Concepts in Calculus II UNIVERSITY PRESS OF FLORIDA Florida A&M University, Tallahassee Florida Atlantic University, Boca Raton Florida Gulf Coast University, Ft. Myers Florida International University, Miami Florida State University, Tallahassee New College of Florida, Sarasota University of Central Florida, Orlando University of Florida, Gainesville University of North Florida, Jacksonville University of South Florida, Tampa University of West Florida, Pensacola Orange Grove Texts Plus   Concepts in Calculus II Mikl6s B6na and Sergei Shabanov University of Florida Department of Mathematics UNIVERSITY PRESS OF FLORIDA Gainesville * Tallahassee * Tampa * Boca Raton Pensacola * Orlando * Miami * Jacksonville * Ft. Myers * Sarasota  Copyright 2012 by the University of Florida Board of Trustees on behalf of the University of Florida Department of Mathematics This work is licensed under a modified Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 Unported License. To view a copy of this license, visit http:// creativecommons.org/licenses/by-nc-nd/3.0/. You are free to electronically copy, distribute, and transmit this work if you attribute authorship. However, all printing rights are reserved by the University Press of Florida (http://www.upf.com). Please contact UPF for information about how to obtain copies of the work for print distribution. You must attribute the work in the manner specified by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). For any reuse or distribution, you must make clear to others the license terms of this work. Any of the above conditions can be waived if you get permission from the University Press of Florida. Nothing in this license impairs or restricts the author's moral rights. ISBN 978-1-61610-161-9 Orange Grove Texts Plus is an imprint of the University Press of Florida, which is the scholarly publishing agency for the State University System of Florida, comprising Florida A&M University, Florida Atlantic University, Florida Gulf Coast University, Florida International University, Florida State University, New College of Florida, University of Central Florida, University of Florida, University of North Florida, University of South Florida, and University of West Florida. University Press of Florida 15 Northwest 15th Street - ~ Flonida Distance Gainesville, FL 32611-2079 " . ./mLearning Consortium http://www.upf.com  Contents Chapter 6. Applications of Integration 1 36. The Area Between Curves 1 37. Volumes 8 38. Cylindrical Shells 17 39. Work and Hydrostatic Force 22 40. Average Value of a Function 28 Chapter 7. Methods of Integration 33 41. Integration by Parts 33 42. Trigonometric Integrals 36 43. Trigonometric Substitution 41 44. Integrating Rational Functions 45 45. Strategy of Integration 51 46. Integration Using Tables and Software Packages 54 47. Approximate Integration 59 48. Improper Integrals 67 Chapter 8. Sequences and Series 77 49. Infinite Sequences 77 50. Special Sequences 85 51. Series 92 52. Series of Nonnegative Terms 99 53. Comparison Tests 106 54. Alternating Series 111 55. Ratio and Root Tests 117 56. Rearrangements 126 57. Power Series 133 58. Representation of Functions as Power Series 139 59. Taylor Series 147 Chapter 9. Further Applications of Integration 157 60. Arc Length 157 Chapter and section numbering continues from the previous volume in the series, Concepts in Calculus I.  vi CONTENTS 61. Surface Area 162 62. Applications to Physics and Engineering 168 63. Applications to Economics and the Life Sciences 176 64. Probability 182 Chapter 10. Planar Curves 193 65. Parametric Curves 193 66. Calculus with Parametric Curves 201 67. Polar Coordinates 207 68. Parametric Curves: The Arc Length and Surface Area 214 69. Areas and Arc Lengths in Polar Coordinates 221 70. Conic Sections 225  CHAPTER 6 Applications of Integration 36. The Area Between Curves 36.1. The Basic Problem. In the previous chapter, we learned that if the function f satisfies f(x) > 0 for all real numbers x in the interval [a, b], then the area of the domain whose borders are the graph of f, the horizontal axis, and the vertical lines x = a and x= b is equal to f[b f(x) dx. If there is no danger of confusion as to what a and b are, then this fact is sometimes informally expressed by the sentence "the integral of f is equal to the area of the domain that is under the graph of f." What can we say about the area of the domain between two curves? There are several ways to ask this question. The easiest version, dis- cussed by the following theorem, differs from the previous situation only in that the horizontal line is replaced by another function g. THEOREM 6.1. Let f and g be two functions such that, for all real numbers x E [a, b], the inequality f (x) > g(x) holds. Then the domain whose borders are the graph of f, the graph of g, and the vertical lines x = a and x = b has area A (f (x) - g(x)) d. See Figure 6.1 for an illustration of the content of Theorem 6.1. The reader is invited to explain why this theorem is a direct conse- quence of the fact that we recalled in the first paragraph of this section. The reader is also invited to explain why the theorem holds even if f and g take negative values. EXAMPLE 6.1. Compute the area A(D) of the domain D whose borders are the graph of the function f (x) = x3+1, the function g(x) x2 + 2, and the vertical lines x = 2 and x = 3. See Figure 6.2 for an illustration of this specific example. Solution: In order to see that Theorem 6.1 is applicable, we must first show that, for all cc E [2, 31, the inequality f(xc) ;> g(xc) holds. 1  2 6. APPLICATIONS OF INTEGRATION fW jb (f (x) - g(x)) dz g(x) z = a x=b FIGURE 6.1. Area enclosed by f(x) and g(x) between x= a and x =b. f(x)= x3+1 40 30 20 10 -=x2 +2 2 3 4 FIGURE 6.2. Area enclosed by the graphs of f(x) x3_ 1 and g(x) x2 + 2 between x 2 and = 3. This is not difficult, since we only need to show that if x E [2, 3], then f(x) > g(x), that is, x3 + 1 >12+2, x3 x2 > x2(x - 1) > 1, and this is clearly true since x > 2, so x2 > 4, and x - 1 > 1, forcing x2(x - 1) > 4.  36. THE AREA BETWEEN CURVES 3 Therefore, Theorem 6.1 applies, and we have A(D) (x3+1)(x2+2) dx (x3-x21) dx (x4 x3 3 3 2 11 8-. 12 D 36.2. Intersecting Curves. Sometimes, mined by the curves themselves, and case, we have to compute them before the points a and b are deter- not given in advance. In that we can apply Theorem 6.1. EXAMPLE 6.2. Find the area A(D) of the domain D whose borders are the graphs of the functions f (x) = x2 + 3x + 5 and g(x) = 2x2 + 7x + 8. See Figure 6.3 for an illustration. Solution: Let us find the points in which the graphs of f and g inter- sect. In these points, we have x2+3x+5= 2x2+7x+8, 0=x2 + 4x + 3, 0 = (x + 3)(x + 1). That is, the two curves intersect in two points, and these points have horizontal coordinates a = -3 and b = -1. Furthermore, if 1 E g(x) =2x2+7x+8 f(x)= x2+3x+5 6 -3 -2 -1 0 FIGURE 6.3. Area enclosed by the graphs of f(x) and g(x) between x = -3 and x = -1.  4 6. APPLICATIONS OF INTEGRATION [-3, -1], that is, if x is between those two intersection points, then f (x) - g(x) (x2 + 3x + 5) - (2x2 + 7x + 8) = (X2 + 4x + 3) -(x + 3)(x + 1) > 0, since c + 3 > 0 and c + 1 < 0. Therefore, if c E [-3, -1], then f (x) > g(x), and Theorem 6.1 applies. So we have A(D) (f(x)-g(x)) dz =-(x2+4x+3) dx =- - 2x2 - 3) \ -3 1 3 The situation becomes slightly more complicated if f > g does not hold throughout the entire interval [a, b]. For instance, it could happen that f (x) > g(x) at the beginning of the interval [a, b], and then, from a given point on, g(x) > f(x). In that case, we split [a, b] up into smaller intervals so that on each of these smaller intervals, either f(x) ;> g(x) or g(x) > f(x) holds. Then we can apply Theorem 6.1 to each of these intervals. As on some of these intervals f(x) ;> g(x) holds, while on some others g(x) - f(x) holds, the application of Theorem 6.1 will sometimes involve the computation of f (f(x) -g(x)) dc and sometimes f(g(x) - f(x)) dc. The following theorem formalizes this idea. THEOREM 6.2. Let f and g be two functions. Then the area of the domain whose borders are the graph of f, the graph of g, the vertical line c = a and the vertical line z = b is equal to bf (x) - g(x)| dc. Note that Theorem 6.1 is a special case of Theorem 6.2, namely, the special case when f(x) - g(x) =If(x) - g(x)| for all c E [a, b]. EXAMPLE 6.3. Let f(cc) =cc3 + 3cc2 + 2cc and let g(cc) =cc3 -+ cc2 Compute the area A(fD) of the domain whose borders are the graphs of f and g and the vertical lines xc= -2 and zc= 1. See Figure 6.4 for an illustration.  36. THE AREA BETWEEN CURVES 5 4 2 f(x) Igx 1 -2 -4 FIGURE 6.4. Graphs of f(x) and g(x) on [-2, 1]. Solution: In order to use Theorem 6.2, we first need to compute |f(x) - g(x) . We have f (x) - g(x) (x3 + 3x2 + 2x) - (x3 + x2) 2x2 + 2x = 0, 2x(x + 1) = 0. That is, there are only two points where these two curves intersect, namely, at x = -1 and x = 0. If x < -1 or if x > 0, then f(x) -g(x) 2x(x+1) > 0, so |f (x) -g(x) = f (x) -g(x) = 2X2+2x. If -1 < <0, then f(x) - g(x) < 0, so |f(x) - g(x) = g(x) - f(x) = -2x2 - 2x. Figure 6.5 shows the behavior of the function |f(x) - g(x)|. We can now directly apply Theorem 6.2. We get A(D) = If (x) - g(x) dx /2 ( -21 +j1 /1 +j1 10 3 f (x) - g(x)) dx + (f(x) - g(x)) dx 0 -1 (g(x) - f (x)) dx 2x2 + 2x) dx + (2x2 + 2x) dx 0 (-2x2 - 2x) dx -1 I + x2) 2x3 3 - x2 0 (2X3 3 + x2 1 0 2 1I  6 6. APPLICATIONS OF INTEGRATION 5 4 3 2 1 -2 -1 0 1 FIGURE 6.5. Graph of |f(x) - g(x) on [-2, 1]. 36.3. Curves Failing the Vertical Line Test. Sometimes we want to com- pute the area between two curves that do not pass the vertical line test; that is, they contain two or more points on the same vertical line. Such curves are not graphs of functions of the variable x. If they pass the horizontal line test, that is, if they do not contain two points on the same horizontal line, then they can be viewed as functions of y. We can then change the roles of x and y in Theorems 6.1 and 6.2 and proceed as in the earlier examples of this section. EXAMPLE 6.4. Compute the area A(D) of the domain between the vertical line x = 4 and the curve given by the equation y2 X. See Figure 6.6 for an illustration. Solution: Neither curve satisfies the vertical line test, but both satisfy the horizontal line test. Therefore, we set f(y) = 4 and g(y) = y2. It is 4 y2 =x A(D) -1 -2 --------------------- FIGURE 6.6. Graph of y2 x andxz= 4.  36. THE AREA BETWEEN CURVES 7 clear that the two curves intersect at the points given by y = -2 and y = 2. Between these two curves, the value of f(y) is larger. Therefore, Theorem 6.1 applies (with the roles of x and y reversed). So we have A(D) =j(f(x) - g(x)) dx /2 = r2 2 = 4y - 2 = 16 - 3 2 = 10-. 3 - Note that the geometric meaning of reversing the roles of x and y is simply reflecting all curves through the x= y line. That reflection does not change the area of any domain, so one can expect analogous methods of computing areas before and after that reflection. 36.4. Exercises. (1) Find the area of the domain whose borders are the vertical line x= 0, the vertical line x= 2, and the graphs of the functions f (x) = x2 + 3 and g(x) =sinx. (2) Find the area of the domain whose borders are the vertical line x = 2, the vertical line x = 4, and the graphs of the functions f (x) = x2 + 3 and g(x) = x. (3) Find the area of the domain between the horizontal axis and the graph of the function f (x) = x2 + 6x. (4a) Find the area of the domain between the horizontal axis and the graph of the function f (x) = 18 - 2x2. (4b) Find the area of the domain between the horizontal axis and the graph of the function f(x) = 16 - x4. (5) Find the area of the domain whose borders are the vertical line cc= 1, the vertical line cc= 3, and the graphs of the functions f(cc) =cc3 and g(cc) =e-2. (6) Find the area of the domain whose borders are the vertical line cc= -1, the vertical line cc= 1, and the graphs of the functions f(cc) =cc3 and g(cc) =-2.  8 6. APPLICATIONS OF INTEGRATION (7) Find the area of the domain between the graphs of the func- tions f(x) = ex and g(x) = x + 1 and the vertical lines x = 0 and x= 1. How do we know that between those two vertical lines, the graph of f is above the graph of g? (8) Find the area of the domain between the graphs of the func- tions f (x) = x2 + 2 and g(x) = 4x - 1. (9) Find the area of the domain between the graphs of the func- tions f (x) = x3 - 3x2 and g(x) = x2. (10) Find the area of the domain between the graphs of the func- tions f (x) = x2 and g(x) = z1/5 (11) Find the area of the domain between the graphs of the func- tions f (x) = x3 and g(x) = x4. (12) Find the area of the domain between the graphs of the func- tions f (x) = x and g(x) cVc. (13) Compute the area between the curves given by the equations x =5y and c =y2+6. (14) Compute the area between the curves y = x - 2 and y = (x - 2)3. (15) Compute the area between the y axis and the curve c = y2 - 7y+6. (16) Compute the area between the y axis and the curve c = y3 - 11y2 + 6y - 6. (17) Compute the area between the curves y = x2 and y = 32- x2. (18) Compute the area between the three curves f(x) = x, g(x) -x, and h(x) = 4. (19) Compute the area between the curves y =cc and y = x2 -12. (20) Compute the area between the curves y =cc and y = 20- x2. 37. Volumes 37.1. Extending the Definition of Volumes. If a solid S can be built up using unit cubes, then we can simply say that the volume V(S) of S is the number of unit cubes used to build S. However, if the borders of S are not planes, then this method will have to be modified. A ball or a cone is an example of this. So we would like to define the notion of volume so that it is applica- ble to a large class of solids, not just to those solids that are bordered by planes. This definition should agree with our intuition. It should also be in accordance with the fact that we can approximate all solids with a collection of very small cubes; therefore, V(S) must be close to the number of unit cubes used in the approximation.  37. VOLUMES 9 With these goals in mind, we recall that we already defined the area of a domain in the plane whose borders are the graphs of continuous functions. Building on that definition, we say that the volume of a prism is its base area times its height. More formally, let S be a solid whose base and cover are identical copies of the plate P, located at distance h from each other, on two parallel planes that are at distance h from each other. Then we define the volume of S to be V(S) = A(P)h, where A(P) is the area of P. See Figure 6.7 for an illustration. In particular, the volume of a cylinder whose base is a circle of radius r and whose height is h is rrh. Now let S be any solid located between the planes given by the equations x = a and x = b. In order to define and compute the volume V(S) of S, we cut S into n parts by the planes x = zx for i = 0, 1, ... , n, where a=zo 2t - t) 2 = 2r 3 7 - - rs3 3 43 =-r37. 3 There is nothing magical about the x axis as far as Theorem 6.3 is concerned. The argument that yielded that theorem can be repeated for the y axis instead of the x axis, yielding the following theorem. THEOREM 6.4. Let S be a solid located between the planes y = a and y = b, and let B(t) be the area of the intersection of S and the plane y = t. Then the volume V(S) of S satisfies the equation V(S) = B(t) dt. EXAMPLE 6.6. Let S be the right circular cone whose symmetry axis is the y axis, whose apex is at y = h, and whose base is a circle in the plane y = 0 with its center at the origin and with radius r. Find the volume of S. Solution: The cone S is between the planes y = 0 and y = h, and B(t) of Theorem 6.4 is easier to compute than A(t) of Theorem 6.3, so we use the former. The intersection of the plane y =t and S is a circle. The radius r+ of this circle, by similar triangles, satisfies re h -t  12 6. APPLICATIONS OF INTEGRATION showing that rt = r(h - t)/h. Therefore, B(t) = r2(h - t)2r/h2, and Theorem 6.4 implies V (S) = r2h- t )\r/h2) dt /h 0 h23 h 2 (h 2- ht +5) d 0 T 127 , h 3 h2 3 = 1-hr2br. 3 See Figure 6.9 for an illustration. D 37.2. Annular Rings. In the examples that we have solved so far, the computation of A(t) or B(t), that is, the computation of the area of the intersection between a solid and a horizontal or vertical plane, was not difficult. That computation could be done directly. There are situations in which the domains whose areas we need to compute are not convex; that is, visually speaking, there is a hole in them. This happens particularly often when S is obtained by rotating a domain D around a line. h h-t t r FIGURE 6.9. Right circular cone.  37. VOLUMES 13 37. VOLUMES 13 EXAMPLE 6.7. Let D be the domain between the two curves y z2 = f(x) and y = 2x = g(x) and let S be the solid obtained by rotating D about the line z = -3. Find the volume of S. Solution: The two curves intersect at the points (0, 0) and (2, 4). The intersection of the horizontal plane y = t with S has the form of an annular ring, which is sometimes informally called a washer. This is simply a smaller circle cut out off the middle of a larger circle, so that the two circles are concentric. If the larger circle has radius r1 and the smaller circle has radius r2, then the annular ring has area7(ri2 - r2). This general recipe enables us to compute B(t) in the example at hand. The points in D satisfy c E [0, 2] and y E [0, 4]. As 0 < g(x) for all x E [a, b], and f and g are continuous functions on [a, b], then the average value of f on [a, b] is at least as large as the average value of g on [a, b]. (13) Which is larger, the average value of f(x) = x2 on [2, 3] or the average value of g(x) = --2 on [1/3, 1/2]? (14) A bicyclist covered a certain distance at an average velocity of 25 miles per hour. Prove that there was a moment at which the instantaneous velocity of the bicyclist was 25 miles per hour. (15) Starting from a standstill, a car needed 10 seconds to reach a traveling velocity of 20 meters per second. Prove that there is a moment at which the instantaneous acceleration of the car is 2 m/s2. (16) Let us assume that fj f(cx) dc = 10. Prove that there is a number x E [0, 4] such that f (4) = 2.5. (17) Let f be an odd function. What is the relation between the av- erage value of f on [a, b] and the average value of f on [-a, -b]? (18) Let f be an odd function. Compute the average value of f for any interval [-R, R]. (19) Let f be an even function. What is the relation between the average value of f on [-R, R] and the average value of f on [0, R]? (20) Let a be a fixed real number and let f be a continuous function. Define g(x) as the average value of f on the interval [a, c]. Compute g(x) for f (x) = 3, f (x) = 2x + 7, and f (x) = x2 + 2x + 1.   CHAPTER 7 Methods of Integration 41. Integration by Parts 41.1. Method of Integration by Parts. Let u and v be two differentiable functions of the variable x. We used the simple product rule (7.1) (uv)'= u'v +UV' to compute the derivative of the product of these two functions. Is there a similar rule for computing the integral of the product of two functions? In general, the answer is no. There is no rule that provides the integral of the product of two functions that would work in every case. However, there are many cases in which a relatively simple way of "reversing" the product rule of differentiation will give us the answer we are trying to obtain. Indeed, integrating both sides of the product rule (7.1) of differen- tiation (7.1) with respect to x, we get the identity u(x)v(x) =J(u'(x)v(x)) dx + (u(x)v'(x)) dx or, after rearrangement, (7.2) J(u'(x)v(x)) dx =u(x)v(x) - J(u(x)v'(x)) dx. Formula (7.2) is very useful if we want to compute the integral of the product of two functions, one of which can play the role of u' and the other one of which can play the role of v. If we can compute u, and f(uv'), then formula (7.2) enables us to compute f(u'v) as well. If we cannot carry out one or both of these computations, then formula (7.2) will not help. EXAMPLE 7.1. Compute f xex dx. Solution: We set u'(x) = ex and v(x) = x. Then formula (7.2) is easy to apply, since v(x) = x and v'(x) = 1. Therefore, (7.2) implies that f ze dz= e. x - fe. 1,d=e. x - eX + C =e(x - 1) +C. 33  34 7. METHODS OF INTEGRATION The reader is encouraged to verify that the obtained solution is correct by computing the derivative of ex(x - 1) and checking that it is indeed equal to ex -cz. At this point, the reader may be asking how we knew that we needed to set u'(x) = ex and v(x) = x, and not the other way around. The answer is that the other distribution of roles, that is, i'(x) = x and v(x) = ex would not have helped. Indeed, if we had chosen u' and v in that way, we would have needed to compute f(uv') dcc f (ezz2)/2 dz. That would have been more complex than the original problem. We should always choose u' and v so that f(uv') dz is easy to compute. That usually means selecting v so that it becomes simpler when differentiated, and to select u' so that u' does not get much more complex when integrated (or at least one of these two desirable outcomes occur). EXAMPLE 7.2. Compute f x cos x dx. Solution: We set u'(x) = coscx and v(x) = x, which means that u(x) = sincx and v'(x) = 1. So formula (7.2) implies fcx cos x dx = x sin x - fsin x dx = x sin x + cos x + C. The technique of integration we have just explained is called inte- gration by parts. 41.2. Advanced Examples. Sometimes the integrand does not seem to be a product, but it can be transformed in to one. The following is a classic example. EXAMPLE 7.3. Compute f ln c dc. Solution: The crucial observation is that writing ln x = 1 ln x helps. Let u'(x) = 1 and v(x) = lncc. Then u(x) = x and v'(x) = 1/c, so, crucially, u(x)v'(x) = 1. Therefore, formula (7.2) yields f ln x dc f 1 -ln x d = xln x- f1 d = xln x - xC +C. Sometimes integration by parts leads to an equation or a system of equations that needs to be solved in order to get the solution to our problem. EXAMPLE 7.4. Compute f ex cos c.  41. INTEGRATION BY PARTS 35 Solution: We set u'(x) = ex and v(x) = cos x. Then u(x) = ex and v'(x) - sin x, and formula (7.2) yields (7.3) f ex cos x dx = ex cos x + fex sin x dx. So we could solve our problem if we could compute the integral f ex sin x dc. We can do that by applying the technique of integration by parts again, setting u'(x) = ex and v(x) =sin c. We obtain (7.4) f exsin x dx = ex sinxc- fexcos x dx. Finally, note that (7.3) and (7.4) is a system of equations with un- knowns f ex cos x d and f ex cos x d. We can solve this system, for instance, by adding these two equations and noting that f ex sin x can- cels. We get the equation ex cos x dx = ex(coscx + sincx) - fexcos x dx or eX cos x dx = -(coscx + sincx) + C. Note that substituting the obtained expression for f ex cos x d into (7.4), we get a formula for f ex sin x d, namely, exdsincdc 2(inx -cosx)+C. 41.3. Definite Integrals. If we evaluate both sides of formula (7.2) from a to b and we apply the fundamental theorem of calculus, we get the identity (7.5)(u/v dcc= (tv) b - jb x EXAMPLE 7.5. Evaluate f2 ln x dc. Solution: As we saw in Example 7.3, we can set u(x) = x and v(x) lncc. Then u'(cc) =1, v'(cc) =1/cc, and formula (7.5) yields flnc =(clnc)-f1 dc= 2ln 2 -1.  36 7. METHODS OF INTEGRATION 41.4. Exercises. (1) Compute fcxsinxdz. (2) Compute f z2ex dz. (3) Compute f x2 cos x dz. (4) Compute fcxlnxdz. (5) Compute fcx2lncz. (6) Compute f e2x sin x dz. (7) Compute f ze-- dc. (8) Compute f(x+cosz)2 dc. (9) Compute f /zlncx dz. (10) Compute fcxtan-1(x) dz. (11) Compute fcxsin-1(x) dz. (12) Compute f(lncz)2 dz. (13) Evaluate fcxccos 2x dz. (14) Evaluate fcx2 sin x dz. (15) Evaluate fczex dz. (16) Evaluate fo ze2 dc. (17) Evaluate f cx3ex dc. (18) Evaluate f23 xclncx dc. (19) Evaluate fj(lncz)2 dz. (20) Use integration by parts to compute f(cos z)2 dc. 42. Trigonometric Integrals 42.1. Powers of sin and cos. In this section, we consider functions of the form f(x) = sinm x cos" c and discuss techniques for their integration. It seems natural to first consider the cases when m or n is 0, that is, when f is just a power of sin or cos. Even these special cases will break up into further subcases. The easiest subcase is when the exponents are even numbers. In that case, we can use the trigonometric identities (7.6) and (7.7) cos 2x = 2 cos2 x - 1 sin 2x = 2 sin x cos x to eliminate high powers of trigonometric functions in the integrand. EXAMPLE 7.6. Compute f cos4 x dc.  42. TRIGONOMETRIC INTEGRALS 37 Solution: Using (7.6), we get that cos2 x = 1+cos2x, and so c4x (1 + cos 2x 2 1 cos 2x cos2 2x Applying (7.6) again, with 2x replacing x, we get that cos2 2x = 1+cos4x, so the previous displayed equation turns into o4 3 cos2x cos4x cosx= + + . 8 2 8 Having eliminated the powers of cos, the integration is easy to carry out as follows: 4 f 3( cos2x cos4x cos' x dz x + dz 8 2 8 3x sin 2x sin 4x + + +C. 8 4 32 The computation is more complex if the integrand is an odd power of sin or cos. In that case, we separate one factor and convert the rest into the other trigonometric function, using the rule cos2 x + sin2 x = 1. EXAMPLE 7.7. Compute f sin3 x dx. Solution: We have sin3 x = sin x" sin2 x = sin x" (1 - cos2x) = sin x - sin xcos2 X. The advantage of this form is that it makes integration by substi- tution easy. Indeed, set u = cos x, then du/dx - sin x, and so f2d u33 cos3x f sinx cos2xdx du = + C = + C. 3 3 Comparing the two displayed equations of this solution and noting that f sin x dx = - cos x, we get 1 3 cos3 x sin3x -cosx+ +C. The methods shown above can be used to compute the integral of products of powers of sin x and cos z. In other words, the method allows us to compute f cosm x sin" x dx as shown below. EXAMPLE 7.8. Compute f cos2 x sin3xdx  38 7. METHODS OF INTEGRATION Solution: Just as in Example 7.7, we separate one sin x factor. This accomplishes two things. It allows us to convert the remaining even number of sin x factors to cos x factors, and it allows us to integrate by substitution. Indeed, fcos2x sin3x dx fcos2x sin2x sin x dx = f cos2 x(1 - cos2 x) sin x = cos2x sinx - cos4 x sin x = f-u2 du + uA du u3 5 =- + + c 3 5 -cOs3 x cos5 x 3 + +C 3 5 where we used the substitution u = cos x. D We can always proceed this way if at least one of m and n in the integrand cosm x sine x is odd. Indeed, in that case, after separating one factor from that odd power, an even power remains, and that can be converted to the other trigonometric function using the identity sin2 x + cos2 x = 1. If both m and n are even, then we can use that identity right away. EXAMPLE 7.9. Compute f cos4 x sin2 x dx. Solution: We have f cos4 x sin2 x dx f cos4 x(1 - cos2 x) dx f cos4x - f cos6 x dx. Now note that we computed f cos4 xdx in Example 7.6. You are asked to compute f cos6 x in Exercise 42.4.4. The difference of these two results then provides the solution of the present example. Q 42.2. Powers of tan and sec. When integrating a product of the form tmn1x sec" x, we will use the identity sec2 x =tan2 x + 1 and the dif- ferentiation rules (t an z)' =sec2 x and (sec x)' =sec x t an z. There are two easy cases, namely, when mn is odd (and n is at least 1) and when nm> 2 is even.  42. TRIGONOMETRIC INTEGRALS 39 In the first case, that is, when m is odd and n;> 1, we separate one factor of tan x sec x and express the remaining factors in terms of sec x by the identity -1 + sec2 x =tan2 x. Then we substitute ut= sec x, which leads to d = tan x secx. EXAMPLE 7.10. Compute f tan3 x sec x dx. Solution: Following the strategy explained above, we have J tan3 xsecxdx= antan2xtanxsecxdx (-1 + sec2x)tanxsecxdx = f(-1+u2)du u3 =-u+ +C 3 sec3 x - secx+ +C. 3 In the second case, that is, when n > 2 is even, we separate one factor of sec2 x, express the remaining factors in terms of tan x using the identity sec2 x = 1 + tan2 x, and substitute ui= tan x, which leads to = sec2 X. EXAMPLE 7.11. Compute f sec4 x dx. Solution: We have f sec4 x dx = sec2 x sec2 x dx = (1 + tan2 x) sec2 x dx = (1 + 2) du u3 -u+ +C 3 tan3 x =tan x + + C. 3 If we are not in these two easy cases, then there is no recipe that will always work. We then need to have a separate strategy for each problem. We will show examples of that in Exercises 42.4.13, 42.4.14, 42.4.15, and 42.4.16.  40 7. METHODS OF INTEGRATION 42.3. Some Other Trigonometric Integrals. If our goal is to compute in- tegrals of the form f cos mx sin nx dx, f cos mx cos nx dx, and f sin mx sin nx dx, then we can often make use of the following identities: (7.8) (7.9) (7.10) 1 sin a cos b =- sin(a 2 1 cos acos b = - cos(a 1 sin a sin b =- cos(a 2 1 b) + - sin(a + b), 2 1 b) + - cos(a + b), 1 b) - - cos(a +b). 2 EXAMPLE 7.12. Compute f cos 3x cos 5x dx. Solution: Using (7.9) with a = 3x and b cos -2x = cos 2x, we get that cos 3x cos 5x = so = 5x and noting that j cos 2x + 2 cos 8x, and I cos 3x cos 5x dx J 1 cos 2x +-cos8x 1dz 1 1 - sin 2x+ -sin 8x +C. 4 16 F- 42.4. Exercises. (1) Compute (2) Compute (3) Compute (4) Compute (5) Compute (6) Compute (7) Compute (8) Compute (9) Compute (10) Compute (11) Compute (12) Compute (13) Compute integrand (14) Compute (15) Compute f cos3 x dx. f sin5 x dx. f cosy x dx. f cos6 x dx. f sin4 x dx. f sin3 x cos2 xdx. f sin x cos3 x dx. f sin3 x cos3 xdx. f sin2 x cos2 x dx. f sin4 x cos4xdx. f tan2 x sec4xdx. f tan3 x secs x dx. f tan5 x dx by separating one factor of tan2 x in the and expressing it in terms of sec2 x. f tan3 x dx using integration by parts. f sec3 x dx using integration by parts, with u'(x) sec2 x and v(x) = secx.  43. TRIGONOMETRIC SUBSTITUTION 41 (16) Compute f sec' x dx. (Hint: Recall the result of Example 7.11.) (17) Compute f sin x cos x dx. (18) Compute f sin 2x cos 4x dx. (19) Compute f sin 3x sin 7x dx. (20) Compute f cos 2x cos 4x dx. 43. Trigonometric Substitution 43.1. Reversing the Technique of Substitutions. Let us assume that we want to compute the area of a circle by viewing one-fourth of that circle as the domain under a curve. Let r be the radius of the circle, and let us place the center of the circle at the origin. Then the northeastern quarter of the circle, shown in Figure 7.1, is just the domain under the graph of the function f(x) = r2 - x2, where x ranges from 0 to r. In other words, we need to compute the integral (7.11) j r2 - x2 dx. In Chapter 5, we presented the technique of integration by substi- tution. This technique worked in situations when the best way to com- pute an integral was to define a simple function of x, such as y(x) = x2 and then continue the integration in terms of that new variable y. In order to compute the integral in (7.11), we use the reverse of the strategy mentioned in the previous paragraph. We define another variable y so that x is a simple function f of y. It is important to define f and y so that f is one-to-one, since that assures that f(y) = x is equivalent to f -1(x) = y. In computing the integral in (7.11), we can set x = r sin y. Then dx/dy = r cosy, and the limits of integration are y = 0 and y = 7/2. -r 0 r FIGURE 7.1. The northeastern quadrant of the unit circle.  42 7. METHODS OF INTEGRATION This yields f/r2 - 2 d rj2 - r2 sin2 y r cos y dy 0 0 = r2 J/ 1 - sin2ydy 0 07r/2 = 7T2 i COS22JCJ 2 y + sin 2y 'r/2 2 0 27F~ 4 Note that we could write cos y for 1 - sin2 y, since 0 < y r/2, and in that interval, cos y is nonnegative. The preceding computation resulted in finding a definite integral. We point out that by converting the indefinite integral r2 + sin 2y back to a function of x using the rule c = r sin y, we 2 get the corresponding indefinite integral, that is 2 rin- The result that we are going to compute in the next example will be useful in the next section, when we will learn a technique to integrate rational functions. EXAMPLE 7.13. Compute the integral f dcz. Solution: We use the substitution x = tan y. Then y =tan-1(x), and so dy/dz = 1/(1 + X2), and hence dy = dc/(1 + X2). This yields (1 + x2)2 d 1 +cX2dY 1+tan2 =fcos2Yd y sin 2y 2 4  43. TRIGONOMETRIC SUBSTITUTION 43 y sin ycos y 2 2 11 1 x =--.tan-(x)+-"- +C. 2 2 x2+1 The last step is justified since x____ tan yt (7.12) x 2an y = tan y cos2 y =sin y cos y. x2+-1 1 +tan y Figure 7.2 illustrates this trigonometric argument. EXAMPLE 7.14. Compute the integral fI . Solution: The denominator reminds us of the trigonometric identity tan2 y = sec2 y - 1, and so, if y E [0, 7/2), then tan y = sec2y - 1. Therefore, we use the substitution x = sec y. Then dz/dy = tan y sec y. Hence, we have f x -i f sejyidX / 11 d v/z2 -f taysecYd_ J tan y = n sec y dy =lnjsecy+tanyl+C =ln x + v/2 - 1| +C. Figure 7.3 illustrates this trigonometric argument. FIGURE 7.2. Some expressions from (7.12).  44 7. METHODS OF INTEGRATION 2-1 FIGURE 7.3. Some expressions occuring in the solution of Example 7.14. 43.2. Summary of the Most Frequently Used Trigonometric Substitutions. The three examples that we have seen so far in this section show the three most frequently used reverse substitutions. That is, (i) To compute f r2 - x2 dv, use the reverse substitution x = r sin y. (ii) To compute integrals involving (r2 + X2) under a root sign or in the denominator of a fraction, use the reverse substitution x = r tan y. (iii) To compute f v2 - r2 dc, use the reverse substitution x = r see y. Finally, a word of caution. The availability of the method of reverse substitution does not mean that this method is always the best one to compute an integral that contains a square root sign. Some of the following exercises can be solved by another method faster (and, no, we are not revealing which ones). 43.3. Exercises. (1) Use the method presented in this section to compute the area of an ellipse determined by the equation X2 2 a2 + b2 = 1. (2) Compute f 1 - 4x2 dc. (3) Compute f c+16 . (4) Compute f x3/25 - x2 dc. (5) Compute f 4 - 36x2 dc. (6) Compute f v1 +cv2dv. (7) Compute f 4+ 2 dcv. (8) Compute f -5dcv. (9) Compute f 22 4dcv.  44. INTEGRATING RATIONAL FUNCTIONS 45 (10) Compute fx x2 - 4dx. (11) Computefx 2 - 2x dx. (12) Compute f 9+x2 dx. (13) Compute f 42-25 dx. (14) Compute f 4x+13dx. (15) Compute f 2+4x02 dx. (16) Compute f 2x - x2 dx. (17) Compute f x2 2 dx. (18) Compute f e2 - 16 dx. (19) Compute fex 4 - e2xdx. (20) Find the average value of 1+9 on the interval [0, 3]. 44. Integrating Rational Functions 44.1. Introduction. Recall that a rational function is the ratio of two polynomials, such as P(x) 3x + 5 Q(x) 2x2 + 4x + 9 Integrating rational functions is relatively simple, because most of these functions can be obtained as sums of even simpler functions. If the degree of P(x) is at least as large as the degree of Q(x), then we can divide P(x) by Q(x), getting a polynomial as a quotient, and possibly a remainder. That is, if the degree of P is at least as large as the degree of Q, then there exist polynomials P1(x) and P2(x) such that the degree of P2(x) is less than the degree of Q(x) and P(x) _ P1Q(x)+P2(x) P2(x) Q(x) Q(x) PiQ) + P(x) As P1(x) is a polynomial, it is easy to integrate. Therefore, the diffi- culty of integrating R(x) lies in integrating R(x), which is a rational function whose denominator is of higher degree than its numerator. For this reason, in the rest of this section, we focus on integrating rational functions with that property, that is, when the degree of the denominator is higher than the degree of the numerator. E XAMPLE 7.15. Let R(x) = %22±+1 Then dividing P(x) by Q(x) using long division, we get P(x) =(x + 1)(x2 - x + 1) + 2x,  46 7. METHODS OF INTEGRATION so _P(x) -+ + 2x R(x)- x + 1+, Q(x) x2 -_X+ 1 and integrating Q(x) boils down to integrating .2x 44.2. Breaking Up the Denominator. In order to decide how to break up a rational function R(x) into the sum of simpler terms, we analyze the denominator Q(x) of R(x). A theorem in complex analysis, some- times called the fundamental theorem of algebra, implies that if q(x) is a polynomial whose coefficients are real numbers, then q(x) can be written as a product of polynomials that are of degree 1 or 2. This decomposition, or factorization, of Q(x) will determine the way in which we break up our rational function into the sum of simpler terms. There are several cases to distinguish, based on the factorization of Q(x). 44.2.1. Distinct Linear Factors. The easiest case is when Q(x) factors into the product of polynomials of degree 1, and each of these terms occurs only once. EXAMPLE 7.16. Compute f X2+3X+2 da. Solution: Note that x2+3x+2 = (x+1)(x+2). Using that observation, we are looking for real numbers A and B such that 1 A B (7.13) 1 = + B x2+3x+2 x+1 x+2 as functions, that is, such that (7.13) holds for all real numbers x. Multiplying both sides by x2 + 3x + 2, we get (7.14) 1 = A(x + 2) + B(x + 1). If (7.14) holds for all real numbers x, it must hold for x = -1 and x = -2 as well. However, if x = -1, then (7.14) reduces to 1 = A, and if x = -2, then (7.14) reduces to 1 = -B. So we conclude that A = 1 and B =-1 are the numbers we wanted to find. It is now easy to compute the requested integral as follows: fx2+3x+2d - f +1dx-f +2dc =ln(c-+1) - ln(cc+2)+ C. The above method can always be applied if Q(cc) factors into a product of linear polynomials, each of which occurs only once. In  44. INTEGRATING RATIONAL FUNCTIONS 47 particular, if Q(x) decomposes as a(x - ai)(x - a2) - -- (x - ak), then we can decompose R(x) into a sum of the form A1 A2 __... _A x-ai x -a2 x-ak After determining the numbers AZ, we can integrate each of the above k summands. 