Exercises and Problems in Calculus
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Abel's Identity, 131 Abel's Test, 131–132 Abel's Theorem, 463–464 Absolute Convergence, 113–114 Implication of Condi
INDEX Abel’s identity, 131 Bolzano-Weierstrass theorem, 66–68 Abel’s test, 131–132 for sequences, 99–100 Abel’s theorem, 463–464 boundary points, 50–52 absolute convergence, 113–114 bounded functions, 142 implication of conditional bounded sets, 42–43 convergence, 114 absolute value, 7 Cantor function, 229–230 reverse triangle inequality, 9 Cantor set, 80–81, 133–134, 383 triangle inequality, 9 Casorati-Weierstrass theorem, 498–499 algebraic properties of Rk, 11–13 Cauchy completeness, 106–107 algebraic properties of continuity, 184 Cauchy condensation test, 117–119 algebraic properties of limits, 91–92 Cauchy integral theorem algebraic properties of series, 110–111 triangle lemma, see triangle lemma algebraic properties of the derivative, Cauchy principal value, 365–366 244 Cauchy sequences, 104–105 alternating series, 115 convergence of, 105–106 alternating series test, 115–116 Cauchy’s inequalities, 437 analytic functions, 481–482 Cauchy’s integral formula, 428–433 complex analytic functions, 483 converse of, 436–437 counterexamples, 482–483 extension to higher derivatives, 437 identity principle, 486 for simple closed contours, 432–433 zeros of, see zeros of complex analytic on circles, 429–430 functions on open connected sets, 430–432 antiderivatives, 361 on star-shaped sets, 429 for f :[a, b] Rp, 370 → Cauchy’s integral theorem, 420–422 of complex functions, 408 consequence, see deformation of on star-shaped sets, 411–413 contours path-independence, 408–409, 415 Cauchy-Riemann equations Archimedean property, 5–6 motivation, 297–300 -
Sequences and Series Pg. 1
Sequences and Series Exercise Find the first four terms of the sequence 3 1, 1, 2, 3, . [Answer: 2, 5, 8, 11] Example Let be the sequence defined by 2, 1, and 21 13 2 for 0,1,2,, find and . Solution: (k = 0) 21 13 2 2 2 3 (k = 1) 32 24 2 6 2 8 ⁄ Exercise Without using a calculator, find the first 4 terms of the recursively defined sequence • 5, 2 3 [Answer: 5, 7, 11, 19] • 4, 1 [Answer: 4, , , ] • 2, 3, [Answer: 2, 3, 5, 8] Example Evaluate ∑ 0.5 and round your answer to 5 decimal places. ! Solution: ∑ 0.5 ! 0.5 0.5 0.5 ! ! ! 0.5 0.5 ! ! 0.5 0.5 0.5 0.5 0.5 0.5 0.041666 0.003125 0.000186 0.000009 . Exercise Find and evaluate each of the following sums • ∑ [Answer: 225] • ∑ 1 5 [Answer: 521] pg. 1 Sequences and Series Arithmetic Sequences an = an‐1 + d, also an = a1 + (n – 1)d n S = ()a +a n 2 1 n Exercise • For the arithmetic sequence 4, 7, 10, 13, (a) find the 2011th term; [Answer: 6,034] (b) which term is 295? [Answer: 98th term] (c) find the sum of the first 750 terms. [Answer: 845,625] rd th • The 3 term of an arithmetic sequence is 8 and the 16 term is 47. Find and d and construct the sequence. [Answer: 2,3;the sequence is 2,5,8,11,] • Find the sum of the first 800 natural numbers. [Answer: 320,400] • Find the sum ∑ 4 5. -
Lesson 2: the Multiplication of Polynomials
NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 2 M1 ALGEBRA II Lesson 2: The Multiplication of Polynomials Student Outcomes . Students develop the distributive property for application to polynomial multiplication. Students connect multiplication of polynomials with multiplication of multi-digit integers. Lesson Notes This lesson begins to address standards A-SSE.A.2 and A-APR.C.4 directly and provides opportunities for students to practice MP.7 and MP.8. The work is scaffolded to allow students to discern patterns in repeated calculations, leading to some general polynomial identities that are explored further in the remaining lessons of this module. As in the last lesson, if students struggle with this lesson, they may need to review concepts covered in previous grades, such as: The connection between area properties and the distributive property: Grade 7, Module 6, Lesson 21. Introduction to the table method of multiplying polynomials: Algebra I, Module 1, Lesson 9. Multiplying polynomials (in the context of quadratics): Algebra I, Module 4, Lessons 1 and 2. Since division is the inverse operation of multiplication, it is important to make sure that your students understand how to multiply polynomials before moving on to division of polynomials in Lesson 3 of this module. In Lesson 3, division is explored using the reverse tabular method, so it is important for students to work through the table diagrams in this lesson to prepare them for the upcoming work. There continues to be a sharp distinction in this curriculum between justification and proof, such as justifying the identity (푎 + 푏)2 = 푎2 + 2푎푏 + 푏 using area properties and proving the identity using the distributive property. -
