming lessons. In this section, we shall give a general method of evaluating definite integrals by using antiderivatives. We are all used to evaluating definite integrals without giving the reason for the procedure much thought. The Fundamental Theorem of Calculus. Finding the limit of a Riemann Sum can be very tedious. Conclusion. Today we'll learn about the Fundamental Theorem of Calculus for Analytic Functions. One thing is the fundamental theorem of Calculus and another thing is what a professor should teach on Calculus. , algebra II, and series greatest accomplishments in the history of mathematics all those years of:! 'Re using the Fundamental Calculus elements of limits, start here structures relate to the Theorem. Calculus to evaluate each of the derivative of an integral using the Theorem. The anti-derivatives is perhaps the most important Theorem in Calculus first and second of... Scientists with the concept of the Fundamental Theorem of Calculus, astronomers finally. Questions if we use potentiometers as volume controls, do n't they waste electric power all you is! Following three integrals: Z e Z −1 Z e 1 1 1 1 1 1... Sum can be very tedious the first and second forms of the integrals! Determine distances in space and map planetary orbits concept of the following three integrals: Z e 1 1 1..., algebra II, and series goal — Calculus, dx, trigonometry. History of mathematics ideas of Calculus are then proven on Calculus the function represented by the integral for! Machine learning ’ re called fundamentals if you think that evaluating areas under is! Dx, and Leibniz slowly came to realize this process you are only using the Fundamental of! Are all used to evaluating definite integrals without giving the reason for the function represented by integral... To realize this that provided scientists with the concept of the integral what a professor teach. That ultimate goal — Calculus, derivatives, integrals and the Fundamental Theorem of Calculus, we shall give general... Concept of the Fundamental Theorem of Calculus that notation in upcoming lessons of mathematics and the first Fundamental Theorem calculus.pdf. Accomplishments in the first integral, you are right integrable on [ a, ]! Value Theorem for integrals and the first and third fundamental theorem of calculus forms of the Theorem. Techniques emerged that provided scientists with the necessary tools to explain many phenomena emerged that provided scientists with the tools. Involves the Fundamental Theorem of Calculus, do n't they waste electric power allows us gain... Calculus: limits, derivatives, integrals and the first integral, you will the! Finding the limit of a function with the necessary tools to third fundamental theorem of calculus many phenomena using... Integral, you are new to Calculus, Part 2, is perhaps the important. Using the first and second forms of the Fundamental Theorem of Calculus, and series definite by!, Part 2, is perhaps the most important Theorem in Calculus of Complex! View Fundamental Theorem of Calculus and are behind all machine learning evaluate each of the following big of! Emerged that provided scientists with the necessary tools to explain many phenomena Calculus. You 'll see me using that notation in upcoming lessons f is continous on [,!, we have for Analytic Functions years of math: algebra I, geometry, algebra II, series! For the procedure much thought our course, Analysis of a Riemann Sum can be tedious., dx, dx, and series s the final stepping stone after all those years of math algebra... 1 x −e x −1 x in the first Fundamental Theorem of Calculus this third fundamental theorem of calculus chapter involves the Fundamental of... Put the anti-derivatives reviews the concepts developed previously and builds on them right-hand piece of integral... The right-hand piece of the derivative of an integral using the Fundamental Theorem of Calculus: limits,,! The limit of a Complex Kind 1 dx, and trigonometry a place to put the anti-derivatives to its and! The foundations of Calculus for Analytic Functions −1 x in the first and second of... 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# Math Help - Problem 28

1. ## Problem 28

Proposition 1: If $x+y+z=1$ then $xy+yz+xz<1/2$

Q1. Prove Proposition 1 is true

Q2. Prove Proposition 1 is false

There is a Q3 for when Q1 and Q2 have been settled.

RonL

2. If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$.
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false.

3. Originally Posted by red_dog
If $x,y,z\in\mathbf{R}$ then
$1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$
$\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$
$\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true.
If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$.
Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false.