44.2.2. Repeated Linear Factors. The next case is when Q(x) factors into linear terms, but some of these terms occur more than once. EXAMPLE 7.17. Compute f x+7 do. Solution: Just as in the previous case, we decompose the integrand into a sum of simpler fractions. We are looking for real numbers A, B, and C such that 2x+7 A B C (x-+-1)2(x -1) x-+-1 (x-+-1)2 x- 1 Multiplying both sides by the denominator of the left-hand side, we get 2x +7= A(x + 1)(x - 1) + B(x - 1) + C(x + 1)2. Substituting x = 1 in the last displayed equation yields 9 = 4C, so C = 2.25. Substituting x = -1 yields 5 = -2B, so B = -2.5. Finally, the coefficient of x2 on the left-hand side is 0, while on the right-hand side, it is A + C. So A + C = 0, yielding A = -2.25. Now we are in a position to compute the requested integral. 2x + 7 -2.25 -2.5 2.25 (x +1)2(x- 1)dx x + 1dx+ (x + )2dx x - 1dx /-2.25 d + -2.5 d + 2.25 d ~Jddx +J x]l2x I ldx X + 1 (X + 1)2 X - 1 2.5 - -2.251n(x + 1) + x 1+ 2.25ln(x -1). S+ 1-1 In general, if a term (x+a)k occurs in Q(x), then the partial fraction decomposition of R(x) will contain one term with denominator (x +a)' for each i E {1, 2, .. . ,k}. For inst ance, if Q(x) =(x+2)3(x+5)2(x-10), then R(x) will have a partial fraction decomposition of the form A1 A2 A3 A4 A5 A6 x +2 (x +2)2 (x +2)3 x +5 (x +5)2 x -10  48 7. METHODS OF INTEGRATION 44.2.3. Distinct Quadratic Factors. The third case is when the factor- ization of Q(x) contains some quadratic factors that are irreducible (i.e., they are not the product of two linear polynomials with real co- efficients), but none of these irreducible quadratic factors occurs more than once. In that case, after obtaining the partial fraction decompo- sition of R(x), we may have to resort to the formulas ddx= tan-1x + C /2+1 and / 1 1 x1 d1 = - tan-(X)-+)C. x+a2 a a EXAMPLE 7.18. Compute the integral f X34X+i1dx. Solution: It is easy to notice that setting x = -1 turns the denomi- nator to 0; hence, the denominator is divisible by x + 1. Dividing the denominator by x + 1, we get x2 + 1, so the denominator factors as (x + 1)(x2 + 1). The factor x2 + 1 is irreducible (it is not divisible by x - b for any real number b, since no real number b satisfies the equation b2 + 1 = 0). Therefore, we are looking for real numbers A, B, and C such that 4x + 2 A B Cx (7.15) + + x3 +z2+ +1 x+1 x2+1 x2+1 The reader is invited to verify that the third summand of the right- hand side is necessary; that is, if the summandXx is removed, then no pair of real numbers (A, B) will satisfy (7.15). In order to find the correct values of A, B, and C, multiply both sides of (7.15) by (x + 1) - (x2 + 1) and rearrange, to get 4x+2 = (A+C)x2+ (B+C)x+A+B. The coefficient of x2 is 0 on the left-hand side, so it has to be 0 on the right-hand side. Therefore, A + C = 0. Similarly, the coefficient of x is 4 on the left-hand side, so it has to be 4 on the right-hand side, forcing B + C = 4. Similarly, the constant terms of the two sides have to be equal, and, consequently, A + B = 2. Solving this system of equations, we get A = -1, B = 3, and C = 1. Therefore, ]x+x+x1dx = - dx + 3 1 +1 dx x3+2+ +1x+1 ,x , x+ =- ln(x +1) + 3tan-1 x + -ln(x2 +1). 2D  44. INTEGRATING RATIONAL FUNCTIONS 49 In general, if x2+ax+b is a quadratic factor in Q(x), then the partial fraction decomposition will contain a summand of the form x2ax+b and a summand of the form 2+x+b. Again, the latter is necessary, since a rational fraction of the form E+a+ will not equal one of the form x2+ x+b for any choice of E ifF#O0. 44.2.4. Repeated Quadratic Factors. Finally, it can happen that the factorization Q(x) contains irreducible quadratic factors, some of which occur more than once. EXAMPLE 7.19. Compute the integral f x3+2x23x++7 dx. Solution: It is easy to see that the denominator factors as (x2 + 1)2. Hence, we are looking for real numbers A, B, C, and D such that x3+2x2+3x+7 A Bx C Dx x4+2x2+ 1 x2 + 1 x2 + 1 (x2+1)2 (x2+1)2 Multiplying both sides by x4 + 2x2 + 1 and rearranging, we get x3 +2x2+3x+7 =Bx3+Ax2+ (B+ D)x + (A +C). For each k, the coefficients of xk must be the same on both sides. Hence, A =2 and B= 1, so C= 5 and D =2. Now we can compute the requested integral using the preceding partial fraction decomposition as follows: S2+ dx x4 +2x2+ 1 f 2 x 5 2x x2 + 1+ x2 + 1 +(x2 +1)2+ (x2 + 1)2dx (1 5x 5 1 =2-tan +-ln(x2+1)+ + -tanx 1- 2 2(x2 +1) 2 x2 +1 1( 41 1 5- =1 ln(x 2 + 1) + 4.5 tan--' z + - 5-2. 2 2 x2+1 Here we used the formula for f x2+1)2 that we computed in the last section, in Example 7.13. D By now, the reader must know what the general version of the technique of the preceding example is. If the factorization of Q(x) contains (x2 + ax + b)k, then, for each integer i such that 1 < i < k, the partial fraction decomposition of R(x) will contain a summand of the form (x22++) and a summand of the form (2Faxbi'  50 7. METHODS OF INTEGRATION 44.3. Rationalizing Substitutions. There are situations when a function that is not a rational function can be turned into one by an appropriate substitution, and then it can be integrated by the methods presented in this section. The most frequent scenario in which this happens is when the integrand contains roots, but if those roots are replaced by another variable, we get a rational function in that other variable. EXAMPLE 7.20. Compute fx-idx. Solution: We use the substitution cc y. Then dy/dc =g-1= g This leads to dx = 2y dy V + 1 y+ 1 2y22 y + = 2(y - 1) + dy /Y+2 = y2 - 2y+ 21n(y+ 1) =x-2 c+2ln( cc+1). Q Note that the computation would have been very similar if the integrand contained some other root of x. Indeed, if the integrand contained c instead of cc, then we would have substituted y = c, and that would have turned the integrand into a rational function of y. Indeed, y = cl/ implies dy x(1/r)-1 dz r ' dy_ y1r dcrc and therefore dc = ry-1 dy. In other words, dc is a equal to dy times a polynomial function of y, so, indeed, the integrand will be a rational function of y. 44.4. Exercises. (1) Compute f .2+5x+4 dc. (2) Compute f i2t~+ dcc. (3) Compute f ,sf2i+2+11-6 dcc (4) Compute f 3±2x1dcc. (5) Compute f 2322_ dcc.  45. STRATEGY OF INTEGRATION 51 (6) Compute f 4+5x2+4 dx. (7) Compute f x4+42+4 dx. (8) Compute f 14 dx. (9) Compute f '4+1'dx. (10) Compute f X3+4xdx. (11) Compute f xidx. (12) Compute fIx3-i dx. (13) Compute f xdx. (14) Compute f 3x±2 dx. (15) Compute f 1+ dx. (16) Compute f +1 dx. (17) Compute f e~-1 dx. (18) Compute I f2_4ex+ 'dx. (19) Compute I csi dx. (20) Compute f 0sin x+8 dx. (Hint: Use the substitution ut tan(x/2) and observe that this implies that sin x 1=2.) Note that the substitution described above is called the Weier- strass substitution, named after the German mathematician Karl Weierstrass. It is a powerful tool for computing the inte- grals of functions of the type f o g, where f is a trigonometric function and g is a rational function. 45. Strategy of Integration We presented various integration techniques in this chapter and in some preceding chapters. The most general ones were integration by parts and integration by substitution. The most frequently studied special cases were related to trigonometric functions and their inverses. Reverse substitution came up in some special cases. We also discussed the integration of rational functions, using the technique of partial fractions. In short, we have learned a decent number of methods. For this very reason, it is sometimes not obvious which method we should use when trying to integrate a function. While there is no general rule, in this section we will provide a few guidelines. So let f be a function that is not equal to one of the functions whose integral we either know offhand or have a deterministic method  52 7. METHODS OF INTEGRATION to compute. That is, f is not a polynomial, f is not a rational function, f is not the function f(x) = ax or f(x) = log cc for some positive real number a, and f is not one of the basic trigonometric functions like sin c or tan c. Let us also assume that simple algebra will not help, that is, that f cannot be transformed into one of these elementary functions by simple algebraic transformations. Then how do we decide which method to use? 45.1. Substitution. The method that needs the least amount of work, when it is available, is a simple substitution, so it is reasonable to try to use that method first. There is a particularly good chance for this approach to work when f is the composition of two functions, one of which has a constant derivative, or when f is of the form f(x) h'(g(x))g'(x), since then f (x) = ((x), and so ff(x) dz= h(x). In the language of substitutions, this means that substituting y = g(x) will work, since f dc =fh'(g(x))g'(x) dc f h'(y)jdcc =fh'(y) dy. In other words, the integral of the composite function f is turned into something simpler, the integral of the function h. EXAMPLE 7.21. Let f(x) sin 2x. Then, using the substitution y = 2x, we get that f sin2xdx= J'sin y dy = - 2cos y + C = -{cos 2x + C. EXAMPLE 7.22. Let f(x) x+1. Then we set y = x2 + 1, so dy/dx = 2x. This leads to J ccdcc f= cdx cc2 + 1 yJ c J x -+ 1 x 3 1 = 2 dy 3 =- Iny + C 2 =-ln(x2 + 1) + C. 2 The reader should compute the integral f sin" x cos x d at this point. 45.2. Integration by Parts. If f is the product of two functions, but substitution does not seem to help, then integration by parts is the logical next step. This technique is particularly useful when one of the two functions whose product is f is made significantly simpler by differentiation.  45. STRATEGY OF INTEGRATION 53 EXAMPLE 7.23. Compute the integral f xe-x dx. Solution: Considering the integrand, we notice that substitution is unlikely to help, since c and e-x are not closely related. On the other hand, the integrand is a product, and one of the terms, x, is made simpler by differentiation. Therefore, we choose the technique of in- tegration by parts, with c = u and v' = e-x. Then u' = 1, while V = -e-x, and we get Jxe-x dx = -ce-x - fe-x dx = -xe-x + e-x + C =-(1- x)e-x + C. 45.3. Radicals. If the two most general methods (substitution and in- tegration by parts) are not helpful, then it is quite possible that there is a root sign in the integrand. In that case, there are two specific meth- ods that we can try, reverse substitution and rationalizing substitution. The easiest way to know when to use each of these two methods is to remember the relatively few cases in which reverse substitution works directly. As we have seen, these are the integrals involving f r2±x2 dc, f r2 - x2 dc, and f cc2 - r2 dc. If the integrand does not contain any of these functions, then it may be simpler to use a rationalizing substitution, such as in computing the integral f <74 dcc. The ex- ercises at the end of this section will ask the reader to decide which method to use for a few specific examples. 45.4. If Everything Else Fails. If none of our methods work, then it may be that an unexpected transformation of the integrand may help, at least with relating the integral to one that is not quite as challenging to compute. EXAMPLE 7.24. Compute f tgcos? 2x dx. Solution: We use the trigonometric identity tan x = sin c/ cos x, to rewrite the integrand. We get sn4 x cos2 2x dx f sin2 x cos2 x cos2 2x dx tan2 cc -f sin2 2cc cos2 2cc dcc f - sin2 4cc dc,  54 7. METHODS OF INTEGRATION which is easy to integrate with the method we learned for powers of trigonometric functions. D It is important to point out that sometimes there is indeed no so- lution; that is, there exist elementary functions f such that there is no elementary function F(x) satisfying F'(x) = f(x). Examples of such functions f include ex2, ex/x, and l. 45.5. Exercises. In all of the exercises below, compute the integral. (1) f x2e-5x dx. (2) fdx. (3) fx+1 do. (4) fesinxcosxdx. (5) f dx. (6) f dx. (7) f dx. (8) f -dx. (9) fIxinx n x) dx. (10) 1 i dx. (11) f+sin xdx. (12) 1 efi dx. (13) f (x + cos x)2 dx. (14) f ex sin 3x dx. (15) fx2lnxdx. (16) f- dx. (17) f 4 dx. (18) f __xdx. (19) f tan4 x sec2 x dx. (20) 1 l+sinxdx. 46. Integration Using Tables and Software Packages 46.1. Tables of Integrals. Tables of integrals can be found in many calculus textbooks and on the Internet. The website www.integral- table.com is a good example, and we will use it as a reference in this subsection. (When we say "the table of integrals," we mean that table.)  46. INTEGRATION USING TABLES AND SOFTWARE PACKAGES 55 No matter how extensive a table of integrals is, it cannot contain all integrals. It is therefore important to know how to use these tables to compute integrals that are not contained in the tables in the same form. The easiest case is when the integral to be computed is a special case of a more general integral that is in the table. EXAMPLE 7.25. Use the table of integrals to compute f ln(x2+9) dc. Solution: Looking at the table of integrals, we find that the integrand is a special case of integral (45), with a = 3. Using the formula given in the table for the general case with a = 3, we get the result f ln(x2 + 9) dx fxln(x2 + 9) + 6 tan-1() - 2x + C. Sometimes, we have to resort to integration by substitution to be able to use the table of integrals. EXAMPLE 7.26. Use the table of integrals to compute f 9 - 42 dc. Solution: Taking a look at the table of integrals, we find that formula (30) provides a formula for f /a2 - x2 dc. In order to be able to use that formula, we set y = 2x, which implies dy/dz= 2. Therefore, f 9 -4x2 dz= 9f9-y2 y Now we can apply formula (30) from the table of integrals, to get 9 -y2dgy - 9_y2 +tan( y +c 2J4 2 tg -y2 X 9 2x = 2 9-4x2+-tan-1( 942)+C. Sometimes we need to carry out some algebraic manipulation before we can use the technique of substitutions in connection with using the table of integrals. EXAMPLE 7.27. Use the table of integrals to compute f 12+2 +dc. Solution: There is no integral in the table of integrals that would im- mediately stand out as one that is very similar to this one. The crucial observation is that the substitution y = x+1 significantly simplifies our integrand. That substitution leads to the integral f " dy, which,  56 7. METHODS OF INTEGRATION in turn, can be directly found in the table of integrals as item (36). Substituting x back into the obtained formula, we get / x2+2x+1 d f ~ dx X-2+ 2x+ 10 x2+2x+10 9 2In x+1+ x2+2x+10 +C. 46.2. Software Packages. Computer software packages such as Maple and Mathematica are very useful tools of integration. These packages will compute the definite or indefinite integrals of a large class of func- tions, then they will present the results in a form that is usually, but not always, in the form the user expected. In this section, we show a few examples of these unexpected results and explain how to interpret them. To start with a very basic example, type int (x^2+3x, x) ; into Maple. We get the answer 33 2. This is the correct answer having constant term 0. Experimenting with other functions, we note that Maple always answers in this way, that is, without the constant C at the end. It is important not to forget this if we will be using the obtained function in some further computation. Maple does not always provide the simplest form for the integral that it computes. For instance, if we ask Maple to compute the indef- inite integral f x(x2 + 3)6 dx, we get the output 114 312 27 10 1358 4056 7294 729 2 (7.16) - + -x + x + x+ x+ x+ x . 14 2 2 2 2 2 2 However, it is very easy to compute f x(x2 +3)6 dx by hand, using the substitution u = x2. That substitution leads to the same solu- tion, but in a much simpler form, namely, 1(x2 + 3)7. If we want to verify that this result indeed agrees with the one given by Maple and displayed in (7.16), we can ask Maple to expand the expression H = 4(x2 + 3)7 using the expand command. We see that the ex- panded expression indeed agrees with the one given in (7.16), up to the constant terms at the end. There are other commands like expand that, are useful if we want to transform the output of an integration software package. The com- mands rationalize and simplify are examples of these.  46. INTEGRATION USING TABLES AND SOFTWARE PACKAGES 57 There is often more than one way to express integrals involving hyperbolic functions. A striking example is the following. If we ask Maple to compute f 1x2dz by typing int (1/ (1-x^2) , x) ; Maple returns the answer tanh-1 c. This is very surprising, since it is not difficult to integrate the integrand as a rational function, with no hyperbolic functions involved. Indeed, 1 0.5 0.5 1-cx2 x+1 x-1' and so (7.17) J 1 1-z2 1 - ln(xc+ 1) 2 1 - ln(xc x+1 1)=In .+1 cc-1 The result obtained by Maple actually agrees with the result given in (7.17), even if that is not obvious. Indeed, if y = tanh-1 x, then, by definition, cc= (e - e-y)/(ey + e-Y). Solving this equation for y is not completely trivial, but at the end, it yields y =ln . (Hint: Multiply both the numerator and the denominator by ey to get X = (e2y _ 1)/(e2y + 1).) Finally, when computing definite integrals, Maple sometimes an- swers by using the acronyms of some rare functions. For instance, if we want to compute f0 sin (x2) dc by typing int(exp(sin(x^2)),x=0..1). then we get the answer (1/2)*FresnelS(sqrt(2)/sqrt(Pi))*sqrt(2)*sqrt(Pi). Here FresnelS refers to the Fresnel sine integral, a concept beyond the scope of this book. If we simply want to know a numerical value for f0 ec2 dc, we can type eva1f(int(sin(x^2),x=0..1)) instead. Maple outputs 0.4596976941 as the answer. 46.3. Exercises. (1) Use the table of integrals to compute f x ln(3x + 5) dc. (2) Use the table of integrals to compute f x cos - c) dc. (3) Use the table of integrals to compute f + dcc. (4) Use the table of integrals to compute f x6+16 dcc' (5) Use the table of integrals to compute f (4g2.9)cc. (6) Use the table of integrals to compute f v8x2 + 3 dc.  58 7. METHODS OF INTEGRATION (7) The table of integrals provides a formula for f ax2+bx+c dx. Describe the values of a, b, and c for which that formula will not work and compute the integral in those exceptional cases. (8) The table of integrals provides a formula for J ln(ax2 + bx+ c) dx. Describe the values of a, b, and c for which that formula will not work and compute the integral in those exceptional cases. (9) Use your favorite software package to compute f xex dx, f x2ex dx, and f x3ex dx. Do you see a pattern? Try to guess what f x4ex dx is, then verify your guess by using your soft- ware package again. (10) Use your favorite software package to compute f ln x dx, f xln x dx, and f x2 ln x dx. Do you see a pattern? Try to guess what f x3 ln x dx is, then verify your guess by using your software package again. (11) Use your favorite software package to compute f ln x dx, f(lnx)2 dx, and f(lnx)3 dx. Do you see a pattern? Try to guess what f(ln x)4 dx is, then verify your guess by using your software package again. (12) Let f (x) 1 and let g(x) = 1+2. Clearly, f(x) = g(x) 1+x2 for all x for which these functions are defined. Compute the integral of both functions with Maple. If the results seem different, show that they are in fact equal. (13) Compute f x(x2 + 1)8 dx first by Maple, then by substitution. Which answer is simpler? (14) Compute f x x2 + 4x + 29 dx using a software package. If the result appears to be different from what you find in the table of integrals, explain why the two results are equivalent. (15) Compute f tan2 x sec4 x dx. If the result appears to be differ- ent from what you get using the methods of Section 42, explain why the two results are equivalent. (16) Compute f sin2 x cos4 x dx. If the result appears to be different from what you get using the methods of Section 42, explain why the two results are equivalent. (17) Use a software package to compute f sin" x dx for = 3, nm 4, and nm 5. (The case of nm 2 was discussed in the text.) Explain the trend you detect. (18) Use a software package to compute f~ cos" x dx for nm 2, nm 3, nm 4, and nm 5. Explain the trend you detect.  47. APPROXIMATE INTEGRATION 59 (19) Use a software package to compute the value of the expression 1fi106ex2/2 (20) Use a software package to compute the value of the expression 1 106 X4 47. Approximate Integration Sometimes it is not possible to find the exact value of a definite integral fa f(x) dx. It could happen that we cannot find the antideriv- ative of f(x) or that the antiderivative of f(x) is not an elementary function. Or it could happen that f itself is not given by a formula, but instead, f is given by its graph, which is plotted by a computer program. In this case, we resort to methods of approximate integration. 47.1. Basic Approximation Methods. The key observation behind the approximation methods is the fact that if f(x) > 0 for x E [a, b], then f/b f(x) dx is equal to the area below the graph of the function f on that interval. More precisely, fa f(x) dx is equal to the area of the domain bordered by the horizontal axis, the vertical lines x = a and x = b, and the graph of f. In order to estimate the area of this domain D, we cut D into small vertical strips. To do so, we choose real numbers a = x0 f(xi_1). Furthermore, for each point x E [Lxi1, xz], we have f(x_1) < f(x) < f(x2). So the left-endpoint method underestimates the area of each strip Si, while the right-endpoint method overestimates it. Therefore, the area of D -and hence the correct value of f ex dx is between the two values of 1.2759 and 1.7055 computed above. Q Replacing the value of n = 4 by some larger number will result in a more precise approximation (and more work). Using the midpoint method will result in an approximation A that is closer to the actual value of the integral fo ex2 dx, but it is not completely obvious from which side A approximates fo ex2 dx, that is, whether A < fo ex2 dx or A > fo ex2 dx. 47.2. More Advanced Approximation Methods. 47.2.1. Trapezoid Method. If the difference between f(x2_1) and f(x2) is large, then estimating the area of Si by using rectangles could lead  62 7. METHODS OF INTEGRATION Xi - xi_1 FIGURE 7.6. Trapezoid method. to large errors. A more refined approach is to estimate the area of Si by computing the area of the trapezoid whose vertices are the points (xi1, ), (xi, ), f(xi), and f(xi_1). We know that the area of this trapezoid is the average length of its parallel sides times the distance of those parallel sides from each other, that is, ('f(x')f(x - 1))(x.-x1) Summing over all possible values of i, we get an estimate for all the area of D, that is, for fa f(x) dx. Indeed, we obtain the formula /b 2 In particular, if the xi are chosen so that they split the interval [a, b] into n equal parts, then the last displayed equation simplifies to b- b f x ) x = b - a ( f x i + f ( i - ) b (f (a) + 2f (xi) + 2f (x2) + ... + 2f (xn_1) + f (b)). 2n Note that f(xo) = f(a) and f(xn) = f(b) occur only once in the sum in the last line since a and b are each part of only one of the intervals [xi-, xi]. EXAMPLE 7.29. Use the trapezoid method with n = 4 to find the approximate value of fo e d .  47. APPROXIMATE INTEGRATION 63 An Approximation of the Integral of f(x) = exp (x2) on the Interval [0, 1] Using the Trapezoid Rule Area: 1.490678862 2.5 2 1.5 1 0.5 0 -0.5 0.2 0.4 0.6 0.8 X Partitions: 4 f(x) FIGURE 7.7. Trapezoid method. Solution: We Example 7.28. j1 will have x1 This yields 1/4, x2 = 1/2, and x3 = 3/4, just as in eC dx 1 (e + 2e1/16 + 2e1/4 + 2e9/16 + e) = 1.4907. The alert reader may have noticed that the result we obtained is precisely the average of the left-endpoint and right-endpoint approx- imations we obtained for the same integral in the previous section. (This means that it is a better approximation than at least one of the two earlier ones.) This is not an accident, and in Exercise 47.4.17, the reader will be asked to prove that, under certain conditions, this phenomenon will always occur. 47.2.2. Simpson's Method. A similar method is Simpson's method, in which we use parabolas instead of straight lines for approximation. For simplicity, let us now assume that the points x2 split the interval [a, b] into n equal parts, that is, x2 - x2_1 = (b - a)/n. In order to simplify the notation, let us set y2 = f(x2). For any integer i E [1, n-1], consider the points (x2_1, yi-1), (xi, y2), and (xi+1, y2+1). There is exactly one parabola p2 of the form y Ax2 + Bx + C that contains these three points. It can then be proved  64 7. METHODS OF INTEGRATION that the area under that parabola more precisely, the area of the domain P bordered by the horizontal axis, the vertical lines x = x2-1 and x = xi+1 and p2 is equal to (7.19) b - a (yi_1 + 4y2 + yi+1) . 3n If we summed (7.19) over all possible values of i, we would not get a good estimate, since most points of the domain under the curve would be part of two of the P. For instance, a point with a horizontal co- ordinate between x and xi+1 is part of both P and P+1. Therefore, we sum the last displayed equation over all even values of i, and, ac- cordingly, we stipulate that n be an even number. This leads to the following estimate. THEOREM 7.1 (Simpson's Method). Let n be an even positive in- teger. Then f (x) dx ~ (yo + 4y1 + 2Y2 + 4Y3 + ...+ 2Yn-2 + 4yn-1 + yn). ab 3n EXAMPLE 7.30. Use Simpson's method with n = 4 to approximate fj sin (x2) dx. An Approximation of the Integral of f(x) = sin(x2) on the Interval [0, 1] Using Simpson's Rule Area: .3102485324 0.8- 0.6- 0.4 0.2- 0- 0.2 0.4 0.6 0.8 1 x -0.2 Partitions: 4 f(x) FIGURE 7.8. Simpson's method.  47. APPROXIMATE INTEGRATION 65 Solution: We have yo= 0, y1i= sin(1/16), y2 = sin(1/4), y3 = sin(9/16), and y4 =sin 1. So Simpson's method yields ]sin(x2) d 12 (4 sin(1/16) + 2 sin(1/4) + 4 sin(9/16) + sin 1) 0.31. Note that this result confirms our intuition in that if c E [0, 1], then sin (x2) < sincc, and so ] sin(x2) dc < ] sin x = 1 - cos 1 0.4597. 47.3. Bounds on the Error Term. The error term E of an approximation is the difference between the number obtained by the approximation and the actual value of the quantity that was approximated. (In this section, that actual value is the value of f f(cx) dc.) It goes without saying that the smaller the absolute value of the error term, the better the approximation is. The field of numerical analysis studies the error terms of approx- imation methods. The techniques of that field yield various bounds on error terms. We collected some of these bounds in the following theorem. THEOREM 7.2. Let f be a twice-differentiable function on [a, b] such that f"(x)| < M if x e [a,b]. Then the following hold for the approxi- mation methods used to compute fb f(z). (a) If ET is the error term of the trapezoid method, then FT N for some real number N. Time has come to formally define fa f(x) dx. DEFINITION 7.1. Let f be a function. exists for all b > a and limb,, fa f(x) dx - If the integral f f(x) dx L exists as a (finite) real number, then we say that the integral fj7 f(x) dx is convergent, and we write ff(x) dx = L. If limb-oo f f(x) dx does not exist or is infinite, then we say that f ' f(x) dx is divergent. Note that if F is an antiderivative of f, then b lim f (x) dx b-oo a a lim (F(b) - F(a)) b )oo (lim F(b)) - F(a). b-*oo Therefore, the integral fj° f(x) dx limb,,, F(b) exists and is finite. is convergent if and only if EXAMPLE 7.32. Let f (x) x--2. Compute f° f(x) dx.  48. IMPROPER INTEGRALS 69 2.0 1.5~ ( x 1 1.0 0.5 0 1 2 3 4 5 6 7 8 9 10 11 12 13 FIGURE 7.10. Area under the curve y = f(x) from 1 to 00. Solution: We have o0 x--2 dx 1 b lim x- b--o~ 1 -2 dx b lim --- b-ooi 1 lim- + 1 b-oo b 1. In particular, f7 f(x) dx is convergent. Encouraged by the simple solution of the last example, we are going to compute the more general integral f xr for any real number r. EXAMPLE 7.33. Let f(x) = xr. Compute f f(x) dx. Solution: Let us first assume that r # -1. Then we have 1 b z' dx = lim z' dx = lim r+1 b~oo r + 1 b 1 If r > -1, then r + 1 > 0 and limxsoo xr+1 = oc, so the limit in the last displayed row is infinite, and hence f °° xr dx is divergent. If r < -1, then r + 1 < 0 and limx~o0 xr+1 = 0, so the limit in the last displayed row is equal to -i, and hence f °° xr dx is convergent.  70 7. METHODS OF INTEGRATION If r = -1, then we need to compute fj z dz differently, since, in that case, f zo dx a . Instead, we have x-1 dz= lim cx-1 dz = boo j b = lim ln x b-oo 1 = lim ln b = oo. boo Therefore, f0° x-1 dc is divergent. Note that the results of the previous example prove the following important theorem. THEOREM 7.3. Let r be a real number. (i) If r;> -1, then f1 x dx is divergent. (ii) If r < -1, then fx" dx is convergent. The following definition is not very surprising. It is the counterpart of Definition 7.1. DEFINITION 7.2. Let f be a function and let b be a real number such that, for all real numbers a < b, the integral fa f(x) dx exists. If L = lima-,-o f (x) dx exists as a (finite) real number, then we say that the integral fb. f (x) dx is convergent, and we write fb. f (x) dx = L. If lima--oo f(x) dx is infinite or if it does not exist, then we say that fb0 f (x) dc is divergent. The following definition makes it clear how and when we can define an integral on the entire line of real numbers. DEFINITION 7.3. Let f be a function and let m be a real number such that both f_" f(x) dx and fj* f(cx) dx are convergent. Then we say that the integral f_ f(cx) dx is convergent and that f (x) dx = f(x) dx + f(x) dx. -oo -o m Otherwise; we say that f_*, f(cc) dcc is divergent. See Figure 7.11 for an illustration. EXAMPLE 7.34. Compute f°° ex dcc.  48. IMPROPER INTEGRALS 71 48. IMPROPER INTEGRALS 71 X = m FIGURE 7.11. fm, f(x) dx is blue, while fm f(x) dx is orange. -2 -1 0 2 FIGURE 7.12. f0 exdx. Solution: We set m = 0 and apply Definition 7.3. We get that f e-x dx is convergent if both of f_. e-x dx and f° e-x dx are con- vergent. However, -0' 0 c-x dx = lim -c-x : a-*-oo a 1+ o0 is divergent and therefore so is fe . Figure 7.12 shows the domain whose area is equal to f_, e- dx. The reader could ask how we knew that we needed to select 0, and not some other real number, for the role of m, that is, to split the real number line into two parts. The answer is that we did not, and other choices of m would have given the same result since the integrand  72 7. METHODS OF INTEGRATION converges to infinity as x goes to negative infinity. We chose m = 0 because it was convenient to do so. Note that all improper integrals discussed in this section are called Type 1 improper integrals. 48.2. Vertical Asymptotes. Sometimes we may want to compute the integral of a function f on a finite interval [a, b] so that in some point c E [a, b], the function f has a vertical asymptote. An example is the function f(x) = 1/(x2 - 4) on the interval [1, 3]. In this case, we use the technique of limits to formally define fja f(x) dx, as we did in the previous section. DEFINITION 7.4. Let f be a function that is continuous on [a, b], except for one point c E [a, b]. Then we set (7.20) and f (x) dz Jf (x) dz = I b t : lim f (x) dx ac jb f(x) dc. lm+ (7.21) Furthermore, if finite, we set both of the two limits displayed above exist and are f(x) dz j f(x)dz + bf(x)d. Note that if the only point c in which f is not continuous is one of the endpoints of [a, b], then we only have to compute one of (7.20) and (7.21), since the other integral is taken over a trivial interval and is hence zero. EXAMPLE 7.35. Compute fjj -1/2 dc. Solution: As the only point in [0, 1] in which f(x) continuous is 0, we use formula (7.21) with c = 0 and b x-1/2 is not :1.  48. IMPROPER INTEGRALS 73 1 x=0 x=1 FIGURE 7.13. f0x-/2 dx We get x-/ d Ji x-71/2dx 1 = rn 2x1/2 t =2 - lrn t1/2 t-*-o ± =2-0 = 2. So the integral fof x-12 dx is convergent. EXAMPLE 7.36. Compute f> 41x2 dx. D Solution: We apply Definition 7.4 since the interval [-1, 4] has one point, c =0, where the integrand is not continuous. Therefore, -4 IX-2 dx i-1x2 dx + X-2 dx lim x-2dx+ li j X-2 dx limn t -x1 + limn 4 -x1 t 00 +00 00.  74 7. METHODS OF INTEGRATION 1 f (x) = 2 x2 -1 0 4 FIGURE 7.14. f41 x2 dx. So the integral in question is divergent. Q Figure 7.14 shows the domain whose area is equal to fo x-2 dx and the correct way of breaking that interval up to two parts. Note that we would have reached the wrong conclusion if we had disregarded the fact that x-2 is not continuous at x = 0 and tried to apply the fundamental theorem of calculus. Indeed, in that case, we 4 would have obtained the wrong result: -x-1 = - - 1 = - . This result is incorrect, and the incorrect step was to apply the fundamental theorem of calculus for a function that is not continuous in the entire interval of integration. The integrals that we have discussed in this section are called Type 2 improper integrals. 48.3. Further Remarks. 48.3.1. Improper Integrals of Mixed Type. There are some integrals that are improper for two reasons. They are taken over an infinite interval, and that interval contains a point in which the function is not contin- uous. In that case, we split up the interval of integration so that now we have two integrals, one of which is of Type 1 and the other of which is Type 2. EXAMPLE 7.37. Compute foG 1()2 - Solution: We break up the interval [0, oo) to the union of the two intervals [0, 2] and [2, oo), getting /*G 1 d 2 1 d o 1 J (x - 2)2 0 (x - 2)2 2 (x - 2)2  48. IMPROPER INTEGRALS 75 3.0 1 2.5 (x - 2)2 2.0 1.5 1.0 0.5 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 FIGURE 7.15. f1 2dx. 0.12 0.10 1 0.08 x2 in x 0.06 0.04 0.02 0 1 2 3 1 x2 4 5 6 7 8 9 10 11 12 13 14 FIGURE 7.16. f3 X2 n ' The first term on the right-hand side is an improper integral of Type 2, and the second term on the right-hand side is an improper integral of Type 1. We can compute both by the methods presented earlier in this section. D 48.3.2. Comparison Test. Comparison tests for improper integrals work very similarly to those for proper integrals. THEOREM 7.4. Let us assume that, for all x > a, the chain of inequalities 0 < f(x) g(x) holds. (i) If f ° f(x) dx is divergent, then so is f' g(x) dx. (ii) If f g(x) dx is convergent, then so is fa7 f(x) dx. EXAMPLE 7.38. Show that f3 x2 in dx is convergent.  76 7. METHODS OF INTEGRATION Solution: If x ;> 3, then lncx > 1, so x21ln x > x2, and therefore the integrand is less than y. See Figure 7.16 for an illustration. On the other hand, we know that f3o 1/c2 dc is convergent, so our claim follows from the comparison test. D 48.4. Exercises. (1) Is flsin x d convergent or divergent? (2) Is f0 X-15 dcc convergent? (3) Is f00 x-2 dc convergent? (4) Is f e- dc convergent? (5) Is f cex2 dc convergent? (6) Is f,° eX+x dc convergent? (7) Is f2xn dcc convergent? (8) Is f x2-8x+7 dc convergent? (9) Compute f7 e-ax dc, where a is a fixed nonnegative real num- ber. (10) Compute f lnxd. (11) Compute f2(lnc)2 dc. (12) Compute fo dc. (13) Compute f1 3>3dc. (14) Compute f000x2+1 dc. (15) Compute f0e ,2 c+9 (16) Compute fo dc. (17) Is there a real number a such that f oca dc is convergent? (18) Prove that if f is an even function, and f0 f(cx) dc is con- vergent, then f° f0(x) dc is convergent and f° f (x) dcc 2 f°°f()d. (19) State and solve the analogous version of the previous exercise for the case of odd functions f. (20) Show an example of a function f for which jfc(x)dc limJ f(x) dc. 0oo R  CHAPTER 8 Sequences and Series 49. Infinite Sequences A sequence can be thought of as an ordered list of numbers ai, a2, ..., an, an+1, .... The subscript n indicates the position of a number an in the sequence; for example, a1 is the first element, an is the nth element, and so on. DEFINITION 8.1 (Sequence). A sequence is a function f defined on the set of all positive integers; that is, it is a rule that assigns a number to each positive integer. If f(n) = an for n = 1, 2,..., it is customary to denote the range off by the symbol {an} or {an}°. So a sequence can be defined by specifying the rule an = f(n) to calculate the nth term from an integer n. For example, n1n-2 23 n+1 1 1 2' 3' 4 (- ) (-1)" ___1 1 1 an -1 -, ... , n n i 2 3 4 an = q" ->{q"} = {1, q, q2q3 Sequences can also be defined recursively, that is, by a relation that al- lows us to find an if am, m < n, are known. For example, the Fibonacci sequence {f,} is defined by the recurrence relation fi=f/2 =1, fm =fa-_1+f-_2, n ;> 3 ->{fn} ={1, 1,2, 3,5,8, 13, ...