INFINITE SEQUENCES AND SERIES

Transcription

1 INFINITE SEQUENCES AND SERIES y T T x y=si x T T The partial sums T of a Taylor series provide better ad better approximatios to a fuctio as icreases. Ifiite sequeces ad series were itroduced briefly i A Preview of Calculus i coectio with Zeo s paradoxes ad the decimal represetatio of umbers. Their importace i calculus stems from Newto s idea of represetig fuctios as sums of ifiite series. For istace, i fidig areas he ofte itegrated a fuctio by first expressig it as a series ad the itegratig each term of the series. We will pursue his idea i Sectio.0 i order to itegrate such fuctios as e x 2. (Recall that we have previously bee uable to do this.) ay of the fuctios that arise i mathematical physics ad chemistry, such as Bessel fuctios, are defied as sums of series, so it is importat to be familiar with the basic cocepts of covergece of ifiite sequeces ad series. Physicists also use series i aother way, as we will see i Sectio.. I studyig fields as diverse as optics, special relativity, ad electromagetism, they aalyze pheomea by replacig a fuctio with the first few terms i the series that represets it. 674

2 . SEQUENCES A sequece ca be thought of as a list of umbers writte i a defiite order: a, a 2, a 3, a 4,..., a,... The umber a is called the first term, a 2 is the secod term, ad i geeral a is the th term. We will deal exclusively with ifiite sequeces ad so each term a will have a successor a. Notice that for every positive iteger there is a correspodig umber a ad so a sequece ca be defied as a fuctio whose domai is the set of positive itegers. But we usually write a istead of the fuctio otatio f for the value of the fuctio at the umber. NOTATION The sequece {,,,...} is also deoted by a a 2 a 3 a or a (a) (b) EXAPLE Some sequeces ca be defied by givig a formula for the th term. I the followig examples we give three descriptios of the sequece: oe by usig the precedig otatio, aother by usig the defiig formula, ad a third by writig out the terms of the sequece. Notice that does t have to start at. 3 a a 3, 2, 3, 4,..., ,... 2, 3, 4, 5,...,, (c) (d) {s 3 } 3 cos 6 0 a s 3, 3 a cos 6, 0 {0,, s2, s3,..., s 3,...} s3, 2, 2, 0,..., cos 6,... V EXAPLE 2 Fid a formula for the geeral term a of the sequece 3, 4 5, , 6 625, 7 325,... assumig that the patter of the first few terms cotiues. SOLUTION We are give that a 3 5 a a a a Notice that the umerators of these fractios start with 3 ad icrease by wheever we go to the ext term. The secod term has umerator 4, the third term has umerator 5; i geeral, the th term will have umerator 2. The deomiators are the powers of 5, so a has deomiator 5. The sigs of the terms are alterately positive ad egative, so 675

3 676 CHAPTER INFINITE SEQUENCES AND SERIES we eed to multiply by a power of. I Example (b) the factor meat we started with a egative term. Here we wat to start with a positive term ad so we use or. Therefore 2 a 5 EXAPLE 3 Here are some sequeces that do t have a simple defiig equatio. (a) The sequece p, where p is the populatio of the world as of Jauary i the year. (b) If we let a be the digit i the th decimal place of the umber e, the a is a welldefied sequece whose first few terms are 7,, 8, 2, 8,, 8, 2, 8, 4, 5,... (c) The Fiboacci sequece f is defied recursively by the coditios f f 2 f f f 2 3 Each term is the sum of the two precedig terms. The first few terms are,, 2, 3, 5, 8, 3, 2,... This sequece arose whe the 3th-cetury Italia mathematicia kow as Fiboacci solved a problem cocerig the breedig of rabbits (see Exercise 7). 0 2 FIGURE a a a a A sequece such as the oe i Example (a), a, ca be pictured either by plottig its terms o a umber lie as i Figure or by plottig its graph as i Figure 2. Note that, sice a sequece is a fuctio whose domai is the set of positive itegers, its graph cosists of isolated poits with coordiates, a 2, a 2 3, a 3..., a... a From Figure or 2 it appears that the terms of the sequece approachig as becomes large. I fact, the differece a are 7 a = FIGURE 2 ca be made as small as we like by takig sufficietly large. We idicate this by writig lim l I geeral, the otatio lim a L l meas that the terms of the sequece a approach L as becomes large. Notice that the followig defiitio of the limit of a sequece is very similar to the defiitio of a limit of a fuctio at ifiity give i Sectio 2.6.

4 SECTION. SEQUENCES 677 DEFINITION A sequece a has the limit L ad we write lim a L l or a l L as l if we ca make the terms a as close to L as we like by takig sufficietly large. If lim l a exists, we say the sequece coverges (or is coverget). Otherwise, we say the sequece diverges (or is diverget). Figure 3 illustrates Defiitio by showig the graphs of two sequeces that have the limit L. a a FIGURE 3 Graphs of two sequeces with lim a =L ` L 0 L 0 A more precise versio of Defiitio is as follows. 2 DEFINITION A sequece a has the limit L ad we write N Compare this defiitio with Defiitio lim a L l a l L as l if for every 0there is a correspodig iteger N such that or if N the a L Defiitio 2 is illustrated by Figure 4, i which the terms a, a 2, a 3,... are plotted o a umber lie. No matter how small a iterval L, L is chose, there exists a N such that all terms of the sequece from oward must lie i that iterval. a a a aˆ a N+ a N+2 a aß a a a FIGURE 4 0 L- L L+ Aother illustratio of Defiitio 2 is give i Figure 5. The poits o the graph of a must lie betwee the horizotal lies y L ad y L if N. This picture must be valid o matter how small is chose, but usually a smaller requires a larger N. a N y L y=l+ y=l- FIGURE N

5 678 CHAPTER INFINITE SEQUENCES AND SERIES If you compare Defiitio 2 with Defiitio 2.6.7, you will see that the oly differece betwee lim l a L ad lim x l f x L is that is required to be a iteger. Thus we have the followig theorem, which is illustrated by Figure 6. 3 THEORE If lim x l f x L ad f a whe is a iteger, the lim l a L. y y=ƒ L FIGURE x I particular, sice we kow that lim x l x r 0 whe r 0 (Theorem 2.6.5), we have 4 If a becomes large as becomes large, we use the otatio lim l a. The followig precise defiitio is similar to Defiitio DEFINITION lim l a meas that for every positive umber there is a iteger N such that if lim l r 0 N the if r 0 a If lim l a, the the sequece a is diverget but i a special way. We say that a diverges to. The Limit Laws give i Sectio 2.3 also hold for the limits of sequeces ad their proofs are similar. LIIT LAWS FOR SEQUENCES If a ad b are coverget sequeces ad c is a costat, the lim a b lim a lim b l l l lim a b lim a lim b l l l lim ca c lim a l l lim c c l lim a b lim a lim b l l l a lim l b lim a l lim b l if lim l b 0 lim a p [lim a ] p if p 0 ad a 0 l l

6 SECTION. SEQUENCES 679 The Squeeze Theorem ca also be adapted for sequeces as follows (see Figure 7). SQUEEZE THEORE FOR SEQUENCES If a for ad lim a, the lim b b c 0 lim c L L. l l l c Aother useful fact about limits of sequeces is give by the followig theorem, whose proof is left as Exercise 75. b a 6 THEORE If lim a 0, the lim a 0. l l 0 FIGURE 7 The sequece b is squeezed betwee the sequeces a ad c. N This shows that the guess we made earlier from Figures ad 2 was correct. EXAPLE 4 Fid lim. l SOLUTION The method is similar to the oe we used i Sectio 2.6: Divide umerator ad deomiator by the highest power of ad the use the Limit Laws. lim l lim l 0 lim l lim lim l l Here we used Equatio 4 with r. l EXAPLE 5 Calculate lim. l SOLUTION Notice that both umerator ad deomiator approach ifiity as l. We ca t apply l Hospital s Rule directly because it applies ot to sequeces but to fuctios of a real variable. However, we ca apply l Hospital s Rule to the related fuctio f x l xx ad obtai l x x lim lim x l x x l 0 Therefore, by Theorem 3, we have l lim l 0 a _ FIGURE 8 EXAPLE 6 Determie whether the sequece a is coverget or diverget. SOLUTION If we write out the terms of the sequece, we obtai,,,,,,,... The graph of this sequece is show i Figure 8. Sice the terms oscillate betwee ad ifiitely ofte, a does ot approach ay umber. Thus lim l does ot exist; that is, the sequece is diverget.

7 680 CHAPTER INFINITE SEQUENCES AND SERIES N The graph of the sequece i Example 7 is show i Figure 9 ad supports our aswer. a 0 EXAPLE 7 Evaluate lim if it exists. l SOLUTION Therefore, by Theorem 6, lim l lim l 0 lim 0 l _ The followig theorem says that if we apply a cotiuous fuctio to the terms of a coverget sequece, the result is also coverget. The proof is left as Exercise 76. FIGURE 9 7 THEORE If lim a L ad the fuctio f is cotiuous at L, the l lim l f a f L EXAPLE 8 Fid lim si. l SOLUTION Because the sie fuctio is cotiuous at 0, Theorem 7 eables us to write lim si silim si 0 0 l l N CREATING GRAPHS OF SEQUENCES Some computer algebra systems have special commads that eable us to create sequeces ad graph them directly. With most graphig calculators, however, sequeces ca be graphed by usig parametric equatios. For istace, the sequece i Example 9 ca be graphed by eterig the parametric equatios ad graphig i dot mode, startig with t ad settig the t-step equal to. The result is show i Figure 0. 0 FIGURE 0 x t y t!t t 0 V EXAPLE 9 Discuss the covergece of the sequece a!, where! 2 3. SOLUTION Both umerator ad deomiator approach ifiity as l, but here we have o correspodig fuctio for use with l Hospital s Rule ( x! is ot defied whe x is ot a iteger). Let s write out a few terms to get a feelig for what happes to a as gets large: 2 3 a a 2 2 a a 2 3 It appears from these expressios ad the graph i Figure 0 that the terms are decreasig ad perhaps approach 0. To cofirm this, observe from Equatio 8 that a 2 3 Notice that the expressio i paretheses is at most because the umerator is less tha (or equal to) the deomiator. So 0 a We kow that l 0 as l. Therefore a l 0 as l by the Squeeze Theorem.

8 SECTION. SEQUENCES 68 V EXAPLE 0 For what values of r is the sequece r coverget? SOLUTION We kow from Sectio 2.6 ad the graphs of the expoetial fuctios i Sectio.5 that lim for ad lim x l a x x l a x a 0 for 0 a. Therefore, puttig a r ad usig Theorem 3, we have It is obvious that lim r l lim l 0 ad if r if 0 r lim l 0 0 If r 0, the, so 0 r lim r lim r 0 l l ad therefore lim by Theorem 6. If, the r l r 0 r diverges as i Example 6. Figure shows the graphs for various values of r. (The case r is show i Figure 8.) a a r> r= 0 _<r<0 FIGURE The sequece a =r 0 0<r< r<_ The results of Example 0 are summarized for future use as follows. 9 The sequece r is coverget if r ad diverget for all other values of r. lim r 0 if r l if r N The right side is smaller because it has a larger deomiator. 0 DEFINITION A sequece a is called icreasig if a a for all, that is, a a 2 a 3. It is called decreasig if a a for all. It is called mootoic if it is either icreasig or decreasig. EXAPLE The sequece 3 5 is decreasig because ad so a a for all.

9 682 CHAPTER INFINITE SEQUENCES AND SERIES EXAPLE 2 Show that the sequece a 2 SOLUTION We must show that a a, that is, 2 is decreasig. 2 This iequality is equivalet to the oe we get by cross-multiplicatio: 2 2 &? &? &? Sice, we kow that the iequality 2 is true. Therefore a a ad so a is decreasig. x SOLUTION 2 Cosider the fuctio f x : x 2 f x x 2 2x 2 x 2 2 x 2 x wheever x 2 Thus f is decreasig o, ad so f f. Therefore a is decreasig. DEFINITION A sequece a is bouded above if there is a umber such that a for all It is bouded below if there is a umber m such that m a for all If it is bouded above ad below, the a is a bouded sequece. a L 0 23 FIGURE 2 For istace, the sequece a is bouded below a 0 but ot above. The sequece a is bouded because 0 a for all. We kow that ot every bouded sequece is coverget [for istace, the sequece a satisfies a but is diverget from Example 6] ad ot every mootoic sequece is coverget a l. But if a sequece is both bouded ad mootoic, the it must be coverget. This fact is proved as Theorem 2, but ituitively you ca uderstad why it is true by lookig at Figure 2. If a is icreasig ad a for all, the the terms are forced to crowd together ad approach some umber L. The proof of Theorem 2 is based o the Completeess Axiom for the set of real umbers, which says that if S is a oempty set of real umbers that has a upper boud ( x for all x i S), the S has a least upper boud b. (This meas that b is a upper boud for S, but if is ay other upper boud, the b.) The Completeess Axiom is a expressio of the fact that there is o gap or hole i the real umber lie.