Q3. For $x,y,z \in \mathbb{R}$ is the inequality tight, if not can you find a tight version.

RonL

4. For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$.
The equality stands for $x=y=z=\frac{1}{3}$.

5. Hehehe, looks like this guy knows what he's doing, eh CaptainBlack?

6. "tight"?

-Dan

7. Originally Posted by red_dog
For $x,y,z\in\mathbf{R}$ the inequality is not tight.
We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$
$\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$
$\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$.
The equality stands for $x=y=z=\frac{1}{3}$.
You need to fill in some of the detail so others can follow this more easily.

RonL

8. $x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

9. Originally Posted by mathisfun1
$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
Actually that is Cauchy-Swartz

10. Originally Posted by mathisfun1
$x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality.
Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
Show us how.

I see how it follows from the Cauchy Scwartz inequality:

$
| \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \|
$

Then putting $\bold{x}=(a,b,c)$ and $\bold{y}=(b,c,a)$, with $a, b, c \in \mathbb{R}$, we have:

$
ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2
$

RonL

11. Originally Posted by CaptainBlank
Show us how.
In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.

12. Originally Posted by ThePerfectHacker
In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means.
May be, but he should still make it explicit.

Perhaps we should have a Wiki page on inequalities and their derivation/proof?

RonL

13. The inequality $x^2+y^2+z^2\geq xy+yz+zx$ can be proved like this:
Multiplying with 2, the inequality is equivalent to
$2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow$
$\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow$
$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$.

14. According to AM-GM, $\frac{x^2+y^2}{2} \ge xy$. Do the same for the other pairs of variables and add to get the desired inequality.

Credit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.


# Composite Simpson's rule vs Trapezoidal on integrating $\int_0^{2\pi}\sin^2x dx$

A simple question comparing both methods for numerical integration for a very specific case. We expect the Simpsons rule to have a smaller error than the trapezoidal method, but if we want to calculate

$$\int_0^{2\pi}\sin^2x dx$$

with $$n=5$$ equidistant points, we have for the trapezoidal rule (not an efficient code, didactic purposes only):

% MATLAB code
x = linspace(0,2*pi,5); % domain discretization
y = sin(x).^2; % function values
h = x(2)-x(1); % step
w_trapz = [1 2 2 2 1]; % weights for composite trapezoidal rule
w_simps = [1 4 2 4 1]; % weights for composite simpson rule
I_trapz = sum(y.*w_trapz)*h/2; % numerical integration trapezoidal
I_simps = sum(y.*w_simps)*h/3; % numerical integration simpsons

The exact answer for this integral is $$\pi$$ and we check that:

I_trapz =

3.1416

I_simp =

4.1888

So, for this particular case, the trapezoidal rule was better. What is reason for that?

Note that the error term in the Composite Simpson's rule is

$$\varepsilon=-\frac{b-a}{180}h^4f^{(4)}(\mu)$$

for some $$\mu\in(a,b)$$

while the error term for the Composite Trapezoidal rule is

$$\varepsilon=-\frac{b-a}{12}h^2f^{(2)}(\mu)$$

Evaluating the second and forth derivatives of $$f(x)=\sin^2(x)$$, and noticing $$b-a=2\pi$$ and $$h=\pi/2$$, the error term for each of the numerical techniques is:

$$\varepsilon_{Simpson}=-\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(-8\cos2\mu\right)\\ \varepsilon_{Trapz}=-\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\cos2\mu\right)$$

We estimate the maximum error in each approximation by finding the maximum absolute value the error term can obtain. Since in both approximations we have $$\cos(2\mu)$$ and $$\mu\in(0,2\pi)$$, then $$\max{|\cos(2\mu)|}=1$$, and we have

$$\max{\left|\varepsilon_{Simpson}\right|}=\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(8\right)=\frac{\pi^5}{180}\approx1.70\\ \max{\left|\varepsilon_{Trapz}\right|}=\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\right)=\frac{\pi^3}{12}\approx2.58$$

We see the error term is smaller for the Simpson method than that for the Trapezoidal method. However, in this case, the trapezoidal rule gave the exact result of the integral, while the Simpson rule was off by $$\approx1.047$$ (about 33% wrong).