} Graphic representation of sequences. A sequence can be pictured sim- ilarly to the graph of a function by plotting points (x, y) = (n, an), n = 1, 2, ..., on the xy plane. For example, the sequence an = n/(n+ 1) is the set of points on the graph y =x/(x + 1) corresponding to all positive integer values of x, that is, x = 1, 2, .... 49.1. Limit of a Sequence. The sequence an = n/(n + 1) has the prop- erty that the values an approach 1 as n becomes larger. Indeed, the difference in 1 1- an=1- in+1 in+1 77  78 8. SEQUENCES AND SERIES 78 8. SEQUENCES AND SERIES 1.0 0.8 0.6 0.4- 0.2- 0 1 2 3 4 5 FIGURE 8.1. Set of points on the graph y = x/(x + 1) corresponding to integer values x = n. For large x, x/(x+1) approaches 1 from below, and hence n/(n+1) 1/(1+1/n) -- 1 as n -- oo. The difference 1-n/(n+1) 1/(n + 1) can be made smaller than any (small) number E > 0 for all n > N and some integer N. decreases with increasing n and hence can be made smaller than any preassigned positive number E for all n > N, where N depends on E. For example, put E = 10-2. Then the condition 1 - a. < E implies that 1/(n + 1) < or 1/E - 1 < n or 99 < n. Thus, 1 - a. < 10-2 for all n > 99. If E = 10-4, then 1 - a. < 10-4 for all n > N = 9999. In other words, no matter how small E is, there is only a finite number of elements of the sequence that lie outside the interval (1 - E, 1 + E). In this case, the sequence is said to converge to the limit value 1. DEFINITION 8.2 (Limit of a Sequence). A sequence {an} has the limit a if, for every E > 0, there is a corresponding integer N such that an - a < E for all n > N. In this case, the sequence is said to be convergent, and one writes lim an = a or an -- a as n -- oo. n-* oc If a sequence has no limit, it is called divergent. One can say that a sequence {an} converges to a number a if and only if every open interval containing a has all but finitely many of the elements of {an}. THEOREM 8.1 (Uniqueness of the Limit). The limit of a convergent sequence is unique: lim an = a and lim an = a' a = a'. n -*oc n -oc  49. INFINITE SEQUENCES 79 49. INFINITE SEQUENCES 79 0 _ _ _ _ _ _ _ __-_ _-_ _ _ _ _ _ _ _ _ _ _ a + 6 FIGURE 8.2. Definition of the limit of a sequence. The dots indicate numerical values an (vertical axis). The integer n increases from left to right (horizontal axis). The convergence of as to a number a means that, for any small e > 0, there is an integer N such that all the numbers an, n > N, lie in the interval (a - e, a + e). It is clear that N depends on e. Generally, a smaller e requires a larger N. PROOF. Fix >R Then, by the definition of the limit, there are numbers NandN'suchthat |an - al ai.fn>N and an-a'| if nm> N'. Hence, both inequalities hold for n> max(N, N') and for all such n: 0 0, there is a corresponding number M such that f(x) - al < for all > M) by noting that the range of f(x) contains the sequence an = f(n).  80 8. SEQUENCES AND SERIES 80 8. SEQUENCES AND SERIES EXAMPLE 8.1. Find the limit of the sequence an = Inn/n if it exists or show that the sequence is divergent. Solution: Consider the function f(x) = in x/x so that an = f(n) for all positive integers. Hence, lnn lnx .1/z lim = lm - = lm = 0, n- n x-oo X x-oo 1 where the indeterminate form * arising from in zx/z as x - o has been resolved by means of l'Hospital's rule. Note that l'Hospital's rule applies not to sequences but to functions of a real variable. Q Following the analogy between the limits of sequences and func- tions, one can select a particular class of divergent sequences. DEFINITION 8.3 (Infinite Limits). The limit limn-, an = 0o means that, for every positive number M, there is a corresponding integer N such that an > M for all n> N. Similarly, the limit limn-o an = -oo means that, for every negative number M, there is a corresponding integer N such that an < M for all n> N. EXAMPLE 8.2. Analyze the convergence of the power sequence an = 1/np, where p is real. Solution: Put f(x) = 1/cp for x> 0. Then an = f(n) and therefore 1 1 0 ifp>0, lim - lim - 1 if p=0, n-o jnp x-o o LXo if p <0. EXAMPLE 8.3. Analyze the convergence of the sequence an = q", n = 0, 1, ..., where q is real. Solution: Suppose q > 0. Put f (x) = qx = exlnq. From the properties of the exponential function, it follows that eax oc if a =ln q > 0, eax = 1 if a = ln q = 0, and eo 0 if a = ln q < 0. Therefore, an- 00 if q> 1, an= 1 - 1 if q =1, and an - 0 if 0 < q < 1. When q = 0, an = 0. Suppose q < 0. Then q = -ql and an = (-1)"ql"n= (-1)"e"l1". If |ql < 1, then even and odd terms of the sequence converge to 0: a2n = e2nInq - 0 and a2n-1 = -e(2n-1)ln q - 0 as n -i oc. When q =-1, the sequence an= (-1)< is divergent because a2n 1 and a24-1 =-1; that is, the sequence oscillates between 1 and -1 for all m and an does not approach any number. Finally, if q < -1, the sequence is divergent, too, because a2n - einql - 00 but a24- - e(2-1) in ql -00o. Moreover, it approaches neither 00 nor  49. INFINITE SEQUENCES 81 49. INFINITE SEQUENCES 81 -oo as it oscillates taking ever-increasing positive and negative values. Thus, 0 if q E (-1, 1), lim q" = 1 if q = 1, L 00 if q > 1, and the sequence does not converge if q < -1. D EXAMPLE 8.4. Find the limit of the sequence an = (1 + q/n)1/". Solution: Let f (x) (1 + q/x)x for x > |ql to make 1 + q/x positive for any q. Then lim an = lim (1 + = lim (1u+ q)/U T-oo x-oo X u-O+ where the substitution x = 1/u has been made. To find the latter limit, consider ln f (x) = ln(1+ qu)1/" = ln(1+ qu)/u. Using l'Hospilal's rule, ln(1 + qu) g/(1 + qu) . q lim = lim = lim q q. u-o+ u u-o+ 1 u-o+ 1 + qu It follows from ln f(x) - q that f(x) - eas x - 00 and, hence, lim 1+ - =eq n-oo n for all real q. 49.2. Subsequences. Given a sequence {an}, consider a sequence {nk} of positive integers such that n1 <1n2 <1n3 < """. Then the sequence { an}, k = 1, 2, ..., is called a subsequence of {an}. Recall that a se- quence {an} converges to a number a if and only if every open interval containing a has all but finitely many of the elements of {an}. There- fore, the following theorem holds. THEOREM 8.3. A sequence {a} converges to a if and only if every subsequence of {an} converges to a. This necessary and sufficient criterion for convergence has already been used in Example 8.3. The sequence an = (-1) does not con- verge because it has two subsequences a2,_ 1 and a2-1 =-1, which converge to different numbers, 1 / -1. 49.3. Limit Laws for Sequences. The limit laws for functions also hold for sequences, and their proofs are similar. If {an} and {b,} converge  82 8. SEQUENCES AND SERIES 82 8. SEQUENCES AND SERIES to numbers a and b, respectively, and c is a constant, then lim (an + bn) = lim an + lim bn = a + b, nh-ooTn-oo Tn-oo lim (can) C lim an = ca, nh-ooTn-oo lim (aba) = lim an lim b= =ab, nh-ooTn-oo Tn-oo lim a_ lim - a - I if b#0, n- o bn limn- o bn b' lim (an)p = (lim an)p = ap if p > 0 and an > 0. nh-ooTn-oo It should be emphasized that the convergence of the sequences an and bn is crucial for these relations to hold. For example, let bm = n and an = q - n where q is a number. Then both the sequences diverge. In particular, b - 00 and an - -oo as n o oc. However, an + bn = q and, hence, an + bn - q as n - oo. Evidently, the relation "q 00 - o" makes no sense. Furthermore, take c = 0 so that the sequence ca = 0 - n = 0 is a constant sequence and, hence, converges to 0. The relation "0 = 0 -oc" would not make any sense either. Similarly, put an = n and bn = q/n. Then an diverges, an - o0, while the sequence bn converges bn - 0 as n - 0. However the sequence of the products is a constant sequence anbn = q and, hence, converges. The relation "q = 0 -oc" makes no sense. If an = pn, q > 0, and bn = n, then an o and b - 0o as n oc, where the sequence of the ratios converges: an/bn = q. The relation "q ="" is meaningless. The squeeze theorem also applies to sequences (see Fig. 8.3). THEOREM 8.4 (Squeeze Theorem). If c < an < b for n> N and lim- o bn = lim-- c = a, then lim--o an = a, where a can also be +oo or -oo. EXAMPLE 8.5. Find the limit of an = sin(7/in4). Solution: Since -x < sin x < x if x > 0, one has c = -w/V n an C /5 = bn, where c 0 and b 0 as n oc (see Example 8.2). By the squeeze theorem, sin(w/\) - 0 as n - oo. D THEOREM 8.5. If lim-- la| = 0, then lim- o an = 0. This theorem follows directly from the definition of the limit of a sequence where a = 0. The next result provides a convenient tool to calculate limits of sequences using continuous functions. T HEORE M 8.6. If an -~ a as in -a 00 and the function f is contin- uous at a; then lim f (an) =f (a).  49. INFINITE SEQUENCES 83 49. INFINITE SEQUENCES 83 ----- ---------- ---- ----- ---- --.a FIGURE 8.3. The squeeze theorem. The dots indi- cate numerical values (vertical axis) of the sequences bn (blue), cn (black), and an (red). The integer n increases from left to right (horizontal axis). The sequences bn and cn converge to a number a. This means that the differ- ences bn -a and cn -a can be made arbitrarily small for all n > N and some integer N. Since cn < a_ < b, the difference an - a is also arbitrarily small for all n > N. By the definition of the limit, the sequence an must con- verge to a, too. This theorem asserts that if a continuous function is applied to the terms of a convergent sequence, the result is also convergent. PROOF. The continuity of f at a means that lim,,a f(x) = f(a) or, by the definition of this limit, for any > 0, there is a corresponding 6 > 0 such that |f(x) - f(a) < E whenever x - al < 6. Having found such 6, put E' = 6 and, by the definition of the limit limn-, an = a, for any such E' > 0, there is a corresponding integer N such that |an - a < E' = b if n > N. Therefore, for any E > 0, one can find a corresponding integer N such that |f(an) - f(a) < E for all n > N, which means that limn-, f(an) = f(a). Q EXAMPLE 8.6. Find the limit of the sequence an = exp(1/n2). Solution: Consider the sequence bn = 1/n2. Then 1 lim bn = lim 2 = 0. nt-oo X-*oo X2 Put f(x) = cx. Then an = f(bn). By continuity of the exponential function, lim an = exp ( lim bn) = eo = 1. n-*+oo n-*+oo  84 8. SEQUENCES AND SERIES 84 8. SEQUENCES AND SERIES 49.4. Exercises. (1) Find a formula for the general term an of the sequences: 1 1 1 1 1 {an =l'3' 5'7' 9'11 1 1 1 1 {a?}{l'2' 4' 8'16 1 2 3 4 5 (2) Let {an} converge to a number a. Prove that the sequences be = an+k, where k is a fixed positive integer, c = an2, and do = a2n converge and find their limits. (3) Show that the sequence an = (2n - 1)/(n + 1) converges to 2. Determine how many terms of an lie outside the interval (2-6,2+E) if = 0.1 and =0.01. (4) Show that the sequence an = (n2 - 2(-1)n + 1)/(n2 + 1) converges to 1. Given 2 > E > 0, find the number of terms of the sequence that lie outside the interval (1 - e, 1 + ,) as a function of E. In the following exercises, determine whether the sequence an converges or diverges for the specified ranges of parameters (if the range of a parameter is not given, assume that it can be any real number). If the sequence converges, find its limit. (5) an = 2n. (6) an = 2" - (-1)"24. (7) an = (3 - 5n2)/(1 + n2). (8) an = Pm(n)/Qk(n), where Pm and Qk are polynomials of de- gree m and k, respectively. (9) an = 1/v/n2 + n + 1. (10) an = (n2 + 1)1/3/(n3 + 1)1/2. (11) an = (n + 2)3/(n6 + n2 + 1)1/2. (12) an = 3/(n3 + 1)/(8n3 _p4n2 - 2n - 1) (13) an = [Pm(n)]4/[Qk(n)]P, where q and p are positive numbers and Pm and Qk are polynomials of degrees m and k, respec- tively. (14) an (2" + 3")e-". (15) a = (2" + 3")e-2n (16) an (2"h+3"h+5")/(2"hp+3"h+5"hp), where q and p are positive numbers. (17) an (2-" + 3-4 + 5-")/(2-"p + 3-"l + 5-"hq), where q and p are positive numbers.  50. SPECIAL SEQUENCES 85 50. SPECIAL SEQUENCES 85 (18) an = Pm(p")/Qk(q"h), where Pm and Qk are polynomials of degree m and k, respectively, and 0 < p, q < 1. (19) an = q" sin(pn). (20) an = tanh(n2). (21) an = tan[n7r/(2 + 4n)]. (22) an = sin2[7r(n2 + 2)/(2n2 + 5)]. (23) an = ln(an)/ln(bn), where a and b are positive numbers. (24) an = ln(Pm(n))/ ln(Qk(n)), where Pm and Qk are strictly pos- itive polynomials of degree m and k, respectively. (25) an = n cos(1/n). (26) an = (lnn)P/n. (27) an = n(lnn)P. (28) an= tan-1(n2). (29) an = n2 + n - n. (30) an= n sin(p/n). (31) an = n2(cos(p/n) - 1). (32) an = n(ql/" - 1), where q > 0. (33) an = n2 e_2 _n. (34) an = bn sin(p/bn), where bn - c as n -- oo. (35) an = (b -tan bn)/(bn)p, where p > 0 and b - 0 as n - oo. (36) an = f(ba)/g(ba), where the functions f and g are twice dif- ferentiable and have a root b such that f'(b) = g'(b) = 0, g"(b) 0, and bn - b as n - oo. (37) Prove that a sequence that is the sum of convergent and di- vergent sequences diverge. 50. Special Sequences THEOREM 8.7 (Special Sequences). Let p and q be real numbers. (8.1) lim 9p/ = 1 if p > 0, T--oo (8.2) lim n = 1, T-oo (8.3) limn-= 0 if p > 1, m,-oo p in! (8.4) lim - =0, n- oo in (8.5) lim = h-ooX in!  86 8. SEQUENCES AND SERIES 86 8. SEQUENCES AND SERIES PROOF. (8.1). If p > 1, put an = n/p - 1. Then an > 0 and, by the binomial theorem, p = (1 + a)" = 1 +na + n(n - 1)an/2 + - -+ na"-i+ a". All terms in the right side are positive. Therefore by retaining only the first two terms, a smaller number is obtained: 1+ nan < (1+ an)Th p. It follows from this inequality that p-1 0 1. (8.2). Put an m= g - 1. Then an > 0 and, by the binomial theorem, n(nm-1) 2 n = (1 + an)"> 2 l 2 a. Hence, for n;> 2, 2 n2-1 By the squeeze theorem, an- 0or n/ = an + 1 > 1 as n > 0. (8.3). Consider the function f(x) =§zqe-cx, where c > 0. By the asymptotic property of the exponential function, f(x) - 0 as x - 00 for any q; the exponential grows faster than any power function (which has been proved in Calculus I). Since an = f(n) for c =lnp > 0 if p > 1, one concludes that lim - = lim n2e-" n"p= = lim z-xl"P = 0. h-oo pn m- oo x-oo (8.4). The following inequality holds: n2! 1-2-3---n 1 2 3 n2 1 1 By the squeeze theorem, an -~ 0 as n -~ oo. (8.5). If q > 0, then there is a positive integer k such that k-i <; q < k,  50. SPECIAL SEQUENCES 87 50. SPECIAL SEQUENCES 87 0 . 0 0 an < an+1 S . an > an+1 0 0 0 FIGURE 8.4. The sequence in the left panel is monoton- ically increasing, and the sequence in the right panel is monotonically decreasing. that is, k is the smallest positive integer such that q/k following inequality holds: < 1. The qfl q q . q q..q k-q an= -~I 1, if it exists lim n nI = lim ( n )q = ( lim n n)q = 1q = 1 n-*+oo n-*+oo n-*+oo by (8.2) and the basic limit laws. Q 50.1. Monotonic Sequences. DEFINITION 8.4 (Monotonic Sequences). A sequence an is said to be monotonically increasing if an < an+1, monotonically decreasing if an > an+1 for all n = 1, 2,.... The class of monotonic sequences consists of the increasing and the decreasing sequences. EXAMPLE 8.8. Show that the sequence an tonically decreasing. n/(n2 + 1) is mono-  88 8. SEQUENCES AND SERIES 88 8. SEQUENCES AND SERIES --------------------------M 0 " . " " * . e 0 0 e . . ------------------------------------m FIGURE 8.5. A bounded sequence. The dots indicate numerical values of a. (vertical axis). The integer n in- creases from left to right (horizontal axis). All the num- bers a. lie in the interval: m < a. < M. Solution: The inequality a. > an+1 must be established. It is equiv- alent to the following inequalities obtained by cross-multiplication: n+2 < n , (n+1)(n2+1) < n[(n+1)2+1] (n + 1)2 + 1 n2 + 1 n3+n2+nm+1 1. Therefore, an+1 < an (in fact, the strict inequality an+1 < an holds as well), and the sequence is monotonically decreasing. D DEFINITION 8.5 (Bounded Sequence). A sequence is said to be bounded above if there is a number M such that an < M for all n > 1. A sequence is said to be bounded below if there is a number m such that m < a for all n > 1. A sequence is said to be bounded if it is bounded above and below: m < an < M for all n > 1. For example, the sequence an 1/n is bounded: 0 < an < 1. The sequence an = en is bounded below, but not above.  50. SPECIAL SEQUENCES 89 50. SPECIAL SEQUENCES 89 Completeness axiom for the set of real numbers. The completeness ax- iom for the set of real numbers says that if S is a nonempty set of real numbers that has an upper bound M (x < M for all x E S), then S has a least upper bound. By definition, the number a is a least upper upper bound of S if, for any E > 0, a - E is not an upper bound of S. The least upper bound is called the supremum of S and denoted sup S. Naturally, sup S < M for any upper bound M of S. If S has a lower bound m, then it also has the greatest lower bound, denoted inf S (the infimum of S). The number inf S is a lower bound of S such that inf S + E is not a lower bound of S for any positive E > 0; that is, m < inf S for any lower bound m of S. The completeness axiom is an expression of the fact that there is no gap or hole in the real number line. THEOREM 8.8 (Monotonic Sequence Theorem). Suppose {an} is monotonic. Then {an} converges if and only if it is bounded. PROOF. Suppose an < an+1 (the proof is analogous in the other case). Let S be the range of {an}. If {a} is bounded, let a = sup S be the least upper bound of S (it exists by the completeness axiom). Then an < a for all n > 1. By the definition of sup S, for every E > 0, the interval (a - ,, a] should contain an element of S (otherwise, a - E would be an upper bound of S). Therefore there should exist an integer N such that a - e < aN a. Since {an} increases, all numbers an, n > N, lie in the interval (a-c, a]: a - E< an < a - |a -an| N. By the definition of the limit of a sequence, this inequality shows that {an} converges to a. D The proof is illustrated in Fig. 8.6. EXAMPLE 8.9. Investigate the convergence of the sequence defined by the recurrence relation a1 = 2 and an+1 - (an + 3). Solution: Let us compute the first few terms of the sequence ai1= 2, a2 = 2.5, a3= 2.75, and so on. The initial terms suggest that the sequence is monotonically increasing, and one can try to prove this property a a +1 for all n. A commonly used technique to do so is mathematical induction. The statement is true for nm 1. Suppose that the statement is true for nm k (the hypothesis of mathematical induction). Then, using this hypothesis, one has to prove that the statement holds for nm k + 1. If the proof goes through, then starting  90 8. SEQUENCES AND SERIES 90 8. SEQUENCES AND SERIES ----------------------------------- a , e * * e * ------------------------------------a - e 0 FIGURE 8.6. Monotonic sequence theorem. A bounded monotonic sequence with numerical values indicated by dots (vertical axis). The integer n increases from left to right. If a = sup{an} is the least upper bound of all an, then, for any number E > 0, a - E is not an upper bound of the sequence. Since a. increases monotonically, there is an integer N such that all the numbers a, n > N, are greater than a - E and hence lie in the interval (a - E, a]. This means that a. converges to a. with n = 1, one can establish the statement for n = 2, n = 3, an so on. This is the basic idea of mathematical induction. Using the recurrence relation, 1 1 ak+1 > ak 1 (ak+1 + 3) > -(ak + 3) - ak+2 > ak+1. 2 2 Thus, the sequence is indeed monotonically increasing. If it happens to be bounded, then it converges. Again, mathematical induction turns out to be helpful. The first terms suggest that a. < 3. This is true for n = 1. Suppose the inequality is true for n = k. Let us try to prove that this hypothesis implies that the inequality holds for n = k + 1. Using the recurrence relation, 1 1 ak < 3 -- (ak + 3) < -( ) -> k1<3 2 2 Thus, the sequence is monotonic and bounded and hence converges. If the sequence a. converges to a, then so does the sequence an+k for any integer k (the sequence b. = an+k is a subsequence of the convergent sequence a. and hence b. converges to a by Theorem 8.3). Since the existence of the limit has been established, the limits of both sides of the recurrence relation must coincide by the basic limit laws: 1 1 lim a.+1 = -( lim a. + 3) a = -(a + 3) a = 3. n-*oo 2 n-*oo 2  50. SPECIAL SEQUENCES 91 50. SPECIAL SEQUENCES 91 Thus, an - 3 as n - oo. EXAMPLE 8.10. Investigate the convergence of the sequence defined by the recurrence relation a1 - 2 and an+1 - 2 + an. Solution: The first few terms of the sequence suggest that the se- quence is increasing: ai = < a2 = 2 + 2. Let us try to prove the inequality an < an+1 by induction. Suppose it is true for n = k. Then, by monotonicity of the square root function and the recurrence relation, ak < ak+1 ak < ak+1 - 2+ak < 2 + ak+1 - ak+1< ak+2. The first terms of the sequence suggest that ai < 3 and a2 < 3. Let us try to prove that an < 3 for all n by induction. Suppose the inequality holds for n = k. Then, by the recurrence relation, ak <3 -4 2 + ak < 5 42 + ak< 5 < 3 - ak+1< 3. Thus, the sequence is monotonic and bounded, and hence it converges. If its limit is a, then using the basic limit laws lim an+1 = lim 2+ an = 2+ liman a = 2+ a a = 2. h-oo h-oo h-oo 50.2. Exercises. In (1)-(15), find the limit of the sequence {an} or show that it does not exist. Assume that the parameters are real numbers. (1) an = 2n2 + 3. (2) an = cos2(n2)/. (3) an = nr". (4) an = np-- (5) an = n"q". (6) an = p" - n"q. (7) an = nup" - n"q". (8) an = Pm(n)q"h, where Pm is a polynomial of degree m. (9) an = (2n - 1)!!/(2n)", where (2n - 1)!!= 1 -"3 - 5 ...(2n - 1). (10) a_= (2n)!!/(2n - 1)!!, where (2n)!! =2 - 4... (2n). (12) an #3" + 5Th. (14) an /qn2 + pn+1, where q > 0.  92 8. SEQUENCES AND SERIES 92 8. SEQUENCES AND SERIES (15) an = /Pm(n), where Pm is positive polynomial of degree m > 1. (16) Prove that every convergent sequence is bounded. In (17)-(22), determine whether the sequence is monotonic or not monotonic. Is the sequence bounded? (17) an = (-2). (18) an = (-)n . (19) an= ne--. (20) an = n + . (21) an = sin(qn)/n, where q is real. (22) an = (1 + q/n)", where q is real. In (23)-(27), find the limit of the sequence or show that it does not exist. (23) ai1= 1 and an+= 4 - an. (24) a1 = 1 and an+1 = 1/(1 + an). (25) ai1= 1 and an+ 1= 3 - 1/an. (26) ai1= 2 and an+1 = 1/(3 - an). (27) a1 = 1 and an+1 = 1 + 1/(1 + an). (28) The size of an undisturbed fish population has been modeled by the formula pn+1 = bpn/(a + pm), where pn is the fish pop- ulation after n years and a and b are positive constants that depend on the species and the environment. Suppose that po > 0. Show that pn+1 < (b/a)pa. Then prove that p > 0 if a > b; that is, the population dies out. Finally, show that pn-- b - a if b > a. Hint: Show that pn is increasing and bounded, 0 b-a, then pn is decreasing and bounded, pn > b - a. 51. Series 51.1. Basic Definitions and Notation. With a sequence {an}, one can associate a sequence {sa}, where n s8n=Zak = al+ a2 +-...+ C. k=1 The symbol oo (8.6) Zam =a1+ a2+ as+ - - n=1  51. SERIES 93 is called an infinite series, or just a series. The numbers sn are called the partial sums of the series (8.6). The limits of summation are often omitted to denote a series. The symbol E an also stands for an infinite series _ an. If {sa} converges to s, then the series is said to converge and one writes 00 n an = s or lim ak = s. n=1 k=1 The number s is called the sum of the series. If the sequence of partial sums {sa} diverges, the series is said to diverge. It should be under- stood that s is the limit of a sequence of sums, and it is not obtained merely by addition. For example, the sequence of partial sums for the series E(-1)" is si = -1, s2 = -1 + 1 = 0, 83 = 82 - 1 = -1, or, generally, sn = ((-1)< - 1)/2. This sequence diverges as it has two subsequences s2n = 0 and s2n-1 = -1, which converge to different numbers, 0 f -1. If one simply uses addition, different values for the sum of the series may be obtained: 0O an #a1+a2) + (a3 +a4) + (a5 +a6) +... 0+0+... 0, n=1 00 an #a1+ a2 + a3) + (a4 + a5 + a6) + --- -1 - 1 - --- - , n=1 00 an -a1+ (a2+-a3) + (a4 +-as) +...- -1+0+0+... --1. n=1 Generally, by grouping terms in the sum in different ways (according to the associativity of addition), the sum is found to be any integer! The reader is advised to verify this. Thus, the addition rules cannot generally be applied to evaluate the sum of a series. 51.2. Geometric Series. Take a piece of rope of length 1 m. Cut it in half. Keep one half and cut the other half in two pieces of equal length. Keep doing this, that is, keeping one half and cutting the other half in two equal-length pieces. The total length of the retained pieces is 1 1 1 1, 1 1 1 00 1 2 48 2\2 42 2 2 This series must converge. The partial sum sa here is the total length of retained pieces. The sequence {sa} is monotonically increasing (after each cut piece of rope is added) and bounded by the total length 1. So  94 8. SEQUENCES AND SERIES it converges. From the geometry, it is also clear that 1 - sn = 1/2, where n is the number of cuts, and hence sn - 1 as one would expect (the total length of the rope). So it is concluded = 2. Z2n2 n=0 This series is an example of the geometric series: 00 1+q+q2+q3+-- -= q" n=0 where q is a number. The geometric series does not converge for any value of q. THEOREM 8.9 (Convergence of a Geometric Series). A geometric series _1 q" converges if q < 1, and, in this case, n=0 1 q and the series diverges otherwise. PROOF. If q = 1, the sequence of partial sums obviously diverges. If q f 1, one has s=1 +q+q2 +- +q"- - qs=q+ q2 +q3-6 ----q". Subtracting these equations, one infers 1 - q" sn - qsn = 1 - q" --> sn =. 1 -q Therefore, 1 - q" 1 1n lim sn = lim = 1 limq". n- To n-o0 1-q 1-q 1-qa- o0 It has been found (Example 8.3) that the sequence an = q" converges only if |ql < 1, and, in this case, q" - 0 as n - oo. If |ql ;> 1, the geometric series diverges. D EXAMPLE 8.11. Analyze the convergence of the series 4 - + 2 - 32 27+ Solut ion: The series can be written in the form 4q0 +4q + 4q2 +4q3 + ---,where q =-2/3. So its partial sums are four times the partial sum of the geometric series with q= -2/3. Therefore, 2 a *_2___ 12 4 3 1 25  51. SERIES 95 51. SERIES 95 When real numbers are presented in decimal from, one often en- counters a situation when a number has a repeated pattern of decimal places. Take, for example, the number 1.2131313...; that is, the com- bination 13 repeats itself in all decimal places starting in the second decimal place. EXAMPLE 8.12. Is the number 1.2131313... rational or irrational? If it is rational, write it as a ratio of integers. Solution: By definition of the decimal representation, 13 13 13 13 1 n 1.2131313... =1.2 + 103 + + 1 + I0 Z(10 13 105 107 0 10 n=0 13 100 12 13 1201 103 99 10 990 990 51.3. Necessary Condition for a Series to Converge. The following the- orem follows from the limit laws applied to the sequences of partial sums. THEOREM 8.10 (Properties of Series). Suppose that the series E an and E bn are convergent and their sums are s and t, respectively. Let c be a number. Then the series E(an + bn) and E can converge and Z(an + bn) = an + bn = s + t Z(can) = c an = cs. Indeed, if {sa} and {ta} are the sequences of partial sums of the series E an and E bn, respectively, then the partial sums of the series E(an + bn) and L(can) are sn + to and csn, respectively. By the limit laws, sn + to - s + t and csn - cs. Note that the convergence of the series L(an + bn) does not imply the convergence of E an and E bn. For example, put an = 1 and bn = -1. Then s m = n and t = -n. The sequence sn + t = 0 converges to 0. Therefore E(an + bn) = 0. However, the sequences sn = n and to = -n diverge and so do the series E an = E 1 and E b = E(-1). The equality E(an + bn) E an + E bn becomes meaningless ("0 = 0- 00"). Recall that the basic limits laws hold only for convergent sequences (Section 49.3). This shows that the rules of algebra for finite sums are not generally applicable to series. Only series from a special class of absolutely convergent series, discussed later, behave pretty much as finite sums.  96 8. SEQUENCES AND SERIES 96 8. SEQUENCES AND SERIES Every theorem about sequences can be stated in terms of series and vice versa. The sequence {an} can be expressed via the partial sums of the series E an by putting a1i = s and an = se- s_1, n > 1. In particular, if the series converges, meaning that sn - s as n -oo, then also sn_1 - s. By the basic limit laws, the limit on both sides of the above recurrence relation may be taken: lim an = lim (s-S - s_-1) = lim s - lim s_1 = s - s = 0 . Thus, for a convergent series E an, the sequence {an} necessarily con- verges to 0. THEOREM 8.11 (Necessary Condition for a Series to Converge). If the series E an converges, then limn-, an = 0. The converse is not generally true. The condition limn-, an = 0 is not sufficient for a series to converge. However, it can still be used as a test for divergence of a series. COROLLARY 8.1 (Test for Divergence of a Series). If the limit limnax an does not exist or if limn-, an 0, then the series E an diverges. EXAMPLE 8.13. Show that the series E n3/(3n3 + 1) diverges. Solution: n3 1 1 lim an = lim = lim 1=- /0, n-oo n-oo3ns3+ 1 n-oo 3 + 3 so the series diverges. D If the necessary condition is satisfied, the series may converge or diverge. The sequence of partial sums has to be analyzed. E XAMPL E 8.14. Find the sum of the series L( if it exists or show that it does not exist.  51. SERIES 97 Solution: The necessary condition for convergence is evidently satis- fied. So the sequence of partial sums has to be analyzed for conver- gence: 1 __1_ s=1k(k+1) _ k k+ =1 2) + 2 3+\3 41+.+ n n+1 1 = 1-- 1 as n -oo. n+1 So the sequence {s,} converges to 1 and hence °° 1 = 1. =1 n(n +1) This example is a particular case of a telescopic series. THEOREM 8.12 (Convergence of a Telescopic Series). A telescopic series L_ L1(an - an+1) converges if limn_00 an = a, and, in this case, 00 (an - an+1) = ai - a, n=1 The proof is analogous to the above example and based on the fact that the sequence of partial sums of a telescopic series s = ai - an+1 converges to ai - a. The details are left to the reader as an exercise. 51.4. Exercises. In (1)-(4), find the sequence of partial sums of the series and investigate its convergence. 1 1 1 (-1)"--1 2 4 8 2--1 ( 123 121 (2) (-+-3)+ (-+-9)+--+ (2n+ +--- 1 2 3 n ()-+-+-+---+ +--- '2 4 8 2 ( 1 3 5 2n-1 (4) -+ 22 +23 + 2 Hint: Put nm 1 +1+.--. + 1 (the sum of n units) in (3) and (4), when calculating the partial sum, then use partial sums of a geometric series.  98 8. SEQUENCES AND SERIES 98 8. SEQUENCES AND SERIES In (5)-(9), determine whether the (geometric) series converges or di- verges (p > 0 and q > 0). Find the sum of the series if it exists. 0O 12 (5) 3n+1 n=0 (8) 0 n p + q)" 00o 00 (- 5) " (6) 3 2n (6) Z n1-1 () 321 n=1 n=0 01+2"n+3"n+44n-5"n (9) 5" In (10)-(14), determine whether the series converges or diverges by expressing it as a telescopic series. Find the sum of the series if it exists. ° 1 (10) =2 2 1 00 (12) Lln n1 n1 (14) L( nr+2 00 3 (11) = 2 +3n+2 (13) (3n - 2)(3n + 1) 2 2+1+/) In (15)-(20), determine whether the series converges or diverges. If it converges, find its sum. Here p is a positive number, p > 0. k2 00o 2 3"00 (15) k2 + k + 1 (16) 5n (17) Lnp n=1 n=1 n=1 (18) (sinp)" (19) n (20) (e2+(41) n=1 n=1 n=1 In (21) and (22), express the number as a ratio of integers (21) 1.23232323.... (22) 1.53525252.... In (23)-(25), find the values of x for which the series converges. Find the sum of the series for those values of x. (23) (24) mL X n=1 n=1 In (26) and (27), solve the equation. (26) (1 +x)-" = 3 n=2 (25) L(x n=1 5)" (27) Le"X n=O 9  52. SERIES OF NONNEGATIVE TERMSM 99 (28) Suppose thatL°1 =a. Let 1 < ki < k2 < - - - < k, <- be a strictly increasing sequence of integers. Let A = akn + akn+1 + - - - + ak±1 1. For example, for ki = 2, k2 = 5, and k3 = 7, A1 = a1, A2 = a2 + a3+ a4, and A3 = a5+ a6, respectively. Prove that L= An= a. In other words, the series obtained from a convergent series by grouping its terms without changing the order of its terms converges and has the same sum. The converse is not true (see Section 56). 52. Series of Nonnegative Terms In many applications, the terms of a series decrease monotonically. It appears that there is a relation between convergence of such series and convergence of improper integrals over an interval [1, oc). This relation allows one to establish a necessary and sufficient condition for series of nonnegative terms to converge. 52.1. The Integral Test. Suppose f(x) is a positive, continuous, mono- tonically decreasing function on [1, oc) such that f(x) -- 0 as x - oo. Suppose also that the improper integral lim jaf(x) dx= If exists. The value If is the area under the graph y = f(x) over the in- terval [1, oc). Consider the series ' f(n). The necessary condition for convergence is fulfilled because f(n) -- 0 as n - 0o. To investigate the convergence of the series, one has to analyze the convergence of its partial sums: sm= f (k) = f (1) + f (2)+- + f (n). k=1 Since the function f(x) monotonically decreases and is continuous on every interval [k, k + 1], it attains its minimal and maximal values f(k + 1) f(x) f(k) on this interval and therefore (see Fig. 8.7) (8.7) f (k+1) j f (k+1 Let us take the sum over k =1, 2, ..., n - 1 of the left inequality in (8.7). In the left side, the summation yields f (2) + f (3) +.-.-. + f (n) s- f(1). By moving f(1) to the other side, an upper bound for s, is  100 8. SEQUENCES AND SERIES 100 8. SEQUENCES AND SERIES obtained: Sn f(1)+ f(x)dx +...+i 21 n Similarly, taking the sum over k (8.7), the lower bound is derived: ,n n f(x)dx= f(1) + f(x)dx. -1 1 1, 2, ..., n of the right inequality in sn f (x) dx + f (x) dx + 1 2 Therefore jn- n+1 f (x) dx = f (x) dx, (8.8) 1 n+1 f (x) dx sn < n f (1) + f (x) for all n ;> 1. 1 This inequality shows that the following theorem holds. THEOREM 8.13 (Integral Test). Suppose f is a continuous, posi- tive, decreasing function on 1, oo) and let an = f(n). Then the series an converges if and only if the improper integral f' f(x) dx con- verges. In other words, 00 f(x) dx converges - 1 f (x) dx diverges - 1 f(n) n1 00 n~1 converges, diverges. PROOF. If the improper integral converges to a number If, then by (8.8) the sequence of partial sums is bounded, sn < f(1) + If, and monotonically increases, sn < sn + f(n + 1) = sn+1. Therefore, it is convergent by Theorem 8.8. If the improper integral diverges, then, for any number M > 0, there is an integer N such that f>+ f(x) dx > M for all n > N. By the left inequality of (8.8), M < sn for all n > N; that is, {sn} is a monotonically increasing, unbounded sequence and hence it diverges. v f(k+1) I k+ 1 | k k+1 k+1 f(x) dx k k |+ | 1 k k+1 f(k) I k k+1 FIGURE 8.7. Integral test. An illustration of inequality (8.7).  52. SERIES OF NONNEGATIVE TERMS 101 Remark. Suppose that an = f(n), where f(x) is a function on [1, oc), such that it is continuous, positive, and decreasing on [N, oo), where N;> 1 is an integer. Then 00 f00 (8.9) an converges <-> f(x) dx converges; n=1 N that is, the integral test applies even if the sequence an becomes mono- tonically decreasing only for n > N > 1. This is easy to understand by isolating the first N - 1 terms in the series an=a1+a2+.+ aN-1+ an= al+a2+ + aN-1+ bn, n=1 n=N n=1 where bn = aN+n-1. Convergence of E br implies convergence of an and vice versa as they differ by a number. Put b = g(n), where g(x) = f(x+N-1), which is a continuous, positive, decreasing function on [1, oo), and g(x) d = f(x +N- 1)d = f(u)du 1 1 N by changing the integration variable = x + N - 1. 52.2. Special Series of Nonnegative Terms. THEOREM 8.14 (Convergence of p-series). The p-series 001 np n=1 converges if p > 1 and diverges if p < 1. PROOF. If p < 0, the series diverges because the necessary condi- tion for convergence is not fulfilled, an o oc if p < 0 and an = 1 / 0 if p = 0. For p > 0, consider the function f (x) = x-p, which is positive, continuous, and decreasing on [1, oc), and fadcf1 (1_1i) if p 1, 1icP lna if p = 1. So, by the integral test, the series converges if p > 1 because the improper integral diverges if 0 < p < 1 and converges if p > 1 (the limit a -a oc exists only if p> 1).D Harmonic series. The series E n-'3 diverges for all 0 < p <; 1 despite that the necessary condition to converge is fulfilled: an =rn-3 -~ 0. In particular, the harmonic series i~4 diverges.  