James Clerk Maxwell
James Clerk Maxwell JAMES CLERK MAXWELL Perspectives on his Life and Work Edited by raymond flood mark mccartney and andrew whitaker 3 3 Great Clarendon Street, Oxford, OX2 6DP, United Kingdom Oxford University Press is a department of the University of Oxford. It furthers the University’s objective of excellence in research, scholarship, and education by publishing worldwide. Oxford is a registered trade mark of Oxford University Press in the UK and in certain other countries c Oxford University Press 2014 The moral rights of the authors have been asserted First Edition published in 2014 Impression: 1 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without the prior permission in writing of Oxford University Press, or as expressly permitted by law, by licence or under terms agreed with the appropriate reprographics rights organization. Enquiries concerning reproduction outside the scope of the above should be sent to the Rights Department, Oxford University Press, at the address above You must not circulate this work in any other form and you must impose this same condition on any acquirer Published in the United States of America by Oxford University Press 198 Madison Avenue, New York, NY 10016, United States of America British Library Cataloguing in Publication Data Data available Library of Congress Control Number: 2013942195 ISBN 978–0–19–966437–5 Printed and bound by CPI Group (UK) Ltd, Croydon, CR0 4YY Links to third party websites are provided by Oxford in good faith and for information only. -
Notes on Calculus II Integral Calculus Miguel A. Lerma
Notes on Calculus II Integral Calculus Miguel A. Lerma November 22, 2002 Contents Introduction 5 Chapter 1. Integrals 6 1.1. Areas and Distances. The Definite Integral 6 1.2. The Evaluation Theorem 11 1.3. The Fundamental Theorem of Calculus 14 1.4. The Substitution Rule 16 1.5. Integration by Parts 21 1.6. Trigonometric Integrals and Trigonometric Substitutions 26 1.7. Partial Fractions 32 1.8. Integration using Tables and CAS 39 1.9. Numerical Integration 41 1.10. Improper Integrals 46 Chapter 2. Applications of Integration 50 2.1. More about Areas 50 2.2. Volumes 52 2.3. Arc Length, Parametric Curves 57 2.4. Average Value of a Function (Mean Value Theorem) 61 2.5. Applications to Physics and Engineering 63 2.6. Probability 69 Chapter 3. Differential Equations 74 3.1. Differential Equations and Separable Equations 74 3.2. Directional Fields and Euler’s Method 78 3.3. Exponential Growth and Decay 80 Chapter 4. Infinite Sequences and Series 83 4.1. Sequences 83 4.2. Series 88 4.3. The Integral and Comparison Tests 92 4.4. Other Convergence Tests 96 4.5. Power Series 98 4.6. Representation of Functions as Power Series 100 4.7. Taylor and MacLaurin Series 103 3 CONTENTS 4 4.8. Applications of Taylor Polynomials 109 Appendix A. Hyperbolic Functions 113 A.1. Hyperbolic Functions 113 Appendix B. Various Formulas 118 B.1. Summation Formulas 118 Appendix C. Table of Integrals 119 Introduction These notes are intended to be a summary of the main ideas in course MATH 214-2: Integral Calculus. -
Math 142 – Quiz 6 – Solutions 1. (A) by Direct Calculation, Lim 1+3N 2
Math 142 – Quiz 6 – Solutions 1. (a) By direct calculation, 1 + 3n 1 + 3 0 + 3 3 lim = lim n = = − . n→∞ n→∞ 2 2 − 5n n − 5 0 − 5 5 3 So it is convergent with limit − 5 . (b) By using f(x), where an = f(n), x2 2x 2 lim = lim = lim = 0. x→∞ ex x→∞ ex x→∞ ex (Obtained using L’Hˆopital’s Rule twice). Thus the sequence is convergent with limit 0. (c) By comparison, n − 1 n + cos(n) n − 1 ≤ ≤ 1 and lim = 1 n + 1 n + 1 n→∞ n + 1 n n+cos(n) o so by the Squeeze Theorem, the sequence n+1 converges to 1. 2. (a) By the Ratio Test (or as a geometric series), 32n+2 32n 3232n 7n 9 L = lim (16 )/(16 ) = lim = . n→∞ 7n+1 7n n→∞ 32n 7 7n 7 The ratio is bigger than 1, so the series is divergent. Can also show using the Divergence Test since limn→∞ an = ∞. (b) By the integral test Z ∞ Z t 1 1 t dx = lim dx = lim ln(ln x)|2 = lim ln(ln t) − ln(ln 2) = ∞. 2 x ln x t→∞ 2 x ln x t→∞ t→∞ So the integral diverges and thus by the Integral Test, the series also diverges. (c) By comparison, (n3 − 3n + 1)/(n5 + 2n3) n5 − 3n3 + n2 1 − 3n−2 + n−3 lim = lim = lim = 1. n→∞ 1/n2 n→∞ n5 + 2n3 n→∞ 1 + 2n−2 Since P 1/n2 converges (p-series, p = 2 > 1), by the Limit Comparison Test, the series converges. -