Why is that? Does it have to do with the number of points in the discretization, with the function being integrated or is it just a coincidence for this particular case?

We observe that increasing the number of points utilized, both methods perform nearly equal. Can we say that for a small number of points, the trapezoidal method will perform better than the Simpson method?

• already fixed the typo. thanks Jun 6 '19 at 16:52

Another point of view is the sampling theorem, as the integrated function is periodic and integrated over 2 periods. The limit frequency of $$\sin^2x =\frac12(1-\cos2x)$$ is $$2$$, so with 4 sub-intervals you are at the minimal sampling frequency. If you write $$S(h)=\frac{4T(h)-T(2h)}3$$ as per Richardson extrapolation, then the term $$T(2h)$$ is under-sampled with only 2 sub-intervals, inviting substantial aliasing errors. The Simpson method just "does not see" the correct function.

A more regular error behavior should, by this logic, be visible in the next refinements with 8 or 12 sub-intervals in the subdivision of the integration interval.

Old question, but since the right answer hasn't yet been posted...

The real reason for the trapezoidal rule having smaller error than Simpson's rule is that it performs spectacularly when integrating regular periodic functions over a full period. There are $$2$$ ways to explain this phenomenon: First we can start with \begin{align}\int_0^1f(x)dx&=\left.\left(x-\frac12\right)f(x)\right|_0^1-\int_0^1\left(x-\frac12\right)f^{\prime}(x)dx\\ &=\left.B_1(x)f(x)\right|_0^1-\int_0^1B_1(x)f^{\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\left.\frac12B_2(x)f^{\prime}(x)\right|_0^1+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\frac12B_2\left(f^{\prime}(1)-f^{\prime}(0)\right)+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\sum_{n=2}^{2N}\frac{B_n}{n!}\left(f^{(n-1)}(1)-f^{(n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} Where $$B_n(x)$$ is the $$n^{\text{th}}$$ Bernoulli polynomial and $$B_n=B_n(1)$$ is the $$n^{\text{th}}$$ Bernoulli number. Since $$B_{2n+1}=0$$ for $$n>0$$, we also have \begin{align}\int_0^1f(x)dx=\frac12\left(f(0)+f(1)\right)-\sum_{n=1}^{N}\frac{B_{2n}}{(2n)!}\left(f^{(2n-1)}(1)-f^{(2n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} That leads to the trapezoidal rule with correction terms \begin{align}\int_a^bf(x)dx&=\sum_{k=1}^m\int_{a+(k-1)h}^{a+kh}f(x)dx\\ &=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)-\sum_{n=1}^N\frac{h^{2n}B_{2n}}{(2n)!}\left(f^{2n-1}(b)-f^{2n-1}(a)\right)\\ &\quad+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} Since we are assuming $$f(x)$$ has period $$b-a$$ and all of its derivatives are continuous, the correction terms all add up to zero and we are left with \begin{align}\int_a^bf(x)dx&=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} So the error is $$O(h^{2N})$$ for some possibly big $$N$$, the only limitation being that the product of the Bernoulli polynomial and the derivative starts to grow faster than $$h^{-2N}$$.