102 8. SEQUENCES AND SERIES 102 8. SEQUENCES AND SERIES Riemann's zeta function. The sum of a p-series ((p) = E n-p depends on the value of p > 1; that is, this series defines a function on (1, oo). This function is called Riemann's zeta function. EXAMPLE 8.15. Investigate the convergence of the series L 1(n + 2)-3/2. Solution: The series can be written as 11001 1 i i ~ 1 3/23/2 (n + 2)3/2 = /2/2 n=1 n=3 n=1 The latter series is a p-series that converges for p = 3/2 > 1. Q THEOREM 8.15. The series 00 1 2n(lnn)P converges if p > 1, and it diverges if p < 1. PROOF. Consider the function g(x) = x(lncz)P for x > 1. Its de- rivative reads g'(x) = (lnx)P1(p + In x). If p > 0, then g'(x) > 0 for all x > 1 and g(x) increases, while its reciprocal f(x) = 1/g(x) should decrease. If p < 0, then g'(x) > 0 for all x > e-p and hence g(x) in- creases, while f (x) = 1/g(x) decreases if x > e-p > 1. Thus, for any p, there is an integer N such that the function f(x) = 1/[x(lnx)P] is con- tinuous, positive, and decreases on [N, oc). By the integral test (8.9), the series in question converges if and only if the improper integral 00 dz du JN x(lnx)P JInN converges, where the integration variable has been changed, u =ln x, du = dz/x. This integral diverges if p < 1 and converges if p > 1, and the conclusion of the theorem follows. D On the use of the comparison test for improper integrals. The improper integral of the function f(x) in the integral test often cannot be ex- plicitly evaluated. The comparison test (Theorem 7.4) may be used to assess the convergence of the improper integral. E XAMPLE 8.16. Investigate the convergence of the series E an where  52. SERIES OF NONNEGATIVE TERMS 103 Solution: Let x+1- x-1 cx+1- c-1 c+1+ c-1 xPc xP x+1+ x-1 2 xp( x+1+ cX-1)>O for x > 1. The idea is to find lower and upper bounds for f(x) as func- tions such that the convergence of their improper integrals is easy to analyze. It follows from from the obvious inequality cz + 1 > z - 1 that 2 2 2xp + 1< f (X) < x x-1 2xcP x+1 2xccx c-i Then a lower bound is obtained as a power function by means of x+1 < 2x. Similarly, an upper bound follows from x - 1 > /2 for all x> 2. Thus, 1 2 cc~~7 2< f(c) < ~±/ 1p+1/2 /G W xp+1/2 for all xc> 2. By Theorem 7.4 this inequality implies that the improper integral of f(x) converges if and only if the improper integral of the power function x-p-1/2 converges. The latter is the case if p + 1/2 > 1 or p > 1/2. Thus, the series in question converges if p > 1/2 and diverges otherwise. D EXAMPLE 8.17. Investigate the convergence of the series En2-" where p > 1. Solution: Put f(x) = 2-p > 0 if x > 1. By monotonicity of the exponential function and by the obvious inequality cpc > c for all cc> 1 and p > 1, the function f is bounded from above by g(x) = 2-x e-x n2. It follows from the inequality 0 < f(x) < g(x) that 0<11 f(x)dz <; gx=n and, hence, the series in question converges. D 52.3. Estimate of the Sum. If a partial sum s, is used to estimate the sum of a convergent series of nonnegative terms E f(n), how good is such an estimate? The remainder s - sa has to be investigated to answer this question. COROLLARY 8.2 (Estimate of Sums). Suppose f is a continuous, positive; decreasing function on [1, oo) and let a, f (n). If the series  104 8. SEQUENCES AND SERIES 104 8. SEQUENCES AND SERIES L an converges to a number s, then J f(x)dz s -sm f f(x)dx, n+1 n where {sn} is the sequence of partial sums. PROOF. The first inequality is obtained by taking the limit n - 00 in (8.8) with the result (8.10) 1 fc(x)dcc Za f(1)+f f(c) d, 11 which is a legitimate operation because (8.8) holds for all n and the series converges (and so does the improper integral by the integral test). The remainder estimate is obtained by subtracting (8.8) from (8.10). Note the value of the improper integral does not coincide with the sum; it only determines an interval (8.10) in which the sum of a series lies. EXAMPLE 8.18. Test the series L_, (n2 + 1)-1 for convergence or divergence. If it converges, estimate its sum. Solution: Put f(x) =_(x2 + 1)-1, which is a continuous, positive, de- creasing function on [1, oc), such that the series in question is E f(n). Therefore, the integral test applies, and the series converges because /*dc1a w 7 w 7 =2+ 1 lim tan- x 1 lim tan-1a- =-- - 1 2 1 a-oo 1 a-oo 4 2 4 4 By (8.10), its sum lies in the interval 4 s G f(1) + 4 = } +4. E EXAMPLE 8.19. Test the series L_°1 ne for convergence or di- vergence. If it converges, estimate its sum. Solution: Consider the function f (x) = ze-x. Since f'(x) = e-x - ze-x = (1 - z)e-x < 0 if x ;> 1, the function decreases on [1, oc), and the integral test applies to assess convergence of the series Ef(n): /d m a - 1 1 2 z dz - zf-+-xde-x limze- + edz= +- 1a1 ao 1 i e e e where the integration by parts has been used to evaluate the integral. The series converges to a number s that lies in the interval 2e-1 < s < f (1) + 2e-1 =3e-1.D E XAMPLE 8.20. Estimate values of Riemann's zeta function C(p). How many terms does one need to retain in the partial sum su to ap- proximate C(p) so that the error is less than 10-N g2  52. SERIES OF NONNEGATIVE TERMS 105 Solution: Riemann's zeta function is defined by the sum of the series ((p) L= i n-P. For p > 1, 1 XP lim = a-oo1 - p 1 lim a-oo 1-p 1 1-p 1 p- 1 Since f (1) = 1, by (8.10), 1 p_ 0, is to be approximated with an error not exceeding 10-N, then the remainder should be less than 10-N, which yields the condition on the number of terms: 1 < 10-N or n > 10N/ Por n > 10N/p/ 1/p. The smallest n satisfying this inequality is 1 plus the integer part of 10N/p/p1/p. For example, take p = 1 and N = 2. Then ((2) - s <0.01 for all n> 101. D 52.4. Exercises. In (1)-(15), determine whether the series converges or diverges. (1) n2-n (2) nil/ (4) n2e- (5 n1 (7n9/8 (8) n=1 (10) 2 1 n=1 00 1 n=1 n(n + 4) =21 00 (3) Ze 1-2 n=1 (6) Z01 (9 1 n(n + 1)(n + 2) (9) 001 - n ln n (9)n2 n=2 00 1 (11) n=1 n 4+ 2n + 1 )m1 (15) n(n + 1) n0 (12)n n=1 4 t an-' n n2+ 1  106 8. SEQUENCES AND SERIES 106 8. SEQUENCES AND SERIES In (16)-(26), determine the values of p for which the series is conver- gent. 00 (16) , p>0 pn +100' n=1 00 n=11 (21) n lnn(ln(lnmn))P n +|Ip| - jn - |p|I (17)j + 1- np n=1 00 a a>1 ( n=1 00 (22) n(1 + n2)p n=1 00 (25) nPn-"" (2 n=1 00 20) plI"nn n=1 00 (23) ZnPe- n=1 6) n-(n"n)P n=1 (24) n=1 1 l n2+ 11 (27) How many terms of the series in Theorem 8.15 does one have to add to approximate its sum with an error less than 10-N? In (28)-(30), estimate the sum of the series and, in (28) and (29), also determine how many terms has to be added to approximate its sum with an error less than 0.01. (28) 2-4 n=1 (29) 212 +5 n1n4+5n2+4 (30) 141 n2-1 (31) Show that the sequence 1 1 1 an=1+-+-+---+- Inn converges. The limit limnoo an = 7 is called the Euler num- ber. Hints: (i) Use (8.8) to show that if sn is the partial sum of the harmonic series, then sn < 1 + ln n and hence an < 1 (i.e., the sequence {an} is bounded). (ii) Interpret an - an+1 as a difference of areas to show that {an} is monotonic. 53. Comparison Tests Consider the series 00 a n=1 n= 1 It has terms smaller than the corresponding terms of the convergent p-series: oo bn n=l 00 1  53. COMPARISON TESTS 107 because an < bn for all n. If s, is the partial sum for E an and to is the partial sum for E bn, then sn < tn. Since to converges to a number t, it is bounded, t, < t, and hence s, < t. The sequence {s,} is monotonic and bounded, and therefore it converges. This line of arguments admits a generalization. THEOREM 8.16 (Comparison Test). Suppose that 1 an and 1 bn are series such that an > 0 and bn > 0 for all n> N and some integer N > 1. Then bnconverges and an br for all n> N = an converges, bndiverges and an br for all n> N = an diverges. PROOF. The series n a1n and L-N an differ by a number a1 + a2 + - - - + aNl. So convergence of L_'N an implies convergence of ° an and vice versa. Therefore, it is sufficient to consider the case N = 1. The sequences of partial sums {sa}, s = a1 + a2 + - - - + an, and {t }, t = b1 + b2 +. -+ bn, are monotonically increasing sequences because an > 0 and bn > 0. If E b converges, then to - t as n - 0o and to < t for all n. By the hypothesis an < b for all n > 1, and therefore s<, to < t, which shows that {sa} is monotonic and bounded and, hence, converges. If E bn diverges, then t - oo. From the hypothesis an > bn, it follows that sn > tn. Thus, s - 0o as n -- oo. D When applying the comparison test, the convergence properties of the series E b should be easy to establish. In many instances, a good choice is a geometric series (Theorem 8.9), a p-series (Theorem 8.14), a telescopic series (Theorem 8.12), and the series in Theorem 8.15. EXAMPLE 8.21. Test the series E'_1(2n + 1)/(3n3+ + 2 + 1) for convergence. Solution: Since an > 0 is a rational function of n, a convenient choice of a series in the comparison test is a p-series: 2n+1 2n+1 2 1+1 1 0< 3 < + . 3ns + n2+ 1 3n3 3 n2 3 n3 The series j 2+1 - 2 1 1 1 T1 h=1 Th=1 converges as the sum of two convergent p-series.D EXAMPLE 8.22. Test the series L'_i(v n + 1- /m) for convergence.  108 8. SEQUENCES AND SERIES 108 8. SEQUENCES AND SERIES Solution: One has 2+1 =(v/n +I1- v/) (v/n +1+ vn) 12n+ 1+ v/2 1 1 1 1 /n+1+ v2 v+ 1+ v2 v2 The p-series E 1//n diverges and so does the series in question by the comparison test. D THEOREM 8.17 (Limit Comparison Test). Suppose that E an and L bn are series with positive terms. Let c = limna (an/bn). " If c = 0 and E bn converges, then E an converges. " Iff0 < c < oo, then E an converges if only if E bn converges. " If c = oo and E bn diverges, then E an diverges. PROOF. If c = 0, then there is an integer N such that an/bn < 1 for all n > N by the definition of the limit. Hence, an < b for all n > N. If E bn converges, then E an converges by the comparison test. If c E (0, oc), then, by the definition of the limit, for any number c > E > 0, there is an integer N such that c - a < -- m=c-n< aN. bn bn Therefore, mbn N. By the comparison test, convergence of E bn implies convergence of E an due to the inequality an < Mba. The divergence of E b implies divergence of E an, again by the comparison test as mbn < an. If c 00, then, for any M> 0, there is an integer N such that an/bn > M when n> N. The inequality an > Mbn shows that divergence of E bn implies divergence of E an by the comparison test. D It is often helpful to investigate the asymptotic behavior of an as n -+ 00 to identify a suitable bn in the limit comparison test (Recall the notion of a slant asymptote of a function f(x) in Calculus I). EXAMPLE 8.23. Test the series Ln1(2n3/2 + 1)7/126 + n4 + 1 for convergence. Solution: Let us find the asymptotic behavior of an as 12 -~ 00. For large 12, the top of the ratio is 13/2(2+121/2) 213/2 because the term 12-1/2 may be neglected as compared to 2. The asymptotic behavior  53. COMPARISON TESTS 109 will be denoted by ~, in particular, 2n3/2 + 1223/2. The bottom of the ratio behaves as ~(n6)1/2 _ n3 . Therefore, 2n3/2 + 12 2n3/2(1 _ 2) 2n3/2 - an= = ~ = 2n-/ = bn n/6+-n4+ 1 3 1 + 1 + 1 n3 n2 26 in the asymptotic region n - oc. This shows that .an 1+ 1 lim = lim2 = 1 = C. n-oo b n-noo 1 + A+ ) By the limit comparison test, the series E an converges because the p-series E b = 2 E n-3/2 converges. EXAMPLE 8.24. Test the series E'_1(n+5 + 3n)/ n3 + 5n for con- vergence. Solution: Recall that the power function increases more slowly than the exponential function, that is, npq- -n 0 as n - oc for any q > 1 and any p. Hence, the asymptotic behavior of an is 3"(1+ n53-") 3" ( 3 - S55/2 (1+ n35-)1/2 5n/2 / 5 This shows that the ratio an/b converges to c = 1 as n - oc. By the limit comparison test, E an diverges because the geometric series E bn diverges (q= 3/v5 > 1). D 53.1. Estimating Sums. Let _1 b be a convergent series of non- negative terms. Suppose that 0 < an < b for all n. Then by the comparison test, the series L001 an converges. The sum of E an can be estimated by comparing remainders for the series E bn. Indeed, put E an = s and E bn = t. Let {ta} and {sa} be the sequences of partial sums for E b1 and E an, respectively. The remainders satisfy the inequality: s - s = an+1+an+2+ - -- bn+1+bn+2+---=t - tn. So the accuracy of the approximation s s, is the same or higher than that of the approximation t ~Lt_. If, for example, one finds that 1 = N is sufficient for the equality t =tN to be COrreCt within a specified error, then s =3N is also COrreCt within the same or even smaller error. The remainder is easy to estimate if b1 = f(12) and the function f is simple enough to integrate, t - t12 f7 f~x) dzv (Corollary 8.2).  110 8. SEQUENCES AND SERIES 110 8. SEQUENCES AND SERIES EXAMPLE 8.25. Determine how many terms are needed to estimate the sum of the series L', tan-1(n2)/(3 + 1) correct within an error not exceeding 10-4. Solution: The function tan-1 x is monotonically increasing for x > 0 approaching asymptotically the value r/2. Therefore, tan-1(n2) 7 1 7 1 an- = 31 - 1 . n 1 2n+12n Hence. s - sC < t w f dx t2 I1x3 10 >n> 2 102 157.1. So the needed accuracy is guaranteed if n = 158 or larger. 53.2. Exercises. In (1)-(15), determine whether the series converges or diverges. D 00 1 (1) (4) +m3+ ns3 2n/3+ 00+1 (2) 1 00 n(lnn)4 - (5) n 22+1 n=2 (3) Z 1210 + 2n2/2 2 n=2 1±+ (-1)" 00 +__2" 7) n1 3/2+ 1 (8)s1 + n +1 9 1 n2 +2 1/ (10) 2n+ 1 (11) Zsin2&) (12) n 1 i n = 1 n= n2 00_1_0 0o n (13) n1+1/ (14) ( 0/0-2)(15) n=1 n=1 n=1 In (16)-(20), determine the values of the parameters for which the series converges. sre co v r e .000(16) in (2n) (17) (np + n + 1)q (18) np (1+ q) n=1 n=1 n=1 (19) ( n + 1 - 1 (20) (vn+1 n=1 /)P in (n +) (21) Assume an = PN(n)/QM(n), where PN and QM are polyno- mials of degree N and M, respectively, and QM(n) > 0 for all n. Investigate the convergence of the series E an.  54. ALTERNATING SERIES 111 In (22) and (23), determine how many terms one needs in the partial sum to estimate the sum of the series correct within an error less than 0.001. Zsins om2 (22) Thn1 n (23) i= n3 +=1n+3 (24) Consider a sequence {an}, where an can take any value from the set {0, 1, 2, ..., p - 1} and p > 1 is an integer. The meaning of the representation 0.aia3a3... of a number with base p is that a1 a2 as (8.11) 0.aia3a3... =a+2 + 3 + -.--, p p p When p = 10, the decimal system is obtained. The binary representation corresponds to p= 2. The Maya used p = 20 (the number of fingers and toes). The Babylonians used p 60. Show that the series (8.11) always converges. (25) Show that if an > 0 and E an converges, then 1 ln(1 + an) converges, too. (26) Prove that if the series E an and E bn converge, then the series E lanbn| and L(an + bn)2 converge. (27) Prove that the convergence of E an, where an > 0, implies the convergence of E a/n. (28) If E an converges and if the sequence {b,} is monotonic and bounded, prove that E anb converges. (29) Prove that if limn-0 na= c / 0, then the series E an di- verges. (30) Let the series of nonnegative terms, E an and E bn, con- verge. What can be said about the convergence of the se- ries E max(an, bn) and E min(an, ba), where min(p, q) and max(p, q) are the smallest and largest numbers in the pair (p, q), respectively, and min(p, p) = max(p, p) = p? 54. Alternating Series DEFINITION 8.6 (Alternating Series). Let {b,} be a sequence of nonnegative terms. The series Z(-1)"-1b, b - b2 + b3 - b4 + b5 is called an alternating series.  112 8. SEQUENCES AND SERIES 112 8. SEQUENCES AND SERIES For example, the series (8.12) 1- 1-+ 1 + 1-+ .-- = 2 3 45 n2 n=1 is an alternating series. It is called the alternating harmonic series. THEOREM 8.18 (Alternating Series Test). If a sequence of positive terms {b,} is monotonically decreasing and limn-,b Q = 0, then the alternating series L*_1(-1)"1 bn converges: (i) bn+1 < bn for all n 1 converges. (ii) imn-, bn =0 1 ecnegs n=1 PROOF. The convergence of the sequence of partial sums {sn} is to be established. Consider a subsequence of even partial sums {s2k}. One has s2 = bi - b2 > 0, s4 =s2 + (b3 - b4) s2, and, in general, s2k - s2(k-1) + (b2k-1 - b2k) s32(k-1) s2(k-2) '... 2 0 by the monotonicity of the sequence {b,}. Thus, the subsequence {s2k} is monotonically increasing. By regrouping the terms in a different way, one can see that 2k =bl - (b2 - b3) - (b4 - b5) - (b2k-2 - b2k-1) - b2ki 1 because all numbers in parentheses are nonnegative by hypothesis (i), which shows that {s2k} is also bounded. Therefore, it converges by the monotonic sequence theorem: lim s2k = s. For the subsequence of odd partial sums s2k+1 =82k + b2k+1, one infers by the limit laws and hypothesis (ii) that lim 52k+1 = lim s2k + lim b2k+1 = s + 0 = s. kwk-ow k-ow The convergence of two particular subsequences of a sequence to the same number s does not generally guarantee that the sequence con- verges to s (all its subsequences should converge to s). By definition, the limits of {s2k} and {s2k+1} mean that, given any number e > 0, there are positive integers N1 and N2 such that 1s2k - s| < c if k > N1 and Is2k+1 - s| < c if k > N2. Put N = max(2N1, 2N2 + 1). Then s - s| <&efor alln > N, which means that s, s as - oo. D By this test, the alternating harmonic series (8.12) converges be- cause the sequence be - 1/in is monotonically decreasing and converges to 0.  54. ALTERNATING SERIES 113 s- b2 85 - b6 02 04 ~ 05-b6 ~ S2 64 06 3 07 05 3 = (J S4 + b5 s2 + b3 FIGURE 8.8. Alternating series test. An illustration of its proof where two subsequences, 82k and 82k-1, of the sequence s. of partial sums are analyzed for convergence. EXAMPLE 8.26. Test the series E" sin(7n/2)/n for convergence. Solution: One has sin(7n/2) = 1, 0, -1, 0, 1, respectively, or, in general, for odd n = 2k - while for even n = 2k, sin(7n/2) = sin(7k) question is an alternating series: ... for n = 1, 2, 3, 4, 5, ..., 1, sin(7n/2) = (-1)-1, = 0. Thus, the series in sin(7n/2) n ° (-1)n-1 n-1 n-1 1 2n - 1 The sequence {bn} is monotonically decreasing and bn -- 0 as n -- oc. So the series converges by the alternating series test. Q Scope of the alternating series test. Theorem 8.18 provides only a suffi- cient condition for an alternating series to converge. There are conver- gent and divergent alternating series that do not satisfy the hypothesis (i) of Theorem 8.18. As an example of a convergent series, consider the alternating series 00 ,(-1)n-lb n=1 Dc 1 _ (-1)-1 1)n n-1 n2 0 ( 1)n -1 n=1 + 1 n2  114 8. SEQUENCES AND SERIES 114 8. SEQUENCES AND SERIES The sequence {b,} is not monotonic because b2k = 0 while b2k+1 > 0. Nevertheless, the series converges by Theorem 8.10 because the series E(-1)"-1/n2 and E 1/n2 converge. The former series converges by the alternating series test, while the latter is the p-series with p = 2. On the other hand, consider the alternating series 001 (000(1)(1)-1 1 (-1)-11) - (-1)- + - n=1 n=1 n=1 Here again the sequence {b,} of nonnegative terms is not monotonic. If sn is a partial sum of this series, then s, = to + hn where to is the partial sum of the alternating harmonic series, which has been shown to converge, and hn is the partial sum of the harmonic series, which is known to diverge. Thus, the sequence {s,} is the sum of convergent and divergent sequences and, hence, it diverges. Remark. Hypothesis (i) of Theorem 8.18 may be weakened (i) bn+±1 N for some integer N > 1. Indeed, 00 00 (-1)"-1b, = bi - b2 + b3 - -- - bN-1 + 1- bn n=1 n=N = bi - b2 + b3 - ... - bN-1 + (1)N-1 Z(1)-cn n=1 where c = bn+N-1. The series L(-1)'-bn and +L(-1)"-1cn differ by a number, and therefore the convergence of one of them implies the convergence of the other. The series E(-1)"-Ic converges by Theorem 8.18 as cn+1 c , for all n and limn-00 cn = lim-o00 bn+N-1 = 0. EXAMPLE 8.27. Test the series L'_1(-1)n nP/(n+l) for conver- gence if p < 1. Solution: Here bn =nP/(n + 1) and, for p < 1, np np-1 lim b = lim = lim 1= limnp-1 = 0. n-00 -oo n + 1 n-oo 1 + n --oo So hypothesis (ii) of Theorem 8.18 is fulfilled. However, the mono- tonicity of {b,} is not obvious. To investigate it, consider the function f(x) = P/(x + 1), where z ;> 1. If f(x) monotonically decreases, then so does the sequence be f(in). The condition f'(xc) 0 has to be  54. ALTERNATING SERIES 115 54. ALTERNATING SERIES 115 verified: f'(x) =pXP (p-1)x+pxp-1 < <> p 1. If 0 < p < 1, then f(x) monotoni- cally decreases for x > p7(1 - p) and one can always find an integer N > p/(1-p) such that bn+1 < bn for all n> N. So the series converges for all p <1. D 54.1. Estimating Sums of Alternating Series. A partial sum s of any convergent alternating series can be used as an approximation of the total sum s, but this is not of much use unless the accuracy of the approximation is assessed. The following theorem asserts that the ab- solute error of the approximation s sn does not exceed the value of bn+1- THEOREM 8.19 (Alternating Series Sum Estimation). If s = L(-1)"-1bn is the sum of an alternating series that satisfies (i) 0 < bn+1 < bn for alln and (ii) lim bn = 0, n-oo then s -sn| <;bn+1- PROOF. In the proof of the alternating series test, it was found that the subsequence {s2k} approaches the limit value s from below, s2k < s. On the other hand, the subsequence {s2k-1} approaches the limit value s from above (see Fig. 8.8). Indeed, si = bi, s3 = si - b2 + b3 < si because b3 b2, and, in general, s2k+1 = s2k-1-b2k+b2k+1 82k-1; that is, {s2k+1} is monotonically decreasing. This shows that the sequence of partial sums s, oscillates around s so that the sum s always lies between any two consecutive partial sums sn and sn+1 as depicted in Figure 8.8. Hence, |s - sn| <; Isn+1 - s-l = bn+1- EXAMPLE 8.28. Estimate the number of terms in a partial sum sn needed to approximate the sum of the alternating harmonic series correct within the absolute error not exceeding 10-N. Solution: Here, be 1/in. Hence, the approximation s ~sa has the needed accuracy if |s - s| <; bn+1 <; 10-N or 1/(m + 1) < 10YN or n > 10N -1  116 8. SEQUENCES AND SERIES 116 8. SEQUENCES AND SERIES If the monotonicity condition bn+1 < bn holds only if n > N, the conclusion of Theorem 8.19 also holds only if n > N. Indeed, in the no- tation from Remark above Example 8.27, put t =L(-1)'-ca, where cn = bn+N-1- Let tn be a partial sum for the series L(-1)'-ca. Then s = SN-1 + (_1)N1t and sn =-SN-1 + (_1)N-ltnN+l for n > N. Therefore, s - tsn| t-tn-N+1 N. 54.2. Exercises. In (1) and (2), prove the convergence of the series and find its sum. 1 1 1 1 1 +2 4 8 16 32 (2) 1--3 + 4_ 8 . In (3)-(21), determine whether the series converges or diverges (here p is real). 00(-1)"n 3)n ln(n + 3) 0(-1)"n n=2 (12) n=1 O ( 1) h (15) z : ( 1) T n_1 + p (4) n - 7 0 -( inn=2 (In n (10) (-1 n=1 (13) Z -'1) (16) z:() np (5 cos(nr/2) (5) n/ n2 ( ) (-1) n3 S(8) 2 n=1 ° (-1)"n2 )"n1( (11) n4+ 1 °O=1 ( /-1)" (14) (-)"hbn n=1 (17) 03 (-1)m(2 -) n=1 (2n + 3)2 (18) j nsin(7rn/4) (19) 0j (-1)sin2(w/2) n=1 n=1 -0 1)"n°O(-1)"n (20) 1)n1 (21) ( C) In (22)-(24), find n for which the approximation by partial sums s n is correct within an error not exceeding 10-N (22) 121 (23) (-1)nn 10n n=1 (24) ° (-1)"hn1/3 122/3 + 6  55. RATIO AND ROOT TESTS 117 (25) Prove that the sum of the alternating harmonic series is (0 1)"-1 S= ln 2. n n=1 Hint: Show that a partial sum of the alternating harmonic series is s2 =h2 - hn, where h = an + lnn and the sequence {an} is defined in Exercise 52.4.16. Then use the result of the latter exercise to prove that s , -- 1n2 as n -- 0o. 55. Ratio and Root Tests 55.1. Absolutely Convergent Series. DEFINITION 8.7 (Absolute Convergence). A series E an is called absolutely convergent if the series of absolute values E |an| is convergent. The absolute convergence is stronger than convergence, meaning that there are convergent series that do not converge absolutely. For example, the alternating harmonic series E an, an = (-1)m-1/, is convergent, but not absolutely convergent because the series of absolute values lan| = 1/n is nothing but the harmonic series E 1/n, which is divergent (as a p-series with p = 1). On the other hand, the absolute convergence implies convergence. THEOREM 8.20 (Convergence and Absolute Convergence). Every absolutely convergent series is convergent. PROOF. For any sequence {an}, the following inequality holds; 0 Gan + |an| 2|an| because lan| is either an or -an. It shows that the series E bn, where bn = an + Ian|, converges by the comparison test because E 2|an| 2 E lan| converges if E an converges absolutely. Hence, the series E an = E b - E lan| converges as the difference of two convergent series. D EXAMPLE 8.29. Test the series L[sin n - 2 cos(2n)]/n3/2 for abso- lute convergence. Solution: Making use of the inequality A + BI |Al +|BI and the properties that |sin x < 1 and |cos x < 1, one infers |sinin - 2cos(2n)| | sinin| + 2| cos(2n)| 3 in| 3/2 i3/23/ The series of absolute values E la| converges by comparison with the convergent p-series 3 L in-3/2 (here p =3/2 > 1). So the series in question converges absolutely.D  118 8. SEQUENCES AND SERIES 118 8. SEQUENCES AND SERIES DEFINITION 8.8 (Conditional Convergence). A series E an is called conditionally convergent if it is convergent but not absolutely conver- gent. Thus, all convergent series are separated into two classes of con- ditionally convergent and absolutely convergent series. The key dif- ference between properties of absolutely convergent and conditionally convergent series is studied in Section 56. 55.2. Ratio Test. THEOREM 8.21 (Ratio Test). Given a series E an, suppose the following limit exists: lim an+1 = c n-°o°a where c > 0 or c = oc. " If c < 1, then an converges absolutely. " If c > 1, then an diverges. " If c = 1, then the test gives no information. PROOF. If c < 1, then the existence of the limit means that, for any E > 0, there is an integer N such that -' < an+1 - c N. an an Note that since c is strictly less than 1, one can always take E > 0 small enough so that the number q = c + E < 1. In particular, put n = N + k - 1, where k > 2. Applying the inequality lan+1| < qan| consecutively k times, aN+k N. The series E lan| converges by comparison with the convergent geo- metric series E j3q= =3 E q because q < 1. So E an converges abso- lutely. If c > 1, then there is an integer N such that lan+1|/lan| > 1 or lan+1| > |an| ;> 0 for all n;> N. Hence, the necessary condition for a series to converge, an - 0 as n - oo does not hold; that is, the series L an diverges. If c = 1, it is sufficient, to give examples of a convergent and divergent series for which c = -1. Consider a p-series E -. One has c =lim an+1i lim "P =lim 1 =1 n-oo as n-oo (n + 1)P n-oo (1 + 1/n)P for any p. But a p-series converges if p > 1 and diverges otherwise. D  55. RATIO AND ROOT TESTS 119 EXAMPLE 8.30. Find all values of p and q for which the series L 1 nPq" converges absolutely. Solution: Here an = nPq". One has .*an+1 (n + 1)P qln+ -I(1 + 1/n)p c~ =km =lm =ql lim =ql. n-oa no P |q| ~ n-oo 1 So, for |ql < 1 and any p, the series converges absolutely by the ratio test. If q =+1, the ratio test is inconclusive, and these cases have to be studied by different means. If |ql = 1, then E lan= E np = E 1/n-p, which is a p-series that converges if -p > 1 or p < -1. Thus, the series converges absolutely for all p if |ql < 1 and for p < -1 if q =+1. Note that, for -1 < p < 0 and q = -1, the series conditionally converges (i.e., it is convergent but not absolutely convergent). In this case, it is a convergent alternating p-series L(-1)"/n-P. D The ratio test is advantageous to test convergence of series whose terms contain products of numbers, e.g., factorials n! = 1 - 2 - 3... -m-n. EXAMPLE 8.31. Test the series =0 lo/n! for convergence where, by definition, 0! = 1 Solution: By the ratio test lim a+I= lim = n limX=0 n-oo l an| n-oo (n + 1)! Iz| n-xoo n + 1 The series converges for all real x. Later it will be shown that the sum of the series is the exponential function ex. D 55.3. Root Test. THEOREM 8.22 (Root Test). Given a series E an, suppose the fol- lowing limit exists: lim --1= =c, Th->oo where c > 0 or c = oo. * If c < 1, then E an converges absolutely. * If c > 1, then E an diverges. * If c = 1, then the test gives no information. PROOF. If c < 1, then, as in the proof of the ratio test, the existence of the limit means that there is an integer N and a number q, c < q < 1, such that a N.  120 8. SEQUENCES AND SERIES 120 8. SEQUENCES AND SERIES This shows that the series lan| converges by comparison with the convergent geometric series E q", 0 < q < 1. So E an converges absolutely. If c > 1, then there exists an integer N such that (an > 1 for all n > N, and hence the necessary condition for convergence, an - 0 as n oc, does not hold. The series E an diverges. If c = 1, consider a p-series: P= ( =( )P - 1-p= 1 by Theorem 8.14. But a p-series converges if p > 1 and diverges if p < 1. The root test is inconclusive. D EXAMPLE 8.32. Test the convergence of the series E an, where an = [(2n2 + 5)/(3n2 + 2)]n. Solution: Here a = an, and the absolute convergence is equivalent to the convergence. One has 2n2+5 2+5/n2 2 lim VS/ a| = lim = lim = - < 1. n-oo n-o3n2+2 -o3+2/n2 3 So the series converges. 55.4. Oscillatory Behavior of Sequences in the Root and Ratio Tests. Con- sider a sequence defined recursively by a1= 1/2 and 1 . xn lan+1| 1 Kn an+1 =- sin -an =-> c a - sin (?) 2I2=|a |i2 2 An attempt to test the absolute convergence of E an by the ratio test fails because the sequence cn oscillates between 0 and 1/2 and, hence, does not converge. On the other hand, the absolute convergence of the series E an can easily be established by comparison with the convergent geometrical series: 1 1 1 |an| l _1 | -an-2 I < - -..< 2 4-2n where the inequality |sin x < 1 has been used. Similarly, the sequence used in the root test may also exhibit oscillatory behavior and be non- convergent. For example, an = (sin(--> cn = { 9 = sin . The series E as converges absolutely by comparison with the geometric series: a| < 2-. The ratio and root tests, as stated in Theorems 8.21 and 8.22, assume the existence of the limit lim-o c, - c, while the above ex- amples show that there are absolutely convergent series for which this  55. RATIO AND ROOT TESTS 121 hypothesis is not fulfilled. What can be said about the absolute conver- gence of a series when this limit does not exist? To answer this ques- tion, recall that, in the proof of the ratio or root test, the existence of lim- o cn = c < 1 has been used only to establish the boundedness of the sequence cn < q < 1 for all n > N, which is sufficient for the series E an to converge. But the boundedness property does not imply the convergence! Evidently, the boundedness condition holds in the above examples, cn < 1 < 1 for all n > 1, while the sequence {c,} does not converge. Similarly, the existence of the limit value c > 1 has only been used to show that ar,> 1 or lan| > 1 for infinitely many n to conclude that the sequence {an} cannot converge to 0 and hence E an diverges. If lan+|i/lan ;> 1 for all n > N, then again {an} cannot converge to 0 (by the proof of the ratio test). Thus, the hypothesis of convergence of {c,} in the root or ratio test can be weakened to obtain wider scopes of the tests. THEOREM 8.23 (Ratio and Root Tests Refined). Given a series Ean,1 put cj =|an+I|/an or c =V . Then cn <; q< 1 for all n;> N -- an converges. { a> 1 for infinitely manyn a+1 >1 for all n > N a diverges l anl for some integer N. 55.5. Wider Scope of the Root Test. The difference in the scopes of the root and ratio tests is elucidated by the following lemma that is proved in more advanced calculus courses. LEMMA 8.1. If the limit of |an+1|/an| exists, then so does the limit of an and in this case (8.14) lim -/ = lim lan+1 n-o n-o lan| If the sequence VJan does not converge, neither does |an+I/|an The converse is not true. For example, put an = 2-41(3-(-)") > 0. Then nh-~o m-o 2n+±1 2nm-oo 2 2  122 8. SEQUENCES AND SERIES 122 8. SEQUENCES AND SERIES because (3 - (-1)")/2 = 1 for even n and (3 - (-1)")/2 = 2 for odd n, while 9/p - 1 for any p > 0. In contrast, the sequence lan+1| 2n+1 3- (-1)1 1 3 + (-1)" 1 , n even an| 2n+2 3-(-1)n 2 3 - (-1)n 14,nodd oscillates between 1 and 1/4 and, hence, does not converge. So the convergence of the sequence {9/a, } does not imply the convergence of { an+1/an|} and the sequence {9/a, } may converge even if the sequence {|an+i/an|} does not. Lemma 8.1 shows that the ratio test has the same predicting power as the root test only if |an+I|/|an converges. The root test as stated in Theorem 8.23 has wider scope, meaning that whenever the ratio test shows convergence, the root test does, too, and whenever the root test is inconclusive, the ratio test is, too. The subtlety to note here is that the converse of the latter statement is not generally true. The inconclusiveness of the ratio test does not imply the inconclusiveness of the root test. The assertion can be illustrated with the following example. Consider a convergent series obtained from the sum of two geometric series in which the order of summation is changed: 1 1 1 1 1 1 an 2 + 22 +32+ 2 +33+-- n=1 1 ° 1 1 ° 1 1 1 1 1 3 2 2k +3 3k 2 1- +3 1- 2 k=0 k=0 2 3 where the sum of a geometric series has been used (Theorem 8.9). Now note that if n = 2k is even, then a2k = (1/3)k, and a2k-1 = (1/2)k if n = 2k - 1 is odd. Take the subsequence of ratios for even n = 2k c2k= a2k+1/a2k =(2/3)k/9. It converges to 0 as k -- o. On the other hand, the subsequence of ratios for odd n = 2k - 1 diverges: c2k-1 (3/2)k- oo as k - oo. So the limit of cn does not exist; moreover, the ratio test (as in Theorem 8.23) fails miserably to detect the convergence because cn is not even bounded. The series converges by the root test. Indeed, c2k k /ak -=1/2/ < 1 and c2k-1 = 2k-i1a2k1 =1/ 3< 1. Although the sequence cn does not converge (it oscillates between 1/v/5 and 1/v/2), it is bounded, c_ < q =1/v/2 < 1 for all n, and hence the series converges by T heorem 8.23. The ratio test appears to be sensitive to the order of summation; while this is not so for the root test.  55. RATIO AND ROOT TESTS 123 55.6. When the Ratio Test Is Inconclusive. THEOREM 8.24 (De Morgan's Test). Let E an be a series in which an+1|/|an - 1 as n - oc. The series converges absolutely if lim n ( an+1 - 1) = b 1), then the series E an converges if b < -1. For a p-series, the ratio test is inconclusive (see the proof of the ratio test). However, De Morgan's test resolves the inconclusiveness. Indeed, for large n, nP 1 " (n + 1)P (1+1/n)P n where the asymptotic behavior has been found from the linearization f(x) = (1 + )--P f(0) + f'(O)x = 1 - px for small x= 1/n. So b = -p and the series converges if b < -1 or p > 1. This illustrates a basic technical trick to applying De Morgan's test. Suppose that there is a function f (x) such that lan+l|/lan|I= f(1/n). If f is differentiable at x = 0, then its linearization L(x) = f(0) + f'(0)x is a good approximation in the sense that lim - L(x)= lim - f(0) f'(0) = f'(0) - f'(0) = 0 In other words, f (x) = f (0) + f'(0)x + ze(x) where e(x) - 0 as x - 0 (the error of the linear approximation decreases to zero faster than x). Therefore lan+1|_ f'(0) &(I) f'(0) &( ) an~ -f(1/n) = f(0) + + 1+ / +E . |an| n n n n By the inconclusiveness of the ratio test, f(0) = 1. Then limin ( +1 - 1 = lim (f'(0) + E()) = f'(0) and the series E a converges absolutely if f'(0) < -1. EXAMPLE 8.33. Investigate the convergence of the series n=1  124 8. SEQUENCES AND SERIES 124 8. SEQUENCES AND SERIES Solution: an+1 |an|I The ratio sequence has the form p(p+1)...(p+n - 1)(p+n)| lp(p + 1) ... (p + n - 1) In+pl n+1 1+ So it converges to 1 and the ratio test is inconclusive. Consider the function f(x) = (1 + px)/(1 + x). As a ratio of linear functions, it is differentiable at x = 0. By the ratio rule , p(1+ x) - (1 +px) (1 + x)2 p-1 (1 + X)2 -=f'(0) p-1. By De Morgan's test, the series converges (absolutely) if f'(0) < p-1 <-1 or p < 0. 1 or 55.7. Exercises. In (1)-(18), use the ratio or root tests to determine whether the series is absolutely convergent (here p is real). (1) (1000)2n n=1! (2) Zn2 (2) n=1 (3) n=1 p?40 (4 ) _ _ (5 ) i= 1 n 2 100 100-101 100-101-102 1+ 1-3 + 1-3-5 + 4 4.7 4.7.10 (9) -+ 2- 6 2)( 2- - 2)...( 2- n=1 (6) n=1 2n+1 2) 00 (12) Z('+ °0 2 (10) ( (13) n=1 (n3 5 (11) Z2 1 (14) 0 n=1 1)12 0n+ 2 2n2 p>0  55. RATIO AND ROOT TESTS 125 00 (15) Z (in fn)n (16) 00 pann! (17) Z5.8"""(3n+2) 002 .4---(2n) Z n! n=1 oc0 _n n(n-1) (18) n+ 1 n=2 (19) For which integers p > 0 is the series ' n1(!)2/(pn)! conver- gent? (20) Consider a geometric series with q = 1/2 in which the order of terms is changed by swapping terms in each consecutive pair: 1 1 1 1 1 1 1 a1+a2+a3+---=-+1+-+-+ + + + + . 2 8 4 32 16 128 64 Test the convergence of this series using the root and ratio tests. In (21)-(24), use De Morgan's test to investigate the convergence of the series. (21) (2)p + (L3)P + (.4.$)p + ... (22) + a(a+c) + a(a+c)(a+2c) +-, a>0, b>0,c >0 b b(b+c) b(b+c) (b+2c) / n=1 n!nq (24) Lo 13.5(2n-1) 1_ n=1 2.4.6...(2n) n (25) (Estimating Sums). Given a series E an with positive terms, put c = an+1/an. Suppose that cn - c < 1, that is, the series converges, E an = s. Let sn be a partial sum. Prove that s - sn < an+1 1 -cn+1 if {c,} is a decreasing sequence, and s - Sn < n 1 -c if {c,} is an increasing sequence. Hint: Use the geometric series as in the proof of the ratio test to estimate the remainder s - s ±=an + an+2 +.... (26) Prove De Morgan's test. Hints: Compare the series Elan | with the convergent p-series E be, where be A/ntP and p= -(1 - b)/2 > 1 if b < -1. Show that n(bn+1/bn - 1) - b as n o o. Next, show that, by choosing the constant A, one can always make lan| < bn for all n.  126 8. SEQUENCES AND SERIES 126 8. SEQUENCES AND SERIES 56. Rearrangements Here characteristic distinct properties of conditionally convergent and absolutely convergent series are established through the concept of rearrangement. DEFINITION 8.9 (Rearrangement). Let {k}, n = 1, 2, ..., be an integer-valued positive sequence in which every positive integer appears only once. Given a series E an, put a, = ak. The series E a, is called a rearrangement of E an. The sequence {k,} is obtained by a permutation of terms of the sequence of all integers {1, 2, 3, 4, ...