10 SECTION. SEQUENCES ONOTONIC SEQUENCE THEORE coverget. Every bouded, mootoic sequece is PROOF Suppose a is a icreasig sequece. Sice a is bouded, the set S a has a upper boud. By the Completeess Axiom it has a least upper boud L. Give 0, L is ot a upper boud for S (sice L is the least upper boud). Therefore a N L for some iteger N But the sequece is icreasig so a a N for every N. Thus if N, we have a L so sice a L. Thus 0 L a L a wheever N so lim l a L. A similar proof (usig the greatest lower boud) works if a is decreasig. The proof of Theorem 2 shows that a sequece that is icreasig ad bouded above is coverget. (Likewise, a decreasig sequece that is bouded below is coverget.) This fact is used may times i dealig with ifiite series. EXAPLE 3 Ivestigate the sequece a defied by the recurrece relatio a 2 a 2a 6 for, 2, 3,... SOLUTION We begi by computig the first several terms: a 2 a a a a a a a a N athematical iductio is ofte used i dealig with recursive sequeces. See page 77 for a discussio of the Priciple of athematical Iductio. These iitial terms suggest that the sequece is icreasig ad the terms are approachig 6. To cofirm that the sequece is icreasig, we use mathematical iductio to show that a a for all. This is true for because a 2 4 a. If we assume that it is true for k, the we have a k a k so ad a k 6 a k 6 2a k 6 2a k 6 Thus a k2 a k

11 684 CHAPTER INFINITE SEQUENCES AND SERIES We have deduced that a a is true for k. Therefore the iequality is true for all by iductio. Next we verify that a is bouded by showig that a 6 for all. (Sice the sequece is icreasig, we already kow that it has a lower boud: a a 2 for all.) We kow that a 6, so the assertio is true for. Suppose it is true for k. The a k 6 so a k 6 2 2a k Thus a k 6 This shows, by mathematical iductio, that a 6 for all. Sice the sequece a is icreasig ad bouded, Theorem 2 guaratees that it has a limit. The theorem does t tell us what the value of the limit is. But ow that we kow L lim l a exists, we ca use the recurrece relatio to write lim a lim l l 2a 6 2( lim a 6 l ) 2 L 6 N A proof of this fact is requested i Exercise 58. Sice a l L, it follows that a l L, too (as l, l too). So we have L 2L 6 Solvig this equatio for L, we get L 6, as predicted.. EXERCISES. (a) What is a sequece? (b) What does it mea to say that lim l a 8? (c) What does it mea to say that lim l a?. 2, 7, 2, 7, {, 2 3, 4 9, 8 27,...} 4. { 4, 2 9, 3 6, 4 25,...} 5,, 5,, 5,, (a) What is a coverget sequece? Give two examples. (b) What is a diverget sequece? Give two examples. 5. List the first six terms of the sequece defied by 3 8 List the first five terms of the sequece. 3. a a 3 5. a ! 7. a, a 8. a, a a 3 2a 4 a 9 4 Fid a formula for the geeral term a of the sequece, assumig that the patter of the first few terms cotiues. 9. {, 3, 5, 7, 9,...} 0. {, 3, 9, 27, 8,...} a 2 Does the sequece appear to have a limit? If so, fid it. 6. List the first ie terms of the sequece cos3. Does this sequece appear to have a limit? If so, fid it. If ot, explai why Determie whether the sequece coverges or diverges. If it coverges, fid the limit. 7. a a 3 3

12 SECTION. SEQUENCES a a e a ta a 3 a a 28. cos2 a cos arcta 2 2! e l l 2 e e ,, 0, 0,, 0, 0, 0,, {, 3, 2, 4, 3, 5, 4, 6,...} 45. a 46. a 3! 2! ; Use a graph of the sequece to decide whether the sequece is coverget or diverget. If the sequece is coverget, guess the value of the limit from the graph ad the prove your guess. (See the margi ote o page 680 for advice o graphig sequeces.) 47. a 2e 48. a s si(s ) a 50. a s a 2 cos a 53. a e 34. cos a cos a 38. a s 2 3 si a 2 4. a l2 2 l a 3 5 2! a l l a 3 a a 32 5 a 9 si 2 s l (a) Determie whether the sequece defied as follows is coverget or diverget: (b) What happes if the first term is a 2? 55. If $000 is ivested at 6% iterest, compouded aually, the after years the ivestmet is worth a dollars. (a) Fid the first five terms of the sequece a. (b) Is the sequece coverget or diverget? Explai. 56. Fid the first 40 terms of the sequece defied by ad a. Do the same if a 25. ake a cojecture about this type of sequece. 57. For what values of r is the sequece r coverget? 58. (a) If a is coverget, show that (b) A sequece a is defied by a ad a a for. Assumig that a is coverget, fid its limit. 59. Suppose you kow that a is a decreasig sequece ad all its terms lie betwee the umbers 5 ad 8. Explai why the sequece has a limit. What ca you say about the value of the limit? Determie whether the sequece is icreasig, decreasig, or ot mootoic. Is the sequece bouded? 60. a 2 6. a a a 4 a for a 2 a 3a 2 3 lim a lim l l a 67. Fid the limit of the sequece if a is a eve umber if a is a odd umber 68. A sequece a is give by a s2, a s2 a. (a) By iductio or otherwise, show that a is icreasig ad bouded above by 3. Apply the ootoic Sequece Theorem to show that lim l a exists. (b) Fid lim l a. 62. a 63. a 64. a e 65. a 66. a 2 {s2, s2s2, s2s2s2,...}

13 686 CHAPTER INFINITE SEQUENCES AND SERIES 69. Show that the sequece defied by is icreasig ad a 3 for all. Deduce that a is coverget ad fid its limit. 70. Show that the sequece defied by satisfies 0 a 2 ad is decreasig. Deduce that the sequece is coverget ad fid its limit. 7. (a) Fiboacci posed the followig problem: Suppose that rabbits live forever ad that every moth each pair produces a ew pair which becomes productive at age 2 moths. If we start with oe ewbor pair, how may pairs of rabbits will we have i the th moth? Show that the aswer is f, where f is the Fiboacci sequece defied i Example 3(c). (b) Let a f f ad show that a a 2. Assumig that a is coverget, fid its limit. 72. (a) Let a a, a 2 f a, a 3 f a 2 f f a,..., a f a, where f is a cotiuous fuctio. If lim l a L, show that f L L. (b) Illustrate part (a) by takig f x cos x, a, ad estimatig the value of L to five decimal places. ; 73. (a) Use a graph to guess the value of the limit (b) Use a graph of the sequece i part (a) to fid the smallest values of N that correspod to 0. ad 0.00 i Defiitio Use Defiitio 2 directly to prove that lim l r 0 whe. r a a Prove Theorem 6. [Hit: Use either Defiitio 2 or the Squeeze Theorem.] 76. Prove Theorem Prove that if lim l a 0 ad b is bouded, the lim l a b Let a. 5 lim l! (a) Show that if 0 a b, the a 3 a a 3 a b a b b a (b) Deduce that b a b a. (c) Use a ad b i part (b) to show that a is icreasig. (d) Use a ad b 2 i part (b) to show that a 2 4. (e) Use parts (c) ad (d) to show that a 4 for all. (f) Use Theorem 2 to show that lim l exists. (The limit is e. See Equatio ) 79. Let a ad b be positive umbers with a b. Let a be their arithmetic mea ad b their geometric mea: Repeat this process so that, i geeral, a a a b 2 a b 2 (a) Use mathematical iductio to show that a a b b (b) Deduce that both a ad b are coverget. (c) Show that lim l a lim l b. Gauss called the commo value of these limits the arithmetic-geometric mea of the umbers a ad b. 80. (a) Show that if lim l a 2 L ad lim l a 2 L, the a is coverget ad lim l a L. (b) If a ad a a fid the first eight terms of the sequece a. The use part (a) to show that lim l a s2. This gives the cotiued fractio expasio s The size of a udisturbed fish populatio has bee modeled by the formula p bp a p b sab b sa b where p is the fish populatio after years ad a ad b are positive costats that deped o the species ad its eviromet. Suppose that the populatio i year 0 is p 0 0. (a) Show that if p is coverget, the the oly possible values for its limit are 0 ad b a. (b) Show that p bap. (c) Use part (b) to show that if a b, the lim l p 0; i other words, the populatio dies out. (d) Now assume that a b. Show that if p 0 b a, the p is icreasig ad 0 p b a. Show also that if p 0 b a, the p is decreasig ad p b a. Deduce that if a b, the lim l p b a. 2

14 SECTION.2 SERIES 687 LABORATORY PROJECT CAS LOGISTIC SEQUENCES A sequece that arises i ecology as a model for populatio growth is defied by the logistic differece equatio p kp p where p measures the size of the populatio of the th geeratio of a sigle species. To keep the umbers maageable, p is a fractio of the maximal size of the populatio, so 0 p. Notice that the form of this equatio is similar to the logistic differetial equatio i Sectio 9.4. The discrete model with sequeces istead of cotiuous fuctios is preferable for modelig isect populatios, where matig ad death occur i a periodic fashio. A ecologist is iterested i predictig the size of the populatio as time goes o, ad asks these questios: Will it stabilize at a limitig value? Will it chage i a cyclical fashio? Or will it exhibit radom behavior? Write a program to compute the first terms of this sequece startig with a iitial populatio p 0, where 0 p 0. Use this program to do the followig.. Calculate 20 or 30 terms of the sequece for p 0 2 ad for two values of k such that k 3. Graph the sequeces. Do they appear to coverge? Repeat for a differet value of p 0 betwee 0 ad. Does the limit deped o the choice of p 0? Does it deped o the choice of k? 2. Calculate terms of the sequece for a value of k betwee 3 ad 3.4 ad plot them. What do you otice about the behavior of the terms? 3. Experimet with values of k betwee 3.4 ad 3.5. What happes to the terms? 4. For values of k betwee 3.6 ad 4, compute ad plot at least 00 terms ad commet o the behavior of the sequece. What happes if you chage p 0 by 0.00? This type of behavior is called chaotic ad is exhibited by isect populatios uder certai coditios..2 SERIES If we try to add the terms of a ifiite sequece a we get a expressio of the form a a 2 a 3 a which is called a ifiite series (or just a series) ad is deoted, for short, by the symbol a or a But does it make sese to talk about the sum of ifiitely may terms? It would be impossible to fid a fiite sum for the series because if we start addig the terms we get the cumulative sums, 3, 6, 0, 5, 2,... ad, after the th term, we get 2, which becomes very large as icreases. However, if we start to add the terms of the series

15 688 CHAPTER INFINITE SEQUENCES AND SERIES Sum of first terms we get 2, 4, 8, 6, 32, 64,..., 2,... The table shows that as we add more ad more terms, these partial sums become closer ad closer to. (See also Figure i A Preview of Calculus, page 7.) I fact, by addig sufficietly may terms of the series we ca make the partial sums as close as we like to. So it seems reasoable to say that the sum of this ifiite series is ad to write We use a similar idea to determie whether or ot a geeral series () has a sum. We cosider the partial sums s a s 2 a a 2 s 3 a a 2 a 3 s 4 a a 2 a 3 a 4 ad, i geeral, s a a 2 a 3 a a i i These partial sums form a ew sequece s, which may or may ot have a limit. If lim l s s exists (as a fiite umber), the, as i the precedig example, we call it the sum of the ifiite series a. 2 DEFINITION Give a series a a a 2 a 3, let s deote its th partial sum: s a i a a 2 a i If the sequece s is coverget ad lim l s s exists as a real umber, the the series a is called coverget ad we write a a 2 a s or a s The umber s is called the sum of the series. Otherwise, the series is called diverget. N Compare with the improper itegral y f x dx lim t l y t f x dx To fid this itegral, we itegrate from to t ad the let t l. For a series, we sum from to ad the let l. Thus the sum of a series is the limit of the sequece of partial sums. So whe we write a s, we mea that by addig sufficietly may terms of the series we ca get as close as we like to the umber s. Notice that a lim l a i i EXAPLE A importat example of a ifiite series is the geometric series a ar ar 2 ar 3 ar ar a 0

16 SECTION.2 SERIES 689 N Figure provides a geometric demostratio of the result i Example. If the triagles are costructed as show ad s is the sum of the series, the, by similar triagles, s a a a ar so s a r Each term is obtaied from the precedig oe by multiplyig it by the commo ratio r. (We have already cosidered the special case where a ad r 2 2 o page 687.) If r, the s a a a a l. Sice lim l s does t exist, the geometric series diverges i this case. If r, we have s a ar ar 2 ar ar# ad rs ar ar 2 ar ar ar@ ar@ ar Subtractig these equatios, we get s rs a ar a-ar ar s 3 s a r r a a If r, we kow from (..9) that r l 0 as l, so FIGURE a lim s a r lim l l r r Thus whe the geometric series is coverget ad its sum is a r. If r or r, the sequece r is diverget by (..9) ad so, by Equatio 3, lim l s does ot exist. Therefore the geometric series diverges i those cases. We summarize the results of Example as follows. a r a r lim r l a r 4 The geometric series ar a ar ar 2 N I words: The sum of a coverget geometric series is first term commo ratio is coverget if r ad its sum is ar a r r If r, the geometric series is diverget. V EXAPLE 2 Fid the sum of the geometric series r 2 3 SOLUTION The first term is a 5 ad the commo ratio is r 2 3. Sice, the series is coverget by (4) ad its sum is ( 2 3)

17 690 CHAPTER INFINITE SEQUENCES AND SERIES N What do we really mea whe we say that the sum of the series i Example 2 is 3? Of course, we ca t literally add a ifiite umber of terms, oe by oe. But, accordig to Defiitio 2, the total sum is the limit of the sequece of partial sums. So, by takig the sum of sufficietly may terms, we ca get as close as we like to the umber 3. The table shows the first te partial sums s ad the graph i Figure 2 shows how the sequece of partial sums approaches 3. s s FIGURE 2 N Aother way to idetify a ad r is to write out the first few terms: EXAPLE 3 Is the series coverget or diverget? SOLUTION Let s rewrite the th term of the series i the form ar : ( 4 3) We recogize this series as a geometric series with a 4 ad r 4 3. Sice r, the series diverges by (4). V EXAPLE 4 Write the umber as a ratio of itegers. SOLUTION After the first term we have a geometric series with a 70 3 ad r 0 2. Therefore TEC odule.2 explores a series that depeds o a agle i a triagle ad eables you to see how rapidly the series coverges whe varies. EXAPLE 5 Fid the sum of the series x, where. SOLUTION Notice that this series starts with 0 ad so the first term is x 0. (With series, we adopt the covetio that x 0 eve whe x 0.) Thus This is a geometric series with a ad r x. Sice, it coverges ad (4) gives 5 0 x x x 2 x 3 x 4 0 x 0 x x r x