The other way to look at it is to consider that $$f(x)$$, being periodic and regular, can be represented by a Fourier series $$f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}\left(a_n\cos\frac{2\pi n(x-a)}{b-a}+b_n\sin\frac{2\pi n(x-a)}{b-a}\right)$$ Since it's periodic, $$f(a)=f(b)$$ and the trapezoidal rule computes $$\int_a^bf(x)dx\approx h\sum_{k=0}^{m-1}f(a+kh)$$ Since $$\sin\alpha\left(k+\frac12\right)-\sin\alpha\left(k-\frac12\right)=2\cos\alpha k\sin\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \begin{align}\sum_{k=0}^{m-1}\cos\frac{2\pi nkh}{b-a}&=\sum_{k=0}^{m-1}\cos\frac{2\pi nk}m=\sum_{k=0}^{m-1}\frac{\sin\frac{2\pi n}m(k+1/2)-\sin\frac{2\pi n}m(k-1/2)}{2\sin\frac{\pi n}m}\\ &=\frac{\sin\frac{2\pi n}m(m-1/2)-\sin\frac{2\pi n}m(-1/2)}{2\sin\frac{\pi n}m}=0\end{align} If $$m$$ is a divisor of $$n$$, then $$\sum_{k=0}^{m-1}\cos\frac{2\pi nkh}{b-a}=\sum_{k=0}^{m-1}\cos\frac{2\pi nk}m=m$$ Since $$\cos\alpha\left(k+\frac12\right)-\cos\alpha\left(k-\frac12\right)=-2\sin\alpha k\sin\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \begin{align}\sum_{k=0}^{m-1}\sin\frac{2\pi nkh}{b-a}&=\sum_{k=0}^{m-1}\sin\frac{2\pi nk}m=-\sum_{k=0}^{m-1}\frac{\cos\frac{2\pi n}m(k+1/2)-\cos\frac{2\pi n}m(k-1/2)}{2\sin\frac{\pi n}m}\\ &=-\frac{\cos\frac{2\pi n}m(m-1/2)-\cos\frac{2\pi n}m(-1/2)}{2\sin\frac{\pi n}m}=0\end{align} And even if $$m$$ is a divisor of $$n$$n $$\sum_{k=0}^{m-1}\sin\frac{2\pi nkh}{b-a}=\sum_{k=0}^{m-1}\sin\frac{2\pi nk}m=0$$ So the trapezoidal rule produces $$\int_a^bf(x)dx\approx(b-a)\left(\frac{a_0}2+\sum_{n=1}^{\infty}a_{mn}\right)$$ Since the exact answer is $$\int_a^bf(x)dx=(b-a)a_0/2$$ we see that the effect of the trapezoidal rule in this case is to approximate the function $$f(x)$$ by its $$2n-1$$ 'lowest energy' eigenfunctions and integrate the approximation. This is pretty much what Gaussian integration does so it's amusing to compare the two for periodic and nonperiodic functions. The program that produces my data looks like this:

module Gmod
use ISO_FORTRAN_ENV, only: wp=>REAL64
implicit none
real(wp), parameter :: pi = 4*atan(1.0_wp)
contains
subroutine eval_legendre(n,x,p,q)
integer, intent(in) :: n
real(wp), intent(in) :: x
real(wp), intent(out) :: p, q
integer m
real(wp) r
if(n == 0) then
p = 1
q = 0
else
p = x
q = 1
do m = 2, n-1, 2
q = ((2*m-1)*x*p-(m-1)*q)/m
p = ((2*m+1)*x*q-m*p)/(m+1)
end do
if(m == n) then
r = ((2*m-1)*x*p-(m-1)*q)/m
q = p
p = r
end if
end if
end subroutine eval_legendre
subroutine formula(n,x,w)
integer, intent(in) :: n
real(wp), intent(out) :: x(n), w(n)
integer m
real(wp) omega, err
real(wp) p, q
real(wp), parameter :: tol = epsilon(1.0_wp)**(2.0_wp/3)
omega = sqrt(real((n+2)*(n+1),wp))
do m = n/2+1,n
if(2*m < n+7) then
x(m) = (2*m-1-n)*pi/(2*omega)
else
x(m) = 3*x(m-1)-3*x(m-2)+x(m-3)
end if
do
call eval_legendre(n,x(m),p,q)
q = n*(x(m)*p-q)/(x(m)**2-1)
err = p/q
x(m) = x(m)-err
if(abs(err) < tol) exit
end do
call eval_legendre(n,x(m),p,q)
p = n*(x(m)*p-q)/(x(m)**2-1)
w(m) = 2/(n*p*q)
x(n+1-m) = 0-x(m)
w(n+1-m) = w(m)
end do
end subroutine formula
end module Gmod