}. For example, {1, 2, 3,4, 5,6, 7,8, 9,...} > {k } = f{1,3, 2, 5,7, 4, 9,11, 6, ...} . A rearrangement of a series is accordingly obtained by a permutation of terms in the sum. For a finite sum, a rearrangement (or a permutation) of its terms does not change the value of the sum. This is not generally so for convergent series. Consider an alternating harmonic series: 00 00(-1)"--1 1 1 1 1 1 (8.15) an 1- -++- +---.. n 2 3 4 5 6 n=1 n=1 The series is convergent but not absolutely convergent (its sum is s = ln 2; see Exercise 54.2.25). One of its rearrangements reads 001 1 1 1 1 1 1 1 (8.16) a'=1+ +-+++-+ +--- " 3 2 5 7 4 9 11 6 n=1 in which two positive terms are always followed by one negative. Let s, and s' be partial sums of (8.15) and (8.16), respectively. Put hn = 1 + 1/2 + -"-- + 1/n (a partial sum of the harmonic series). Then 1 1 1 1 1 s2n = 1- - +- -+.--+ 2 3 4 2n - 1 2n 1 1 1 1 i 1 1 1 1 3 5 2n- 1 2 k\2 3 1 J 1 1 1 1 =-h2m- h 2 4 2  56. REARRANGEMENTS 127 In a similar fashion, 1 1 1 1 1 1 1 s'n=1+-+-+-+---+ 3 5 7 4n-1 2 4 2n 1 1 1 1 1 1 =h4n- - hn= h4n - - h2n - -hn 2 4 4n 2 2 2 1 1 = (h4n - h2n) + -(h2n - hn) = s4n + -s2n. 2 2 Since the sequence {sa} converges to s, its subsequence {s4,} and {s2m} also converge to s. By basic limit laws for sequences, the sequence {s'3} converges and by taking the limit n - 00 in the above equality, lim s'3 =lim (s4 + s2n) = limn- o s4 + 1 limn-o 32s h-oo n-oo = s + s/2 = 3s/2. If the rearranged series (8.16) converges, then the subsequence {s'3} of the sequence of partial sums {s'} must converge to the sum of the series and the latter must be s' = 3s/2. Thus, a rearrangement of the series may change its sum! This fact is not specific to the example considered but inherent in all conditionally convergent series. Terms of a conditionally convergent series occur with different signs (positive and negative). By regrouping positive and negative terms, it will be proved that the sum of a conditionally convergent series can be made any number or too. The analysis begins by studying the properties of sums of positive and negative terms of a conditionally convergent series. Given a number x, put x+= (x + Iz|)/2. The number x+ = x if x > 0 and x+ = 0 otherwise. Similarly, x = x if x < 0 and x- = 0 otherwise. LEMMA 8.2. Given a series E an, consider two series E an[ and L a-, where an = (an + |a|)/2 (the series of positive and negative terms). Then (i) If E an converges absolutely, then E a+ and E a- converge. (ii) If E an is conditionally convergent, then E a+ and E a- diverge. PROOF. Let L an = s < o and E lan|j= t, where t c > sk1+mi or |c - sk1+mi l< |qm1 l. This can always be done because partial sums of E Pm can be larger than any number, while partial sums of E qn can be smaller than any number owing to the divergence of these series. So S1 Sm C Ski, 1 n Ck1, where |c- skil < pk1, Sk1 Sn Sk1+mi , 1k n k1 + i1, where |c - ski+mi l< |qk1 Next, take k2 next terms pm, where k2 is the smallest integer such that Sk1+m1+k2 > c, and take m2 next terms qn, where m2 is the smallest integer for which ski+m1+ki+m2 < c, and so on. At the nth step of the procedure, let n1 be the integer for which the last term in sm1 is Pk and let n2 be the integer for which the last term in sm2 is qm , that is, n2 = n1 + mm. The partial sums of the constructed rearrangement oscillate about c, reaching local minima sm1 and local maxima s82: m1 S S3mn2, 121 n 1n22, (8.17) |c - s1 < Pk , |c - s2 < |q By convergence of the series L am, am -~ 0 as n -~ 0. Hence, Pm and q, also converge to 0 and so do the subsequences Pk, -a 0 and qm, > 0. Thus, all local maxima and minima of the sequence of partial sum {sm} converge to c by (8.17), which shows that sm -a c. Finally, if c =+oo,  56. REARRANGEMENTS 129 one can take any divergent sequence c - 0o (or -oo) and construct a rearrangement such that sk1 overshoots ci and ski+mi undershoots ci, ski+m1+k2 overshoots c2 and ski+m1+k2+m2 undershoots c2, and so on. Obviously, this sequence of partial sum diverges. D Absolutely convergent series have a drastically different property. THEOREM 8.26 (Rearrangement and Absolute Convergence). If a series E an converges absolutely, then every rearrangement of E an converges, and they all converge to the same sum PROOF. Let tm = la1| + |a2| + - - - + |aml be a partial sum of the series of absolute values. The sequence {tm} increases monotonically converging to a number t by the hypothesis; that is, for any E > 0, there is an integer N such that t - tm < E for all m > N. Therefore, ak I=tm-tN+1= tm t + t-tN+1 k=N+1 < |tm-t|+|t -tN+1 < 2g. So, by taking N large enough, the sum of any number of terms lak|, k > N, can be made smaller than any preassigned positive number. Let sn and s' be partial sums of E an and its rearrangement E a'. One can take n > N large enough such that s' contains a1, a2,...,aN (i.e., the integers 1, 2, ..., N are in the set of integers k, k2, ..., kn in the notations of Definition 8.9). Then the difference s, - sn contains only terms ak with k > N (the terms ai, a2,..., aN are cancelled). Let m be the largest integer amongst {k1, k2, ..., k1} such that am is not cancelled in s' - sn. Then the difference contains some of the terms aN+1, aN+2, ..., am with possibly reversed sings (the terms of s, that are not cancelled contribute to s' - sn with an opposite sign). Applying the inequality |b t c l < lb +|cl to the sum of all terms in s' - sn, it is concluded that m Is', -s| < l-ak| < 2E. k=N+1 for all n exceeding some integer. If s' - s' and sn - s as n oc, then Is' - s| < 2E, which shows that s' = s because Ec> 0 is arbitrary. Q T hus, an absolutely convergent series is much like a finite sum. The sum does not depend on the order in which the summation is carried out. In contrast; the sum of a conditionally convergent series depends on the summation order. This is the characteristic difference between these two classes of convergent series.  130 8. SEQUENCES AND SERIES 130 8. SEQUENCES AND SERIES 56.1. Strategy for Testing Series. It would not be wise to apply tests for convergence in a specific order to find one that finally works. Instead, a proper strategy, as with integration, is to classify the series according to its form. One should also keep in mind that a conclusion about the convergence of a series can be reached in different ways. 1. Necessary condition for convergence. It is is always easier to check first the condition an - 0 as n - oc than it is to investigate the series E an for convergence. If the condition does not hold, the series diverges. 2. Special series. A series E an coincides with (or is a combina- tion of or is equivalent to) special series such as a p-series, alternating p-series, geometric series, telescopic series, and so on. Their conver- gence properties are known. 3. Series similar to special ones. If a series E an has a form that is similar to one of the special series, then one of the comparison tests should be considered. For example, if an is a rational or alge- braic function (contains roots of polynomials), then the series should be compared with a p-series. 4. Alternating series. If an = (-1)Thb, b ;> 0, then the alter- nating series test is an obvious possibility. 5. Ratio and root tests. Absolute convergence implies conver- gence. So, if the ratio or root test shows convergence, then the series in question converges absolutely. If these tests show divergence, then the series in question may still converge but not absolutely, and a further investigation is required. The root test is convenient for series of the form L(ba)". The ratio test is convenient when an involves the facto- rial n! or similar products of integers. The root test has a wider scope, but it is more difficult to use. The ratio test is often inconclusive if an is a rational or algebraic function (c =|an+1|/lan| - 1). In this case, the asymptotic behavior of cn is rather easy to find, cn ~ 1 + b/n as n - oc, and then use De Morgan's test. 6. Series of nonnegative terms. If an = f(n) > 0 and the integral fjf (x) dx is easy to evaluate, then the integral test is effective. Also, it can be used in combination with the comparison test: an < f (n) andj f(x) dx converges and so is E an, or f(n) < an and f f(x) dx diverges and so is E an. EXAMPLE 8.34. Test the series L(n + 1)/(n2 + in + 1) for conrver- genice. Solution: For large in, the leading terms of the top and bottom of the ratio are in and in2, respectively. So a, 1/in asymptotically for  56. REARRANGEMENTS 131 large n. The series resembles the harmonic series, which diverges. It is natural, then, to try to prove the divergence of the series by comparing it with the harmonic series: n+1 n n 1 n>2+n2+1 n2+n +1n2+1n 2+n2 n 3n Thus, the series indeed diverges by comparison with the harmonic se- ries. Q EXAMPLE 8.35. Test the series E 3"/(2- 4. 6---(2n)) for conver- gence. Solution: Each term an involves a factorial-like product of integers, which suggests the use of the ratio test: an+1 3n+1 2 .4 ... (2n) 3 an 2.4...(2n) - (2n+2) 3n 2n+2 So, the series converges. EXAMPLE 8.36. Test the series E sin(n2)e - for convergence. Solution: One has lan = | sin(n2)e m 0 n=11 (In ln 2 (12) (p2 n=1 =10) n=1 sin(1/n) 12p 1)" (13) n n=1 (14) < 1p Sp -(I - / )n (15) Z(nPji Cp 1 1), p > 0 (16) ( n1p + n (17) = n(- 1 1 1)" n + 1C > 0 (19) n=1 (18) (a'/" - b'/") , a > 0, b n=1  57. POWER SERIES 133 In (20) and (21), use the sum of the alternating harmonic series 1(-1)-1/n = ln 2 to find the sums of its rearrangement. 1 1 1 1 1 (20) 1+--+ - +--+ - - (2) +3 2 5 7 4 1 1 1 1 1 (21) +1 ++-- - 2 4 3 6 8 In (22) and (23), find two rearrangements of the conditionally conver- gent series that converge to +oo and to -oc. 0(-1)"-n-1O 1)n-1 (22) (23) n=1 n-1 (24) Let all positive terms of the alternating harmonic series be di- vided into groups of p terms in the order in which they appear in the series and let all negative terms be divided into groups of q terms in the order in which they appear. Consider the rearrangement in which the first group of positive p terms is followed by the first group of negative q terms, which, in turn, is followed by the second group of p positive terms and the latter is followed by the second group of q negative terms, and so on. If the sum of the alternating harmonic series is ln 2, prove that the sum of this rearrangement is ln 2 + 2 ln(p/q). (25) Let {a}° be the sequence of positive roots of the equation tan x = x in the increasing order. Test the convergence of the series LE',(an)-2. (26) If E an is conditionally convergent, show that 8+n lim = ±, s = (l|+a)1 n~oo 2Z(akj +ak). k=1 57. Power Series DEFINITION 8.10 (Power Series). Given a sequence {c,}, the series 00 caz" = c=O+ c1x + c2x2 +c3x3 _+.. . n=0 is called a power series in the variable x. The numbers cn are called the coefficients of the series. In general, the series will converge or diverge, depending on the choice of x. The power series always converges for x =0 to the num- ber co.  134 8. SEQUENCES AND SERIES 134 8. SEQUENCES AND SERIES EXAMPLE 8.38. For what values of x does the power series L Ox"/n converge? Solution: By the root test, |x| as n-oo. So the series converges for all -1 < x < -1 and diverges as x > 1 or x < -1. The root test is inconclusive for x = +1. These values have to be investigated by different means. For x = 1, the power series becomes the harmonic series E 1/n, which is divergent. For x = -1, the power series becomes the alternating harmonic series L(-1)"/n, which is convergent. Thus, the power series converges if x E [-1, 1) and diverges otherwise. D Given a number a, consider a power series in the variable y = x - a: 00 00 cny" = c (x - a)". n=0 n=0 It is also called a power series centered at a or a power series about a. Let S be the set of all values of x for which a power series in x con- verges and let Sa be the set of all values of x for which the corresponding power series in (x - a) converges. What is the relation between S and Sa? Since the series are obtained from one another by merely shifting the value of the variable by a number a, cv c - a, the set Sa is therefore obtained by adding the number a to every element of S: (8.18) ESa < - aES - Sa={x lc-aES}. For example, the series E" (x - 2)n/n converges if x - 2 E [-1, 1) or c E [1, 3) and diverges otherwise by Example 8.38. Thus, the prob- lem of finding the set Sa is equivalent to the problem of finding the set S. 57.1. Power Series as a Function. Suppose that a power series in x converges on a set S. Then it defines a function on S: f(x)=Zc cv", cE S. n=0 The set S is called the domain of such a function. Functions defined by power series are most common in applications. In what follows, it will be shown that familiar elementary functions such as sin cv, cos cv, and exp cv, etc can also be represented as power series. There are many other (special) functions that are defined by power series.  57. POWER SERIES 135 57. POWER SERIES 135 EXAMPLE 8.39. Find the domain of the Bessel function of order 0 that is defined by the power series Jo(x) Z 22n(n!)2 n=0o where, by common convention, 0! = 1. Solution: Since an = cnc2n contains the factorial, the ratio test is more convenient: an+1| 2 |cn+1 2 22n(n!)2 x2 |an| | Icn-l 22(n+1)((n + 1)!)2 22(n+ 1)2 as n - oc. So the series converges for all x. D Values of a function defined by a power series can be estimated by partial sums that are polynomials in the variable x: n f (X) ~fn(x) = ckza =co + clx + c2x2 + -.-.-+ cnz". k=O Thus, partial sums define a sequence of polynomials that converges to the function on S, f1(x) - f(x) for all x E S. The accuracy of the approximation is determined by the remainder R(x) = f(x) - f(x). The accuracy assessment is discussed in Section 8.59. Since the remainder R1(x) is a function on S, the error of the approximation is not generally uniform; that is, it depends on x. In this regard, recall also a useful result given in Exercise 55.7.25 57.2. Radius of Convergence. The set S on which a power series is convergent is an important characteristic and its properties have to be studied. LEMMA 8.3 (Properties of a Power Series). (i). If a power series L cnx" converges when x = b 0, then it converges whenever |x < |b|. (ii). If a power series E cnx"n diverges when x = d 0, then it diverges whenever |x| > |d|. PROOF. If E cnb" converges, then, by the necessary condition for convergence, cnb"h - 0 as n - oc. This means, in particular, that, for = 1, there exists an integer N such that |cnb"| N. Thus, for n> N, b"b b which shows that the series E c12z" converges by comparison with the geometric series E q"h, where q =cz/b and Iz/b| < 1 or cc| < lbl.  136 8. SEQUENCES AND SERIES 136 8. SEQUENCES AND SERIES Suppose that E cnd" diverges. If x is any number such that Ix > |d|, then E cx" cannot converge because, by part (i) of the lemma, the convergence of E cnz" implies the convergence of E cnd" Therefore, L cnz" diverges. This lemma allows us to establish the following description of the set S. THEOREM 8.27 (Convergence Properties of a Power Series). For a power series E cnz", there are only three possibilities: (i) The series converges only when x = 0. (ii) The series converges for all x. (iii) There is a positive number R such that the series converges if x < R and diverges if |x| > R. PROOF. Suppose that neither case 1 nor case 2 is true. Then there are numbers b / 0 and d / 0 such that the power series converges for x = b and diverges for x = d. By Lemma 8.3, the set of convergence S lies in the interval Ix < dl for all x E S. This shows that |dl is an upper bound for the set S. By the completeness axiom, S has a least upper bound R = sup S. If Ix > R, then x V S, and E cnx" diverges. If Ix < R, then Ix is not an upper bound for S, and there exists a number b E S such that b > I. Since b E S, E cxz" converges by Lemma 8.3. D Theorem 8.27 shows that a power series converges in a single open interval (-R, R) and diverges outside this interval. The set S may or may not include the points x =+R. This question requires a special investigation just like in Example 8.38. So the number R is character- istic for convergence properties of a power series. DEFINITION 8.11 (Radius of Convergence). The radius of conver- gence of a power series E cz" is a positive number R > 0 such that the series converges in the open interval (-R, R) and diverges outside it. A power series is said to have a zero radius of convergence, R = 0, if it converges only when x =-0. A power series is said to have an infinite radius of convergence, R =oco, if it converges for all values of x. The ratio or root test can be used to determine the radius of con- vergence.  57. POWER SERIES 137 COROLLARY 8.3 (Radius of Convergence of a Power Series). Given a power series E cnz", if lim|cn+1|=a --> R n- oo cl a 1 if lim /c|=a -- R=- n-ooa where R= 0 if a =oo and R =oo if a =0. PROOF. Put an = cxn in the ratio test (Theorem 8.21). Then lan+1/lan| cIllcn+l|/lcnl - |la. The series converges if |xla < 1, which shows that R = 1/a. Similarly, using the root test (Theorem 8.22), S/ja_ n j I| /cnI - |xla < 1, which shows that R = 1/a. D Remark. If the sequences in Corollary 8.3 do not converge, then Theorem 8.23 from Section 55.4 should be applied to an = cxn to determine the radius of convergence. Also, one should keep in mind that the root test has wider scope (recall Lemma 8.1 in Section 55.5) Once the radius of convergence has been found and 0 < R < 00, the cases x = +R have to be investigated by some other means (as the root or ratio test is inconclusive in this case) to determine the interval of convergence S of a power series. EXAMPLE 8.40. Find the radius of convergence and the interval of convergence of the power series E cnz", where c = (-q)"/ n+ 1 and q>0. Solution: |cn+1| q+l n+1 n 1+1 1+1/n |cn| n+2 q" n+2 1+2/n Therefore, R =1/a = 1/q. If x = -1/q, then cnz" = (-1)"//n + 1 = (-1)"ba. The sequence bn converges monotonically to 0 so that L(-1)"b converges by the alternating series test. If x = -1/q, then cncz" 1/ n + 1 > 1/v21, n2> 1. The p-series E 1/ni/2 diverges (p = 1/2 < 1) so that E 1/ n+ 1 diverges by the comparison test. Thus, the interval of convergence is S = [-1/q, 1/q). D EXAMPLE 8.41. Find the radius of convergence and the interval of convergence of the power series E 122(cc + 1)n/qn, where q > 0. Solution: Put y cc x+1. If S is the interval of convergence of E ay where c, n22/q, then the interval of convergence in question is ob- tained by adding -1ito all numbers in S according to the rule (8.18).  138 8. SEQUENCES AND SERIES 138 8. SEQUENCES AND SERIES By Corollary 8.3, lim VI | n-o - m m-oo q -(urn l n)2 q n-o 1 q So R = 1/a = q. If y = q, then cny" = n2, and the series E n2 diverges (an = n2 does not converge to 0). If y = -q, then cmz" = (-1)n2, and the series diverges because an = (-1)n122, does not converge to 0. The series converges only if ly| x + 1| < q, and hence the interval of convergence is x E (-q - 1, q - 1) (the interval (-q, q) shifted by -1). 57.3. Exercises. In (1)-(19), find the radius of convergence and the interval of conver- gence of the power series. (1) xnm" m~O (2) n=1 2"0 3x (3) Z2-1) n=1 (5) (x - 2)" p n=1 (4) (~1)"2m 1nn" 2=2ln1 n=2 (6) v/n--(X + 1)"n n=o0 (g 0(4x + 1)" (8) 2 n2 n=1 (10) 2 n=1 24"(n o gnn (12) Zn!(x+3) n=O ( 00 7) (-l m=1 -1)m3 +3(x 1)" 00 x m (9) 1 - 1 3.-5"-""-(2n - 1) (11)~(nk) ! x" , k > 0 (integer) n= (1+)o (13) 1'+)- x n=1 (14) p(p-1)---(p-n+l1) cct (16) n!y (16) p, p > 0, (17) 1 )" l l n n (1)e x" (19) n! e n=1 (15)Z a2n+ b2n n=1a+b 2.-4"-""-(2n) n x" 0[3+ (-1)] ct n n=1 (20) Let p < q be real numbers. Give examples of power series whose intervals of convergence are (p, q), [p, q], (p, q], and [p, q).  58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 139 (21) The Airy function is defined by the power series X3 6X9 A(x)=1+ -+++-. 2-3 +2-3-56 2.3.5.6.8.9 Find its domain. (22) A function f is defined by the power series f(x) =p+qx+px2+q-x3 +pi+qx5+ +-- that is, its coefficients C2k = p and c2k-1 = q, where p and q are real. Find the domain of f and an explicit expression of f(x) (the sum of the series). (23) If f(x) =E cazc", where cn+4 = cn for all n > 0, find the domain of f and a formula for f(x). (24) The power series E cx" and E bnx have the radii of conver- gence R1 and R2, respectively. What is the radius of conver- gence of E(ca + bn)x"? (25) Suppose that the radius of convergence of E cx" is R. What is the radius of convergence of E cmzkn, where k > 0 is an integer? 58. Representation of Functions as Power Series Consider a power series 00 1 2 _ _ 9 .2nh n=0 It is a geometric series with q = -z2, and therefore it converges for all |ql = x2 < 1 or x E (-1, 1). Using the formula for the sum of a geometric series, one infers that 1 =Z1 - X 2 + X4 +"-""-= (-1)"zX2n for all -1G< <1, 1+X x2 n=0 This shows that the function 1/(1 + x2) can be represented as a power series in the open interval (-1, 1). Note that this representation is valid only in the interval of convergence of the power series despite the fact that the function 1/(1 + X2) is defined on the entire real line. In general, one can construct a representation of a function by a power series in (xc - a) for some a. The interval of validity of this representation depends on the choice of a. E XAMPLE 8.42. Find a representation of 1/zc as a power series in (xc - a), a > Q, and determine the interval of its validity.  140 8. SEQUENCES AND SERIES 140 8. SEQUENCES AND SERIES Solution: Put y = x - a. The function can be rewritten in a form that resembles the sum of a geometric series: 1 11 Thy~< n (xl-)a)"Th, cE(0, 2a). x a(l+y/a) anan+1 The geometric series converges if |ql= - y/a= lyl/a < 1, and hence this representation is valid only if -a < y < a or -a < c - a < a or 0 0, then the function f defined by oo f(cc) =co + c1(cc - a) + c2(cc - a)2 + - --3 ca(cc -a) Th~O  58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 141 is differentiable (and therefore continuous) on the interval (a-R, a+R) and 00 f'(x) =ci + 2c2(x- a) + 3c3(x-a)2 +...- - = nc(x -a)"-1, n=1 f f(x)dx=C+co(x-a)+ci 2 +---=C+ cn + n=0 The radii of convergence of these power series are both R. Thus, for power series, the differentiation or integration and the summation can be carried out in any order: d c(x - a)" =3 d[c(x - a)"], cn(x - a)"h) dx =3 J[cn(x - a)"] dx. Remark. Theorem 8.28 states that the radius of convergence of a power series does not change after differentiation or integration of the series. This does not mean that the interval of convergence does not change. It may happen that the original series converges at an endpoint, whereas the differentiated series diverges there. EXAMPLE 8.43. Find the intervals of convergence for f, f', and f" if f (x) = _ 1nx /n2. Solution: Here c, 1/in2 and hence 9/5| 1/9 =n (1/ n i)2 -+ 1 = a. So the radius of convergence is R = 1/a = 1. For x = +1, the series is a p-series E 1/n2 that converges (p = 2 > 1). Thus, f(x) is defined on the closed interval x E [-1, 1]. By Theorem 8.28, the derivatives f'(x) = 1 x"-1/n and f"(x) L -2(n -1)x-2/n have the same radius of convergence R = 1. For x = -1, the series f'(-1) - L(-1)-1/n is the alternating harmonic series that converges, whereas the series f"(-1) L (-1)"(n -1)/n diverges because the sequence of its terms does not converge to 0: |(-1)(n - 1)/nl = 1 - 1/n - 1 -/ 0. For x = 1, the series f'(1) = E 1/n is the harmonic series and hence diverges. The series f"(1) =E(n - 1)/n also diverges ((n - 1)/n does not converge to 0). Thus, the intervals of convergence for f, f', and f" are, respectively, [-1, 1], [-1, 1), and (-1, 1).D The term-by-term integration of a power series can be used to ob- tain a power series representation of antiderivatives. EXAMPLE 8.44. Find a power series representation for tan-1 x.  142 8. SEQUENCES AND SERIES 142 8. SEQUENCES AND SERIES Solution: tan-1x = 1 f 2 = ( -2) dz = C+ (-1)n + 1+Zn=o n=o2n +1 Since tan-10 = 0, the integration constant C satisfies the condition 0 = C + 0 or C = 0. The geometric series with q = -x2 converges if ql < 1 or x2 < 1 or Iz| < 1. Hence, the radius of convergence of the series for tan-1 xis R = 1. D In particular, the number 1/v/ is less than the radius of conver- gence of the power series for tan-1 x. So the number tan-1(1/v/3) 7r/6 can be written as the numerical series by substituting x = 1// into the power series for tan-1 x. This leads to the following represen- tation of the number 7: w=2 n=o (2n + 1) 3n 58.2. Continuity at endpoints of the interval of convergence. Consider the geometric series with q = -x 00 1 (-1)"z" = n=0o By integrating the series term-by-term, the power series for the loga- rithm function is deduced: =x ln(1 +x) = C + ( )x+ 1+xn n+ which holds for cc| < 1 by Theorem 8.28. The constant C is determined by setting x = 0, which yields C =ln 1 = 0. Thus, 00 (_1)n-1 (8.19) ln(1+ cc) - ( x" n=1 This series diverges at x = -1 as a harmonic series, but it converges at the other endpoint x = 1 as the alternating harmonic series. In Exercise 54.2.25 it has been shown that the sum of the alternating harmonic series is ln 2. It follows then that the power series representation (8.19) also holds at cc= 1 lim x"(1)h =c ln 2 x-1-12  58. REPRESENTATION OF FUNCTIONS AS POWER SERIES 143 Here x - 1- means the left limit (x approaches 1 from the left). This observation is not specific to the example considered but rather is of general nature. THEOREM 8.29 (Abel's theorem). If the power series f(x) Ln=0 cnz", |x| 0): 00 f(x) co + ci(x - a)+ c2(x - a)2+ ... - cm(x - a)". n=O where Ix - al < R. By Theorem 8.28, its derivatives f(k)(z) can ob- tained by the term-by-term differentiation of the series, and the result- ing series has the same convergence radius R. Evidently, f(a) = co. What is the significance of the other coefficients cn? The derivative f' is given by f'(x) =ci+ 2c2(x -a) + 3c3(x -a)2 + + kci(x -a)k-l+- -, which shows that f'(a) = c1. The second derivative is f"(x) =2c2+3.2c3(x-a)+...+k(k -1)ck(x-a)k-2+...- Therefore, f"(a) = 2c2. After k such steps, f(k)(z) =k(k - ) ---2-1 c + (k + 1)k(k - ) ---2ck+(x -a) + .-..- and hence f(k)(a) = k!ck or Ck = f(k)(a)/k!. This proves the following theorem. THEOREM 8.30 (Significance of Power Series Coefficients). If f has a power series representation 00 f(x)= c(x-a)", I - al 0, then its coefficients are _f(m) (a) cn =. n!1 DEFINITION 8.12 (Real Analytic Functions). A function f on an open interval I is said to be analytic if, for any a E I, it has a power series representation f(cc) = c,(cc -a<" that converges in some open interval (a - 5, a + 5) c I, where 5 > 0. The class of analytic functions plays a significant role in applica- tions. Their properties are discussed next.  148 8. SEQUENCES AND SERIES 148 8. SEQUENCES AND SERIES THEOREM 8.31 (Power Series Representation of Analytic Func- tions). A function f that is analytic on an open interval I has the power series representation (8.21) f(x) z f()(a) ( a) n=0 for any a E I that converges in an open subinterval of I that includes a. This theorem follows from Definition 8.12 and Theorem 8.30. In Example 8.42 it was found that (8.22)1 00(-1)< (8.22) 1 - Z= an+1 (x-a)", x E (0, 2a). n=0 This shows that the function f(x) = 1/x is analytic for all x > 0 because a can be any positive number; that is, the function has a power series representation that converges in an open subinterval of (0, oc) containing any a > 0. Similarly, the analyticity of f(x) = 1/x can be established for all x < 0. It is important to emphasize that a power series for an analytic function does not necessarily converge on the entire domain of the function. But an analytic function can always be represented by a convergent power series in a neighborhood of every point of its domain. Equation (8.22) illustrates the point. THEOREM 8.32 (Properties of Analytic Functions). (i) The sums and products of analytic functions are analytic. (ii) The reciprocal 1/f of an analytic function f is analytic if f is nowhere zero. (iii) The composition f(g(x)) of analytic functions f and g is analytic. (iv). Analytic functions are differentiable infinitely many times. A proof of properties (i)-(iii) is given in more advanced calculus courses. Property (iv) follows from Theorem 8.28. Its converse is not generally true. There are functions that are differentiable infinitely many times at a point, but they cannot be represented by a power se- ries that converges in an open interval that includes this point. As an example, consider the function f (x) = e-1x if x / 0 and f(0) = 0.  59. TAYLOR SERIES 149 59. TAYLOR SERIES 149 The function is continuous at x = 0 because limx-o e-1/x2= limsom e-" = 0 = f(0). It is differentiable at x = 0 because f () - f (0) e-/r2n fx(0) = lim = lim=0. The first equality is the definition of f'(0). The last limit is established by investigating the left and right limits x - 0* with the help of the substitution x = 1/u - +oo as x - 01; the left and right limits coincide because ue-"2 - 0 as u - +too (the exponential function decreases faster than any power function). In a similar fashion, it can be proved that f(")(0) = 0 for all n (see Exercise 59.5.24). Thus, f(x) has no power series representation E cnx in a neighborhood of x = 0 because, if it did, then, by Theorem 8.31 the function should have been identically 0 in some interval (-b, 8), 5 > 0, (as f(n)(0) = 0 for all n), which is not true (f(x) / 0 for all x / 0). Hence, the function is not analytic at x = 0. 59.2. Taylor and Maclaurin Series. DEFINITION 8.13 (Taylor and Maclaurin Series). The series in (8.21) is called the Taylor series of a function f at a (or about a, or centered at a). The special case of the Taylor series when a = 0 is called the Maclaurin series of a function f. It is important to know Maclaurin series of elementary functions. In particular, Maclaurin series for the power function (1+ zt)P, where p is a real number, is called the binomial series. The numbers p _p(p- 1) ---(p- n+ 1) n n! are called binomial coefficients. THEOREM 8.33 (Maclaurin series of elementary functions). For any real x, the following power series representations hold 2 3 n ex=1+x+2 +3 +---=t ex1++2! 3! n n=1 3 5 7 00 _7n2n+1 3! 5! 7!(2+1) 2 4 6 m _ n 2 cos x =1- + - + - - 2! 4! 6! 2 n (m)  150 8. SEQUENCES AND SERIES For |x| < 1, the following power series representations hold x2 X3 x4 ° O(-i-1 ln(1+) = x-- + 34 +-X 2 3L4 n=1 (1+z)p== 1+ x+c22+---=cx" PROOF. The Maclaurin series of the exponential function ex and the logarithm function ln(1+ zx) have been already established in Sec- tions 58.3 and 58.2, respectively. Let f(x) = sin x. One has f'(x) (sin x)' = coscx and f"(x) = (cos z)' = - sinx. Hence, f(2n)(z) = (-1)Thsincx, f(2n+1)(z) =(-1)"hcoscx, and f(2n)(0) = 0, f(2+1)(0) = (-1)n-1. By Theorem 8.30, the non-zero coefficients of the Maclaurin series are f(2n+1)(0) (-1)< c2n+1 = (2n+1)! (2n +1)! n = 0, 1, 2, ... By the ratio test, c2n+3x2n+3| X22 n+-1)1X 2 lim = lim (= lim = 0 n-o |c2n+i +1 n->oo (2n + 3)! n-oo (2n + 2)(2n + 3) the power series converges for any x. The Maclaurin series for f(x) cos x = (sin x)' is obtained by differentiating the series for sin x term- by-term. By Theorem 8.28, it also converges on the entire real line. Finally, put f (x) = (1 + z)P. The derivatives are f'(x) = p(1 + z)p-1, f"(x) = p(p - 1)(1 + z)p-2, and, in general, f (")(X) = p(p - 1) -.-. (p - n + 1) (1 + z)"-". So the coefficients of the power series coincide with the binomial coef- ficients introduced earlier: f(n)(0) p(p - 1) ...(p-n+l) p n! n! (n) They satisfy the recurrence relation cn+1 =c1(p-n)/(n-+1). Therefore, by Corollary 8.3, the radius of convergence R is determined by lim |c+|=lim =-lim 1 = 1 - n- c - + 1 1+o 1 +  59. TAYLOR SERIES 151 59. TAYLOR SERIES 151 59.3. Taylor Series of Analytic Functions. Every analytic function in a neighborhood of a point is represented by the Taylor series about that point. If the Taylor series converges on the entire real line, then the function is analytic everywhere. In particular, the exponential ex and trigonometric functions sin x and cos x are analytic everywhere. More- over, the properties of analytic functions stated in Theorem 8.32 allows us to add, multiply, and make a composition of the Taylor series (on the common intervals of their convergence) just like ordinary sums to obtain the Taylor series representation of the sums, products, and com- positions of analytic functions. These are extremely useful properties in applications. EXAMPLE 8.47. Find first four terms of the Taylor series for the function f (x) = exp(tan-1 x) about x = 0. Solution: Calculation of the derivatives of such a function is rather tedious. Instead, note that ex and tan-1 x are both analytic in a neigh- borhood of x = 0. So the composition of their Taylor series (see (8.20) and Example 8.44) gives the sought-after Taylor series. Only monomi- als 1, x, x2, and x3 have to be retained when calculating the compo- sition. This implies that it is sufficient to retain two leading terms in the Taylor series tan-1x = x - x3/3+ -"-- and four leading terms in the Taylor series (8.20) of the exponential function: eta"- 1 + tan- zc + 2(tan-i z)2 + 6(tan-i z)3 + X3 1 X3 2 1 x3 3 X 3 2 X 3 + ) + 6 X 3+ =1+ ( - $) + - x2+ - x3- - - 13 2 3 ~321= 1+ x+2-x2 -6-x3+_..._ 1+c2 6x 59.4. Approximations by Taylor Polynomials. An analytic function f can be approximated by a partial sum of the Taylor series: f(x)~ fk! (-a)k = Tn(x). k=0 The polynomial T,(xc) is called a Taylor polynomial of degree n about a. The convergence of the Taylor series guarantees that the remainder converges to 0: R,(x)=f(xc)-Tn(xc)-0 as n-ooo for cc-a| 0 and f(cc) =0 if cc <; 0. Is this function analytic everywhere? (26) Use the identity w/6 =sin-1(1/2) to calculate the number w within the error not exceeding 10-4.  156 8. SEQUENCES AND SERIES (27) Find cos(1°) within the error not exceeding 10-6. (28) Use the first three terms of the binomial series to approximate 9 and estimate the error. In (29)-(32), use various methods to find the Maclaurin series of the function and investigate its convergence radius. (29) Xe-dt (30)t So 1-t4 (31) t (32) fdt t aft (33) Let f(x) = E ocxn, where In!cl < M for n = 1, 2, 3, ... and some number M. Prove that (i) f(x) is differentiable infinitely many times at any point x = a and (ii) the Taylor series representation of f(x) about a holds for Iz| < 00.  CHAPTER 9 Further Applications of Integration 60. Arc Length 60.1. The Length of a Curve. We have seen various applications of in- tegration to the computation of the area of a domain and to the com- putation of the volume of a solid. It is perhaps more surprising that we can also use integration to compute the length of a curve between two given points. This may sound counterintuitive at first, since in the applications we have seen so far, integration was used to compute some parameter of an object that existed in a higher dimension than the function that was being integrated. Let f be a function so that, on the interval [a, b], the derivative f' of f exists and is a continuous function. We would like to know the length of the curve of f, starting at the point A = (a, f(a)) and ending at the point B = (b, f(b)). Intuitively, we can imagine that we lay a rope over the graph of f between the two endpoints, mark A and B on the rope, then straighten that rope out, and measure the distance between them. A more formal definition, which is useful in the actual computation of the length of the curve, is the following. Cut the interval [a, b] into n equal parts, using points a =0zo < x1 < - - - < xn = b. Let Pi = (x, f (x)) = f(x, ys). Let |P_1Pl denote the length of the straight line segment from Pi_1 to P. Then the sum (9.1) Kn = |P-_1Pl i=1 is a little bit smaller than the length of the curve since the points Pi_1 and Pi are on the curve and the straight line is the shortest path between them. If we keep refining the subdivision of the interval [a, b] by having n go to infinity, then it can be proved that limn-, K,1 exists. We define that limit to be the length of the curve of f from A to B. See Figure 9.1 for an illustration. Note that in the case when the graph of f is a straight line segment between A and B, this definition is just the length of that segment, so our definition extends our previous notion of length. 