18 SECTION.2 SERIES 69 EXAPLE 6 Show that the series is coverget, ad fid its sum. SOLUTION This is ot a geometric series, so we go back to the defiitio of a coverget series ad compute the partial sums. s i ii We ca simplify this expressio if we use the partial fractio decompositio ii i i N Notice that the terms cacel i pairs. This is a example of a telescopig sum: Because of all the cacellatios, the sum collapses (like a pirate s collapsig telescope) ito just two terms. N Figure 3 illustrates Example 6 by showig the graphs of the sequece of terms a [ ] ad the sequece s of partial sums. Notice that a l 0 ad s l. See Exercises 62 ad 63 for two geometric iterpretatios of Example 6. (see Sectio 7.4). Thus we have ad so Therefore the give series is coverget ad V s i ii i i lim s lim 0 l l i EXAPLE 7 Show that the harmoic series s a is diverget. SOLUTION For this particular series it s coveiet to cosider the partial sums s 2, s 4, s 8, s 6, s 32,... ad show that they become large. s FIGURE 3 s 2 2 s 4 2 ( 3 4 ) 2 ( 4 4) 2 2 s 8 2 ( 3 4 ) ( ) 2 ( 4 4) ( ) s 6 2 ( 3 4 ) ( 5 8) ( 9 6 ) 2 ( 4 4) ( 8 8) ( 6 6 )

19 692 CHAPTER INFINITE SEQUENCES AND SERIES Similarly, s, s , ad i geeral s 2 2 N The method used i Example 7 for showig that the harmoic series diverges is due to the Frech scholar Nicole Oresme ( ). This shows that s 2 l as l ad so s is diverget. Therefore the harmoic series diverges. 6 THEORE If the series a is coverget, the lim a 0. l PROOF Let s a a 2 a. The a s s. Sice a is coverget, the sequece s is coverget. Let lim l s s. Sice l as l, we also have lim l s s. Therefore lim a lim s s lim s lim l l l s s 0 l s NOTE With ay series a we associate two sequeces: the sequece s of its partial sums ad the sequece a of its terms. If a is coverget, the the limit of the sequece s is s (the sum of the series) ad, as Theorem 6 asserts, the limit of the sequece a is 0. NOTE 2 The coverse of Theorem 6 is ot true i geeral. If lim l a 0, we caot coclude that a is coverget. Observe that for the harmoic series we have a l 0 as l, but we showed i Example 7 that is diverget. 7 THE TEST FOR DIVERGENCE If lim a does ot exist or if lim a 0, the the series l l a is diverget. The Test for Divergece follows from Theorem 6 because, if the series is ot diverget, the it is coverget, ad so lim l a 0. EXAPLE 8 Show that the series SOLUTION diverges. lim a 2 lim lim l l l So the series diverges by the Test for Divergece. NOTE 3 If we fid that lim l a 0, we kow that a is diverget. If we fid that lim l a 0, we kow othig about the covergece or divergece of a. Remember the warig i Note 2: If lim l a 0, the series a might coverge or it might diverge.

20 SECTION.2 SERIES THEORE If a ad b are coverget series, the so are the series ca (where c is a costat), a b, ad a b, ad (i) (iii) ca c a a b a b (ii) a b a b These properties of coverget series follow from the correspodig Limit Laws for Sequeces i Sectio.. For istace, here is how part (ii) of Theorem 8 is proved: Let s a i i s The th partial sum for the series a b is ad, usig Equatio 5.2.0, we have a u i lim u lim l l a i b i lim i l a i i i lim l a i lim i l b i i t a i b i b i i b i t b lim l s lim l t s t Therefore a b is coverget ad its sum is a b s t a b EXAPLE 9 Fid the sum of the series. 3 2 SOLUTION The series 2 is a geometric series with a ad r 2 2, so I Example 6 we foud that So, by Theorem 8, the give series is coverget ad NOTE 4 A fiite umber of terms does t affect the covergece or divergece of a series. For istace, suppose that we were able to show that the series 4 3

21 694 CHAPTER INFINITE SEQUENCES AND SERIES is coverget. Sice it follows that the etire series 3 is coverget. Similarly, if it is kow that the series N a coverges, the the full series is also coverget. a N a a N.2 EXERCISES. (a) What is the differece betwee a sequece ad a series? (b) What is a coverget series? What is a diverget series? 2. Explai what it meas to say that a 5. ; 3 8 Fid at least 0 partial sums of the series. Graph both the sequece of terms ad the sequece of partial sums o the same scree. Does it appear that the series is coverget or diverget? If it is coverget, fid the sum. If it is diverget, explai why ta s s 9. Let a 2. 3 (a) Determie whether a is coverget. (b) Determie whether a is coverget. 0. (a) Explai the differece betwee (b) Explai the differece betwee 20 Determie whether the geometric series is coverget or diverget. If it is coverget, fid its sum a i i a i i ad ad a j j a j i ( 2) Determie whether the series is coverget or diverget. If it is coverget, fid its sum k s l arcta e Determie whether the series is coverget or diverget by expressig s as a telescopig sum (as i Example 6). If it is coverget, fid its sum k l 3 k kk 2 k k cos k k e 2 (s2 ) e

22 SECTION.2 SERIES 695 CAS Express the umber as a ratio of itegers Fid the values of x for which the series coverges. Fid the sum of the series for those values of x x (e e ) cos 2 x 3 0 cos x We have see that the harmoic series is a diverget series whose terms approach 0. Show that l is aother series with this property Use the partial fractio commad o your CAS to fid a coveiet expressio for the partial sum, ad the use this expressio to fid the sum of the series. Check your aswer by usig the CAS to sum the series directly If the th partial sum of a series a is a fid ad. a cos s x 4 x If the th partial sum of a series a is s 3 2, fid ad. a a 57. Whe moey is spet o goods ad services, those who receive the moey also sped some of it. The people receivig some of the twice-spet moey will sped some of that, ad so o. Ecoomists call this chai reactio the multiplier effect. I a hypothetical isolated commuity, the local govermet begis the process by spedig D dollars. Suppose that each recipiet of spet moey speds 00c% ad saves 00s% of the moey that he or she receives. The values c ad s are called the margial propesity to cosume ad the margial propesity to save ad, of course, c s. (a) Let S be the total spedig that has bee geerated after trasactios. Fid a equatio for S. (b) Show that lim l S kd, where k s. The umber k is called the multiplier. What is the multiplier if the margial propesity to cosume is 80%? Note: The federal govermet uses this priciple to justify deficit spedig. Baks use this priciple to justify ledig a large percetage of the moey that they receive i deposits. 58. A certai ball has the property that each time it falls from a height h oto a hard, level surface, it rebouds to a height rh, where 0 r. Suppose that the ball is dropped from a iitial height of H meters. (a) Assumig that the ball cotiues to bouce idefiitely, fid the total distace that it travels. (Use the fact that the ball falls 2 tt 2 meters i t secods.) (b) Calculate the total time that the ball travels. (c) Suppose that each time the ball strikes the surface with velocity v it rebouds with velocity kv, where 0 k. How log will it take for the ball to come to rest? 59. Fid the value of c if c Fid the value of c such that e c I Example 7 we showed that the harmoic series is diverget. Here we outlie aother method, makig use of the fact that e x x for ay x 0. (See Exercise ) If s is the th partial sum of the harmoic series, show that e s. Why does this imply that the harmoic series is diverget? ; 62. Graph the curves y x, 0 x, for 0,, 2, 3, 4,... o a commo scree. By fidig the areas betwee successive curves, give a geometric demostratio of the fact, show i Example 6, that

23 696 CHAPTER INFINITE SEQUENCES AND SERIES 63. The figure shows two circles C ad D of radius that touch at P. T is a commo taget lie; C is the circle that touches C, D, ad T; C 2 is the circle that touches C, D, ad C ; C 3 is the circle that touches C, D, ad C 2. This procedure ca be cotiued idefiitely ad produces a ifiite sequece of circles C. Fid a expressio for the diameter of C ad thus provide aother geometric demostratio of Example A right triagle ABC is give with A ad. CD is draw perpedicular to AB, DE is draw perpedicular to BC, EF AB, ad this process is cotiued idefiitely, as show i the figure. Fid the total legth of all the perpediculars 65. C i terms of b ad. B C C CD DE EF FG What is wrog with the followig calculatio? H (Guido Ubaldus thought that this proved the existece of God because somethig has bee created out of othig. ) 66. Suppose that a a 0 is kow to be a coverget series. Prove that a is a diverget series. 67. Prove part (i) of Theorem If a is diverget ad c 0, show that ca is diverget. P F G C D E C A b D AC b T 69. If a is coverget ad b is diverget, show that the series a b is diverget. [Hit: Argue by cotradictio.] 70. If a ad b are both diverget, is a b ecessarily diverget? 7. Suppose that a series a has positive terms ad its partial sums s satisfy the iequality s 000 for all. Explai why a must be coverget. 72. The Fiboacci sequece was defied i Sectio. by the equatios Show that each of the followig statemets is true. (a) (b) (c) f f f f f f 2 2 f, f 2, f f f f f 2 f f f The Cator set, amed after the Germa mathematicia Georg Cator (845 98), is costructed as follows. We start with the closed iterval [0, ] ad remove the ope iterval ( 3, 2. That leaves the two itervals [0, ad [ 2 3 ) 3 ] 3, ] ad we remove the ope middle third of each. Four itervals remai ad agai we remove the ope middle third of each of them. We cotiue this procedure idefiitely, at each step removig the ope middle third of every iterval that remais from the precedig step. The Cator set cosists of the umbers that remai i [0, ] after all those itervals have bee removed. (a) Show that the total legth of all the itervals that are removed is. Despite that, the Cator set cotais ifiitely may umbers. Give examples of some umbers i the Cator set. (b) The Sierpiski carpet is a two-dimesioal couterpart of the Cator set. It is costructed by removig the ceter oe-ith of a square of side, the removig the ceters of the eight smaller remaiig squares, ad so o. (The figure shows the first three steps of the costructio.) Show that the sum of the areas of the removed squares is. This implies that the Sierpiski carpet has area 0.

24 SECTION.3 THE INTEGRAL TEST AND ESTIATES OF SUS (a) A sequece a is defied recursively by the equatio a 2 a a 2 for 3, where a ad a 2 ca be ay real umbers. Experimet with various values of a ad a 2 ad use your calculator to guess the limit of the sequece. (b) Fid lim l a i terms of a ad a 2 by expressig a a i terms of a 2 a ad summig a series. 75. Cosider the series! (a) Fid the partial sums s, s 2, s 3, ad s 4. Do you recogize the deomiators? Use the patter to guess a formula for s. (b) Use mathematical iductio to prove your guess. (c) Show that the give ifiite series is coverget, ad fid its sum. 76. I the figure there are ifiitely may circles approachig the vertices of a equilateral triagle, each circle touchig other circles ad sides of the triagle. If the triagle has sides of legth, fid the total area occupied by the circles..3 THE INTEGRAL TEST AND ESTIATES OF SUS s i i I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series because i each of those cases we could fid a simple formula for the th partial sum s. But usually it is t easy to compute lim l s. Therefore, i the ext few sectios, we develop several tests that eable us to determie whether a series is coverget or diverget without explicitly fidig its sum. (I some cases, however, our methods will eable us to fid good estimates of the sum.) Our first test ivolves improper itegrals. We begi by ivestigatig the series whose terms are the reciprocals of the squares of the positive itegers: There s o simple formula for the sum s of the first terms, but the computer-geerated table of values give i the margi suggests that the partial sums are approachig a umber ear.64 as l ad so it looks as if the series is coverget. We ca cofirm this impressio with a geometric argumet. Figure shows the curve y x 2 ad rectagles that lie below the curve. The base of each rectagle is a iterval of legth ; the height is equal to the value of the fuctio y x 2 at the right edpoit of the iterval. So the sum of the areas of the rectagles is y y= x FIGURE area= 2@ area= 3@ area= 4@ area= 5@

25 698 CHAPTER INFINITE SEQUENCES AND SERIES If we exclude the first rectagle, the total area of the remaiig rectagles is smaller tha the area uder the curve y x 2 for x, which is the value of the itegral x x 2 dx. I Sectio 7.8 we discovered that this improper itegral is coverget ad has value. So the picture shows that all the partial sums are less tha Thus the partial sums are bouded. We also kow that the partial sums are icreasig (because all the terms are positive). Therefore the partial sums coverge (by the ootoic Sequece Theorem) ad so the series is coverget. The sum of the series (the limit of the partial sums) is also less tha 2: y 2 2 dx 2 x s i si [The exact sum of this series was foud by the Swiss mathematicia Leohard Euler ( ) to be, but the proof of this fact is quite difficult. (See Problem 6 i the Problems Plus followig Chapter 5.)] Now let s look at the series The table of values of s suggests that the partial sums are t approachig a fiite umber, so we suspect that the give series may be diverget. Agai we use a picture for cofirmatio. Figure 2 shows the curve y sx, but this time we use rectagles whose tops lie above the curve. y 2 6 s s s2 s3 s4 s5 y= œ x FIGURE area= area= area= area= œ œ 2 œ 3 œ 4 x The base of each rectagle is a iterval of legth. The height is equal to the value of the fuctio y sx at the left edpoit of the iterval. So the sum of the areas of all the rectagles is s s2 s3 s4 s5 This total area is greater tha the area uder the curve y sx for x, which is equal to the itegral x (sx ) dx. But we kow from Sectio 7.8 that this improper itegral is diverget. I other words, the area uder the curve is ifiite. So the sum of the series must be ifiite; that is, the series is diverget. The same sort of geometric reasoig that we used for these two series ca be used to prove the followig test. (The proof is give at the ed of this sectio.) s