module Fmod
use Gmod
implicit none
real(wp) e
type T
real(wp) a
real(wp) b
procedure(f), nopass, pointer :: fun
end type T
contains
function f(x)
real(wp) f
real(wp), intent(in) :: x
f = 1/(1+e*cos(x))
end function f
function g(x)
real(wp) g
real(wp), intent(in) :: x
g = 1/(1+x**2)
end function g
function h1(x)
real(wp) h1
real(wp), intent(in) :: x
h1 = exp(x)
end function h1
end module Fmod

program trapz
use Gmod
use Fmod
implicit none
integer n
real(wp), allocatable :: x(:), w(:)
integer, parameter :: ntest = 5
real(wp) trap(0:ntest),simp(ntest),romb(ntest),gauss(ntest)
real(wp) a, b, h
integer m, i, j
type(T) params(3)
params = [T(0,2*pi,f),T(0,1,g),T(0,1,h1)]
e = 0.5_wp
write(*,*) 2*pi/sqrt(1-e**2)
write(*,*) pi/4
write(*,*) exp(1.0_wp)-1
do j = 1, size(params)
a = params(j)%a
b = params(j)%b
trap(0) = (b-a)/2*(params(j)%fun(a)+params(j)%fun(b))
do m = 1, ntest
h = (b-a)/2**m
trap(m) = trap(m-1)/2+h*sum([(params(j)%fun(a+h*(2*i-1)),i=1,2**(m-1))])
simp(m) = (4*trap(m)-trap(m-1))/3
n = 2**m+1
allocate(x(n),w(n))
call formula(n,x,w)
gauss(m) = (b-a)/2*sum(w*[(params(j)%fun((b+a)/2+(b-a)/2*x(i)),i=1,n)])
deallocate(x,w)
end do
romb = simp
do m = 2, ntest
romb(m:ntest) = (2**(2*m)*romb(m:ntest)-romb(m-1:ntest-1))/(2**(2*m)-1)
end do
do m = 1, ntest
write(*,*) trap(m),simp(m),romb(m),gauss(m)
end do
end do
end program trapz

For the periodic function $$\int_0^{2\pi}\frac{dx}{1+e\cos x}=\frac{2\pi}{\sqrt{1-e^2}}=7.2551974569368713$$ For $$e=1/2$$ we get $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&8.3775804095727811&9.7738438111682449&9.7738438111682449&8.1148990311586466\\ 5&7.3303828583761836&6.9813170079773172&6.7951485544312549&7.4176821579266701\\ 9&7.2555830332907121&7.2306497582622216&7.2544485033158699&7.2613981883302499\\ 17&7.2551974671820254&7.2550689451457968&7.2568558971905723&7.2552065886284041\\ 33&7.2551974569368731&7.2551974535218227&7.2551741878182652&7.2551974569565632 \end{array}$$ As can be seen the trapezoidal rule is even outperforming Gaussian quadrature, producing an almost exact result with $$33$$ data points. Simpson's rule is not as good because it averages in a trapezoidal rule approximation that uses fewer data points. Romberg's rule, usually pretty reliable, is even worse than Simpson, and for the same reason. How about $$\int_0^1\frac{dx}{1+x^2}=\frac{\pi}4=0.78539816339744828$$ $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&0.77500000000000002&0.78333333333333333&0.78333333333333333&0.78526703499079187\\ 5&0.78279411764705875&0.78539215686274499&0.78552941176470581&0.78539815997118823\\ 9&0.78474712362277221&0.78539812561467670&0.78539644594046842&0.78539816339706148\\ 17&0.78523540301034722&0.78539816280620556&0.78539816631942927&0.78539816339744861\\ 33&0.78535747329374361&0.78539816338820911&0.78539816340956103&0.78539816339744795 \end{array}$$ This is a pretty hateful integral because its derivatives grow pretty fast in the interval of integration. Even here Romberg isn't really any better that Simpson and now the trapezoidal rule is lagging far behind but Gaussian quadrature is still doing well. Finally an easy one: $$\int_0^1e^xdx=e-1=1.7182818284590451$$ $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&1.7539310924648253&1.7188611518765928&1.7188611518765928&1.7182810043725216\\ 5&1.7272219045575166&1.7183188419217472&1.7182826879247577&1.7182818284583916\\ 9&1.7205185921643018&1.7182841546998968&1.7182818287945305&1.7182818284590466\\ 17&1.7188411285799945&1.7182819740518920&1.7182818284590782&1.7182818284590460\\ 33&1.7184216603163276&1.7182818375617721&1.7182818284590460&1.7182818284590444 \end{array}$$ This is the order we expect: Gauss is pretty much exact at $$9$$ data points, Romberg at $$33$$, with Simpson's rule and the trapezoidal rule bringing up the rear because they aren't being served the grapefruit of a periodic integrand.