157  158 9. FURTHER APPLICATIONS OF INTEGRATION (X1iY1 (x2, Y 2) - - ( x 3 , Y 3 ) I I I II x0 = a x1 X2 x3 = b a b FIGURE 9.1. Arc length as a limit. Let us now return to (9.1) in order to compute limn-Ka. Let (b - a)>/n = (x - x-1)/n = Ax. Note that then |P -iPil- = 2 -z- _1)2 +(yj - yi-1)2 = (Az)2 + (A)2y - Yi-1)2 V) (x2 - x _1) = x 1+ (.y - Yi-i)2 (x-x i1)2 Now observe that since f'is continuous, the Mean Value Theorem (Theorem 6.9) implies that there is a real number x E [xi_1, x] such that Y-Yi= f'(x). Hence, the previous chain of equalities yields |PPl= Axz'\1 + f'(zi)2. Summing over all i, we get Kn = Ox 1 + f'/(zi)2. i=1 As n goes to infinity, the left-hand side, by definition, converges to the length of the curve of f between A and B, while the right-hand side, being a Riemann sum, converges to ff 21 + f'(x)2 dx. Hence, we have proved the following theorem. T HEOREM 9.1. If f' is a continuous function on the interval [a, b], then the length of the graph of f (x) from the point (a, f (a)) to the point  60. ARC LENGTH 159 60. ARC LENGTH 159 0.8 0.6 f(x) = 2x3/2 3 0.4 0.2 0.8 1.0 ,,, % ,1,1 -0.2 0.2 0.4 0.6 1.2 FIGURE 9.2. The curve of f(x) x3/2 (b, f(b)) is equal to L = /1 +f'(cx)2 dz. EXAMPLE 9.1. Find the length of the curve of f (z) (0,0) to (1, 2/3). See Figure 9.2 for an illustration. 2x3/2 from Solution: We have f'(x) and therefore Theorem 9.1 vcc, so f'is a continuous function on [0, 1], applies. Using that theorem, we obtain L=f 1+(v c)2d 0 0 1 + x dc 1 2(1 +z)3/2/ 0 4v/2 3 2 3. Note that the result is remarkably close to the length of the straight line that connects the two points in question, which is 13/3. We can use our new technique to verify a classic formula. EXAMPLE 9.2. Use Theorem 9.1 to compute the circumference of a circle of radius 1. Solution: Let us place the center of the unit circle at the origin. Then the boundary of the circle is the set of points satisfying x2 + y2 1.  160 9. FURTHER APPLICATIONS OF INTEGRATION I, I I I 1 I I 2 2 FIGURE 9.3. One quarter of the unit circle. We want to use Theorem 9.1, so we need a part of the circle where that satisfies the vertical line test (so y is a function of f) and where the tangent line to the circle is never vertical (so that f'(x) exists). For instance, we can choose the quarter of the circle that starts in the point x = (-a) and ends in the point x = (a). See Figure 9.3 for an illustration. On that part of the curve, f'(x) =- Xis continuous, so Theorem 9.1 implies L = 1 + f'(x)2 dz -v/2 rh/2 x2 ~1122 1+ 1 dzc -12/2 1-2 _-2/2 v1 - x2 12 = sin-1 x =r/2. This implies that the circumference of the full circle is four times this much, that is, 27r.D  60. ARC LENGTH 161 60.2. Remarks. Recall that in the first paragraph of this section, we discussed why it may seem counterintuitive that integration plays a role in the computation of are lengths. Now we can see that the purported contradiction explained there is resolved by the fact that the integrand in Theorem 9.1 contains f', not f. Compared to other formulas we learned in our earlier studies of integration, it is relatively rare that the formula given by Theorem 9.1 can be explicitly computed, since f 1 + f'(x)2 dc is often difficult to handle. Therefore, we must often resort to approximate integration while computing are lengths. 60.3. Exercises. (1) Find the length of the curve f(x) =x2/2 between the points given by c = 0 and c= 1. (2) Find the length of the curve f(x) = 3x2/2 between the points given by c = 0 and c= 1. (3) Find the length of the curve f(x) = 4- x x between the points given by c = 2 and c =4. (4) Find the length of the curve f(x) = ln(cos c) between the points given by c = 0 and cc =w/4. (5) Find the length of the curve f(x) = ln(sin c) between the points a =w7/4 and b = 37/4. (6) Find the length of the curve f(x) = lncx - $ from x = 4 to c=5. (7) Verify that Theorem 9.1 provides the correct value for the are length of f when f is a linear function. (8) Let f = (x2 + 2x)3/2 - 2 ln(c+1+ cc2 + 2x). Find the length of the curve of f from x = 1 to c = 2. (9) Let f =(2x+4) c2+4xc+3-jln(x+2+ c2+4x+3). Find the length of the curve of f from x = 1 to c = 2. (10) Let f (x) = 2xcc2 - 1 - 2 ln(x + cc2 - 1). Find the length of the curve of f from x = 1 to c = 2. (11) Let f(x) = 2xcc2 + 1 + 2 ln(x + cc2 + 1). Find the length of the curve of f from x = 1 to c = 2. (12) Let f and g be two functions that have continuous derivatives on the interval [a, b] and assume that f'(x) -g'(x) for all cc E [a, b]. Prove that the curves of f and g have the same length between cc= a and a =b. Try to find two different arguments. (13) Use a method of approximate integration to estimate the length of the curve f(cc) =ex as from (0, 1) to (1, e).  162 9. FURTHER APPLICATIONS OF INTEGRATION (14) Use a method of approximate integration to estimate the length of the curve f(x) = ex as from (0, 1) to (1, e). Compare the result with that of the previous exercise. (15) Compute the exact are length in the previous exercise by either using a software package or an appropriate substitution to find the needed indefinite integral. (16) Use a method of approximate integration to estimate the length of the curve f(x) = sin x from (0, 0) to (7, 0). (17) Use a method of approximate integration to estimate the length of f(x) = x3 from (0, 0) to (1, 1). (18) Use the result of the previous exercise to estimate the length of f (x) = cos x from x = -7 to w= . (19) Find the length of the curve f(x) = lncx from x = 1 to xc 2. You can use a software package to compute the needed indefinite integral. (20) Use a method of approximate integration to estimate the length of f(x) = tancx from x = 0 to w= /4. 61. Surface Area 61.1. The Definition of Surface Area. In the last section, we defined the length of a curve, and deduced a formula for the computation of that length. Let us now take a curve, say of a function f(x) = y, where x E [a, b] and f'is continuous on [a, b]. Let us rotate this curve around the horizontal axis, as shown in Figure 9.4. What is the area of the obtained surface of revolution? The definition of the area in question, and its computation, will be quite similar to what we have discussed in the previous section for the are length. Cut the interval [a, b] into n equal parts, using points a =zo, x2] at 1g, the objects at the endpoints of the interval will balance. See Figure 9.8 for an illustration. We assume that the interval [i>, x2], or the stick representing it, has negligible mass. If mi = m2, then we clearly have 1g = (X1 + x2)/2. Otherwise, we make use of the well-known fact of physics that the interval will balance if the moments on the two sides of the fulcrum are equal, that is, when (9.5) m1(x9 - xi) = m2(x2 - x9) holds. Solving (9.5) for xg, we get (9.6) 1g = .+ m22 mI + m2 The point og of the real line is called the center of mass or center of gravity of the system described above, that is, the system of an object of mass mi at x1 and an object of mass m2 at x2. The moment of an object with respect to a point P is the mass of the object times the distance of the object from P. In particular, in the above system, the two objects had moments m1i and m2x2 with respect to the origin. So the total system had moment m11 + m2x2. Note that if we replace the two ob- jects by a simple object of mass m1 +m2 placed at og, then the moment of the system about the origin does not change. This is an important property that only the center of mass has, and therefore we repeat it. If we concentrate the total mass of the system at the center of mass, the moment of the system with respect to the origin will not change. If we consider a system of k distinct objects of mass m1, m2, ... , mk placed at points x1, x2, ... , Ik along the horizontal axis, then we can use an analogous argument to show that the center of mass of the system is at x1 xg 22 | |Xg ml m2 FIGURE 9.8. Center of mass.  170 9. FURTHER APPLICATIONS OF INTEGRATION (9.7) Xg=- E2=1 mn 62.2. Two-Dimensional Systems. 62.2.1. Discrete Two-Dimensional Systems. Let us now consider the more general case when the k objects of mass m1, m2, ... ,mk are placed in points (x1, y1), (x2, 2), ..., (Xk, Yk) of the plane. We would like to find the center of mass (xg, yg) of this system. In other words, we assume that a plate of negligible mass is placed under our system, and we want to find the point (xg, yg) with the property that if we place a fulcrum under the plate at that point, the plate will balance. Using methods similar to the one-dimensional case, it can be proved that the plate will balance if the fulcrum is placed at (xg, yg) with iEk mizi Ek myy (9.8) Xg = and yg = =. E~i1 m2 Li1 mi This corresponds to the intuitively appropriate concept that the plate will balance if it balances both "horizontally" and "vertically." The sum MM =_E 1 myy is called the moment of the system with respect to the x axis. This name is due to the fact that if we tried to balance the system on the x axis, the larger the number M2, the more would the weights of the system rotate the plate. Similarly, the sum Mg = Ek1rryz is called the moment of the system with respect to the y axis. 62.2.2. Symmetry Lines. Now let us consider the continuous version of the problem. Let P be a plate and let us try to find the center of mass of P. (We no longer assume that the mass of the plate is negligible; in fact, that mass is the object of our study now.) Let us assume, for the rest of this chapter, that the mass of P is uniformly distributed over P. Let us also assume that the density of the material of which P is made is 1. That is, the mass of a unit square within P is 1. Once these assumptions are made, it is clear that the center of mass of P is determined by the purely geometric characteristics of P. In order to emphasize this, if the assumptions of this paragraph hold, we will call the center of mass of P the centroid of P. Sometimes we can find the centroid of P without computation. A symmetry line of P is a straight line t such that the image of P when reflected through t is P itself. That implies that the two parts into which t cuts P are congruent, and the plate balances on the line t. Consequently, the centroid C of P must be on t, since if we concentrate the entire mass of P in C, it still has to balance on the line P.  62. APPLICATIONS TO PHYSICS AND ENGINEERING 171 The argument of the previous paragraph shows that the centroid of P must be on every symmetry line of P. So if P has more than one symmetry line, then these symmetry lines must all intersect in one point, namely, in the centroid of P. In this case, we obtain the center of mass of P as the intersection of any two symmetry lines of P. For example, we can find the center of mass of a circle, ellipse, rectangle, or rhombus in this way. 62.2.3. A Formula for Continuous Two-Dimensional Systems. Let us keep the conditions from the previous section and let us impose the new con- dition that P is a "domain under a curve"; that is, the borders of P are the vertical lines x = a and x = b, the horizontal axis, and the graph of the continuous function f(x) = y. We would like to use formula (9.8) to find the approximate location of the centroid of P. Let us cut the interval [a, b] into n equal parts, using the intermediate points a x=zo < x1 < x2 < "< xn = b, and let Ax = (b - a)/n. The vertical lines x = x cut P into n vertical stripes. Let Si be the ith such stripe. The area, and hence the mass, of Si is close to AOx f(x), where x is the midpoint (x21 + x)/2 of the interval [x-_, x]. So we are approximating Si by a rectangle Ri. Let us concentrate the entire mass of RZ in the centroid of R2, that is, at (x,f(x)/2). Now we can compute the moment of the obtained system of n ob- jects with respect to the x axis. Note that the mass of RZ is equal to the area of R2, that is, Axf(x). So we have n n MJ(n) = m)Z y 3= A x- f(x2) -.f(x )/2. i=1 i=1 The right-hand side is a Riemann sum, so, as n goes to infinity, it will converge to the corresponding integral, while the left-hand side will converge to the moment MM of the original plate with respect to the x axis. This yields (9.9) M Xf(x)=2dx. A similar argument using horizontal stripes instead of vertical ones shows that (9.10) M = f(x) dx. Finally, now that the moments of P are known, it is straightforward to compute the coordinates of the center of mass of P. Indeed, the  172 9. FURTHER APPLICATIONS OF INTEGRATION center of mass is the unique point (x9, yg) with the property that if the entire mass A= fj f(x) dc of P is placed in that point, then the moments of this one-object system are identical to the moments of P. In other words, Mx = ygA and My =czgA. Therefore, zc=g Ms/A and yg = M/A, which means that formulas (9.9) and (9.10) imply the following theorem. THEOREM 9.6. Let f be a function such that f(x) ;> 0 if x E [a, b]. Let D be a domain whose borders are the vertical lines x = a and x = b, the horizontal axis, and the curve of the function f(x) = y. Let A(D) denote the area of D. Let xg and yg be the coordinates of the centroid of D. Then we have f (x)dx f [f(x)]2 d xc= and yg A(D f A ( D )a d9A ( D ) EXAMPLE 9.6. Find the coordinates of the centroid of the quarter of the unit circle that is in the northeastern quadrant. Note that if we asked the same question for the entire unit circle, the answer would obviously be that zg = yg = 0, since the centroid of any domain must be on all symmetry lines. If we asked the same question for the half of the unit circle that is in the northern half-plane, then zcc= 0 would clearly hold, since the vertical axis is a symmetry line of that semicircle. Solution: (of Example 9.6): Note that the domain D in question has a symmetry line, namely, the line determined by the equation x = y. So the centroid of D is on that line, that is, zg = yg. Therefore, it suffices to compute one of zg and yg. We have f(x) = v1 -cx2=y, so yg is somewhat easier to compute. Theorem 9.6 yields f .(1-x2)d yg = gr/4 1 2(cc3 7r 3 0 2 2 73 4 So the center of gravity of the quarter of the unit circle in the north- eastern quadrant is at (e, A) See Figure 9.9 for an illustration. D  62. APPLICATIONS TO PHYSICS AND ENGINEERING 173 Note that 4/(37) ~ 0.424. This makes perfect sense since this shows that the centroid of D is closer to the horizontal axis (the bottom of the quarter circle) than to the y = 1 line (the top of the quarter circle). That is reasonable, since the bottom of D is wider than the top of D, so it constitutes a larger portion of the total weight of D than the top of D. EXAMPLE 9.7. Find the centroid of the domain D whose borders are the vertical lines x = 0 and x = 1, the horizontal axis, and the graph of the function f(x) = x2=y. Solution: The domain D in question does not have a symmetry line, so we must use Theorem 9.6 to compute both of zc and yg. We have fa of (x) dz cg A(D) f x-"x2 dc 1 x4/4 0 1 x3/3 0 3 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 FIGURE 9.9. The centroid of a quarter of the unit circle.  174 9. FURTHER APPLICATIONS OF INTEGRATION and lfa j[f(x)]2 dx ygz A(D) f o(X4/2) dx 1/3 1 x5/10 __0 1/3 3 10 So the centroid of D is at (0.75, 0.3). This agrees with our intuition, since the bottom of D is larger than its top, and the left-hand side of D is smaller than the right-hand side of D. illustration. See Figure 9.10 for an 62.3. Exercises. (1) Find the centroid of the unit semicircle that lies in the northern half-plane. (2) Find the centroid of the unit semicircle that lies in the western half-plane. (3) Find the centroid of the plate whose borders are the lines x 0 and x = r/2, the graph of the function f(x) horizontal axis. sin x, and the (4) Find the centroid of the plate whose borders are the vertical lines x = 1 and x = 2, the horizontal axis, and the graph of the function f(x) = In X. 1.0 0.8 0.x2 0.6 y 0.4 0.2 0.2 0.4 0.6 0.8 1.0 FIGURE 9.10. Center of mass of the area defined in Ex- ample 9.7.  62. APPLICATIONS TO PHYSICS AND ENGINEERING 175 (5) Find the centroid of the plate whose borders are the vertical lines x = 1 and x= 2, the horizontal axis, and the graph of the function f (x) = ex. (6) Find the centroid of the plate whose borders are the vertical lines x = 0 and x= 1, the horizontal axis, and the graph of the function f (x) = x3. (7) Find the centroid of the plate whose borders are the vertical lines x = 0 and x = 1, the horizontal axis, and the graph of the function f (x) = x. (8) Find the centroid of the plate whose borders are the vertical lines x = 0 and x = 1, the horizontal axis, and the graph of the function f(x) =V/. (9) Find the centroid of the plate whose vertices are at (0, 0), (10, 10), and (20, 0). (10) Find the centroid of the (10, 10), and (50, 0). (11) Find the centroid of the (15, 0), (1, 1), and (8, 1). (12) Let ABC be any triangle. plate whose vertices are at (0, 0), plate whose vertices are at (0, 0), Let us assign objects of equal mass to A, B, and C. Let A1 denote the midpoint of the segment BC, let B1 denote the midpoint of the segment AC, and let C1 denote the midpoint of the segment AB. Prove that the lines AA1, BB1, and CC1 intersect in one point, centroid of the system ABC. (13) Keep the notation of the preceding exercise. Prove that the center of mass of ABC is the same as the centroid of A1B1C1, if we assign objects of equal weight to A1, B1, and C1. (14) An object consists of two squares. The first is the square with vertices (0, 0), (0, 2), (2, 0), and (2, 2), and the other is the square with vertices (0, 2), (1, 2), (0, 3), and (1, 3). The density of the material of the small square is twice the density of the material of the large square. Where is the centroid of this object? (15) Let n > 1 be a positive integer. Find the centroid M(n) of the plate whose borders are the vertical lines x = 1 and x = n, the horizontal axis, and the graph of the function f(x) = x-3. (16) Keep the notation of the preceding exercise. Describe the be- havior of the point M(n) as n goes to infinity. (17) How does the answer to the two preceding exercises change if f is replaced by g, where g(x) = x-2?  176 9. FURTHER APPLICATIONS OF INTEGRATION (x2(,1,pxx1)) . (x3,p(x3)) Number of Units x FIGURE 9.11. The demand function p(x) of a commodity. (18) Solve exercise 15 with f unchanged, but with the two vertical lines changed to x= 1/n and x= 1. Try to find an answer without additional integration. (19) Find the centroid of the plate whose borders are the horizontal lines y = 0 and y = 1, the vertical axis, and the graph of the function x= y. (20) Find the centroid of the plate whose borders are the horizontal lines y = 0 and y = 1, the vertical axis, and the graph of the functionx= =y + 1. 63. Applications to Economics and the Life Sciences 63.1. Consumer Surplus. Let us consider the problem of pricing some merchandise whose value is highly subjective; that is, it is worth more to some customers than to others. Examples of this could be tickets for various sporting events, air line tickets to vacation destinations, or popular books. Let p(x) be the demand function of this commodity. That is, p(x) is the price that will result in selling x units of the commodity. Lower prices usually lead to higher sales, therefore p(x) is usually a decreasing function as illustrated in Figure 9.11. The area under the graph of p represents the total revenue the company could possibly have, if it managed to charge each customer the maximum price that that customer is willing to pay. Indeed, if the highest amount anyone is willing to pay for one unit is p(zi), and x1 customers are willing to pay that price, then the revenue coming from these most enthused customers is xip(zi), which is the area of the domain under the graph of p that is between the vertical lines x =0  63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 177 and y = xi. We could continue in this way, noting that if the second highest price that some customers are willing to pay is p(x2), and there are x2 - x1 people who are willing to pay this price (not including those who are willing to pay even p(xi)), then the revenue from them will be (x2 - x1)p(x2). This is the area of the domain under the graph of p that is between the lines x = x1 and X = x2, and so on. If the seller decides to set one fixed price p(z), then the seller will sell z items, for a total revenue of zp(z) (the area of the rectangle R bordered by the two coordinate axes and the lines x = z and y = p(z)). This means that the customers who would have paid an even higher price for these goods have saved money. Besides losing that potential revenue, the seller also loses revenue by not getting any purchases from customers who were willing to pay some amount, but not p(z), for one unit. Let xc be the number of items that the seller can sell at the lowest price at which the seller is still willing to sell these items. It is a direct consequence of the above discussion that the total amount saved by all customers who bought the item at p(z) dollars is the area under the curve of p but above the rectangle R, that is, n (9.11) (p(Xi) - p(z))(Xi - zi_1) i=1 If the number n of prices at which various customers are willing to buy goes to infinity, then the Riemann sum in (9.11) approaches the definite integral (9.12) CS (p(x)- p(z)) dx. In economics, CS is called the customer surplus for the given commod- ity. Similarly, the integral p(x) dx is the amount of missed revenue, that is, the money the company could have received from buyers who found the product too expensive. Note that this is the area of the domain under the graph of p, but on the right of R. E XAMPLE 9.8. Tickets for a certain flight are normally priced at $300, and in an average month, 500 tickets are sold. Research shows that, for every $10 that the price is reduced, the number of tickets sold  178 9. FURTHER APPLICATIONS OF INTEGRATION goes up by 20. Find the demand curve and compute the consumer surplus for these tickets if the price is set at $240. Solution: If the airline wants to sell c tickets, then the price that the airline needs to charge is 030010 c-500 x -500 p(x) = 300 - 10 "-50= 300 - . 0 20 2 Indeed, in order to sell x-500 extra tickets, the airline needs to decrease its price by $10 for each 20-pack of extra tickets. If the price is set at p(z) = $240, then the last displayed equation shows that z = 60. Now we can apply formula (9.12) to compute that the customer surplus is CS = (p(x) - 240) dz f60 x - 500 f 60( - dz 0 2 /60x = 0 310 - -dz = 17, 700. So customers would save a total of $17,700 in an average month if the price of the tickets were set at $240. D 63.2. Survival and Renewal. EXAMPLE 9.9. Let us assume that there are currently 30, 000 people in the United States who have a certain illness. Let us also assume that we know that the fraction of that population who will still have the illness t months from now is given by the function f(t) = e-0.051. We also know that every month 1000 new patients will get the illness. How many people in the United States will have the illness in 20 months? Solution: Clearly, f(20) = e-1 = 0.368 of the people who currently have the illness will still have it 20 months from now. Now we have to compute the number of people who will get the illness between now (t = 0) and 20 months from now (t = 20) and will still be ill 20 months from now. Subdivide the interval [0, 20] into n equal subintervals using the points Set 20/in =At. Then, for all i, there are 1000 - At people who will get the illness during the time period [ti_1, ti). That means that 20 months  63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 179 from now, in other words, approximately 20 -ti_1 months from getting the illness, the fraction of them who will still have the illness will be f(20 - ti_1). So their number will be about 1000 - At.- f(20 - ti_1). Summing over all allowed values of i, we get that the total number of people in the United States who will have the illness 20 months from now is 30, 000- f(20) + 1000. At- f(20 - ti_1). i=1 We recognize that the above sum is a Riemann sum, so, infinity, the above expression converges to D = 30, 000 - f (20) + 1000f (20 - t) dt 020 = 30, 000 - e-1 + 1000] e.05t-1 dt 02 as n goes to 11, 036.38 + 12, 642.41 23, 679. So 20 months from now, 23,679 people in the United States will still have the illness. D Note that the result of the previous example shows that the number of people in the United States who have the illness will decrease during the next 20 months. Try to find an intuitive explanation for that fact that does not involve integration. 63.3. Exercises. (1) How much money would customers save in total if the flights discussed in Example 9.8 were sold for $210 each? (2) Consider the flights discussed in Example 9.8 and assume that the tickets are sold at their original price of $300. Compute the missed revenue for the airline. (3) Consider the flights discussed in Example 9.8 and assume that the tickets are sold at a reduced price of $270. Compute the missed revenue for the airline. (4) A demand curve is given by p(x) = 600/(x + 10). Find the consumer surplus when the selling price is $20. (5) A demand curve is given by p(x) = 1000/(x + 40). Find the missed revenue when the selling price is $25. (6) We deposit P dollars into a bank account at an annual interest rate of r%. Let us assume that the interest is compounded n times a year. Find a formula for our account balance after t  180 9. FURTHER APPLICATIONS OF INTEGRATION years. Then use that result to compute our account balance after t years if the interest is compounded continuously. (7) We deposit $100,000 into a bank account, where it will earn an annual interest of 5%. The interest is compounded contin- uously. Each year, we deposit $2000 into this same account in a continuous manner. What will our account balance be in 15 years? (8) Consider the bank account of the preceding exercise, with the sole difference that, after 10 years, we increase our annual deposit to $3000. What will our account balance be in 15 years? (9) We deposit $100,000 into a bank account, where it will earn an annual interest of 5%. The interest is compounded continu- ously. Each year, we withdraw a total of $4000 in a continuous manner. What will our account balance be in 20 years? (10) A country currently has a population of 80 million people and a natural growth rate of 1.5%. The natural growth g of the population in a given year is computed as the difference be- tween the number of births and the number of deaths in that year, while the natural growth rate for that year is g divided by the size of the population at the beginning of that year. Let us assume that each year 1.1 million people emigrate from this country. If the current trends continue, how large will the population of this country be in 20 years? (11) A country currently has a population of 100 million people and a natural growth rate of 1%. See the preceding exercise for the definition of natural growth rate. Let us assume that each year 0.5 million people emigrate from this country. If the current trends continue, how large will the population of this country be in 25 years? (12) A country currently has a population of 80 million people and a natural growth rate of -0.5%. Let us assume that each year 0.35 million people immigrate to this country. If the current trends continue, how large will the population of this country be in 20 years?  63. APPLICATIONS TO ECONOMICS AND THE LIFE SCIENCES 181 (13) Tickets to a certain section of the arena for a basketball game usually cost $50. This results in the sales of 1000 tickets. For every dollar that the price is dropped, the number of tickets sold goes up 1%. Find the demand function for these tickets. (14) Compute the consumer surplus for the tickets discussed in the previous exercise if the tickets are sold at $40. (15) Let S(x) be the supply function for a certain commodity. That is, S(x) is the price that one unit of the commodity has to cost in order to attract enough sellers to provide x units for sale. Note that S(x) is an increasing function, since a higher price is needed to attract more sellers. Let us assume that the units are sold at a fixed price T - S(t). That means that the sellers who would be willing to sell at a lower price are making a profit. The total amount of the profit made by all sellers is sometimes called the producer surplus. Prove that the producer surplus for this commodity can be computed by the formula (T - S(x)) dx. (16) Let us assume that the supply function for a certain commod- ity is given by S(x) = x2+2x. Compute the producer surplus (which we defined in the preceding exercise) if the selling price is set at $24. (17) How does the producer surplus change in the situation de- scribed in the previous exercise if the selling price is raised to $35? (18) Let us assume that the supply function for tickets to flights between two given cities is given by S(x) = x2 + 10x + 5. Compute the producer surplus if the selling price is set at $205. (19) The gates of a football stadium open at 3 p.m. After t hours, fans enter the stadium at a rate of f (t) = -t3 + 20t2 + 30t + 64 people per hour, where t = 0. How many fans enter the stadium between 5 p.m. and 6 p.m.? (20) There are currently 50,000 people in the United States who take a certain medication. The fraction of that population who will still take that medication t months from now is given by the function f(t) e e-o.0 Every month, the medication will be taken by 500 new patients. How many people in the United states will take the medication a year from now?  182 9. FURTHER APPLICATIONS OF INTEGRATION 64. Probability The word "probability" is often used in informal conversations, even if it is sometimes not clear what the speaker means by that word. It turns out that there are two distinct concepts of probability. These two concepts complement each other in that they are applicable in different circumstances, and use very different methods. 64.1. Discrete Probability. Let us say that we are tossing a fair coin four times. What is the probability that we will get at least three heads? This is a situation in which the event that we study, that is, the sequence of four coin tosses, has only a finite number of outcomes. Indeed, there are 24 possible outcomes, since each coin toss has two outcomes (heads or tails), and the coin is tossed four times. Among these 16 possible outcomes, five are favorable outcomes, namely, HHHH, HHHT, HHTH, HTHH, and THHH. Further- more, each single outcome (favorable or not) is equally likely to occur, since the coin is fair, and the result of each coin toss is equally likely to be heads or tails. In this situation, that is, when the number of all possible outcomes is finite, and each outcome is equally likely to occur, define Number of favorable outcomes (9.13) Probability of event = Number of al outcomes Number of all outcomes' which, in our example, shows that the probability of getting at least three heads is 5/16. Probabilities defined by formula (9.13) are called discrete proba- bilities. The formula is applicable only when the number of possible outcomes is finite. If we want to apply this formula in complicated situations, we need advanced techniques to count the number of all outcomes and the number of favorable outcomes. The fascinating dis- cipline studying those techniques is called enumerative combinatorics, and will not be discussed in this book. 64.2. Continuous Probability. Let us say we want to know the probabil- ity that during the next calendar year the city of Gainesville, Florida, will have more than 40 inches of precipitation, or we want to know the probability that Gainesville will have less than 50 inches of precipita- tion, or that Gainesville will have at least 42 but at most 48 inches of precipitation. In this case, formula (9.13) is not applicable, since both the number of favorable outcomes and all outcomes is infinite. Indeed, the amount  64. PROBABILITY 183 of precipitation in Gainesville next year can be any nonnegative real number. Furthermore, not all outcomes are equally likely. Receiving very little or very much precipitation is far less likely than receiving close to the usual amount. We need a totally different approach. Our approach, while different from the one in the previous section, shares some of the most important features of that approach. For instance, the probability of an event that is certain to happen will be 1, while the probability of an event that never happens will be 0. Similar situations occur when we want to know the probability that a certain device will work for more than t years, or that a randomly selected person weighs more than p pounds but less than q pounds, or that the blood pressure of a randomly selected person is below a given value. The quantities mentioned here are called random variables. We would like to define the probability F(a) that the amount X of precipitation in Gainesville next year will be at most a inches. This probability will sometimes be denoted by P(X < a). While we do not yet know how to compute F(a), we know that the function F has to satisfy the following requirements: (1) We will have F(ai) F(a2) if ai < a2. In other words, F is increasing. Indeed, if X < ai, then X < a2 since ai < a2. (2) We will have lima- F (a) = 1, since the amount of precipita- tion is always finite. (3) The function F(a) is a continuous function, since a little bit of change in a will only mean a little bit of change in F(a) P(X < a). We sometimes refer to this fact by saying that X is a continuous random variable. Note that the function F is called the distribution of the random variable X. It can be proved that if F has all these properties, then there exists a unique function f : R - R that has the following properties: (a) For all a E R, we have f_", f (x) dc = F(a) = P(X < a). (b) The equality f f (x) dc = 1 holds. (c) For all real numbers x, the inequality f(x) ;> 0 holds. If f is the unique function described by the three properties above, then f is called the probability density function, or simply density func- tion, of the continuous random variable X. Note that property (a) above implies that, for all real numbers a < b, the equality (9.14) P(a 5X b) =ff (c) d  184 9. FURTHER APPLICATIONS OF INTEGRATION y = f (x) a b FIGURE 9.12. The probability P(a < X < b) as an area. holds. In other words, P(a X b) is equal to the area of the domain between the graph of the density function f, the horizontal axis, and the vertical lines x = a and x = b. See Figure 9.12 for an illustration. Indeed, we have P(a X 1. Verify that f is indeed a density function and compute the proba- bility P(0.3 < X < 0.6). Solution: In order to see that f is indeed a density function, we must verify that its definite integral, taken over the entire line of real numbers, is equal to 1. This is not difficult, since f(x) interval [0, 1]. This leads to 0 outside the f(x) dx 6 (- 2) )dx 1 0 6 1 1. So f is indeed a valid density function.  64. PROBABILITY 185 1.4 1.2 \ 1.0 \4 0.8 0.6- 0.4 0.2 -1.0 -0.5 0.5 1.0 1.5 2.0 FIGURE 9.13. The graph of the function f in Example 9.10. We can use formula (9.14) to compute the requested probability. We get /0.6 P(0.3 < X < 0.6) =f(x c) dxc 0.3 =6(cc-cc2) dcc 0.6 00.3 = 6(0.108 - 0.036) = 0.432. 64.2.1. Exponential Distribution. Consider the following density func- tion. Let A be a positive real number and let {0 ifx <0, (9.15) e- if 0 x . We see that f is a decreasing function on the interval [0, oc). Figure 9.14 shows how the speed at which f decreases depends on the param- eter A. It turns out that this density function is a very frequently occurring one. Therefore, it has a name. It is called the exponential density func- tion with parameter A. Using the right constant A and under the right circumstances, it can be used in many scenarios, typically connected to waiting times. For instance, it could be used to measure the proba- bility that, given a starting moment, a given cell phone will ring in less than t minutes, or that, at a given location, it will start raining in less  186 9. FURTHER APPLICATIONS OF INTEGRATION A = 2.0 2.0 A = 1.5 1.5 A = 1.0 1.0 0.5 -1 0 1 2 3 FIGURE 9.14. A = 1 (red), A = 1.5 (blue), A = 2 (orange). than h hours, or that, given a random store, a customer will enter in s seconds. The exponential density function will give a good approxi- mation to compute these probabilities if the mentioned processes take place at a roughly constant rate. That is, we should choose a part of the day when that given cell phone receives calls at roughly constant frequency, a season when it rains at that location at roughly constant time periods, or a time of day when customers enter that store at a roughly constant rate. EXAMPLE 9.11. The probability that a certain kind of new refrig- erator will need a major repair in x years is given by the exponential density function with parameter A = 1/9. What is the probability that a new refrigerator will not need a major repair for 10 years? Solution: First, we compute the probability that the refrigerator will need a major repair in 10 years. Let X denote the number of years passing before the first major repair is needed. Then that probability is P(X < 10) = f (x) dx /10 1 0 1 - 1 0 -- 1 __ e-1l0/9 = 0.671.  