26 SECTION.3 THE INTEGRAL TEST AND ESTIATES OF SUS 699 THE INTEGRAL TEST Suppose f is a cotiuous, positive, decreasig fuctio o, ad let a. The the series f a is coverget if ad oly if the improper itegral f x dx is coverget. I other words: y (i) If f x dx is coverget, the is coverget. y x a a (ii) If f x dx is diverget, the is diverget. NOTE Whe we use the Itegral Test, it is ot ecessary to start the series or the itegral at. For istace, i testig the series 4 Also, it is ot ecessary that f be always decreasig. What is importat is that f be ultimately decreasig, that is, decreasig for x larger tha some umber N. The N a is coverget, so a is coverget by Note 4 of Sectio.2. EXAPLE Test the series we use for covergece or divergece. SOLUTION The fuctio f x x 2 is cotiuous, positive, ad decreasig o, so we use the Itegral Test: y 4 x 3 2 dx y dx lim x 2 t l yt lim t l ta t t dx lim x] x 2 t l ta 4 Thus x is a coverget itegral ad so, by the Itegral Test, the series x 2 dx 2 is coverget N I order to use the Itegral Test we eed to be able to evaluate x f x dx ad therefore we have to be able to fid a atiderivative of f. Frequetly this is difficult or impossible, so we eed other tests for covergece too. EXAPLE 2 For what values of is the series V p coverget? p SOLUTION If p 0, the lim. If, the lim l p l p p 0. I either case, lim l p 0, so the give series diverges by the Test for Divergece (.2.7). If p 0, the the fuctio f x x p is clearly cotiuous, positive, ad decreasig o,. We foud i Chapter 7 [see (7.8.2)] that y p dx coverges if p ad diverges if p x It follows from the Itegral Test that the series p coverges if p ad diverges if 0 p. (For p, this series is the harmoic series discussed i Example 7 i Sectio.2.) The series i Example 2 is called the p-series. It is importat i the rest of this chapter, so we summarize the results of Example 2 for future referece as follows.

27 700 CHAPTER INFINITE SEQUENCES AND SERIES The p-series is coverget if p ad diverget if p. p EXAPLE 3 (a) The series is coverget because it is a p-series with p 3. (b) The series 3 s 3 s 3 2 s 3 3 s 3 4 is diverget because it is a p-series with p 3. NOTE We should ot ifer from the Itegral Test that the sum of the series is equal to the value of the itegral. I fact, whereas y 2 dx x Therefore, i geeral, a y f x dx V EXAPLE 4 Determie whether the series coverges or diverges. SOLUTION The fuctio f x l xx is positive ad cotiuous for x because the logarithm fuctio is cotiuous. But it is ot obvious whether or ot f is decreasig, so we compute its derivative: Thus f x 0 whe l x, that is, x e. It follows that f is decreasig whe x e ad so we ca apply the Itegral Test: y l x x f x xx l x x 2 l l x x 2 dx lim t l y t l x l x 2 t dx lim x t l 2 l t 2 lim t l 2 Sice this improper itegral is diverget, the series l is also diverget by the Itegral Test. ESTIATING THE SU OF A SERIES Suppose we have bee able to use the Itegral Test to show that a series a is coverget ad we ow wat to fid a approximatio to the sum s of the series. Of course, ay partial sum s is a approximatio to s because lim l s s. But how good is such a approximatio? To fid out, we eed to estimate the size of the remaider R s s a a 2 a 3

28 SECTION.3 THE INTEGRAL TEST AND ESTIATES OF SUS 70 y y=ƒ The remaider R is the error made whe s, the sum of the first terms, is used as a approximatio to the total sum. We use the same otatio ad ideas as i the Itegral Test, assumig that f is decreasig o,. Comparig the areas of the rectagles with the area uder y f x for x i Figure 3, we see that a + a x FIGURE 3 Similarly, we see from Figure 4 that R a a 2 y f x dx y y=ƒ R a a 2 y So we have proved the followig error estimate. f x dx a + a x FIGURE 4 2 REAINDER ESTIATE FOR THE INTEGRAL TEST Suppose f k a k, where f is a cotiuous, positive, decreasig fuctio for x ad a is coverget. If R s s,the y f x dx R y f x dx V EXAPLE 5 (a) Approximate the sum of the series 3 by usig the sum of the first 0 terms. Estimate the error ivolved i this approximatio. (b) How may terms are required to esure that the sum is accurate to withi ? SOLUTION I both parts (a) ad (b) we eed to kow x f x dx. With f x x 3, which satisfies the coditios of the Itegral Test, we have (a) y 3 dx lim x t l s Accordig to the remaider estimate i (2), we have R 0 y 0 2x 2 3 dx x So the size of the error is at most (b) Accuracy to withi meas that we have to fid a value of such that R Sice R y t lim t l 2t dx x 2 2 we wat

29 702 CHAPTER INFINITE SEQUENCES AND SERIES Solvig this iequality, we get or s We eed 32 terms to esure accuracy to withi If we add s to each side of the iequalities i (2), we get 3 s y f x dx s s y f x dx because s R s. The iequalities i (3) give a lower boud ad a upper boud for s. They provide a more accurate approximatio to the sum of the series tha the partial sum does. s EXAPLE 6 Use (3) with 0 to estimate the sum of the series. SOLUTION The iequalities i (3) become s 0 y From Example 5 we kow that x dx s s 3 0 y 0 x dx 3 y 3 dx x so s s s Usig s , we get s If we approximate s by the midpoit of this iterval, the the error is at most half the legth of the iterval. So with error If we compare Example 6 with Example 5, we see that the improved estimate i (3) ca be much better tha the estimate s s. To make the error smaller tha we had to use 32 terms i Example 5 but oly 0 terms i Example 6. PROOF OF THE INTEGRAL TEST We have already see the basic idea behid the proof of the Itegral Test i Figures ad 2 for the series 2 ad s. For the geeral series a, look at Figures 5 ad 6. The area of the first shaded rectagle i Figure 5 is the value of f at the right edpoit of, 2,

30 SECTION.3 THE INTEGRAL TEST AND ESTIATES OF SUS 703 y y=ƒ that is, f 2 a 2. So, comparig the areas of the shaded rectagles with the area uder y f x from to, we see that a a a a x FIGURE 5 y y=ƒ a a - a a a a x FIGURE 6 4 (Notice that this iequality depeds o the fact that f is decreasig.) Likewise, Figure 6 shows that 5 (i) If y f x dx sice f x 0. Therefore a 2 a 3 a y f x dx y f x dx a a 2 a is coverget, the (4) gives i2 s a a i y f x dx y f x dx i2 a i a y f x dx, say Sice s for all, the sequece s is bouded above. Also s s a s sice a f 0. Thus s is a icreasig bouded sequece ad so it is coverget by the ootoic Sequece Theorem (..2). This meas that a is coverget. (ii) If is diverget, the x x f x dx f x dx l as l because f x 0. But (5) gives y f x dx a i s i ad so s l. This implies that s l ad so a diverges..3 EXERCISES. Draw a picture to show that 2 What ca you coclude about the series? 2. Suppose f is a cotiuous positive decreasig fuctio for x ad a f. By drawig a picture, rak the followig three quatities i icreasig order: y 6 f x dx y.3 5 a i i dx.3 x 6 a i i2 3 8 Use the Itegral Test to determie whether the series is coverget or diverget s e 8. 5 s 4 2

31 704 CHAPTER INFINITE SEQUENCES AND SERIES 9 26 Determie whether the series is coverget or diverget s l s2 3s3 4s4 5s Fid the values of p for which the series is coverget l p 2 3 e l p The Riema zeta-fuctio is defied by x ad is used i umber theory to study the distributio of prime umbers. What is the domai of? 32. (a) Fid the partial sum s of the series 4 0. Estimate the error i usig s 0 as a approximatio to the sum of the series. (b) Use (3) with 0 to give a improved estimate of the sum. (c) Fid a value of so that is withi of the sum. s x e l 2 4 l ll p l p CAS 33. (a) Use the sum of the first 0 terms to estimate the sum of the series 2. How good is this estimate? (b) Improve this estimate usig (3) with 0. (c) Fid a value of that will esure that the error i the approximatio s s is less tha Fid the sum of the series 5 correct to three decimal places. 35. Estimate 2 6 correct to five decimal places. 36. How may terms of the series 2 l 2 would you eed to add to fid its sum to withi 0.0? 37. Show that if we wat to approximate the sum of the series.00 so that the error is less tha 5 i the ith decimal place, the we eed to add more tha terms! 38. (a) Show that the series l 2 2 is coverget. (b) Fid a upper boud for the error i the approximatio s s. (c) What is the smallest value of such that this upper boud is less tha 0.05? (d) Fid s for this value of. 39. (a) Use (4) to show that if s is the th partial sum of the harmoic series, the s l 0,30 (b) The harmoic series diverges, but very slowly. Use part (a) to show that the sum of the first millio terms is less tha 5 ad the sum of the first billio terms is less tha Use the followig steps to show that the sequece t 2 3 l has a limit. (The value of the limit is deoted by ad is called Euler s costat.) (a) Draw a picture like Figure 6 with f x x ad iterpret t as a area [or use (5)] to show that t 0 for all. (b) Iterpret t t l l as a differece of areas to show that t t 0. Therefore, t is a decreasig sequece. (c) Use the ootoic Sequece Theorem to show that t is coverget. 4. Fid all positive values of b for which the series b l coverges. 42. Fid all values of c for which the followig series coverges. c

32 SECTION.4 THE COPARISON TESTS THE COPARISON TESTS I the compariso tests the idea is to compare a give series with a series that is kow to be coverget or diverget. For istace, the series 2 remids us of the series, which is a geometric series with a ad r ad is therefore coverget. Because the series () is so similar to a coverget series, we have the feelig that it too must be coverget. Ideed, it is. The iequality 2 2 shows that our give series () has smaller terms tha those of the geometric series ad therefore all its partial sums are also smaller tha (the sum of the geometric series). This meas that its partial sums form a bouded icreasig sequece, which is coverget. It also follows that the sum of the series is less tha the sum of the geometric series: 2 Similar reasoig ca be used to prove the followig test, which applies oly to series whose terms are positive. The first part says that if we have a series whose terms are smaller tha those of a kow coverget series, the our series is also coverget. The secod part says that if we start with a series whose terms are larger tha those of a kow diverget series, the it too is diverget. THE COPARISON TEST Suppose that a ad b are series with positive terms. (i) If b is coverget ad a b for all, the a is also coverget. (ii) If b is diverget ad a b for all, the a is also diverget. N It is importat to keep i mid the distictio betwee a sequece ad a series. A sequece is a list of umbers, whereas a series is a sum. With every series a there are associated two sequeces: the sequece a of terms ad the sequece s of partial sums. Stadard Series for Use with the Compariso Test PROOF (i) Let s a i i t b i i t b Sice both series have positive terms, the sequeces s ad t are icreasig s s a s. Also t l t, so t t for all. Sice a i b i, we have s t. Thus s t for all. This meas that s is icreasig ad bouded above ad therefore coverges by the ootoic Sequece Theorem. Thus a coverges. (ii) If b is diverget, the t l (sice t is icreasig). But a i b i so s t. Thus s l. Therefore a diverges. I usig the Compariso Test we must, of course, have some kow series b for the purpose of compariso. ost of the time we use oe of these series: N A p-series [ p coverges if p ad diverges if p ; see (.3.)] N A geometric series [ ar coverges if ad diverges if ; see (.2.4)] r r

33 706 CHAPTER INFINITE SEQUENCES AND SERIES V EXAPLE Determie whether the series coverges or diverges. SOLUTION For large the domiat term i the deomiator is 2 2 so we compare the give series with the series Observe that because the left side has a bigger deomiator. (I the otatio of the Compariso Test, is the left side ad is the right side.) We kow that a b is coverget because it s a costat times a p-series with p 2. Therefore is coverget by part (i) of the Compariso Test. NOTE Although the coditio a b or a b i the Compariso Test is give for all, we eed verify oly that it holds for N, where N is some fixed iteger, because the covergece of a series is ot affected by a fiite umber of terms. This is illustrated i the ext example. V EXAPLE 2 Test the series for covergece or divergece. SOLUTION This series was tested (usig the Itegral Test) i Example 4 i Sectio.3, but it is also possible to test it by comparig it with the harmoic series. Observe that l for 3 ad so We kow that is diverget ( p-series with p ). Thus the give series is diverget by the Compariso Test. NOTE 2 The terms of the series beig tested must be smaller tha those of a coverget series or larger tha those of a diverget series. If the terms are larger tha the terms of a coverget series or smaller tha those of a diverget series, the the Compariso Test does t apply. Cosider, for istace, the series The iequality l l b ( 2 ) is useless as far as the Compariso Test is cocered because is coverget ad a. Noetheless, we have the feelig that 2 b ought to be coverget because it is very similar to the coverget geometric series ( 2). I such cases the followig test ca be used.