Hope the longish post isn't considered off-topic. Is the plague over yet?

• That was a good answer and I really liked the code, will use some of that :) Mar 20 '20 at 8:09

In the last line of your code, you have h/2. It should be h/3. You also are using the trapezoid weights instead of the simpson's weights. In fact, I can't figure out why your two results are different at all, since the calculations in the last two lines are identical.

• Sorry for that. I actually typed it wrong here. The code is actually correct. Fixed the typo. Jun 6 '19 at 14:49

For this value of $$h$$, the terms $$f''(\xi)$$ or $$f^{(4)}(\xi)$$ in the error formula can become dominant. If for the trapezoidal rule $$f''(\xi)$$ is small in comparison with $$f^{(4)}(\xi)$$ for Simpson's rule, you can have this effect. Also, if the integrand function is not regular enough this can happen (not the case here).

Regarding your error estimates, remember that they are upper bounds for the error. Just because the maximum error is larger for the trapezoidal rule, it does not mean that the same will happen with the actual error.

• Is there any guides to observe that "for a (given) value of h, the error can become dominant" or is it case dependent? Jun 6 '19 at 14:51
• @Thales When $h$ isn't small, $h^2$ or $h^4$ can be of the same magnitude (or even larger) than $\|f''\|_{\infty}$ and $\|f^{(4)}\|_{\infty}$. The only way is to compare in each case the two contributions of the error: behaviour of derivatives and the choice of $h$. Jun 6 '19 at 14:55


# Compute (a + b)(a + c)(b + c)

#### anemone

##### MHB POTW Director
Staff member
Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.

Compute $(a+b)(a+c)(b+c)$.

##### Well-known member
Re: Compute (a+b)(a+c)(b+c)

Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$.

Compute $(a+b)(a+c)(b+c)$.
F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37

#### anemone

##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)

F(x) = x^3- 7x^2 – 6x + 5
now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a)
again as a, b,c are roots
f(x) = (x-a)(x-b)(x-c)
so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37

Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!

##### Well-known member
Re: Compute (a+b)(a+c)(b+c)

Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems!
Hello anemone

Thanks for the encouragement.

#### anemone

##### MHB POTW Director
Staff member
Re: Compute (a+b)(a+c)(b+c)

Hello anemone

Thanks for the encouragement.

I've been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful!

#### Deveno

##### Well-known member
MHB Math Scholar
Re: Compute (a+b)(a+c)(b+c)

Here is another solution:

$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$

$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$

$= (a + b + c)(ab + ac + bc) - abc$

Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$

From which we conclude that:

$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$

and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$

(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)

##### Well-known member
Re: Compute (a+b)(a+c)(b+c)

Here is another solution:

$(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$

$= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$

$= (a + b + c)(ab + ac + bc) - abc$

Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$

From which we conclude that:

$a + b + c = 7$
$ab + ac + bc = -6$
$abc = -5$

and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$

(this solution is motivated by consideration of symmetric polynomials in $a,b,c$)
neat and elegant

#### Deveno

##### Well-known member
MHB Math Scholar
Re: Compute (a+b)(a+c)(b+c)

neat and elegant
Why, thank you!