An Introduction to Mathematical Modelling
An Introduction to Mathematical Modelling Glenn Marion, Bioinformatics and Statistics Scotland Given 2008 by Daniel Lawson and Glenn Marion 2008 Contents 1 Introduction 1 1.1 Whatismathematicalmodelling?. .......... 1 1.2 Whatobjectivescanmodellingachieve? . ............ 1 1.3 Classificationsofmodels . ......... 1 1.4 Stagesofmodelling............................... ....... 2 2 Building models 4 2.1 Gettingstarted .................................. ...... 4 2.2 Systemsanalysis ................................. ...... 4 2.2.1 Makingassumptions ............................. .... 4 2.2.2 Flowdiagrams .................................. 6 2.3 Choosingmathematicalequations. ........... 7 2.3.1 Equationsfromtheliterature . ........ 7 2.3.2 Analogiesfromphysics. ...... 8 2.3.3 Dataexploration ............................... .... 8 2.4 Solvingequations................................ ....... 9 2.4.1 Analytically.................................. .... 9 2.4.2 Numerically................................... 10 3 Studying models 12 3.1 Dimensionlessform............................... ....... 12 3.2 Asymptoticbehaviour ............................. ....... 12 3.3 Sensitivityanalysis . ......... 14 3.4 Modellingmodeloutput . ....... 16 4 Testing models 18 4.1 Testingtheassumptions . ........ 18 4.2 Modelstructure.................................. ...... 18 i 4.3 Predictionofpreviouslyunuseddata . ............ 18 4.3.1 Reasonsforpredictionerrors . ........ 20 4.4 Estimatingmodelparameters . ......... 20 4.5 Comparingtwomodelsforthesamesystem . ......... -
Example 10.8 Testing the Steady-State Approximation. ⊕ a B C C a B C P + → → + → D Dt K K K K K K [ ]
Example 10.8 Testing the steady-state approximation. Å The steady-state approximation contains an apparent contradiction: we set the time derivative of the concentration of some species (a reaction intermediate) equal to zero — implying that it is a constant — and then derive a formula showing how it changes with time. Actually, there is no contradiction since all that is required it that the rate of change of the "steady" species be small compared to the rate of reaction (as measured by the rate of disappearance of the reactant or appearance of the product). But exactly when (in a practical sense) is this approximation appropriate? It is often applied as a matter of convenience and justified ex post facto — that is, if the resulting rate law fits the data then the approximation is considered justified. But as this example demonstrates, such reasoning is dangerous and possible erroneous. We examine the mechanism A + B ¾¾1® C C ¾¾2® A + B (10.12) 3 C® P (Note that the second reaction is the reverse of the first, so we have a reversible second-order reaction followed by an irreversible first-order reaction.) The rate constants are k1 for the forward reaction of the first step, k2 for the reverse of the first step, and k3 for the second step. This mechanism is readily solved for with the steady-state approximation to give d[A] k1k3 = - ke[A][B] with ke = (10.13) dt k2 + k3 (TEXT Eq. (10.38)). With initial concentrations of A and B equal, hence [A] = [B] for all times, this equation integrates to 1 1 = + ket (10.14) [A] A0 where A0 is the initial concentration (equal to 1 in work to follow). -
11.3-11.4 Integral and Comparison Tests
11.3-11.4 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1; 1). Let an 1 P R 1 be defined by an = f(n). Then, the series an and the improper integral 1 f(x) dx either BOTH n=1 CONVERGE OR BOTH DIVERGE. Notes: • For the integral test, when we say that f must be decreasing, it is actually enough that f is EVENTUALLY ALWAYS DECREASING. In other words, as long as f is always decreasing after a certain point, the \decreasing" requirement is satisfied. • If the improper integral converges to a value A, this does NOT mean the sum of the series is A. Why? The integral of a function will give us all the area under a continuous curve, while the series is a sum of distinct, separate terms. • The index and interval do not always need to start with 1. Examples: Determine whether the following series converge or diverge. 1 n2 • X n2 + 9 n=1 1 2 • X n2 + 9 n=3 1 1 n • X n2 + 1 n=1 1 ln n • X n n=2 Z 1 1 p-series: We saw in Section 8.9 that the integral p dx converges if p > 1 and diverges if p ≤ 1. So, by 1 x 1 1 the Integral Test, the p-series X converges if p > 1 and diverges if p ≤ 1. np n=1 Notes: 1 1 • When p = 1, the series X is called the harmonic series. n n=1 • Any constant multiple of a convergent p-series is also convergent. -