34 SECTION.4 THE COPARISON TESTS 707 N Exercises 40 ad 4 deal with the cases c 0 ad c. THE LIIT COPARISON TEST Suppose that a ad b are series with positive terms. If a lim c l b where c is a fiite umber ad c 0, the either both series coverge or both diverge. PROOF Let m ad be positive umbers such that m c. Because a b is close to c for large, there is a iteger N such that m a b whe N ad so mb a b whe N If b coverges, so does b. Thus a coverges by part (i) of the Compariso Test. If b diverges, so does mb ad part (ii) of the Compariso Test shows that a diverges. EXAPLE 3 Test the series 2 SOLUTION We use the Limit Compariso Test with for covergece or divergece. ad obtai a 2 b 2 a 2 lim lim lim l b l 2 l Sice this limit exists ad 2 is a coverget geometric series, the give series coverges by the Limit Compariso Test. EXAPLE 4 Determie whether the series 2 lim 2 l s coverges or diverges. SOLUTION The domiat part of the umerator is 2 2 ad the domiat part of the deomiator is s This suggests takig a 22 3 s5 5 b a lim lim lim l b l s5 5 2 l 2s5 5 lim l s0

35 708 CHAPTER INFINITE SEQUENCES AND SERIES b 2 2 Sice is diverget ( p-series with p 2 ), the give series diverges by the Limit Compariso Test. Notice that i testig may series we fid a suitable compariso series b by keepig oly the highest powers i the umerator ad deomiator. ESTIATING SUS If we have used the Compariso Test to show that a series a coverges by compariso with a series b, the we may be able to estimate the sum a by comparig remaiders. As i Sectio.3, we cosider the remaider For the compariso series b we cosider the correspodig remaider Sice a b for all, we have R T. If b is a p-series, we ca estimate its remaider T as i Sectio.3. If b is a geometric series, the T is the sum of a geometric series ad we ca sum it exactly (see Exercises 35 ad 36). I either case we kow that R is smaller tha. V EXAPLE 5 Use the sum of the first 00 terms to approximate the sum of the series 3. Estimate the error ivolved i this approximatio. SOLUTION Sice T the give series is coverget by the Compariso Test. The remaider T for the compariso series 3 was estimated i Example 5 i Sectio.3 usig the Remaider Estimate for the Itegral Test. There we foud that Therefore the remaider R R s s a a 2 T t t b b 2 T y dx x 2 2 for the give series satisfies R T 2 2 With 00 we have R Usig a programmable calculator or a computer, we fid that with error less tha

36 SECTION.4 THE COPARISON TESTS EXERCISES. Suppose a ad b are series with positive terms ad b is kow to be coverget. (a) If a b for all, what ca you say about a? Why? (b) If a b for all, what ca you say about a? Why? 2. Suppose a ad b are series with positive terms ad b is kow to be diverget. (a) If a b for all, what ca you say about a? Why? (b) If a b for all, what ca you say about a? Why? 3 32 Determie whether the series coverges or diverges s arcta s cos s si 0 0 s s Use the sum of the first 0 terms to approximate the sum of the series. Estimate the error s The meaig of the decimal represetatio of a umber 0.d d 2d 3...(where the digit d i is oe of the umbers 0,, 2,...,9) is that 0.d d 2d 3d 4... d 0 Show that this series always coverges. 38. For what values of p does the series 2 p l coverge? 39. Prove that if ad coverges, the 2 a 0 a a also coverges. 40. (a) Suppose that a ad b are series with positive terms ad b is coverget. Prove that if the a is also coverget. (b) Use part (a) to show that the series coverges. (i) l 3 d2 0 a lim 0 l b (ii) d3 d4 2 3 si l s e s (a) Suppose that a ad b are series with positive terms ad b is diverget. Prove that if s s ! si e e! s a lim l b the a is also diverget. (b) Use part (a) to show that the series diverges. (i) (ii) l 2 l 42. Give a example of a pair of series a ad b with positive terms where lim l a b 0 ad b diverges, but a coverges. (Compare with Exercise 40.) 43. Show that if a 0 ad lim l a 0, the a is diverget. 44. Show that if a 0 ad a is coverget, the l a is coverget. 45. If a is a coverget series with positive terms, is it true that sia is also coverget? 46. If a ad b are both coverget series with positive terms, is it true that a b is also coverget?

37 70 CHAPTER INFINITE SEQUENCES AND SERIES.5 ALTERNATING SERIES The covergece tests that we have looked at so far apply oly to series with positive terms. I this sectio ad the ext we lear how to deal with series whose terms are ot ecessarily positive. Of particular importace are alteratig series, whose terms alterate i sig. A alteratig series is a series whose terms are alterately positive ad egative. Here are two examples: We see from these examples that the th term of a alteratig series is of the form a b where b is a positive umber. (I fact, b a.) The followig test says that if the terms of a alteratig series decrease toward 0 i absolute value, the the series coverges. or a b THE ALTERNATING SERIES TEST If the alteratig series b b b 2 b 3 b 4 b 5 b 6 b 0 satisfies (i) (ii) b b lim b 0 l for all the the series is coverget. Before givig the proof let s look at Figure, which gives a picture of the idea behid the proof. We first plot s b o a umber lie. To fid s 2 we subtract b 2, so s 2 is to the left of s. The to fid s 3 we add b 3, so s 3 is to the right of s 2. But, sice b 3 b 2, s 3 is to the left of s. Cotiuig i this maer, we see that the partial sums oscillate back ad forth. Sice b l 0, the successive steps are becomig smaller ad smaller. The eve partial sums s 2, s 4, s 6,... are icreasig ad the odd partial sums s, s 3, s 5,... are decreasig. Thus it seems plausible that both are covergig to some umber s, which is the sum of the series. Therefore we cosider the eve ad odd partial sums separately i the followig proof. b -b +b -b +b -bß FIGURE 0 s s sß s s s s

38 SECTION.5 ALTERNATING SERIES 7 PROOF OF THE ALTERNATING SERIES TEST We first cosider the eve partial sums: I geeral s 2 b b 2 0 s 4 s 2 b 3 b 4 s 2 s 2 s 22 b 2 b 2 s 22 sice b 2 b sice b 4 b 3 sice b 2 b 2 Thus 0 s 2 s 4 s 6 s 2 But we ca also write s 2 b b 2 b 3 b 4 b 5 b 22 b 2 b 2 Every term i brackets is positive, so s 2 b for all. Therefore the sequece s 2 of eve partial sums is icreasig ad bouded above. It is therefore coverget by the ootoic Sequece Theorem. Let s call its limit s, that is, lim s 2 s l Now we compute the limit of the odd partial sums: lim s 2 lim s 2 b 2 l l lim l s 2 lim l b 2 s 0 s [by coditio (ii)] N Figure 2 illustrates Example by showig the graphs of the terms a ad the partial sums s. Notice how the values of s zigzag across the limitig value, which appears to be about 0.7. I fact, it ca be proved that the exact sum of the series is l (see Exercise 36). Sice both the eve ad odd partial sums coverge to s, we have lim l s s [see Exercise 80(a) i Sectio.] ad so the series is coverget. V EXAPLE The alteratig harmoic series s satisfies (i) (ii) b b because lim b lim l l 0 a 0 so the series is coverget by the Alteratig Series Test. V EXAPLE 2 The series 3 4 is alteratig but FIGURE 2 lim b 3 3 lim lim l l 4 l 4 3 4

39 72 CHAPTER INFINITE SEQUENCES AND SERIES so coditio (ii) is ot satisfied. Istead, we look at the limit of the th term of the series: lim a 3 lim l l 4 This limit does ot exist, so the series diverges by the Test for Divergece. EXAPLE 3 Test the series 2 3 for covergece or divergece. SOLUTION The give series is alteratig so we try to verify coditios (i) ad (ii) of the Alteratig Series Test. Ulike the situatio i Example, it is ot obvious that the sequece give by b 2 3 is decreasig. However, if we cosider the related fuctio f x x 2 x 3, we fid that f x x2 x 3 x 3 2 N Istead of verifyig coditio (i) of the Alteratig Series Test by computig a derivative, we could verify that b b directly by usig the techique of Solutio of Example 2 i Sectio.. Sice we are cosiderig oly positive x, we see that f x 0 if 2 x 3 0, that is, x s 3 2. Thus f is decreasig o the iterval (s 3 2, ). This meas that f f ad therefore b b whe 2. (The iequality b 2 b ca be verified directly but all that really matters is that the sequece b is evetually decreasig.) Coditio (ii) is readily verified: lim b 2 lim lim l l 3 l 3 0 Thus the give series is coverget by the Alteratig Series Test. ESTIATING SUS A partial sum s of ay coverget series ca be used as a approximatio to the total sum s, but this is ot of much use uless we ca estimate the accuracy of the approximatio. The error ivolved i usig s s is the remaider R s s. The ext theorem says that for series that satisfy the coditios of the Alteratig Series Test, the size of the error is smaller tha, which is the absolute value of the first eglected term. b N You ca see geometrically why the Alteratig Series Estimatio Theorem is true by lookig at Figure (o page 70). Notice that s s s 4 b 5, s5 b 6, ad so o. Notice also that s lies betwee ay two cosecutive partial sums. ALTERNATING SERIES ESTIATION THEORE If s b is the sum of a alteratig series that satisfies the (i) 0 b b ad (ii) R s s b lim b 0 l PROOF We kow from the proof of the Alteratig Series Test that s lies betwee ay two cosecutive partial sums ad. It follows that s s s s s s b

40 SECTION.5 ALTERNATING SERIES 73 EXAPLE 4 Fid the sum of the series V correct to three decimal places. 0! (By defiitio, 0!.) SOLUTION We first observe that the series is coverget by the Alteratig Series Test because (i)!!! (ii) 0 so as l! l 0! l 0 To get a feel for how may terms we eed to use i our approximatio, let s write out the first few terms of the series: s 0!! 2! 3! 4! 5! 6! 7! Notice that ad b s N I Sectio.0 we will prove that e x 0 x! for all x, so what we have obtaied i Example 4 is actually a approximatio to the umber e. By the Alteratig Series Estimatio Theorem we kow that s s 6 b This error of less tha does ot affect the third decimal place, so we have s correct to three decimal places. NOTE The rule that the error (i usig s to approximate s) is smaller tha the first eglected term is, i geeral, valid oly for alteratig series that satisfy the coditios of the Alteratig Series Estimatio Theorem. The rule does ot apply to other types of series..5 EXERCISES. (a) What is a alteratig series? (b) Uder what coditios does a alteratig series coverge? (c) If these coditios are satisfied, what ca you say about the remaider after terms? 2 20 Test the series for covergece or divergece s2 s3 s4 s5 s l 4 s cos l si ! 34 e l si2! cos 5 s 2s

41 74 CHAPTER INFINITE SEQUENCES AND SERIES ; 2 22 Calculate the first 0 partial sums of the series ad graph both the sequece of terms ad the sequece of partial sums o the same scree. Estimate the error i usig the 0th partial sum to approximate the total sum Show that the series is coverget. How may terms of the series do we eed to add i order to fid the sum to the idicated accuracy? Approximate the sum of the series correct to four decimal places ! e 5 ( error ) ( error 0.000) ( error ) ( error 0.0) 3 8 3! 3. Is the 50th partial sum s 50 of the alteratig series a overestimate or a uderestimate of the total sum? Explai For what values of p is each series coverget? 32. p p 35. Show that the series b, where b if is odd ad b 2 if is eve, is diverget. Why does the Alteratig Series Test ot apply? 36. Use the followig steps to show that h s Let ad be the partial sums of the harmoic ad alteratig harmoic series. (a) Show that s 2 h 2 h. (b) From Exercise 40 i Sectio.3 we have h l l as l ad therefore l 2 h 2 l2 l Use these facts together with part (a) to show that s 2 l l 2 as l. p l 2 as l.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS Give ay series a, we ca cosider the correspodig series a a a 2 a 3 whose terms are the absolute values of the terms of the origial series. N We have covergece tests for series with positive terms ad for alteratig series. But what if the sigs of the terms switch back ad forth irregularly? We will see i Example 3 that the idea of absolute covergece sometimes helps i such cases. DEFINITION A series a is called absolutely coverget if the series of absolute values is coverget. a a a Notice that if a is a series with positive terms, the ad so absolute covergece is the same as covergece i this case. EXAPLE The series

42 SECTION.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 75 is absolutely coverget because 2 is a coverget p-series ( p 2). EXAPLE 2 We kow that the alteratig harmoic series is coverget (see Example i Sectio.5), but it is ot absolutely coverget because the correspodig series of absolute values is which is the harmoic series ( p-series with p ) ad is therefore diverget. 2 DEFINITION A series a is called coditioally coverget if it is coverget but ot absolutely coverget. Example 2 shows that the alteratig harmoic series is coditioally coverget. Thus it is possible for a series to be coverget but ot absolutely coverget. However, the ext theorem shows that absolute covergece implies covergece. 3 THEORE If a series a is absolutely coverget, the it is coverget. 0.5 FIGURE s a 0 a N Figure shows the graphs of the terms ad partial sums s of the series i Example 3. Notice that the series is ot alteratig but has positive ad egative terms. PROOF Observe that the iequality is true because is either a or a. If a is absolutely coverget, the is coverget, so 2 a is coverget. Therefore, by the Compariso Test, is coverget. The is the differece of two coverget series ad is therefore coverget. V a EXAPLE 3 Determie whether the series is coverget or diverget. 0 a a 2 a a (a a ) a cos cos cos 2 cos SOLUTION This series has both positive ad egative terms, but it is ot alteratig. (The first term is positive, the ext three are egative, ad the followig three are positive: The sigs chage irregularly.) We ca apply the Compariso Test to the series of absolute values cos 2 cos 2 a (a a )