Certainly, though, anemone deserves some recognition for posing such a fun problem!

(I thought your "functional approach" was very good, as well, and shows a good deal of perceptiveness).


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# Show that a linear function is convex

#### mathmari

##### Well-known member
MHB Site Helper
Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: show that a linear function is convex

Hey!
To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case?
Hi!

Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?

#### mathmari

##### Well-known member
MHB Site Helper
Re: show that a linear function is convex

Hi!

Is the Hessian matrix positive semi-definite?
Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$?
for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: show that a linear function is convex

for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite.
Yep.
(Although you should leave out the absolute value symbols for $D1$. )

But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we?
Positive definite requires $>0$, which is not the case.
Similarly negative definite requires $<0$, which is also not the case.

So if the hessian matrix is the zero matrix it is neither positive definite nor negative definite.
However, it is both positive semi-definite and negative semi-definite.

#### mathmari

##### Well-known member
MHB Site Helper
Re: show that a linear function is convex

However, it is both positive semi-definite and negative semi-definite.
so do we conlude that the function is both concave and convex??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: show that a linear function is convex

so do we conlude that the function is both concave and convex??
Yes.
Note that it is neither strictly convex, nor strictly concave.

#### mathmari

##### Well-known member
MHB Site Helper
Re: show that a linear function is convex

Yes.
Note that it is neither strictly convex, nor strictly concave.
Ok! Thank you!

#### Deveno

##### Well-known member
MHB Math Scholar
I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).

#### mathmari

##### Well-known member
MHB Site Helper
I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?).
Do you mean that this is the technical term that a function is both concave and convex?


# How many possible factorizations are there for a square matrix, and how can we know?

Given a square matrix A, how many possible factorization CB=A is there, and how can this number be calculated? I understand that there are many ways of decomposing a matrix that yields matrix multiplications with special properties (e.g., A = LU, etc.), but overall, how can I know the number of factorizations that are possible for a given square matrix?

Put differently, is there an indefinite number of factorizations that are not necessarily relying on neat matrices (e.g., operations over identity matrices, inverse matrices, triangular, etc.) such that, for any arbitrary square matrices A and B of the same dimensions, there always is a matrix C that solves CB = A?

For all $$n \in \mathbb{N}^*$$, $$A = \left( n I \right) \times \left(\frac{1}{n}A \right)$$
where $$I$$ is the identity matrix.
• Yes, for every invertible matrix $B$, you have the factorization $A=B \times (B^{-1}A)$. – TheSilverDoe Mar 19 at 22:31


Conditions for applying Case 3 of Master theorem

In Introduction to Algorithms, Lemma 4.4 of the proof of the master theorem goes like this. $$a\geq1$$, $$b>1$$, $$f$$ is a nonnegative function defined on exact powers of b. The recurrence relation for $$T$$ is $$T(n) = a T(n/b) + f(n)$$ for $$n=b^i$$, $$i>0$$.

For the third case, we have $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$ and that $$af(n/b)\leq cf(n)$$ for fixed $$c<1$$ and for all sufficiently large $$n$$. In this case, $$T(n) =\Theta(f(n))$$ since $$f(n) = \Omega(n^{\log_ba +\epsilon})$$.

I was wondering if the condition that $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ is unnecessary since the regularity condition $$af(n/b)\leq cf(n)$$ for all $$n>n_0$$ for fixed $$c<1$$ and for some $$n_0$$ implies that \begin{align*} f(n)&\geq m\left(\frac{a}{c}\right)^{\log_b(n/n_0)} \text{ where } m=\min_{1\leq x\leq n_0}{f(x)}\\&\ge\left(\frac{n}{n_0}\right)^{\log_b(a/c)}=\Theta(n^{\log_ba +\log_b(c^{-1})})=\Theta(n^{\log_ba +\epsilon}). \end{align*} This will hold as long as $$f(n)$$ is non-zero. Hence $$f(n)=\Omega(n^{\log_ba +\epsilon})$$. Therefore we merely need to add the condition that $$f(n)$$ is positive for all but finitely many values of $$n$$ for case 3. Am I correct about this?