64. PROBABILITY 187 Therefore, the probability that the refrigerator will not need a major repair in 10 years is P(X > 10) = 1 - 0.671 = 0.329. D 64.2.2. Mean. If we want to compute the average weight of a per- son selected from a given population of n people, we can simply take the weights ai, a2, ... , an of those people and compute their arithmetic mean, or average, that is, the real number A=ai+a2+---+anh n This could take a very long time if n is a very large number. If the data are given in a more organized form, we may be able to save some time. In particular, if we know that there are b1 people in the population whose weight is x1, there are b2 people whose weight is x2, and so on, then we can compute the average weight of the population as (9.16) A b1x1 + b2x2 + ... + bkxk bl+b2+---+bk since this fraction is the total weight of the population divided by the number of people in the population. Now note that p = b2/(bi + b2+- -"-+ bk) is just the probability that a randomly selected person of this population has weight x2. Therefore, (9.16) can be written as (9.17) P1x1+P2x2+---+Pkk. Theoretically, the weight of a person can take infinitely many val- ues since the measuring scale can be always be more precise. It is not difficult to prove that, as k goes to infinity, the sum in (9.17) will turn into a Riemann sum, and the weights x2 will be measured by a contin- uous random variable X, and the probabilities pi will be expressible by the definite integrals of a density function. This leads to the following definition. DEFINITION 9.1. Let f be the density function of the continuous random variable X. Then the value of ,p(X) = t f (t) dt is called the average value or mean or expected value of X. E XAMPLE 9.12. Let X be the continuous random variable whose density function is the exponential density function with parameter A that we defined in (9.15). Then pu(X ) =1/A.  188 9. FURTHER APPLICATIONS OF INTEGRATION Solution: Using Definition 9.1, we have p(X) f tf(t)dt =fAte-^tdt 00 =0-te-At - e 0 1 A In other words, the parameter and the mean of an exponential dis- tribution are reciprocals of each other. In view of this, we can reformu- late the result of Example 9.11 as follows. If the average time before a new refrigerator needs a major repair is 9 years, then the probability that a refrigerator will not need a major repair for 10 years is 0.329. 64.2.3. Normal Distribution. Let p be a real number and let o be pos- itive real number. Consider the density function 1 (x _ )2 f W(-)2= exp (-2 . The distribution defined by this density function is called the nor- mal distribution with parameters y and o. This distribution is denoted by N(p, o-). In particular, if p,= 0 and o-= 1, then the obtained distribution N(0, 1) is called the standard normal distribution. Plotting the graph of f for various values of p and o, we see that the graph has a bell curve; its highest point is reached when x = p, and it increases on the left of that and decreases on the right of that. The smaller the value of o, the steeper is the rise and fall of the graph of f. See Figure 9.15 for an illustration. It can be proved that p is precisely the mean of N(p, o-). The constant o is called the standard deviation of N(p, o-). It measures how spread out the values of our variable X are. (The precise definition is that o is the square root of the mean of (X - p)2.) Many scenarios are modeled by a normal distribution, such as test scores, athletic results, or annual snowfall at a given location. E XAMPLE 9.13. In an average year, Northtown gets 10 feet of snow, with a standard deviation of 2 feet. What is the probability that, in a random year; Northtown gets between 9 and 12 feet of snow if snowfall is modeled by a normal distribution?  64. PROBABILITY 189 64. PROBABILITY 189 FIGURE 9.15. or (green), and o = = 1 (red), 2.5 (blue). or 1.5 (orange), ou 2 Solution: Let X denote the snowfall in a random year in Northtown. We need to find the probability P(9 < X < 12). As snowfall is modeled by a normal distribution, be the distribution N(10, the given parameters imply that that must 2). Therefore, by formula (9.14), we have P(9 < X < 12) I exp J9 2 2 0.5328, ( S ) dx where the definite integral has to be computed by some approximation method (or a software package) since e-- has no antiderivative among elementary functions. D 64.3. Exercises. (1) For which value of c will f(x) [0, 1]? (2) For which value of c will f(x) on [-oo, oo]? cx4 be a density function on S1+4x2 be a density function (3) Consider the value of c that makes the function f of the pre- ceding exercise a density function. Using that value c, find P(-2 < x < 2). (4) Prove that the exponential density function, which we defined in (9.15), is indeed a density function. (5) Let X be a random variable whose density function is 0 outside the interval [0, 1] and satisfies f(x) 2x for x E [0, 1]. Prove that f is indeed a density function, and compute the mean of X.  190 9. FURTHER APPLICATIONS OF INTEGRATION (6) Let us say that the lifetime of a bicycle tire (measured in months) has an exponential distribution with A = 7. What is the probability that a tire will last between 5 and 8 months? (7) What is the probability that the tire of the preceding exercise will last more than 14 months? (8) Find the number a of months such that there is exactly 1/2 probability that the tire of the previous exercise will last at least a months. (9) Let us assume that the lifetime of a given product, measured in months, has an exponential distribution with parameter A. What is the probability that the product will last for more than twice its expected lifetime? (10) Considering the product of the preceding exercise, what is the probability that it will last at least k times its expected life- time, where k is any positive real number? (11) Considering the product of the preceding exercise, what is more likely, that its lifetime will be at most A - 1 months or that it will be at least A + 1 months? (12) The average score on an exam is 100 points. In order to pass, a student cannot be more than 2 standard deviations below the average. If the scores have a normal distribution with a standard deviation of 6, how large a fraction of the students will pass the exam? (13) Using the conditions of Example 9.13, what is the probability that Northtown will get less than 5 feet of snow in a given year? (14) Use a software package to prove that the standard normal distribution function is indeed a distribution function. (15) The height of the adult male population of the Netherlands has a normal distribution with a mean of 73 inches and a standard deviation of 3 inches. What percentage of the adult male population is more than 75 inches tall? (16) Consider the adult male population of the Netherlands given in the preceding exercise. What percentage of that population is between 70 and 75 inches tall? (17) Prove that the mean of the standard normal distribution is indeed 0. (18) Let X be the random variable that counts the goals scored by an offensive soccer player of a certain elite league during an entire season. An offensive player is considered exceptional if the number of goals he scores exceeds the average of all  64. PROBABILITY 191 offensive players by at least 3 standard deviations. Let us say that X has distribution N(33, 3). What percentage of offensive players is considered exceptional? (19) For the standard normal distribution, find the probability that the value of the random variable is within 3 standard devia- tions of the mean. (20) For any normal distribution, find the probability that the value of the random variable is within 3 standard deviations of the mean.   CHAPTER 10 Planar Curves 65. Parametric Curves Every point in a plane can be defined as an ordered pair of real num- bers (x, y) called the rectangular or Cartesian coordinates. A graph of a function f is the set points in a plane whose coordinates satisfy the condition y = f(x). The graph gives a simple example of a planar curve. More generally, a planar curve can be defined as the set of points whose coordinates satisfy the condition F(x, y) = 0 called the Carte- sian equation of a curve. In many instances, an equation F(x, y) = 0 has multiple solutions for every given x. For example, consider the circle of unit radius: (10.1) 2 + y2 1 _-> y=+ 1 -cc2, E[-1, 1]. The two solutions represent two semicircles. The graph y = 1 - x2 is the semicircle above the x axis, while the graph y= - 1 - x2 is the semicircle below the x axis. The union of the two graphs is the full circle. This example shows a deficiency in describing planar curves by the graph of a function because the curves cannot always be represented as the graph of a single function. On the other hand, (10.1) admits a different solution: (10.2) X2 + y2 1 == =c cost , y = sint, t E [0,27r], which immediately follows from the trigonometric identity cos2 t + sin2 t = 1 for all values of t. This representation means that a point of the coordinate plane is assigned to every value of t E [0, 27] by the rule (x, y) = (cost, sin t). The coordinates of points of the circle are functions of a third variable called a parameter. As t changes, the point (cos t, sin t) traces out the circle of unit radius centered at the origin in the plane. The parame- ter t has a simple geometrical interpretation. It is the angle counted counterclockwise from the positive x axis to a ray from the origin on which the point (cos t, sin t) lies. This observation admits a natural generalization. DEFINITION 10.1 (Parametric curves). Let cc(t) and y(t) be contin- uous functions on [a, b]. A parametric curve in the coordinate plane is 193  194 10. PLANAR CURVES 194 10. PLANAR CURVES FIGURE 10.1. y Circle: x(t) cos t, y(t) sin t. y(b) y(a) y(t) x x(a) x(t) x(b) FIGURE 10.2. Parametric curve. As t increases from a to b, the point (x(t), y(t)) traces out a curve in the xy plane. the set of points satisfying the conditions, called the parametric equa- tions, X = x(t), y = y(t) , t c [a, b]. The points (x(a), y(a)) and (x(b), y(b)) are called the initial and ter- minal points of the curve, respectively. The graph of a function f is a particular example of a parametric curve: x = t, y = f(t).  65. PARAMETRIC CURVES 195 Y P2 t = ,r P4 t =27r 0 37r NN P3 t= 2 FIGURE 10.3. The spiral x = t cost, y = t sin t, t E [0, 27]. The distance from the origin R = /2 + y2 = t increases linearly as the angle t, counted counterclock- wise from the positive x axis, increases from 0 to 27r. Parametric curves are common in everyday life. The position of a particle in a plane is defined by its rectangular coordinates (x, y) in the plane. When the particle moves, its coordinates become functions of time t so that the parametric curve x = x(t), y = y(t) is the trajectory of the particle. EXAMPLE 10.1. Sketch the curve with the parametric equations x = t cos t, y = t sin t, t E [0,27]. Solution: A basic approach to visualize the shape of a parametric curve is to plot points (z(tk), y(tk)), k = 1, 2, ..., n, corresponding to successive values of t: ti < t2 < ... < tn. For n large enough, a fairly good picture of the curve emerges. This approach can be followed here, and the reader is advised to do so, for example, tk = 27k/n, k = 0, 1, ..., n. However, there is another way in part specific to this very problem. Note that x2 + y2 = t2 so that the equations may be written in the form of the parametric equations of the circle cc= R cos t, y =R sin t, where the radius increases linearly with t, R =R(t) t. The parameter t can be viewed as the angle between a ray from the origin and the positive cc axis counted counterclockwise. Thus, the curve has the following interpretation. As the point (cc(t), y(t))  196 10. PLANAR CURVES 196 10. PLANAR CURVES rotates about the origin, the distance R between it and the origin increases linearly with the rotation angle. Such a motion occurs along an unwinding spiral. In the interval t E [0, 27], the spiral makes one full turn from the initial point (0, 0) to the terminal point (2w, 0). Q 65.1. Parametric Curves and Curves as Point Sets. If a curve is defined as a point set in the coordinate plane, for example, by the Cartesian equation F(x, y) = 0, then there are many parametric equations that describe it. For example, the circle (10.1) may also be described by the following parametric equations: (10.3) X 2 + y2 _-> zX=cos(3T) , y=-sin(3T) , TE [0, 2w]. What is the difference between (10.2) and (10.3)? First, note that, as the parameter t in (10.2) increases, the point (cost, sin t) traces out the circle counterclockwise (the initial point (1, 0) moves upward as y sint > 0 for 0 < t < w/2). In contrast, the point (cos(3r), - sin(3T)) does so clockwise with increasing T (the initial point (1, 0) moves down- ward as y = - sin(3T) for 0 < T < w/6). Second, as t ranges over the interval [0, 2w], the point (cos t, sin t) traces out the circle only once, while the point (cos(3r), - sin(3T)) winds about the origin three times because the period of the trigonometric functions involved is 27/3, so the point returns to the initial point (r = 0) three times when T= 2w/3, T = 4w/3, and T = 6/3 = 2w. Third, there is a relation between the parameters t and T: t = -3T. This example illustrates the main differences between curves defined as a point set by the Cartesian equation F(x, y) = 0 and parametric curves. " A parametric curve C is oriented; that is, the point (x(t), y(t)) traces out C in a particular direction (from the initial to the terminal point). " A parametric curve may repeat itself (or some of its parts) multiple times. " Parametric equations describing the same point set in the plane differ by the choice of parameter; that is, if (x(t), y(t)) and (X(T), Y(T)) trace out the same point set C in the plane, then there is a function g(T) such that X(T) = x(g(T)) and Y(T) =y(g(T)). The change of the parameter t =g(T) is called a reparameterization of a curve C. A good mechanical analogy is the motion of cars along a road. The road is a point set. A parametric curve describes the actual motion of a  65. PARAMETRIC CURVES 197 particular car along the road. Naturally different cars moves differently along the very same road. EXAMPLE 10.2. Suppose a curve C is described by the parametric equations x = x(t), y = y(t) if t E [a, b]. Find the parametric equations of C such that the curve is traced out backward, that is, from the point (x(b), y(b)) to (x(a), y(a)) (the initial and terminal points are swapped). Solution: One has to find a new parameter T, t = g(r), such that g(b) = a and g(a) = b. When T increases from a to b, the parameter t decreases from b to a, and the sought-after parametric equations are obtained by the composition X(T) = x(g(T)) and Y(T) = y(g(T)). The simplest possibility is to look for a linear relation between t and T, g(T) = c + dT. The coefficients c and d are fixed by the conditions g(a) = b or b = c + da and g(b) = a or a = c + db. Therefore, by subtracting these equations, b - a = (c + da) - (c + db) = -(b - a)d or d = -1. By adding these equations with d = -1, b+a = (c-a)+(c-b) or c = a + b. Hence, t = (a + b) - T, so that the parametric equations of C with reversed orientation are Xz= x(a + b - r), y = y(a + b - r), TFEc[a, b]. For example, if C is the circle oriented counterclockwise as in (10.2), then the same circle oriented clockwise is described by x= cos(27 - T) = cos T, y = sin(27r-T) = -5sin T, T E [0,27r]. D 65.2. The Cycloid. The curve traced by a fixed point on the circum- ference of a circle as the circle rolls along a straight line is called a cycloid (see Figure 10.4). To find its parametric equations, suppose that the circle has a radius R and it rolls along the x axis. Let the fixed point P on the circumference be initially at the origin so that the center of the circle is positioned at the point (0, R) (on the y axis). Let CP denote the straight line segment between the center of the circle C and P. Initially, CP is perpendicular to the x axis. As the circle rolls, the segment CP rotates about the center of the circle. Therefore, it is natural to choose the angle of rotation 0 as a parameter. The coordi- nates of P are functions of 0 to be found. If the circle rolls a distance D so that its center is at (D, R), then the are length RB of the part of the circle between F and the touch point T has to be equal to D, that is, D =RO. Let Q be a point on the segment CT such that FQ and CT are perpendicular. Consider the right-angled triangle CFQ. Its hypotenuse CF has length |CPF R, and the lengths of its catheti are |CQ| |CFP cosO0 R cos06 and |PQ| |CFP sinO0 R sin 0. Let  198 10. PLANAR CURVES C(R G,R) R 1 p ......................... Q T >x 00 OR FIGURE 10.4. Definition of a cycloid. A disk of radius R is rolling along the x axis. A curve traced out by a fixed point on its edge is called a cycloid. FIGURE 10.5. Overall shape of a cycloid. (x, y) be coordinates of the point P. The parametric equations of the cycloid are x= D - PQ RO-RsinO =R(O-sin). y= R - CQ =R R-Rcosoz= R(1-cos). It looks like an upward arc over the interval 0 < x < 27R, with max- imal height Ymax = 2R (0 = r/2), and the arc repeats itself over the next interval of the length of circumference 27R and so on. Remark. In 1696, the Swiss mathematician Johann Bernoulli posed the brachistochrone problem: Find the curve along which a parti- cle will slide (without friction) in the shortest time (under the influence of gravity) from a point A to a lower point B not directly beneath A. The particle will take the least time sliding from A to B if the curve is a part of an inverted arch of a cycloid.  65. PARAMETRIC CURVES 199 65.3. Families of Curves. Different values of R define different cycloids. In general, if the parametric equations contain a numerical parameter, then the parametric equations define a family of curves; each family member corresponds to a particular value of the numerical parameter. EXAMPLE 10.3. Investigate the family of curves with the parametric equations x = acost, y = bsin(2t), t E [0,2w], where a and b are positive numbers. Solution: Consider first the simplest case a = b = 1. The function x(t) = cos t has a period of 2w, and y(t) = sin(2t) has a period of w. The initial point is (x(0), y(0)) = (1, 0). As t increases, the point moves upward so that x(t) decreases (becomes less than 1), while y(t) increases, reaching its maximum value 1 at t =w7/4. After that, y(t) begins to decrease, while x(t) continues to decrease. At t =w7/2, the point arrives at the origin and passes through it into the third quadrant so that x(t) and y(t) continue to decrease. When t = 3w/4, y(t) attains the minimum value -1 and begins to increase for t > 3w/4, while x(t) =cost is still decreasing toward its minimal value -1, which is reached at t =w7, and the curve crosses the x axis moving into the second quadrant. In the second quadrant w < t < 3w/2, x(t) increases toward 0, while y(t) first reaches its maximum value 1 at t =w7 + w/4 (the curve touches the horizontal line y = 1) and then decreases to 0. The curve passes the origin again at t = 3w/2 and moves into the fourth quadrant, where it again touches the horizontal line y = -1 at t = 3w/2 + w/4 and at t = 2w it arrives at the initial point. The shape of the curve resembles the infinity sign (oc) embedded into the square bounded by the lines y =+1 and x= +1 so that it touches each of the horizontal sides y =+1 twice and each of the vertical sides x =+1 once. If a and b are arbitrary, the transformation x - ax stretches (a > 1) or compresses (a < 1) any geometrical set horizontally in the coordinate plane. The transformation y -- by does the same but in the vertical direction. So the family of curves consists of curves of the o0 shape stretched to fit into the rectangle bounded by the lines x = ta andy =+b. 65.4. Exercises. In (1)-(9), sketch the curve by plotting its points. Include the arrow showing the orientation of the curve. Eliminate the parameter to find a Cartesian equation of the curve.  200 10. PLANAR CURVES 200 10. PLANAR CURVES (1) x =1+ 2t , y = 3 - t (2) x = 1 t, y = 2 - t (3) x =t2 , y= t3 (4) x = 1+2et , y = 3 - et (5) x = cosh t , y =sinht (6) x = 2 sin t, y = 3 cos t (7) x = 2 - 3 cos t , y = -1+ sint (8) x = cos t, y = sin(4t) (9) x = t2 sin t , y = t2 cos t In (10)-(12), sketch the family of parametric curves a2 -b2 a2 -b2 (10) x= a cos3 t, yb sinst ab (11) x= a cosh2 t, y= b sinh2 t (12) x = a(sinht - t) , y = a(cosht - 1) In (13)-(16), find parametric equations of the curve defined by a Carte- sian equation and sketch it. Investigate the dependence of the shape of the curve on the parameter a. (13) x2/3 _+ 2/3-_ a2/3 (14) x2 + y2 atan-(y/x) (15) x3 +y3 = 3axy (16) (x2 + y2)2 = a2(x2 -Y2) Hint: Put y = tz in (15) and y= xtant in (16). (17) The curves x = a sin(nt), y = b cos t, where n is a positive inte- ger, are called Lissajous figures. Investigate how these curves depend on a, b, and n. (18) Consider a disk of radius R. Let P be a point on the disk at a distance b from its center. Find the parametric equations of the curve traced out by the point P as the disk rolls along a straight line. The curve is called a trochoid. Are the equations well defined if b > R. Sketch the curve for b < R, b = R, and b > R. (19) The swallowtail catastrophe curves are defined by the para- metric equations x = 2ct - 4t3, y -ct2 + 3t4. Sketch these curves for a few values of c. What features do the curves have in common? How do they change when c increases? (20) A hypocycloid is the curve traced out by a point on the circle that is rolling along a fixed circle so that it remains within the latter. Find parametric equations of the hypocycloid if the radius of the moving circle is a > b, where b is the radius of the fixed circle. Sketch the curve if the ratio b/a is an integer nm 2 and nm 4. (21) An epicycloid is the curve traced out by a point on the circle that is rolling along a fixed circle so that it remains out of  66. CALCULUS WITH PARAMETRIC CURVES 201 the latter. Find parametric equations of the epicycloid if the radius of the moving circle is a and b is the radius of the fixed circle. Sketch the curve if the ratio a = b. 66. Calculus with Parametric Curves 66.1. Tangent Line to a Parametric Curve. Consider a parametric curve x = x(t), y = y(t), where the functions x(t) and y(t) are continu- ously differentiable and the derivatives x'(t) and y'(t) do not vanish simultaneously for any t. Such parametric curves are called smooth. THEOREM 10.1 (Tangent Line to a Smooth Curve). A smooth para- metric curve x = x(t), y = y(t) has a tangent line at any point (xo, yo), and its equation is (10.4) x'(to)(y - yo) - y'(to)(x - zo) = 0, where (xo, yo) - (x(to),y(to)). PROOF. Take a point of the curve (xo, yo) = (x(to), y(to)) corre- sponding to a particular value t = to. Suppose that x'(to) / 0. Then, by the continuity of x'(t), there is a neighborhood IS = (to - b, to + b) for some b > 0 such that x'(t) / 0 for all t E Ib; that is, the de- rivative is either positive or negative in Is. By the inverse function theorem (studied in Calculus I), there is an inverse function t = f(x) that is differentiable in some open interval that contains xo. Substi- tuting t = f(x) into the second parametric equation y = y(t), one obtains that near the point (xo, yo) the curve can be represented as a part of the graph y = F(x) such that yo = F(xo). The function F is differentiable as the composition of two differentiable functions. The derivative F'(xo) determines the slope of the tangent to the graph, and the equation of the tangent line reads (10.5) y = yo + F'(zo)(X - zo). By construction, y = F(x) -- y(t) = F(x(t)) for all t E Is. Differentiation of this equation with respect to t by means of the chain rule yields x,(t)Yx (to) Substituting this equation into (10.5) , the latter can be written in the form (10.4). If z'(to) =0, then y'(to) / 0 by the definition of a smooth curve so that there is a differentiable inverse t =g(y) and  202 10. PLANAR CURVES 202 10. PLANAR CURVES hence x = G(y) = x(g(y)). Similar arguments lead to the conclusion that the tangent line to the graph x = G(y) has the form (10.4). The details are left to the reader as an exercise. D The rule for calculating the slope of the tangent line can also be obtained by means of the concept of the differential. Recall that the differentials of two related quantities y = F(x) are proportional: dy = F'(x) dx. On the other hand, x= x(t), y = y(t) and therefore dx= x'(t) dt and dy =y'(t) dt. Hence, , dg dy y'(t) F'(x)=dx-=dam= y'(t) dt These manipulations with differentials are based on a tacit assumption that, for a smooth curve x= x(t), y = y(t), there exists a differentiable function F such that y = F(x). In the proof of the tangent line the- orem, this has been shown to be true as a consequence of the inverse function theorem. The use of the differentials establishes the following helpful rules to calculate the derivatives: d d 1 dd d 1 d =da =and -=- = - dx and(t)ddttdgd y'(t) dt 66.2. Concavity of a Parametric Curve. The concavity of a graph y = F(x) is determined by the sign of the second derivative F"(x). If F"(x) > 0, the graph is concave upward, and it is concave downward if F"(x) < 0. If y(t) and x(t) are twice differentiable, then the concavity of the curve can be determined by the sign of the second derivative d2y ddy 1 d {dy /) y"x' - x"y' d2 x dx dx x' dt \dxJ x' - (' EXAMPLE 10.4. A curve C is defined by the parametric equations x = t2, y = t3 - 3t. (i) Show that C has two tangent lines at the point (3,0). (ii) Find the points on C where the tangent line is horizontal or vertical. (iii) Determine where the curve is concave upward or downward. Solution: (i) Note that y(t) t(t2 - 3) 0 has three solutions t 0 and t = tv/. But the curve has only two points of intersection with the x axis, (0, 0) and (3, 0), because x(+v/5) =3; that is, the curve is self-intersecting at the point (3, 0). This explains why the curve may have two tangent lines. One has z'(t) =2t and y'(t) =3t2 - 3 so that  66. CALCULUS WITH PARAMETRIC CURVES 203 z'(tv/3) = +29/3 and y'(tv/3) = 6. So the slopes of the tangent lines are (y'/x')(t/53) = v5, and the equations of the lines read y v3(x-3) and y=-v/ 3(x-3). (ii) The tangent line becomes horizontal when y'(t) = 3t2 - 3 = 0 (see Eq. (10.4)). This happens when t =+1. Thus, the tangent line is horizontal at the points (1, +2). The tangent line is vertical if x'(t) = 2t = 0 or t = 0. So the tangent line is vertical at the origin (0, 0). (iii) The second derivative is d2y 1 d dy 1 d 3t2 - 3 3 d( 1l 3 ( 1 d2x z'dtdz 2tdt 2t 4tdt tJ 4t1+t2 This equation shows that the curve is concave upward if t > 0 (the second derivative is positive) and the curve is concave downward if t < 0 (the second derivative is negative). D 66.3. Cusps of Planar Curves. Consider a curve defined by the Carte- sian equation x2 - y3= 0. This equation can be solved for y, y - X2/3 such that dy/dz = -1/3. For x > 0, the slope of the tangent line diverges, y'(x) - oc as x -- 0+ (as x approaches to 0 from the right). For x < 0, it also diverges, y'(x) -- -oc as x -- 0- (as x approaches 0 from the left). The two branches of the curve (x > 0 and x < 0) are joined at x = 0 and have a common tangent line, which is the vertical line x = 0 (the y axis) in this case, but the slope suffers a jump dis- continuity (from -oc to oc). So the curve is not smooth at x = 0 and exhibits a horn like shape near x = 0. Such a point of a planar curve is called a cusp. A parametric curve x = x(t), y = y(t) may have cusps even though both derivatives z'(t) and y'(t) are continuous for all t. For example, consider the parametric curve xc= t3, y = t2. For all values of t, X2 _ y3 = 0. So this curve coincides with that discussed above and has a cusp at the origin (t = 0). The derivatives z'(t) = 3t2 and y'(t) = 2t are continuous everywhere, and, in particular, x'(0) = y'(0) = 0 at the cusp point. Despite the continuity of the derivatives, the slope of the curve is not defined since dy/dz= y'/x' is an undetermined form . A closer investigation shows that the slope y'(t)/z'(t) = 3t-1 suffers a jump discontinuity (from -oo to +oo as t changes from negative to positive). The definition of a smooth parametric curve requires that the derivatives c'(t) and y'(t) are continuous and do not vanish simultaneously at any t. This condition eliminates possible cusps that may occur at points where both derivatives vanish.  204 10. PLANAR CURVES 1.0 0.8 0.6 0.4-y,=X2/ .2/ -1.0 -0.5 0.5 1.0 FIGURE 10.6. Plot of y = x2/3. The curve has a cusp at the origin. Furthermore, consider the curve x = t2, y = t3. The slope dy/dzc y'/x' = 2t is continuous everywhere and, in particular, at t = 0, where X'(0) = y'(0) = 0. Nevertheless, the curve has a cusp at the origin. To see this, let us investigate the Cartesian equation of this curve x3 _ 2 0, which can be solved for x, z = y2/3. Therefore, the derivative dx/dy = 2y-1/3 exhibits a jump discontinuity from -oc to oc as y changes from negative to positive. The two branches of the curve (y > 0 and y < 0) have a common tangent line (the horizontal line y = 0), but at their joining point a cusp is formed. Note also that the rate dx/dy = z'/y' = 3t-1 suffers a familiar infinite jump discontinuity, thus indicating a cusp. This example shows that both rates dy/dz = y'/x' and dx/dy = z'/y' must be studied to determine whether there is a cusp at the point where y' = z' = 0. EXAMPLE 10.5. Find the tangent line to the astroid defined by the parametric equation x = a cos3 t, y = a sin3 t, t E [0, 27] at the points t = 7/4. Determine the points where the tangent line is horizontal and vertical. Is the curve smooth? Specify the regions of upward and downward concavity. Use the results to sketch the curve. Solution: The slope of the tangent line at a generic point is dy -- y1 3a si2tcost = -tant. dc ccz' cit 3a cos2 tsin ta The value t = /4 corresponds to the point cc= a/23/2, y - 2/ because sin(w/4) =cos(w/4) =1/v/2, and the slope at this point is  66. CALCULUS WITH PARAMETRIC CURVES 205 1.0- 0.5- 0.2 0.4 0.6 0.8 1.0 -.5 FIGURE 10.7. The curve x= x'(0) =y'(0) =0 vanish at t: at the point (x(0), y(0)) =_(0, t2, y = t. The derivatives = 0. The curve has a cusp 1. So the tangent line is a y 2v"2 ( x a) 2d V a or y = x. The slope dy/dt = - tan t vanishes at t = 0 and t =w7 so the tangent line is horizontal (y = 0) at the points (ta, 0). However, the derivatives x'(t) -3a cos2 t sin t and y'(t) = 3a sin2 t cos t vanish simultaneously at t = 0 and t =w7. The inverse slope dz/dy = 1/(dy/dx) =_- cott exhibits an infinite jump discontinuity at t = 0 and t =w7, so the curve has cusps at (ta, 0) and hence it is not smooth at these points. The slope dy/dz is infinite at t =w7/2 and t = 37/2. Therefore, the curve has a vertical tangent line (x = 0) at (0, ta). However, the slope dy/dz = - tan t has an infinite jump discontinuity at t =w7/2 and t = 37/2. So the curve has cusps and is not smooth at (0, ta). Note also that both derivatives x' and y' vanish at these points. Thus, the curve consists of four smooth pieces, and the curve has cusps at the joining points of its smooth pieces. The second derivative d2y z' dtdy d2cc cc' dt dcx 1 (tan t)' 3a cos2 t sin t 1 3a sin t cos4 t  206 10. PLANAR CURVES 206 10. PLANAR CURVES is positive if sin t > 0 (or y > 0) and negative if sin t < 0 (or y < 0). So the two branches of the curve above the x axis are concave upward, while the two branches below it are concave downward. The curves look like a square with vertices (ta, 0), (0, ta) whose sides are bent inward toward the origin. D 66.4. Exercises. In (1)-(4), find an equation of the tangent line(s) to the curve at the given point. Sketch the curve and the tangent(s). (1)x=t2 +t,y=4sint, (0, 0) (2) x = sin t + sin(2t) , y cost + cos(2t) , (1, -1) (3) x2 + 2y2 = 3x , (2, 1), (2, -1) 2 (4) + 2 () -+ y 1, (V2 ,1/ V2) , (-2 ,1/ V2) In (5)-(8), investigate the concavity of the curve. (5)x=t3-_ntyt2+-l1 (6)x =sin(2t),y cost (7) x = t2 - ln t , y = t2 + ln t (8) x = 3 sin t3 , y = 2 cos t3 (9) Investigate the slope of the trochoid x= R# - b sin #, y - R - b cos # in terms of #. Find the condition on the parameters R and b such that the trochoid has vertical tangent lines. (10) At what points on the curve x= 2t3, y - 1 + 4t - t2 does the tangent line have slope 1? (11) Find equations of the tangents to the curve x= 2t3 + 1, y 3t2 + 1 that pass through the point (3, 4). (12) Find points on the curve x2+4y2 = 2 at which it has a tangent line parallel to the line y - x = 1. In (13)-(18), investigate whether the curve has cusps or not. If it does, find their position. Sketch the curve. (13) x=t3 (4t3 1) x -t5, y -t2 (15) x = (t2-1)3' Y(t3_112 (16) ()+( = 1, a >0,b>0,n>0 (17) (X2 + y2)2 = a2(y2 - X2) (18) c3 + Y3 = 3azy In (19) and (20), find the points of intersection of two curves C1 and C2. At each point of intersection, find the tangent lines to C1 and C2 and determine the angle between these lines.  67. POLAR COORDINATES 207 (20) C1 : x2 + y2 = a2 ; C2: x2 + y2 = 2ay 67. Polar Coordinates A point on a plane is described by an ordered pair of numbers (zo, Yo) in the rectangular coordinate system. This description implies a geometrical procedure to obtain the point as the intersection of two mutually perpendicular lines x = zo and y =_Yo. The set of vertical and horizontal lines form a rectangular grid in a plane. There are other possibilities to label points on a plane by ordered pair of numbers. Here the polar coordinate system is introduced, which is more convenient for many purposes. Fix a point 0 on a plane. A horizontal ray from 0 is called the polar axis, and the point 0 is called the origin or pole. Consider a generic point P on the plane. Let 0 be the angle between the polar axis and the ray OP from 0 through P. The angle 0 is counted counterclockwise from the polar axis. The position of the point P on the ray OP is uniquely determined by the distance r = |OPl. Thus, any point P on a plane is uniquely associated with the ordered pair (r, 0), and r, 0 are called the polar coordinates of P. The coordinate r is called the radial variable, and 0 is called the polar angle. All points on the plane that have the same value of the radial vari- able form a circle of radius r centered at the origin (all points that have P Ox FIGURE 10.8. Definition of the polar coordinates in a plane. r is the distance |OPF, and 0 is the angle counted counterclockwise from the horizontal ray outgoing from O to the right. The rectangular coordinates of a point P are related to the polar ones as xc= r cosO6, y =r sinO6.  208 10. PLANAR CURVES 208 10. PLANAR CURVES the same distance from the origin). All points on the plane that have the same value of the polar angle form a ray (a half-line bounded by the origin). So a point P with polar coordinates (r, 0) is the intersection of the circle of radius r and the ray that makes the angle 0 with the polar axis. Concentric circles and rays originating from the center of the circles form a polar grid in a plane (see Figure 10.9). To represent all points of a plane, the radial variable has to range over the interval r E [0, oc), while the polar angle takes its values in the interval [0, 27) because any ray from the origin does not change after rotation about the origin through the angle 2w. It is convenient, though, to let 0 range over the whole real line. Positive values of 0 correspond to rotation angles counted counterclockwise, while negative values of 0 are associated with rotation angles counted clockwise. All pairs (r, 0) with a fixed value of r and values of 0 different by integer multiples of 2w represent the same points of the plane. For example, the ordered pairs (r, 0) = (1, -w) and (1, w) correspond to the same point. Indeed, both points are on the circle of unit radius. The ray O = w is obtained from the polar axis by counterclockwise rotation of the latter through the angle w. But the same ray is obtained by rotating the polar axis through the angle w clockwise; that is, the rays O = w and 0= -w coincide. Furthermore, the meaning of the radial variable r can be extended to the case in which r is negative by agreeing that the pairs (-r, 0) and (r, 0 + w), r > 0, represent the same point. Geometrically, the points (tr, 0) lie on a line through the origin at the same distance Ir l from the origin but on the opposite sides of the origin. With this agreement on extending the meaning of the polar coordinates, each point on a plane may be represented by countably many pairs: (10.6) (r, 0) < (r, 0 + 2wn) or (-r, 0 + (2n + 1)w), where n is an integer. 