43 76 CHAPTER INFINITE SEQUENCES AND SERIES Sice cos for all, we have cos 2 2 We kow that 2 is coverget ( p-series with p 2) ad therefore is coverget by the Compariso Test. Thus the give series cos 2 cos 2 is absolutely coverget ad therefore coverget by Theorem 3. The followig test is very useful i determiig whether a give series is absolutely coverget. THE RATIO TEST (i) If, the the series lim a a is absolutely coverget l a L (ad therefore coverget). (ii) If or, the the series lim a lim a l a a l a L is diverget. (iii) If lim a, the Ratio Test is icoclusive; that is, o coclusio ca be l a draw about the covergece or divergece of a. PROOF (i) The idea is to compare the give series with a coverget geometric series. Sice L, we ca choose a umber r such that L r. Sice lim a ad L r l a L the ratio such that a a a a 4 or, equivaletly, will evetually be less tha r; that is, there exists a iteger N r wheever N Puttig successively equal to N, N, N 2,... i (4),we obtai ad, i geeral, a a r wheever N a N a N r a N2 a N r a N r 2 a N3 a N2 r a N r 3 5 a Nk a N r k for all k

44 SECTION.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 77 Now the series is coverget because it is a geometric series with 0 r. So the iequality (5), together with the Compariso Test, shows that the series is also coverget. It follows that the series is coverget. (Recall that a fiite umber of terms does t affect covergece.) Therefore a is absolutely coverget. (ii) If a a or a a, the the ratio a l l L a will evetually be greater tha ; that is, there exists a iteger N such that wheever N This meas that wheever N ad so a Therefore diverges by the Test for Divergece. NOTE Part (iii) of the Ratio Test says that if iformatio. For istace, for the coverget series 2 we have a a k a N r k a N r a N r 2 a N r 3 a N a a 2 2 whereas for the diverget series we have k a Nk a N a N2 a N3 a a a lim a 0 l lim l a a 2 2 l 2, the test gives o as l a a l as l Therefore, if lim l a a, the series a might coverge or it might diverge. I this case the Ratio Test fails ad we must use some other test. N ESTIATING SUS I the last three sectios we used various methods for estimatig the sum of a series the method depeded o which test was used to prove covergece. What about series for which the Ratio Test works? There are two possibilities: If the series happes to be a alteratig series, as i Example 4, the it is best to use the methods of Sectio.5. If the terms are all positive, the use the special methods explaied i Exercise 34. EXAPLE 4 Test the series for absolute covergece. SOLUTION We use the Ratio Test with a 3 3 : a a l 3 3 3

45 78 CHAPTER INFINITE SEQUENCES AND SERIES Thus, by the Ratio Test, the give series is absolutely coverget ad therefore coverget. V EXAPLE 5 Test the covergece of the series. SOLUTION Sice the terms a!! are positive, we do t eed the absolute value sigs. a a!!!! l e as l (See Equatio ) Sice e, the give series is diverget by the Ratio Test. NOTE Although the Ratio Test works i Example 5, a easier method is to use the Test for Divergece. Sice a! 2 3 it follows that a does ot approach 0 as l. Therefore the give series is diverget by the Test for Divergece. The followig test is coveiet to apply whe th powers occur. Its proof is similar to the proof of the Ratio Test ad is left as Exercise 37. THE ROOT TEST (i) If lim L, the the series is absolutely coverget (ad therefore coverget). (ii) If lim L or lim, the the series is diverget. (iii) If lim l s a l s a l s a l s a a, the Root Test is icoclusive. a If lim l s a, the part (iii) of the Root Test says that the test gives o iformatio. The series a could coverge or diverge. (If L i the Ratio Test, do t try the Root Test because L will agai be. Ad if L i the Root Test, do t try the Ratio Test because it will fail too.) EXAPLE 6 Test the covergece of the series 2 3 V. 3 2 SOLUTION a s a l 2 3 Thus the give series coverges by the Root Test.

46 SECTION.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS 79 REARRANGEENTS N Addig these zeros does ot affect the sum of the series; each term i the sequece of partial sums is repeated, but the limit is the same. The questio of whether a give coverget series is absolutely coverget or coditioally coverget has a bearig o the questio of whether ifiite sums behave like fiite sums. If we rearrage the order of the terms i a fiite sum, the of course the value of the sum remais uchaged. But this is ot always the case for a ifiite series. By a rearragemet of a ifiite series a we mea a series obtaied by simply chagig the order of the terms. For istace, a rearragemet of a could start as follows: It turs out that if a is a absolutely coverget series with sum s, the ay rearragemet of a has the same sum s. However, ay coditioally coverget series ca be rearraged to give a differet sum. To illustrate this fact let s cosider the alteratig harmoic series 6 (See Exercise 36 i Sectio.5.) If we multiply this series by, we get Isertig zeros betwee the terms of this series, we have 7 Now we add the series i Equatios 6 ad 7 usig Theorem.2.8: 8 a a 2 a 5 a 3 a 4 a 5 a 6 a 7 a l l l l 2 Notice that the series i (8) cotais the same terms as i (6), but rearraged so that oe egative term occurs after each pair of positive terms. The sums of these series, however, are differet. I fact, Riema proved that if a is a coditioally coverget series ad r is ay real umber whatsoever, the there is a rearragemet of a that has a sum equal to r. A proof of this fact is outlied i Exercise EXERCISES. What ca you say about the series a i each of the followig cases? (a) lim (b) lim a l a a 0.8 l a 8 (c) lim a l a 2 28 Determie whether the series is absolutely coverget, coditioally coverget, or diverget ! s e k k( 2 3 ) k si 4 4! 00 s !

47 720 CHAPTER INFINITE SEQUENCES AND SERIES arcta l ! The terms of a series are defied recursively by the equatios Determie whether a coverges or diverges. 30. A series a is defied by the equatios Determie whether a coverges or diverges. 3. For which of the followig series is the Ratio Test icoclusive (that is, it fails to give a defiite aswer)? (a) (c) 3 2 cos3! s 3 5 5! a 2 a ! ! 2! a a 32. For which positive itegers k is the followig series coverget?! 2 k! 33. (a) Show that 0 x! coverges for all x. (b) Deduce that lim l x! 0 for all x. 34. Let a be a series with positive terms ad let r a a. Suppose that lim l r L, so a coverges by the (b) (d) 3 cos 23 2! ! a 2 cos s 2 5 l s 2 a Ratio Test. As usual, we let R be the remaider after terms, that is, R a a 2 a 3 (a) If r is a decreasig sequece ad r, show, by summig a geometric series, that (b) If r is a icreasig sequece, show that 35. (a) Fid the partial sum s of the series 2 5. Use Exercise 34 to estimate the error i usig s 5 as a approximatio to the sum of the series. (b) Fid a value of so that s is withi of the sum. Use this value of to approximate the sum of the series. 36. Use the sum of the first 0 terms to approximate the sum of the series 2 Use Exercise 34 to estimate the error. 37. Prove the Root Test. [Hit for part (i): Take ay umber r such that L r ad use the fact that there is a iteger N such that wheever N.] s a r 38. Aroud 90, the Idia mathematicia Sriivasa Ramauja discovered the formula 2s2 4! ! William Gosper used this series i 985 to compute the first 7 millio digits of. (a) Verify that the series is coverget. (b) How may correct decimal places of do you get if you use just the first term of the series? What if you use two terms? 39. Give ay series a, we defie a series a whose terms are all the positive terms of a ad a series a whose terms are all the egative terms of a. To be specific, we let a a a 2 R r R L a a a a a 2 Notice that if a, the a ad a 0 a 0, whereas if a, the a ad a 0 a 0. (a) If a is absolutely coverget, show that both of the series a ad a are coverget. (b) If a is coditioally coverget, show that both of the series a ad a are diverget. 40. Prove that if a is a coditioally coverget series ad r is ay real umber, the there is a rearragemet of a whose sum is r. [Hits: Use the otatio of Exercise 39. Take just eough positive terms a so that their sum is greater tha r. The add just eough egative terms a so that the cumulative sum is less tha r. Cotiue i this maer ad use Theorem.2.6.]

48 SECTION.7 STRATEGY FOR TESTING SERIES 72.7 STRATEGY FOR TESTING SERIES We ow have several ways of testig a series for covergece or divergece; the problem is to decide which test to use o which series. I this respect, testig series is similar to itegratig fuctios. Agai there are o hard ad fast rules about which test to apply to a give series, but you may fid the followig advice of some use. It is ot wise to apply a list of the tests i a specific order util oe fially works. That would be a waste of time ad effort. Istead, as with itegratio, the mai strategy is to classify the series accordig to its form.. If the series is of the form p, it is a p-series, which we kow to be coverget if p ad diverget if p. 2. If the series has the form ar or ar, it is a geometric series, which coverges if r ad diverges if r. Some prelimiary algebraic maipulatio may be required to brig the series ito this form. 3. If the series has a form that is similar to a p-series or a geometric series, the oe of the compariso tests should be cosidered. I particular, if a is a ratioal fuctio or a algebraic fuctio of (ivolvig roots of polyomials), the the series should be compared with a p-series. Notice that most of the series i Exercises.4 have this form. (The value of p should be chose as i Sectio.4 by keepig oly the highest powers of i the umerator ad deomiator.) The compariso tests apply oly to series with positive terms, but if a has some egative terms, the we ca apply the Compariso Test to a ad test for absolute covergece. 4. If you ca see at a glace that lim l a 0, the the Test for Divergece should be used. 5. If the series is of the form b or b, the the Alteratig Series Test is a obvious possibility. 6. Series that ivolve factorials or other products (icludig a costat raised to the th power) are ofte coveietly tested usig the Ratio Test. Bear i mid that a a l as l for all p-series ad therefore all ratioal or algebraic fuctios of. Thus the Ratio Test should ot be used for such series. 7. If is of the form b, the the Root Test may be useful. 8. If a f, where f x dx is easily evaluated, the the Itegral Test is effective (assumig the hypotheses of this test are satisfied). I the followig examples we do t work out all the details but simply idicate which tests should be used. V a EXAPLE x 2 Sice a l 2 0 as l, we should use the Test for Divergece. EXAPLE 2 Sice a s is a algebraic fuctio of, we compare the give series with a p-series. The

49 722 CHAPTER INFINITE SEQUENCES AND SERIES compariso series for the Limit Compariso Test is b, where b s V EXAPLE 3 e 2 Sice the itegral x dx is easily evaluated, we use the Itegral Test. The Ratio Test xex2 also works. EXAPLE Sice the series is alteratig, we use the Alteratig Series Test. EXAPLE 5 2 k V k k! Sice the series ivolves k!, we use the Ratio Test. EXAPLE Sice the series is closely related to the geometric series 3, we use the Compariso Test..7 EXERCISES 38 Test the series for covergece or divergece sl 2 k 9. k 2 e k l ! 5.! l s 2 2 k s k 2 2 k k! k 2! 2 e 3 si si 2 2 k 5 5 k ta 24.! e k l k 28. k 3 k cosh k k 4 k 33. si 34. s k 2 (s 2 ) si 2 5 e 2 j j! 4 s cos 2 l l sj j 5 (s 2 )

50 SECTION.8 POWER SERIES POWER SERIES A power series is a series of the form N TRIGONOETRIC SERIES A power series is a series i which each term is a power fuctio. A trigoometric series is a series whose terms are trigoometric fuctios. This type of series is discussed o the website Click o Additioal Topics ad the o Fourier Series. N a cos x b si x 0 Notice that! ! where x is a variable ad the c s are costats called the coefficiets of the series. For each fixed x, the series () is a series of costats that we ca test for covergece or divergece. A power series may coverge for some values of x ad diverge for other values of x. The sum of the series is a fuctio whose domai is the set of all x for which the series coverges. Notice that f resembles a polyomial. The oly differece is that f has ifiitely may terms. For istace, if we take c for all, the power series becomes the geometric series which coverges whe x ad diverges whe (see Equatio.2.5). ore geerally, a series of the form is called a power series i x a or a power series cetered at a or a power series about a. Notice that i writig out the term correspodig to 0 i Equatios ad 2 we have adopted the covetio that x a 0 eve whe x a. Notice also that whe x a, all of the terms are 0 for ad so the power series (2) always coverges whe x a. V SOLUTION We use the Ratio Test. If we let the a!x. If x 0, we have lim a l a lim l coverget?, as usual, deote the th term of the series, By the Ratio Test, the series diverges whe x 0. Thus the give series coverges oly whe x 0. V 2 EXAPLE For what values of is the series x!x 0 EXAPLE 2 For what values of does the series x 3 x SOLUTION Let a x 3. The a a c x c 0 c x c 2 x 2 c 3 x 3 0 f x c 0 c x c 2 x 2 c x x x x 2 x 0 c x a c 0 c x a c 2 x a 2 0 x 3 a!x!x x 3 x 3 l x 3 x lim l x coverge? as l