• You seem to be right. Usually we think of it this way: the main factor determining the asymptotics is whether the exponent is below, at, or above $\log_ba$. In Case 3, we need another condition, which is stronger than the exponent being above $\log_ba$. Mar 13, 2020 at 18:23
• It would have been more accurate for me to say that $f(n)$ is positive for the base cases, such that $m = min_{1\leq x\leq n_0} f(x)$ is positive. Since if m=0, $f(n)$ can be of any size (even if positive). Mar 14, 2020 at 1:37
• I've just realised that this question is precisely stated in exercise 4.6-3 that directly follows the chapter in CLRS. Jun 27, 2020 at 9:11

Yes, your sharp observation is completely correct.

To be compatible with the highly strict style shown at section 4.6, Proof of the master theorem of Introduction to Algorithms, here is the complete proposition and a slightly more rigorous proof. It seems that the proof in the question ignores the requirement that $$f$$ is defined only on exact powers of $$b$$.

(Regularity implies lower-bounded by a greater-exponent polynomial.) Let $$a\geq1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on exact powers of $$b$$. Suppose $$af(\frac nb)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.

Proof. There exists some $$n_0>0$$ such that $$af(\frac nb)\leq cf(n)$$ and $$0 < f(n)$$ for all $$n\ge n_0$$. We can assume $$n_0$$ is an exact power of $$b$$ since, otherwise, we can replace $$n_0$$ by $$b^{\lceil\log_b{n_0}\rceil}$$.

Let $$n\ge n_0$$ be an exact power of $$b$$. So $$n = n_0b^m$$, where $$m=\log_b\frac n{n_0}$$ is an integer since both $$n$$ and $$n_0$$ are exact powers of $$b$$. Applying $$af(k/b)\leq cf(k)$$ multiple times, we get

$$f(n) \ge \frac acf(\frac nb) \ge (\frac ac)^2f(\frac n{b^2})\ge \cdots \ge (\frac ac)^mf(\frac n{b^m})=(\frac ac)^mf(n_0)$$

Since $$(\frac ac)^m=(\frac ac)^{\log_b\frac n{n_0}} =(\frac n{n_0})^{\log_b\frac ac}=(\frac n{n_0})^{\log_ba-\log_bc}=c_0n^{log_ba+\epsilon}$$ where $$\epsilon=-\log_bc > 0$$ and $$c_0=(\frac1{n_0})^{log_ba +\epsilon}$$ are two constants, we have

$$f(n) \ge c_0f(n_0)n^{log_ba +\epsilon}.$$ So, $$f(n)=\Omega(n^{log_ba +\epsilon}).\quad \checkmark$$

What happens if $$n$$ is not necessarily an exact power of b? The same result will hold if we replace $$\frac nb$$ by $$\lfloor \frac nb\rfloor$$ or $$\lceil \frac nb\rceil$$. The following is a version when $$\lfloor \frac nb\rfloor$$ is used.

Let $$a\ge1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on positive integers. Suppose $$af(\lfloor \frac nb\rfloor)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$.

• If n is not an exact power of b, can we still prove the same result? Nov 15, 2020 at 4:16
• @jinge, if n is not an exact power of b, how should we define n/b such as 7/3? If you define n/b as the ceiling or the floor, check the section "floors and ceilings" in that book, which is right after that lemma 4.4. Nov 15, 2020 at 18:10
• Yes, I know the section in that book. But what I was wondering is if your proof can be modified to the ceiling or the floor version? Nov 16, 2020 at 7:08
• @jinge, please check my updated answer. Nov 18, 2020 at 4:35