67.1. Rectangular and Polar Coordinates. Suppose that the polar axis is set so that it coincides with the positive x axis of the rectangular coordinate system. Every point on the plane is either described by the rectangular coordinates (x, y) or the polar coordinates (r, 0). It is easy to find the relation between the polar and rectangular coordinates of a point P by examining the rectangle with the diagonal OP. Its horizontal and vertical sides have lengths x and y, respectively. The length of the diagonal is r. The angle between the horizontal side and  67. POLAR COORDINATES 209 the diagonal is 0. Therefore, cos 0 = z/r and sin 0 = y/r, or x=rcosO, y=rsinO < r2 =2+y2, tan O= . These relations allow us to convert the polar coordinates of a point to rectangular coordinates and vice versa (see Fig. 10.8). EXAMPLE 10.6. Find the rectangular coordinates of a point whose polar coordinates are (2, 7/6). Find the polar coordinates of a point with rectangular coordinates (-1,1). Solution: For r = 2 and w= /6, one has x = 2 cos(7/6) = 2v/3/2 = vand y = 2 sin(7/6) = 2/2 = 1, so (x, y) = (V3, 1). For x = -1 and y =1, one has r2= 2 orr =v/2 and tan =-1. The point (-1, 1) lies in the second quadrant, that is, r/2 0 < wr. Therefore, 0 = 37/4. Alternatively, one can take 0 = 37/4 - 27r= -57/4. D 67.2. Polar Graphs. A polar graph is a curve defined by the equation r = f(0) or, more generally, F(r, 0) = 0. It consists of all points that have at least one polar representation (r, 0) that satisfies the equation. Here polar coordinates are understood in the extended sense of (10.6) when they are allowed to take any value. FIGURE 10.9. Polar grid. Coordinate curves of the po- lar coordinates. The curves of constant values of r are concentric circles. The curves of constant values of 0 are rays outgoing from the origin.  210 10. PLANAR CURVES 210 10. PLANAR CURVES The simplest polar graph is defined by a constant function r = a, where a is real. Since r represents the distance from the origin, the pairs (|al, 0) form a circle of radius |al centered at the origin. Similarly, the graph 0 = b, where b is real, is the set of all points (r, b), where r ranges over the real axis, which is the line through the origin that makes an angle b radians with the polar axis. Notice that the points (r, b), r > 0, and (r, b), r < 0, lie in the opposite quadrants relative to the origin as the pairs (r, b) and (-r, b + w) represent the same point. In general, the shape of a polar graph can be determined by plotting points (f(Ok), Ok), k = 1, 2, ..., n, for a set of successively increasing values of 0, 01 < 02 < - - - < On; that is, one takes a set of rays 0 = 64 and marks the point on each ray at a distance rk= f(Ok) from the origin. EXAMPLE 10.7. Describe the curve r = 2 cos 0. Solution: By converting the polar graph equation to rectangular co- ordinates, one finds: r = 2 cosO B r2 = 2r cos0 X2 + y2 =2x (x - 1)2+ y2 . The latter equation is obtained by completing the squares. It represents a circle with center (1, 0) and radius 1. Note also that by looking at the graph of the cosine function, one can see that the point (2 cos 0, 0) gets closer to the origin when 0 changes from 0 to 7/2 (the first quad- rant), reaching the origin at w= /2. This gives the upper part of the circle. A similar behavior is observed when 0 changes from 0 to -w/2 (the lower part of the circle in the fourth quadrant). In the in- tervals (-7, -7/2) and (7/2, 7), the radial variable is negative. The representation (r, 0) is equivalent to (-r, 0 + w). Therefore, the points (2cos0,0) and (-2cos0,0B+w) = (2cos(0 +w7),08+w7) are the same for 0 E [-o, -7/2]. But the latter set can also be described by the pairs (2 cos 0, 0) if 0 E [0, 7/2]. Similarly, the set traced out by the pair (2 cos 0,08) for 0 E [7/2,w7] is the same as when 0 E [-7/2, 0]. So the pair (2 cos 0, 0) traces out the same set (the circle) each time 0 ranges an interval of length 7. D EXAMPLE 10.8. Describe the shape of the curve r =08, 0 > 0. Solution: The point (0, 0) lies on the ray that makes an angle 0 with the polar axis and is a distance r 0 from the origin. As the ray rotates counterclockwise about the origin with increasing 0, the distance of the point from the origin increases proportionally. So the curve is a spiral unwinding counterclockwise (see Fig. 10.10).D  67. POLAR COORDINATES 211 5 5 K4 f FIGURE 10.10. Polar curve r =0. It is a spiral because the distance from the origin r increases with the angle 0 as the point rotates about the origin through the angle 0. 67.3. Symmetry of Polar Graphs. When sketching polar graphs, it it is helpful to take advantage of symmetry, just like when plotting graphs y = f(x) for symmetric (f(-x) = f(x)) or skew-symmetric (f(-x) -f(x)) functions. (i) If a polar equation is unchanged when 0 is replaced by -0, the curve is symmetric about the polar axis. Note that the transformation (r, 0) - (r, -0) means that (x, y) - (x, -y), which is the reflection about the x axis (or the polar axis). (ii) If a polar equation is unchanged when (r, 0) is replaced by (-r, 0) or by (r, 0 + 7), the curve is symmetric about the ori- gin. Again, these transformations are equivalent to (x, y) - (-x, -y), which is the reflection about the origin. (iii) If the equation is unchanged under the transformation (r, 0) - (r,w7 - 0), then the curve is symmetric about the vertical line 0 = w/2. In the rectangular coordinates, this transformation is (x, y) - (x, -y), which is the reflection about the y axis. EXAMPLE 10.9. Describe the cardioid r = 1+ sin 0. Solution: The equation is unchanged under 0 - w - 0 so the curve is symmetric about the vertical axis (the y axis). It is sufficient to investigate the curve in the interval 0 E [-7/2, 7/2] (in the fourth and first quadrants). Consider a ray that rotates counterclockwise from 0 = -w/2 to 0 w= /2 (from the negative y axis to the positive y  212 10. PLANAR CURVES 1.5 =1+sinO 1.0 0.5 0 0.5 0.5 10 FIGURE 10.11. The cardioid r = 1 + sinO0. axis). When 0 = -7r/2, r = 0. As 0 increases from -7r/2 to 0 (the fourth quadrant), the distance from the origin r = 1+ sin 0 increases monotonically from 0 to 1 (r = 1 on the polar axis). In the interval [0, 7r/2] (the first quadrant), the distance from the origin r continues to increase monotonically and reaches its maximal value 2 on the vertical axis. The cardioid is shown in Fig. 10.11. D 67.4. Tangent to a Polar Graph. To find a tangent line to a polar graph r = f(0), the polar angle is viewed as a parameter so that the para- metric equations of the graph are x=rcosO =f()cosO, y=rsin=f()sinO. By the product rule for the derivative, dy f'(O)sinO+f(O)cosO6 dx -f'()cosO-f(O)sinO6 In particular, if the curve passes through the origin, r = 0, the equation for the slope at the origin is simplified d= ~tanO6 if = f'(O)#/0. dzc dO Note that if f'(0) =0, then the slope is an undetermined form0 because z'(O) =y'(O) =0 for any value of 0 such that f'(0) =f (0) =0. This means that the curve may have a cusp at the origin and hence is not smooth.  67. POLAR COORDINATES 213 EXAMPLE 10.10. Find the slope of the cardioid r = 1 + sinO in terms of 0. Investigate the behavior of the cardioid near the origin. Solution: Here f (O) = 1 + sinOB and f'() = cosO8. This leads to the slope dy cos 8 sin0B + (1 + sin0B) cos0 cos 0(1 + 2 sin08) dxc cos2O - (1 + sin8) sinO 6 (1 + sinOB)(1 - 2 sinB) where the identity cos2 0 = 1-sin2 0 has been used to transform the de- nominator. The cardioid passes through the origin as 0 passes through -7/2. The slope dy/dz is undetermined because the numerator and denominator of the ratio vanish at 0 = -7/2 (both derivatives dz/dO and dy/dO vanish). The left and right limits have to be investigated to see if the slope has a jump discontinuity thus indicating a cusp. The numerator vanishes because of the factor cos 0, while the denominator vanishes because of the factor (1 + sin 0). Hence, dy 1 cosO _ 1 -sinO8 O-(-w/2)+ dz 3 O-(-w/2)+ 1 + sin 0 3 0-(-7/2)+ cos 0 ' where l'Hospital's rule has been used to resolve the undetermined form hand the property that tanO 8- +oo as 0 -- (-7/2)* has been invoked to find the limit. The cardioid has a vertical tangent line at the origin. The slope has an infinite jump discontinuity, meaning that the cardioid has a cusp at the origin (see Figure 10.11). D 67.5. Exercises. (1) Find rectangular coordinates of a point whose polar coordi- nates are given: (r, 0) = (1, -w/3), (-1, 57/6), (4, 10/3) (2) Find polar coordinates of a point whose rectangular coordi- nates are given: (x, y) = (1, 1), (1, -1), (3, 4), (-2, 0), (0, -2) In (3)-(5), convert the polar graph equation to a Cartesian equation and sketch the curve. (3)r =4 sinO6 (4) r =tan 6secO6 (5)r =2 sinO6- 4cosO6  214 10. PLANAR CURVES 214 10. PLANAR CURVES In (6)-(14), sketch the curve with the given polar equation. (6)r ( 0,9<0 (7)r=ln 0,;>1 (8) r2-3r+2=0 (9) r = 4 cos(60) (10) r2 = 9 sin(20) (11) r = 1 + 2 cos(20) (12) r = 2 + sin(30) (13) r = 1 + 2 sin(30) (14) r20 = 1 (15) Sketch the curve (x2 + y2)2 = 4x2y2. Hint: Use polar coordi- nates. (16) Investigate the dependence of the shape of the curve r cos(nO) as the integer n increases. What happens if n is not an integer? (17) Show that the curve r = 1 + a sin 0 has an inner loop when |al > 1 and find the range of 0 that corresponds to the inner loop. (18) For what values of a is the curve r = 1+ a sin 0 smooth? In (19) and (20), find the slope of the tangent line to the given curve at the point specified by the value of 0 and give an equation of the tangent line. (19) r = 2sin0, 08= 7/3 (20) r = 1 - 2cos0, 0B= 7/6 (21) Show that the curves r = asin 0 and r = a cos 0 intersect at right angles. In (22)-(24), investigate the concavity of the polar graph: (22) r = a > 0 (23) r = a0 (24) r =a(1l+ cos08) Hint: Use differentials to express the derivative d2y/dz2 in polar coor- dinates similarly to the calculation of the slope in Section 67.4. 68. Parametric Curves: The Arc Length and Surface Area 68.1. Arc Length of a Smooth Curve. Let C be a smooth curve defined by the parametric equations x = x(t), y = y(t), where t E [a, b]. Suppose that C is traversed exactly once as t increases from a to b and consider a partition of the interval [a, b] such that to = a and tk = to + k At, k = 0,1, 2, ..., n, are the endpoints of the partition intervals of width At = (b - a)/n. Then the points P with coordinates (z(tk), y(tk)) lie on the curve so that Po and Pn are the initial and terminal points, respectively. The curve C can be approximated by a polygonal path with vertices Pk. By definition, the length L of C is the limit of the lengths of these approximating polygons as n -~ 00: (10.7) L =iimZ Pk_1Pk, k=1 provided the limit exists.  68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 215 By the mean value theorem, when applied to the functions x(t) and y(t) on the interval [tk_1, tk], there are numbers t* and t** in (tk_1, tk) such that Axk x(tk)-x(tkl_) =_(t ) At, Ay k= y(tk)-y(tkl_) =y(t *) At. Therefore, Pk-lPk (Axk)2 + (Ayk)2 (x'(t*))2 k + (yIQt*))2 At. The sum in (10.7) resembles a Riemann sum for the function F(t) - (z'(t))2 + (y'(t))2. It is not exactly a Riemann sum because t* t ** in general. However, if x'(t) and y'(t) are continuous, it can be shown that the limit (10.7) is the same as if t* and t** were equal, namely, L is the integral of F(t) over [a, b]. THEOREM 10.2 (Arc Length of a Curve). If a curve C is described by the parametric equations x = x(t), y = y(t), t E [a, b], where x'(t) and y'(t) are continuous on [a, b] and C is traversed exactly once as t increases from a to b, then the length of C is L= ( dtC2 (dY)2dt. Pi > x FIGURE 10.12. The arc length of a smooth parametric curve is approximated by the length of n straight line segments connecting points on the curve. The arc length is defined in (10.7) as the limit n -~ oo.  216 10. PLANAR CURVES 216 10. PLANAR CURVES If C is a graph y= f(cx), then x = t, y = f(t), and dc = dt, and the length is given by the familiar expression L = (d + 2 It is convenient to introduce the are length of an infinitesimal segment of a curve (the differential of the arc length) ds =/(dz)2 + (dy)2 > L = ds dt. The symbol fC means the summation over infinitesimal segments of the curve S (the integral along a curve C) and expresses a simple fact that the total length is the sum of the lengths of its (infinitesimal) pieces. 68.2. Independence of Parameterization. By its very definition, the are length is independent of the parameterization of the curve. If a curve C is defined as a point set, then any parametric equations can be used to evaluate the are length. Let C be traced out only once by c = x(t), y = y(t), where t E [a, b], and by cc= X(T), y = Y(T), where T E [a,,3]. As noted, there is a relation between the parameters t and T, r = g(t), such that g(t) increases from a to /3 as t increases from a to b, that is, j = g'(t) > 0, such that x(t) = X(g(t)) and y(t) = Y(g(t)). Therefore, the integrals (10.7) corresponding to different parametric equations of the same curve are related by a change of the integration variable: b d x 2 y 2 jb d c 7)2 d Y )7 2 L = a V d Ct / 2+Cd t 2 dt = ab v()dCT dt 2/+CdC dt / 2dt / b 2 d 2 d7 d-F 23d 2 a r/ + dCT/Cdt Cdt = ,+ dr.) Thus, the arc length is independent of the curve parameterization and can be computed in any suitable parameterization of the curve. A circle of radius R is described by the parametric equations x = R cost, y = R sint, t E [0, 2,]. Then dzc= -R sint dt and dy = R cost dt. Hence, ds2 = (R sint dt)2+(R cost dt)2 = R2(sin2t+cos2t) dt2 = R2 dt2, or ds = R dt, and L ds =fRdt =Rfdt =2,wR. E XAMPLE 10.11. Find the length of one arch of the cycloid  68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 217 Solution: According to the description of the cycloid, one arch cor- responds to the interval # E [0, 2w]. The are length differential ds is found as follows: dx= R(1 - cos#) d#o, dy = R sin#d#, ds2 dx2 + dy2 = [(1 - cos #)2 + sin2 #]R2 d02 =[1 - 2 cos # + cos2 # + sin2 # ]R2 d02 = (2 - 2 cos #)R2 d02. ds = 2(1 -cos #) R d#. To evaluate the integral of 2(1 - cos #), the double-angle identity is invoked, sin2(#/2) =_(1-cos #)/2. Since 0 <#/2 w7when E [0, 2w], the sinus is nonnegative, sin(#/2) > 0, in the integration interval, and hence, after taking the square root (v = lul), the absolute value can be omitted. Thus, L=RJ 4sin2(#/2) d#o= 2Rfsin(#/2) d# 0 i0 2,r = 2R[-2 cos(/2)] = 8R. D 0 68.3. Area of a Planar Region. The area under the curve y = f(x) and above the interval x E [a, b] is given by A =1fb f(x) dx, where f(x) > 0. Suppose that the curve is also described by parametric equations x = x(t), y = y(t), so that the function x(t) is one-to-one. Then, by changing the integration variable, dx = z'(t) dt and A fydz fy(t)x'(t) dt. /b a The new integration limits are found as usual. When x = a, t is either a or 3, and when x = b, t is the remaining value. EXAMPLE 10.12. Find the area under one arch of the cycloid x= R(# - sin #), y = R(1 - cos #). Solution: When # E [0, 2w], x E [0, 2wR] for one arch of the cycloid, and y(#) ;> 0. Using the differential dx found in the previous example, / 27rR 2r A j= RIy dz= R2 f 1 - cos #)2 d =:R2f (1 -2 cos#+ cos25) i5 R2r  218 10. PLANAR CURVES where fo" cos # d# = 0 and fj" cos(2#) d# = 0 by the 27 periodicity of the cosine function. Q 68.4. Surface Area of Axially Symmetric Surfaces. An axially symmetric surface is a surface symmetric relative to rotations about a line. Such a line is called the symmetry axis. For example, a cylinder is symmetric relative to rotations about its axis, a sphere is symmetric relative to rotations about its diameter, and so on. An axially symmetric surface is swept by a planar curve when the latter is rotated about a line. A cylinder of radius R and height h is obtained by revolving a straight line segment of length h about a line parallel to the segment at a distance R. A sphere of radius R is obtained by revolving a circle of radius R about its diameter. dA FIGURE 10.13. A surface is obtained by rotation of a smooth curve about a vertical line. If ds is the arc length of an infinitesimal segment of the curve at a point P and R is the distance of the point P from the rotation axis, then the surface area swept by the curve segment is dA = 27R ds (the surface area of a cylinder of radius R and height ds).  68. PARAMETRIC CURVES: THE ARC LENGTH AND SURFACE AREA 219 Let ds be the are length of an infinitesimal segment of a smooth curve C positioned at a point (x, y). If the distance between the point (x, y) and the symmetry axis is R(x, y), then the area dA of the part of the surface swept by the curve segment when the latter is rotated about the symmetry axis is the area of a cylinder of radius R(x, y) and height ds: dA = 2wR(x, y) ds. The total surface area is the sum of areas of all such parts of the surface (10.8) A = 27fR(x, y) ds= 2w fR(z(t), y(t)) t dt, where x = x(t), y = y(t), a < t < b are parametric equations of C. Here it is again assumed that the point (x(t), y(t)) traces out the curve C only once as t increases from a to b. In particular, if the symmetry axis coincides with the x axis, then R(x, y) = ly| (the distance of the point (x, y) to the x axis) and A = 2f lyj ds y=2t)I(d)2 + (dY)2 dt. EXAMPLE 10.13. Find the area of the surface obtained by revolving one arch of the cycloid x = R(# - sin#0), y = R(1 - cos #) about the z axis. Solution: The differential of the are length of the cycloid has been computed in Example 10.11. Since y(t) > 0 here, the absolute value may be omitted and A = 2wfy ds = 2wf R(1 - cos#) 2(1 -cos #) R d# C i0 28wR2 fnsin3(#/2) d# = 16 R2 f sin3 u du 0 F 0 = 16R2 f(1 - cos2 u) sin u du = 16xR2 f(1 - z2) dz =l16R2(z - z/3) 1. -13 where the double-angle identity has been used again, sin2(#S/2) =(1 - cos #f)/2, and then two successive changes of the integration variable have been done to evaluate the integral, u = /2 E [0, w] and z cos u E [-1, 1].D  220 10. PLANAR CURVES 220 10. PLANAR CURVES 68.5. Exercises. In (1)-(6), find the are length of the curve. (1) (2) (3) (4) (5) (6) (7) x = 2+ 3t2, y = 1 - 2t3 between the points (2, 1) and (5, -1). x = 3 sin t - sin(3t), y = 3 cos t - cos(3t), 0 < t < 7. x = t/(1 + t), y = ln(1 + t) between the points (0, 0) and (2/3, 1n 3). x = a cos3 t, y = bsin3t c2 -=a2 - b2. x = cos4t, y = sin4 t. x =a(sinht - t), y = a(cosht - 1), 0 t a (about the x axis) (c/a)2 + (y/b)2 = 1, 0 < b < a (about the c axis) (c/a)2 + (y/b)2 = 1, 0 < b < a (about the y axis) Let V be the volume a solid bounded by an axially symmetric surface. Show that V =r fc[R(x, y)]2 ds, where C is the curve whose revolution about the symmetry axis gives the boundary surface and R(x, y) is defined in (10.8). Find the volume of the solid bounded by the surface described in Example 10.13.  69. AREAS AND ARC LENGTHS IN POLAR COORDINATES 221 69. Areas and Arc Lengths in Polar Coordinates 69.1. Area of a Planar Region. THEOREM 10.3 (Area of a Planar Region in Polar Coordinates). Let a planar region D be bounded by two rays from the origin 0 = a, O = b and a polar graph r = f(O), where f(O) ;> 0, that is, D = {(r,)|0 < r < f() a g() ;> 0 if 0 E [a, b] and 0 < b - a < 27; that is, D is the set of points whose polar coordinates satisfy the inequalities: D = {(r, 0)|0< g() < r < f(0), a < 0 < b}. Then the area of D is given by A ~ j[f (0)]2 dO - 1j()2 dO fb ([.f(0)]2 - [g(0)]2) dO. EXAMPLE 10.15. Find the area of a region D bounded by the car- dioid r = 1+ sin 0 and the circle r = 3/2 that lies above the polar axis (in the first and second quadrants). Solution: The polar graphs r = 1+ sine0= f(0) and r = 3/2 = g(0) are intersecting when f(0) = g(0) or 1 + sine0= 3/2 or sine0= 1/2. Since the region D lies in the first two quadrants, that is, 0 < 0 < <, the values of 0 for the points of intersection have to be chosen as 0 = 7/6 = a and w= - 7/6 = b. Therefore, D ={(r,08)|3/2 c r 1+ sin08, 7/6 0 55/6}, and hence the area of D is A = [(1 + sin0)2 - ] dO ] [- + 2sin08+ sin20]dO [-= +2sin0+ (1 -cos(2))]dB [- + 2 sin0B- j cos(2)] dB 2[- 0 - 2cos - 4sin(20)] 5./6 _ 9 -2 w 4/68 D-  69. AREAS AND ARC LENGTHS IN POLAR COORDINATES 223 Remark. When finding points of intersection of two polar graphs, r = f(O) and r = g(O), by solving the equation f(O) = g(O), one has to keep in mind that a single point has many representations as described in (10.6). So some of the pairs (f(0), 0), where 0 ranges over solutions of the equation f(0) = g(0), may correspond to the same point. To select distinct points, all pairs (f(0), 0) satisfying the intersection condition can be transformed by means of (10.6) so that r E [0, oc) and 8 E [0, 27). In this range of polar coordinates, there is a one-to-one correspondence between points on a plane and pairs (r, 0) with just one exception when r = 0; all the pairs (0, 0) correspond to the origin of the polar coordinate system. 69.2. Arc Length. Suppose that a curve C is traversed by the point (r, 0) = (f(0), 0) only once as 0 increases from a to b. Choosing 0 as a parameter, the curve is described by the parametric equations x = r cosO8, y = r sinO8, where r = f(0). To find the are length of C, one has to find the relation between the are length differential ds and dB. One has dx= =( cos 0B- r sin0idB, dy= ( dBsin 0B+ r cos0)cd Therefore, dsE2dx2 + dy2 = (cos0- rsin0)2 + (dsino+ r cos)2] dO2 os2 0 + sin2 0) + r2(cos2 0 + sin2 0)] dO2 [ (d) 2 ] d The are length of the curve C is L=/ ds=f dd= r2+ 2 where r = f(0) and b > a. EXAMPLE 10.16. Find the length of the cardioid r =-1+ sin0. Solution: One has r2+=(1+sn2 +( 2( sn)+ (cos0)2 =2(1 +sin06),  224 10. PLANAR CURVES 224 10. PLANAR CURVES where the trigonometric identity sin2 0+ cos2 0B= 1 has been used. The cardioid is traversed once if 0 E [-w, w]. Therefore, the length is /7r sr/2 v/ /2- - 1 + sin 0 d8 = 2 v/1 + sin 0 d8, since the cardioid is symmetric about the vertical This integral can be evaluated by the substitution ui so that du = cos B dB, where cos 1= - sin2 08= u(2-u). Hence, L=2 2f du = 2 f = - Jon u(2-u) Jo0v/2-u line (the y axis). = 1+ sinO E[0, 2] 1 - (u - 1)2 2 2-u2 0 8. 69.3. Surface Area. If a surface is obtained by rotating a polar graph r = f(0) about a line, the (10.8) can be used to find the area of the surface where the distance R(x, y) and the are length differential ds have to be expressed in the polar coordinates with r = f(0). EXAMPLE 10.17. Find the area of the surface obtained by rotating the cardioid r = 1+ sin 0 about its symmetry axis. Solution: The symmetry axis of the cardioid is the y axis. So the distance from the y axis to a point (x, y) is R(x, y) = Iz|. The surface can be obtained by rotating the part of the cardioid that lies in the fourth and first quadrants, that is, x > 0 or 0 E [-w/2, w/2]. Since x = r cos 0, the surface area is A =27 f|x| ds 2w / ds c2 / J ,oO r2( r cos 0 r2 + ( de d. F The derivative ds/dO has been calculated in the previous example. Therefore, 7l/2 f A = 2v 2] (1 + sin 0)3/2 cos dO = 2 J ](1 + u)3/2 du (1 + u)5/2 1 5/2 -1 where the substitution u= integral. 327 5, sin 6 has been made to evaluate the D-  70. CONIC SECTIONS 225 69.4. Exercises. In (1)-(4), sketch the curve and find the area that it encloses. (1) r = 4 cos(2) (2) r =a(1 + cosO8) (3) r = 2 - cos(20) (4) r2 = 4 cos(20) In (5)-(7), sketch the curve and find the area of one loop of the curve. (5) r = 9 sin(30) (6) r = 1+2 sinOB (inner loop) (7) r = 2 cosOB-secOB In (8) and (9), find the area of the region that lies inside the first curve and outside the second curve. Sketch the curves. (8) r= 2sin, r =1 (9) r= 3cos, r =1+cosO In (10)-(13), find the area of the region bounded by the curves. Sketch the region. (10) r = 28 , r = B , B E [0, 27r] (11) r2 = sin(26) , r2 = cos(26) (12)r=2asinO,r=2bcos6,a,b>0 (13)r=3+2cosO, r=3+2sinO (14) Find the area inside the larger loop and outside the smaller loop of the limagon r = 1/2 - cosOB. In (15)-(17), sketch the curve and find its length. (15) r = 2a sinO8 (16) r = 0,O E [0,27r] (17) r = a + cosOB , a > 1 In (18)-(21), find the area of the surface obtained by rotating the curve about the specified axis. Sketch the surface. (18) r = a > 0, about a line through the origin. (19) r = 2a cosOB, a > 0, (i) about the y axis and (ii) about the x axis. (20) r2 = cos(20) about the polar axis. (21) 0 - a and 0 0, about the polar axis. 70. Conic Sections Consider two intersecting lines in space, L1 and L2. A surface swept by the line L2 when it is rotated about the line L1 is a circular double cone. The line L1 is the symmetry axis of the cone. The point of inter- section of the lines is called the vertex of a cone. Any plane that does not pass through the vertex intersects the cone along a curve. It ap- pears that all such curves fall into three types as shown in Figure 10.14. If the curve of intersection is a loop, then it is an ellipse. If the plane is parallel to the line L2, then the curve is a parabola. If the plane  226 10. PLANAR CURVES 226 10. PLANAR CURVES is parallel to the axis of the cone, then the curve is a hyperbola. The curves of intersection of a plane and a cone are called conic sections, or conics. They have a pure geometrical description, which will be presented here. Remark. A trajectory of any massive object in the solar system (e.g., comet, asteroid, planet) is a conic section-that is, a parabola, hyperbola, or ellipse. This fact follows from Newton's Law of Gravity and will be proved in Calculus III. 70.1. Parabolas. A parabola is the set of points in a plane that are equidistant from a fixed point F (called the focus) and a fixed line (called the directrix). Let P be a point in a plane. Consider the line through P that is perpendicular to the directrix and let Q be the point of their intersection. Then P lies on a parabola if FPI =|QP. This condition is used to derive the equation of a parabola. A particularly simple equation of a parabola is obtained if the co- ordinate system is set so that the y axis coincides with the line through the focus and perpendicular to the directrix. The origin 0 is chosen so that F = (0, p) and hence the parabola contains the origin 0, while the directrix is the line y = -p parallel to the x axis (the origin is Circle . Parabola Hyperbola ..... Ellipse FIGURE 10.14. Conic sections are curves that are in- tersections of a cone with various planes. The shape of a conic section depends on the orientation of the plane relative to the cone symmetry axis.  70. CONIC SECTIONS 227 70. CONIC SECTIONS 227 Q FIGURE 10.15. Left: Geometrical description of a parabola as a set of points P in a plane that are equidis- tant from a fixed point F, called the focus, and a fixed line called the directrix (a horizontal line in the figure). Right: A circular paraboloid is the surface obtained by rotating a parabola about the line through its focus and perpendicular to its directrix. at distance p from F and from the directrix). If P = (x, y), then FP = x2 + (y - p)2, the point Q has the coordinates (x, -p), and PQ| = (y + p)2. An equation of the parabola with focus (0, p) and directrix y = -p is FP|2 PQ2 __ 2+(y-p)2 =(y+p)2 _ 2 = 4py. In the 16th century, Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. The surface obtained by rotating a parabola about its symmetry axis is called a paraboloid. If a source of light is placed at the focus of a paraboloid mirror, then, after the reflection, the light forms a beam parallel to the symmetry axis. This fact is used to design flashlights, headlights, and so on. Conversely, a beam of light parallel to the symmetry axis of a paraboloid mirror will be focused to the focus point after the reflection, which is used to design reflecting telescopes. 70.2. Ellipses. An ellipse is the set of points in a plane, the sum of whose distances from two fixed points F1 and F2 is a constant. The fixed points are called foci (plural of focus). Let P be a point on a plane. Then P belongs to an ellipse if PF + PF2 = 2a, where a > 0  228 10. PLANAR CURVES F1 F2 FIGURE 10.16. Left: An ellipse is the set of points in a plane, the sum of whose distances from two fixed points F1 and F2 (the foci) is a constant. Right: A circular ellipsoid is the surface obtained by rotating an ellipse about the line through its foci. is a constant (the factor 2 is chosen for convenience to be seen later). Evidently, F11F2 < 2a; otherwise, no ellipse exists. A particularly simple equation of an ellipse is obtained when the coordinate system is set so that the foci lie on the x axis and have the coordinates F1 = (-c, 0) and F2 = (c, 0), where c < a. Let P = (x, y) be a point in a plane. Then PFil = f(x + c)2 + y2 and PF2 = f(x - c)2 + y2. The point P is on an ellipse if PF1 + PF2 = 2a < PF2 = 2a - PF1|, (10.9) PF2|2 = (2a - PF1|)2 = 4a2 - 4a|PF1 + PF1|2 16a2 P 1|2 = (4a2 + P7 2 - PF2 2)2. These transformations serve only one purpose, that is, to get rid of the square roots. Note that now all the distances are squared. So PF1 2 - pF2 2 = (x + c)2 + y2 - (x - c)2 - 2 = 4cx. The substitution of the latter into the condition (10.9) yields 16a2(x+c)2+y2] = (4a2+4cx)2 _ (a2-c2)x2+a2y2 = a2(a2-c2). By dividing both sides of this equation by a2(a2 - c2), an equation of an ellipse with foci (±a, 0) becomes a2+ =1 where b2 = a2 - c2 so that a > b > 0. The ellipse intersects the x axis at (±a, 0) and the y axis at (0, ±b) (called the vertices of an ellipse). The line segment joining the points (±a, 0) is called the major axis. If the foci of an ellipse are located on the y axis, then x and y are swapped in this equation, and the major axis lies on the y axis. This shows that the restriction a > b can be dropped in the ellipse equation. In particular, an ellipse becomes a circle of radius a if a = b.  70. CONIC SECTIONS 229 One of Kepler's laws is that the orbits of the planets in the solar system are ellipses with the Sun at one focus. 70.3. Hyperbolas. A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points F1 and F2 (the foci) is a constant. For any point P on a hyperbola, |PFl - PF2|_= +2a (as the difference of the distances can be negative). Let the foci be at (+c, 0). Following the same procedure used to derive an equation of an ellipse, an equation of a hyperbola with foci (+c, 0) is found to be x2 y2 a2 b2 where c2 = a2 + b2. The details are left to the reader as an exercise. This equation shows that x2/a2 > 1 for any y, that is, c ;> a or x < -a. A hyperbola therefore has two branches. The branch in x < -a intersects the c axis at x = -a, while the branch in x ;> a does so at x = a. The points (ta, 0) are called vertices. Furthermore, in the asymptotic region c| - 0oc, a hyperbola has slant asymptotes y = +(b/a). Indeed, -x2 ba1| a2 bax|ca2x bx| y=+b -1=c1- bc1-c a2 a xc2 aX 2xc2 a as Icc - oc. Here the linearization 1+ u 1 + u/2 has been used to obtain the asymptotic behavior for small u = -a2/2 -- 0. If the foci of a hyperbola are on the y axis, then, by reversing the roles of x and y, it follows that the hyperbola y2 x2 y=cc a2 -b2= has foci (0, +c), where c2 = a2 + b2, vertices (0, +a), and slant asymp- totes y +(a/b)z. 70.4. Shifted Conics. Consider a curve defined by a quadratic Carte- sian equation Ay2+ Bc2+ ay +#xc+>= 0. Suppose that A / 0 and B / 0. By completing the squares, this equation can be transformed to the standard form a_ 2__ 2 _ /32- d Ay-2A)+Bc- 2B) 4A 4B or (y -yo)2 +(cc-cco)2 A/d B/d  230 10. PLANAR CURVES 230 10. PLANAR CURVES 4F2 P - F1 F2 FIGURE 10.17. Left: A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points F1 and F2 (the foci) is a constant. Top right: A circular hyperboloid of one sheet is the surface obtained by rotating a hyperbola about the line through the midpoint of the segment F1F2 and perpendicular to it (the vertical line in the left panel). Bottom right: A circular hyperboloid of two sheets is the surface obtained by rotating a hyperbola about the line through its foci (the horizontal line in the left panel). where xo = ,/(2B) and yo = a/(2A), provided d $ 0. Depending on the signs of A/d and B/d, this equation describes either an ellipse or a hyperbola as if the origin was moved to the point (xo, yo). If A/d and B/d are both negative, then the equation has no solution. If either A or B vanishes, but not both, then the quadratic Cartesian equation describes a parabola (the details are left to the reader as an exercise). If A = B = 0, the the equation describes a straight line. If d = 0, solutions of the equation form a set of two straight lines, y - Yo = -(B/A)(x - xo), through the point (xo,yo), provided AB < 0. When solutions of the Cartesian equation form a hyperbola (d $ 0, AB < 0), these lines are its slant asymptotes.  70. CONIC SECTIONS 231 70.5. Conic Sections in Polar Coordinates. The following theorem of- fers a uniform description of conic sections. THEOREM 10.4 (Conic Sections). Let F be a fixed point (called the focus) and L be a fixed line (called the directrix) in a plane. Let e be a fixed positive number (called the eccentricity). The set of points P in the plane whose the ratio of the distance from F to the distance from L is the constant e is a conic section. The conic is (1) an ellipse if e < 1 F (2) aparabola if e 1 e= . (3) a hyperbola if e > 1 PLI PROOF. Set the coordinate system so that F is at the origin and the directrix is parallel to the y axis and d units to the right. Thus, the directrix has the equation x = d > 0 and is perpendicular to the polar axis. If the point P has polar coordinates (r, 0) and rectangular coordinates (x, y), then |PFI = r = z2+ y2 and PL|= d - x d-r cos 0. The condition |PF = e|PL| yields the equation r = e(d-x). By squaring it, one infers a quadratic Cartesian equation x2 +y2 e2(d - X)2 (1-e2)X2 + y2 + 2e2 dx- e2d=0, which has been investigated in the preceding section. If e = 1, then the equation describes the shifted parabola y2 = -2d(x - 1/2). When e f 1, by completing the squares, this equation is brought to the standard form e2d l2 2y2 e24 2 (x+1-e2+-e2 (1_2 If e < 1, then all the coefficients are positive, and the equation describes a shifted ellipse (z - z0)2 y2 2 b 2 e2d2 e2d a2 + 1 aoe21-e2 x e2l -c where c is the distance from the origin to the foci of the ellipse, c2 - a2 - b2. The eccentricity is then e = c/a. Similarly, if e > 1, then the coefficients have opposite signs, and the equation describes a shifted hyperbola (x-cco)2 y 12 c2 2b2 a2 b2a In the beginning of the proof, the polar equation for conic sections was given as r =e(d - r cos 0). If the directrix is chosen to be to the left of the focus as cc= -d, then cos 0 is replaced by - cos 0 in the polar  232 10. PLANAR CURVES 232 10. PLANAR CURVES equation. If the directrix is chosen to be parallel to the polar axis as y = td, then the conic sections are r = e(d t y) = e(d t r sinO8). These equations can be solved for r to obtain conic sections as polar graphs. COROLLARY 10.5 (Conics in Polar Coordinates). A polar equation of the form ed ed r 1t e cos or r 1 + e sin 0 represents a conic section of eccentricity e. The conic section is an ellipse if e < 1, a parabola if e = 1, and a hyperbola if e > 1. 70.6. Exercises. In (1)-(9), classify the conic section. Find the vertices, foci (or focus), directrix, and asymptotes (if the curve is a hyperbola). Sketch the curve. (1) y2 = 16x (2) x2 = -4y (3) y + 12x - 2x2 = 18 (4) x2 +4y2 = 16 (5) 9x2 - 18x + 4y2 = 27 (6) x2 +3y2 + 2x - 12y + 10 = 0 (7) 4x2 -9y2 = 36 (8)y2-2y=4x2+3 (9)y2-4x2+2y+16x+3=0 (10) A long-range radio navigation system uses two radio stations, located at points A and B along the coastline, that trans- mit simultaneous signals to a ship located at point P in the sea. The onboard computer converts the time difference in receiving these signals into a distance difference |PAl -|PBl. This locates the ship on one branch of a hyperbola. Suppose that station B is located D miles from station A. A ship re- ceives the signal from B T microseconds (,us) before it receives the signal from A. The signal travels with the speed of light, c = 980 ft/ps. How far off the coastline is the ship? If the coordinate system is set so that the line AB coincides with the x axis and A is at the origin, find the coordinates of the ship as functions of T. In (11)-(13), classify the conic section. Find the eccentricity, an equa- tion of the directrix, and sketch the conic. (11) r 8=(12 r 1= (13) r 1 4+sin0 ( )T2 -5cosO 3+3cos0 (14) Show that the conic sections r =a/(1 - cos 0) and r =b/(1 + cos 0) intersect at right angles.  70. CONIC SECTIONS 233 In (15)-(20), find the polar equations for the curve. (15) x2 +4y2 = 16 (16) x2-4Y2=9 (17) y2 + 6x = 0 (18) x = -4 cos t, y = 9 sin t (19) x = ta cosh t, y = b sinh t (20) x = a sect , y = b tan t (21) The orbit of Halley's Comet, last seen in 1986 and due to re- turn in 2062, is an ellipse with eccentricity 0.97 and one focus at the Sun. The length of its major axis is 36.18 AU. An as- tronomical unit (AU) is the mean distance between the Earth and the Sun, about 93 million miles. Find a polar equation for the orbit of Halley's Comet. What is the maximal and minimal distance from the comet to the Sun?