51 724 CHAPTER INFINITE SEQUENCES AND SERIES By the Ratio Test, the give series is absolutely coverget, ad therefore coverget, whe ad diverget whe. Now x 3 x 3 x 3 &? x 3 &? 2 x 4 so the series coverges whe 2 x 4 ad diverges whe x 2 or x 4. The Ratio Test gives o iformatio whe x 3 so we must cosider x 2 ad x 4 separately. If we put x 4 i the series, it becomes, the harmoic series, which is diverget. If x 2, the series is, which coverges by the Alteratig Series Test. Thus the give power series coverges for 2 x 4. Natioal Film Board of Caada We will see that the mai use of a power series is that it provides a way to represet some of the most importat fuctios that arise i mathematics, physics, ad chemistry. I particular, the sum of the power series i the ext example is called a Bessel fuctio, after the Germa astroomer Friedrich Bessel ( ), ad the fuctio give i Exercise 35 is aother example of a Bessel fuctio. I fact, these fuctios first arose whe Bessel solved Kepler s equatio for describig plaetary motio. Sice that time, these fuctios have bee applied i may differet physical situatios, icludig the temperature distributio i a circular plate ad the shape of a vibratig drumhead. EXAPLE 3 Fid the domai of the Bessel fuctio of order 0 defied by J 0 x 0 x 2 2 2! 2 N Notice how closely the computer-geerated model (which ivolves Bessel fuctios ad cosie fuctios) matches the photograph of a vibratig rubber membrae. SOLUTION Let a x 2 2 2! 2. The a a x 2 2 2! 2 2! 2 2 x 2 x ! 22! 2 2 x 2 x l 0 for all x Thus, by the Ratio Test, the give series coverges for all values of x. I other words, the domai of the Bessel fuctio is,. Recall that the sum of a series is equal to the limit of the sequece of partial sums. So whe we defie the Bessel fuctio i Example 3 as the sum of a series we mea that, for every real umber x, J 0 x lim l s x The first few partial sums are J 0 where s x i0 i x 2i 2 2i i! 2 s 0 x s x x 2 4 s 2 x x 2 4 x 4 64 s 3 x x 2 4 x 4 64 x s 4 x x 2 4 x 4 64 x x 8 47,456

52 SECTION.8 POWER SERIES 725 y s 0 x FIGURE Partial sums of the Bessel fuctio J y s s s y=j (x) s J Figure shows the graphs of these partial sums, which are polyomials. They are all approximatios to the fuctio J 0, but otice that the approximatios become better whe more terms are icluded. Figure 2 shows a more complete graph of the Bessel fuctio. For the power series that we have looked at so far, the set of values of x for which the series is coverget has always tured out to be a iterval [a fiite iterval for the geometric series ad the series i Example 2, the ifiite iterval, i Example 3, ad a collapsed iterval 0, 0 0 i Example ]. The followig theorem, proved i Appedix F, says that this is true i geeral. 3 THEORE For a give power series c x a, there are oly three 0 possibilities: (i) The series coverges oly whe x a. (ii) The series coverges for all x. (iii) There is a positive umber R such that the series coverges if ad diverges if. x a R x a R _0 FIGURE x The umber R i case (iii) is called the radius of covergece of the power series. By covetio, the radius of covergece is R 0 i case (i) ad R i case (ii). The iterval of covergece of a power series is the iterval that cosists of all values of x for which the series coverges. I case (i) the iterval cosists of just a sigle poit a. I case (ii) the iterval is,. I case (iii) ote that the iequality x a R ca be rewritte as a R x a R. Whe x is a edpoit of the iterval, that is, x a R, aythig ca happe the series might coverge at oe or both edpoits or it might diverge at both edpoits. Thus i case (iii) there are four possibilities for the iterval of covergece: a R, a Ra R, a Ra R, a Ra R, a R The situatio is illustrated i Figure 3. covergece for x-a <R FIGURE 3 a-r a a+r divergece for x-a >R We summarize here the radius ad iterval of covergece for each of the examples already cosidered i this sectio. Series Radius of covergece Iterval of covergece Geometric series x 0 R, Example! x 0 R 0 0 Example 2 x 3 R 2, 4 Example 3 x ! 2 R,

53 726 CHAPTER INFINITE SEQUENCES AND SERIES I geeral, the Ratio Test (or sometimes the Root Test) should be used to determie the radius of covergece R. The Ratio ad Root Tests always fail whe x is a edpoit of the iterval of covergece, so the edpoits must be checked with some other test. EXAPLE 4 Fid the radius of covergece ad iterval of covergece of the series 0 3 x s SOLUTION Let a 3 x s. The a a By the Ratio Test, the give series coverges if ad diverges if. Thus it coverges if x ad diverges if x 3 x 3 x. This meas that the radius of covergece is R We kow the series coverges i the iterval ( 3, 3), but we must ow test for covergece at the edpoits of this iterval. If x 3, the series becomes 3 ( 3 ) 0 s 0 3 x s s 2 3 x s s s2 s3 s4 which diverges. (Use the Itegral Test or simply observe that it is a p-series with p.) If x 2 3, the series is 3 ( 3) 0 s 0 3x x l 3 x as l s which coverges by the Alteratig Series Test. Therefore the give power series coverges whe, so the iterval of covergece is ( 3, 3 x 3 3]. V EXAPLE 5 Fid the radius of covergece ad iterval of covergece of the series x SOLUTION If a x 2 3, the a a x x x 2 Usig the Ratio Test, we see that the series coverges if 3 ad it diverges if. So it coverges if x 2 3 ad diverges if x 2 3. Thus the radius of covergece is R 3. x 2 3 x 2 l x 2 3 as l

54 SECTION.8 POWER SERIES 727 x 2 3 The iequality ca be writte as 5 x, so we test the series at the edpoits 5 ad. Whe x 5, the series is which diverges by the Test for Divergece [ does t coverge to 0]. Whe x, the series is which also diverges by the Test for Divergece. Thus the series coverges oly whe 5 x, so the iterval of covergece is 5,..8 EXERCISES. What is a power series? 2. (a) What is the radius of covergece of a power series? How do you fid it? (b) What is the iterval of covergece of a power series? How do you fid it? 3 28 Fid the radius of covergece ad iterval of covergece of the series. 3. x x x 2. s l x 4 8. s x , 22. x a b 0 b x s x 0! 3 2 x x x 0 s x x 0 x 3 3x 2 3 x x 2 2! x x 4 x x !2x (a) c 2 (b) c Suppose that 0 c x coverges whe x 4 ad diverges whe x 6. What ca be said about the covergece or divergece of the followig series? (a) (c) x 3 5 2!x If c 4 is coverget, does it follow that the followig series are coverget? c 0 c 3 0 (b) (d) 2 c 8 0 c If k is a positive iteger, fid the radius of covergece of the series! k 0 k! x 2 x x 2 l Let p ad q be real umbers with p q. Fid a power series whose iterval of covergece is (a) p, q (b) p, q (c) p, q (d) p, q 33. Is it possible to fid a power series whose iterval of covergece is 0,? Explai. 24.

55 728 CHAPTER INFINITE SEQUENCES AND SERIES ; 34. Graph the first several partial sums s x of the series 0 x, together with the sum fuctio f x x, o a commo scree. O what iterval do these partial sums appear to be covergig to f x? 35. The fuctio defied by is called the Bessel fuctio of order. (a) Fid its domai. ; (b) Graph the first several partial sums o a commo scree. CAS (c) If your CAS has built-i Bessel fuctios, graph J o the same scree as the partial sums i part (b) ad observe how the partial sums approximate. 36. The fuctio A defied by Ax x 3 J J x 0 x 2!! x x is called the Airy fuctio after the Eglish mathematicia ad astroomer Sir George Airy (80 892). (a) Fid the domai of the Airy fuctio. ; (b) Graph the first several partial sums o a commo scree. J CAS (c) If your CAS has built-i Airy fuctios, graph A o the same scree as the partial sums i part (b) ad observe how the partial sums approximate A. 37. A fuctio f is defied by f x 2x x 2 2x 3 x 4 that is, its coefficiets are c 2 ad c 2 2 for all 0. Fid the iterval of covergece of the series ad fid a explicit formula for f x. 38. If f x 0 c x, where c 4 c for all 0, fid the iterval of covergece of the series ad a formula for f x. 39. Show that if lim l s c c, where c 0, the the radius of covergece of the power series c x is R c. 40. Suppose that the power series c x a satisfies c 0 for all. Show that if lim l cc exists, the it is equal to the radius of covergece of the power series. 4. Suppose the series c x has radius of covergece 2 ad the series d x has radius of covergece 3. What is the radius of covergece of the series c d x? 42. Suppose that the radius of covergece of the power series c x is R. What is the radius of covergece of the power series c x 2?.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES I this sectio we lear how to represet certai types of fuctios as sums of power series by maipulatig geometric series or by differetiatig or itegratig such a series. You might woder why we would ever wat to express a kow fuctio as a sum of ifiitely may terms. We will see later that this strategy is useful for itegratig fuctios that do t have elemetary atiderivatives, for solvig differetial equatios, ad for approximatig fuctios by polyomials. (Scietists do this to simplify the expressios they deal with; computer scietists do this to represet fuctios o calculators ad computers.) We start with a equatio that we have see before: N A geometric illustratio of Equatio is show i Figure. Because the sum of a series is the limit of the sequece of partial sums, we have lim x sx l where s x x x 2 x is the th partial sum. Notice that as icreases, s x becomes a better approximatio to f x for x. x x x 2 x 3 x 0 We first ecoutered this equatio i Example 5 i Sectio.2, where we obtaied it by observig that it is a geometric series with a ad r x. But here our poit of view is differet. We ow regard Equatio as expressig the fuctio f x x as a sum of a power series. y s sˆ f s x s FIGURE ƒ= 0 x -x ad some partial sums _

56 SECTION.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES 729 V EXAPLE Express x 2 as the sum of a power series ad fid the iterval of covergece. SOLUTION Replacig x by x 2 i Equatio, we have x 2 x 2 0 Because this is a geometric series, it coverges whe, that is, x 2, or x. Therefore the iterval of covergece is,. (Of course, we could have determied the radius of covergece by applyig the Ratio Test, but that much work is uecessary here.) EXAPLE 2 Fid a power series represetatio for x 2. SOLUTION I order to put this fuctio i the form of the left side of Equatio we first factor a 2 from the deomiator: 2 x x2 0 2 x 2 x 2 x 2 x 2 x 4 x 6 x x 2 0 x x x 2 x 2 This series coverges whe, that is,. So the iterval of covergece is 2, 2. N It s legitimate to move x 3 across the sigma sig because it does t deped o. [Use Theorem.2.8(i) with c x 3.] EXAPLE 3 Fid a power series represetatio of x 3 x 2. SOLUTION Sice this fuctio is just x 3 times the fuctio i Example 2, all we have to do is to multiply that series by x 3 : x 3 x 2 x 3 x 2 x x 0 3 x 2 2 x 3 4 x 4 8 x 5 6 x 6 Aother way of writig this series is as follows: x 3 x x As i Example 2, the iterval of covergece is 2, 2. DIFFERENTIATION AND INTEGRATION OF POWER SERIES The sum of a power series is a fuctio f x 0 c x a whose domai is the iterval of covergece of the series. We would like to be able to differetiate ad itegrate such fuctios, ad the followig theorem (which we wo t prove) says that we ca do so by differetiatig or itegratig each idividual term i the series, just as we would for a polyomial. This is called term-by-term differetiatio ad itegratio.

57 730 CHAPTER INFINITE SEQUENCES AND SERIES 2 THEORE If the power series c x a has radius of covergece R 0, the the fuctio f defied by f x c 0 c x a c 2 x a 2 c x a is differetiable (ad therefore cotiuous) o the iterval a R, a R ad 0 (i) f x c 2c 2 x a 3c 3 x a 2 c x a N I part (ii), x c 0 dx c 0x C is writte as c 0x a C, where C C ac 0, so all the terms of the series have the same form. (ii) y f x dx C c 0 x a c C c 0 x a x a2 2 c 2 x a3 3 The radii of covergece of the power series i Equatios (i) ad (ii) are both R. NOTE (iii) Equatios (i) ad (ii) i Theorem 2 ca be rewritte i the form d dx 0 c x a 0 d dx c x a (iv) y 0 c x a dx 0 y c x a dx We kow that, for fiite sums, the derivative of a sum is the sum of the derivatives ad the itegral of a sum is the sum of the itegrals. Equatios (iii) ad (iv) assert that the same is true for ifiite sums, provided we are dealig with power series. (For other types of series of fuctios the situatio is ot as simple; see Exercise 36.) NOTE 2 Although Theorem 2 says that the radius of covergece remais the same whe a power series is differetiated or itegrated, this does ot mea that the iterval of covergece remais the same. It may happe that the origial series coverges at a edpoit, whereas the differetiated series diverges there. (See Exercise 37.) NOTE 3 The idea of differetiatig a power series term by term is the basis for a powerful method for solvig differetial equatios. We will discuss this method i Chapter 7. EXAPLE 4 I Example 3 i Sectio.8 we saw that the Bessel fuctio J 0 x 0 x 2 2 2! 2 is defied for all x. Thus, by Theorem 2, J 0 is differetiable for all x ad its derivative is foud by term-by-term differetiatio as follows: J 0 x 0 d dx x 2 2 2! 2 2x 2 2 2! 2

58 SECTION.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES 73 V EXAPLE 5 Express x 2 as a power series by differetiatig Equatio. What is the radius of covergece? SOLUTION Differetiatig each side of the equatio x x x 2 x 3 x 0 we get x 2 2x 3x 2 x If we wish, we ca replace by ad write the aswer as Accordig to Theorem 2, the radius of covergece of the differetiated series is the same as the radius of covergece of the origial series, amely, R. EXAPLE 6 Fid a power series represetatio for l x ad its radius of covergece. SOLUTION We otice that, except for a factor of, the derivative of this fuctio is x. So we itegrate both sides of Equatio : l x y x x 2 x 2 To determie the value of C we put x 0 i this equatio ad obtai l 0 C. Thus C 0 ad 0 2 x 3 3 C 0 x x dx y x x 2 dx x C x C x l x x x 2 2 x 3 3 x x The radius of covergece is the same as for the origial series: R. Notice what happes if we put x i the result of Example 6. Sice l 2 2 l 2, we see that l V EXAPLE 7 Fid a power series represetatio for f x ta x. SOLUTION We observe that f x x 2 ad fid the required series by itegratig the power series for x 2 foud i Example. ta x y x 2 dx y x 2 x 4 x 6 dx C x x 3 3 x 5 5 x 7 7 2