3.3 Convergence Tests for Infinite Series
3.3 Convergence Tests for Infinite Series 3.3.1 The integral test We may plot the sequence an in the Cartesian plane, with independent variable n and dependent variable a: n X The sum an can then be represented geometrically as the area of a collection of rectangles with n=1 height an and width 1. This geometric viewpoint suggests that we compare this sum to an integral. If an can be represented as a continuous function of n, for real numbers n, not just integers, and if the m X sequence an is decreasing, then an looks a bit like area under the curve a = a(n). n=1 In particular, m m+2 X Z m+1 X an > an dn > an n=1 n=1 n=2 For example, let us examine the first 10 terms of the harmonic series 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + : n 2 3 4 5 6 7 8 9 10 1 1 1 If we draw the curve y = x (or a = n ) we see that 10 11 10 X 1 Z 11 dx X 1 X 1 1 > > = − 1 + : n x n n 11 1 1 2 1 (See Figure 1, copied from Wikipedia) Z 11 dx Now = ln(11) − ln(1) = ln(11) so 1 x 10 X 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + > ln(11) n 2 3 4 5 6 7 8 9 10 1 and 1 1 1 1 1 1 1 1 1 1 1 + + + + + + + + + < ln(11) + (1 − ): 2 3 4 5 6 7 8 9 10 11 Z dx So we may bound our series, above and below, with some version of the integral : x If we allow the sum to turn into an infinite series, we turn the integral into an improper integral. -
The Mean Value Theorem Math 120 Calculus I Fall 2015
The Mean Value Theorem Math 120 Calculus I Fall 2015 The central theorem to much of differential calculus is the Mean Value Theorem, which we'll abbreviate MVT. It is the theoretical tool used to study the first and second derivatives. There is a nice logical sequence of connections here. It starts with the Extreme Value Theorem (EVT) that we looked at earlier when we studied the concept of continuity. It says that any function that is continuous on a closed interval takes on a maximum and a minimum value. A technical lemma. We begin our study with a technical lemma that allows us to relate 0 the derivative of a function at a point to values of the function nearby. Specifically, if f (x0) is positive, then for x nearby but smaller than x0 the values f(x) will be less than f(x0), but for x nearby but larger than x0, the values of f(x) will be larger than f(x0). This says something like f is an increasing function near x0, but not quite. An analogous statement 0 holds when f (x0) is negative. Proof. The proof of this lemma involves the definition of derivative and the definition of limits, but none of the proofs for the rest of the theorems here require that depth. 0 Suppose that f (x0) = p, some positive number. That means that f(x) − f(x ) lim 0 = p: x!x0 x − x0 f(x) − f(x0) So you can make arbitrarily close to p by taking x sufficiently close to x0. -
Antiderivatives and the Area Problem
Chapter 7 antiderivatives and the area problem Let me begin by defining the terms in the title: 1. an antiderivative of f is another function F such that F 0 = f. 2. the area problem is: ”find the area of a shape in the plane" This chapter is concerned with understanding the area problem and then solving it through the fundamental theorem of calculus(FTC). We begin by discussing antiderivatives. At first glance it is not at all obvious this has to do with the area problem. However, antiderivatives do solve a number of interesting physical problems so we ought to consider them if only for that reason. The beginning of the chapter is devoted to understanding the type of question which an antiderivative solves as well as how to perform a number of basic indefinite integrals. Once all of this is accomplished we then turn to the area problem. To understand the area problem carefully we'll need to think some about the concepts of finite sums, se- quences and limits of sequences. These concepts are quite natural and we will see that the theory for these is easily transferred from some of our earlier work. Once the limit of a sequence and a number of its basic prop- erties are established we then define area and the definite integral. Finally, the remainder of the chapter is devoted to understanding the fundamental theorem of calculus and how it is applied to solve definite integrals. I have attempted to be rigorous in this chapter, however, you should understand that there are superior treatments of integration(Riemann-Stieltjes, Lesbeque etc..) which cover a greater variety of functions in a more logically complete fashion.