59 732 CHAPTER INFINITE SEQUENCES AND SERIES N The power series for ta x obtaied i Example 7 is called Gregory s series after the Scottish mathematicia James Gregory ( ), who had aticipated some of Newto s discoveries. We have show that Gregory s series is valid whe x, but it turs out (although it is t easy to prove) that it is also valid whe x. Notice that whe x the series becomes This beautiful result is kow as the Leibiz formula for. To fid C we put x 0 ad obtai C ta 0 0. Therefore ta x x x 3 3 x 5 5 x 7 7 Sice the radius of covergece of the series for x 2 is, the radius of covergece of this series for ta x is also. EXAPLE 8 (a) Evaluate x x 7 dx as a power series. (b) Use part (a) to approximate x 0.5 x 7 dx correct to withi SOLUTION (a) The first step is to express the itegrad, x 7, as the sum of a power series. As i Example, we start with Equatio ad replace x by x 7 : 0 x 2 2 x 7 x x 7 x 7 x 7 x 4 N This example demostrates oe way i which power series represetatios are useful. Itegratig x 7 by had is icredibly difficult. Differet computer algebra systems retur differet forms of the aswer, but they are all extremely complicated. (If you have a CAS, try it yourself.) The ifiite series aswer that we obtai i Example 8(a) is actually much easier to deal with tha the fiite aswer provided by a CAS. Now we itegrate term by term: y x 7 dx y 0 C x x 8 x 7 x 7 dx C x This series coverges for, that is, for. (b) I applyig the Fudametal Theorem of Calculus, it does t matter which atiderivative we use, so let s use the atiderivative from part (a) with C 0: 8 x x x 7 7 y This ifiite series is the exact value of the defiite itegral, but sice it is a alteratig series, we ca approximate the sum usig the Alteratig Series Estimatio Theorem. If we stop addig after the term with 3, the error is smaller tha the term with 4: So we have x dx x x x x y x dx

60 SECTION.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES EXERCISES. If the radius of covergece of the power series 0 c x is 0, what is the radius of covergece of the series c x? Why? 2. Suppose you kow that the series 0 b x coverges for. What ca you say about the followig series? Why? 3 0 Fid a power series represetatio for the fuctio ad determie the iterval of covergece. 3. f x 4. x 5. x 2 f x 2 3 x 0 b x 6. f x 3 x 4 f x x Fid a power series represetatio for the fuctio ad determie the radius of covergece. 5. f x l5 x x f x 8. x 2 2 f x x 2 2x 2 f x arctax3 ; 9 22 Fid a power series represetatio for f, ad graph f ad several partial sums s x o the same scree. What happes as icreases? x 9. f x 20. f x lx 2 4 x f x l x 22. f x ta 2x x 7. f x x 9 x 2 9. f x x 0. x 8. f x x 2x 2 f x x 2 a 3 x Evaluate the idefiite itegral as a power series. What is the radius of covergece? 23. y t t 8 dt 24. y l t t dt 2 Express the fuctio as the sum of a power series by first usig partial fractios. Fid the iterval of covergece. 3. f x 2. f x x 2 x 2 3. (a) Use differetiatio to fid a power series represetatio for f x What is the radius of covergece? (b) Use part (a) to fid a power series for f x (c) Use part (b) to fid a power series for f x x 2 x 3 x 2 x 3 x 2 2x 2 x 4. (a) Fid a power series represetatio for f x l x. What is the radius of covergece? (b) Use part (a) to fid a power series for f x x l x. (c) Use part (a) to fid a power series for f x lx y x ta x dx 26. y ta x 2 dx x Use a power series to approximate the defiite itegral to six decimal places. 27. y x dx 5 y x arcta3x dx Use the result of Example 6 to compute l. correct to five decimal places. 32. Show that the fuctio is a solutio of the differetial equatio 33. (a) Show that J 0 (the Bessel fuctio of order 0 give i Example 4) satisfies the differetial equatio (b) Evaluate x 2 J 0x xj 0x x 2 J 0x 0 x 0 f x 0 x 2 2! f x f x 0 y 0.4 l x 4 dx 0 y x 2 x 4 dx J0x dx correct to three decimal places.

61 734 CHAPTER INFINITE SEQUENCES AND SERIES 34. The Bessel fuctio of order is defied by (a) Show that (b) Show that J 0x J x. 35. (a) Show that the fuctio satisfies the differetial equatio is a solutio of the differetial equatio (b) Show that f x e x. 36. Let f x si x 2. Show that the series f x coverges for all values of x but the series of derivatives f x diverges whe x 2, a iteger. For what values of x does the series f x coverge? 37. Let J J x 0 x 2!!2 2 x 2 J x xj x x 2 J x 0 f x 0 f x x! f x f x x 2 Fid the itervals of covergece for f, f, ad f. 38. (a) Startig with the geometric series 0 x, fid the sum of the series (b) Fid the sum of each of the followig series. (i) x, (ii) (c) Fid the sum of each of the followig series. (i) x, (ii) (iii) 39. Use the power series for ta x to prove the followig expressio for as the sum of a ifiite series: 40. (a) By completig the square, show that (b) By factorig x 3 as a sum of cubes, rewrite the itegral i part (a). The express x 3 as the sum of a power series ad use it to prove the followig formula for : s3 x x 2s3 y x 0 x dx x 2 x s3.0 TAYLOR AND ACLAURIN SERIES I the precedig sectio we were able to fid power series represetatios for a certai restricted class of fuctios. Here we ivestigate more geeral problems: Which fuctios have power series represetatios? How ca we fid such represetatios? We start by supposig that f is ay fuctio that ca be represeted by a power series f x c 0 c x a c 2 x a 2 c 3 x a 3 c 4 x a 4 x a R Let s try to determie what the coefficiets c must be i terms of f. To begi, otice that if we put x a i Equatio, the all terms after the first oe are 0 ad we get f a c 0 By Theorem.9.2, we ca differetiate the series i Equatio term by term: 2 f x c 2c 2 x a 3c 3 x a 2 4c 4 x a 3 ad substitutio of x a i Equatio 2 gives x a R f a c

62 SECTION.0 TAYLOR AND ACLAURIN SERIES 735 Now we differetiate both sides of Equatio 2 ad obtai 3 f x 2c 2 2 3c 3 x a 3 4c 4 x a 2 Agai we put x a i Equatio 3. The result is x a R f a 2c 2 Let s apply the procedure oe more time. Differetiatio of the series i Equatio 3 gives 4 f x 2 3c c 4 x a 3 4 5c 5 x a 2 ad substitutio of x a i Equatio 4 gives x a R f a 2 3c 3 3!c 3 By ow you ca see the patter. If we cotiue to differetiate ad substitute x a, we obtai f a c!c Solvig this equatio for the th coefficiet c, we get c f a! This formula remais valid eve for 0 if we adopt the covetios that 0! ad f 0 f. Thus we have proved the followig theorem. 5 THEORE If f has a power series represetatio (expasio) at a, that is, if f x c x a 0 x a R the its coefficiets are give by the formula c f a! Substitutig this formula for c back ito the series, we see that if f has a power series expasio at a, the it must be of the followig form. 6 f x 0 f a! f a f a! x a x a f a 2! x a 2 f a 3! x a 3

63 736 CHAPTER INFINITE SEQUENCES AND SERIES TAYLOR AND ACLAURIN The Taylor series is amed after the Eglish mathematicia Brook Taylor (685 73) ad the aclauri series is amed i hoor of the Scottish mathematicia Coli aclauri ( ) despite the fact that the aclauri series is really just a special case of the Taylor series. But the idea of represetig particular fuctios as sums of power series goes back to Newto, ad the geeral Taylor series was kow to the Scottish mathematicia James Gregory i 668 ad to the Swiss mathematicia Joh Beroulli i the 690s. Taylor was apparetly uaware of the work of Gregory ad Beroulli whe he published his discoveries o series i 75 i his book ethodus icremetorum directa et iversa. aclauri series are amed after Coli aclauri because he popularized them i his calculus textbook Treatise of Fluxios published i 742. The series i Equatio 6 is called the Taylor series of the fuctio f at a (or about a or cetered at a). For the special case a 0 the Taylor series becomes 7 f x 0 f 0! x f 0 f 0! x f 0 2! x 2 This case arises frequetly eough that it is give the special ame aclauri series. NOTE We have show that if f ca be represeted as a power series about a, the f is equal to the sum of its Taylor series. But there exist fuctios that are ot equal to the sum of their Taylor series. A example of such a fuctio is give i Exercise 70. V EXAPLE Fid the aclauri series of the fuctio f x e x ad its radius of covergece. SOLUTION If f x e x, the f x e x, so f 0 e 0 for all. Therefore the Taylor series for f at 0 (that is, the aclauri series) is 0 f 0! x 0 x! x! x 2 2! x 3 3! To fid the radius of covergece we let a x!. The so, by the Ratio Test, the series coverges for all x ad the radius of covergece is R. The coclusio we ca draw from Theorem 5 ad Example is that if series expasio at 0, the has a power So how ca we determie whether e x does have a power series represetatio? Let s ivestigate the more geeral questio: Uder what circumstaces is a fuctio equal to the sum of its Taylor series? I other words, if f has derivatives of all orders, whe is it true that As with ay coverget series, this meas that f x is the limit of the sequece of partial sums. I the case of the Taylor series, the partial sums are T x i0 a a f ia i! f a f a! x!! x x a i f x e x 0 0 f a! x a f a 2! x! x l 0 x a x a 2 f a! e x x a

64 SECTION.0 TAYLOR AND ACLAURIN SERIES 737 y y= y=t (x) y=t (x) y=t (x) (0, ) y=t (x) 0 x y=t (x) FIGURE N As icreases, T x appears to approach e x i Figure. This suggests that e x is equal to the sum of its Taylor series. Notice that T is a polyomial of degree called the th-degree Taylor polyomial of f at a. For istace, for the expoetial fuctio f x e x, the result of Example shows that the Taylor polyomials at 0 (or aclauri polyomials) with, 2, ad 3 are The graphs of the expoetial fuctio ad these three Taylor polyomials are draw i Figure. I geeral, f x is the sum of its Taylor series if If we let T x x R x f x T x T 2 x x x 2 f x lim l T x so that T 3 x x x 2 f x T x R x 2! x 3 3! the R x is called the remaider of the Taylor series. If we ca somehow show that lim l R x 0, the it follows that 2! lim T x lim f x R x f x lim R x f x l l l We have therefore proved the followig. 8 THEORE If f x T x R x, where T is the th-degree Taylor polyomial of f at a ad lim R x 0 l x x a a R R for, the f is equal to the sum of its Taylor series o the iterval. I tryig to show that lim l R x 0 for a specific fuctio f, we usually use the followig fact. f x x a d 9 TAYLOR S INEQUALITY If for, the the remaider R x of the Taylor series satisfies the iequality R x! x a for x a d To see why this is true for, we assume that f x, so for a x a d we have f x. I particular, we have y x a f t dt y x dt A atiderivative of f is f, so by Part 2 of the Fudametal Theorem of Calculus, we have f x f a x a or f x f a x a a

65 738 CHAPTER INFINITE SEQUENCES AND SERIES N As alteratives to Taylor s Iequality, we have the followig formulas for the remaider term. If f is cotiuous o a iterval I ad x I, the R x! yx x t f t dt a This is called the itegral form of the remaider term. Aother formula, called Lagrage s form of the remaider term, states that there is a umber z betwee x ad a such that R x f z x a! This versio is a extesio of the ea Value Theorem (which is the case 0). Proofs of these formulas, together with discussios of how to use them to solve the examples of Sectios.0 ad., are give o the website Thus f x f a f ax a 2 But R x f x T x f x f a f ax a. So R x 2 A similar argumet, usig f x, shows that y x f t dt y x f a t a dt a a f x f a f ax a R x 2 x a2 x a2 x a2 x a2 2 Click o Additioal Topics ad the o Formulas for the Remaider Term i Taylor series. So R x 2 x a 2 Although we have assumed that x a, similar calculatios show that this iequality is also true for x a. This proves Taylor s Iequality for the case where. The result for ay is proved i a similar way by itegratig times. (See Exercise 69 for the case 2.) NOTE I Sectio. we will explore the use of Taylor s Iequality i approximatig fuctios. Our immediate use of it is i cojuctio with Theorem 8. I applyig Theorems 8 ad 9 it is ofte helpful to make use of the followig fact. 0 x lim l! 0 for every real umber x This is true because we kow from Example that the series x! coverges for all x ad so its th term approaches 0. V EXAPLE 2 Prove that is equal to the sum of its aclauri series. SOLUTION If f x e x, the f x e x for all. If d is ay positive umber ad x d, the f x e x e d. So Taylor s Iequality, with a 0 ad e d, says that R x Notice that the same costat e d works for every value of. But, from Equatio 0, we have lim l e x e d! x e d! x e d lim l for x d x! 0

66 SECTION.0 TAYLOR AND ACLAURIN SERIES 739 It follows from the Squeeze Theorem that lim l R x 0 for all values of x. By Theorem 8, aclauri series, that is, lim l R x 0 e x ad therefore is equal to the sum of its e x 0 x! for all x N I 748 Leoard Euler used Equatio 2 to fid the value of e correct to 23 digits. I 2003 Shigeru Kodo, agai usig the series i (2), computed e to more tha 50 billio decimal places. The special techiques employed to speed up the computatio are explaied o the web page umbers.computatio.free.fr I particular, if we put x i Equatio, we obtai the followig expressio for the umber e as a sum of a ifiite series: 2 e 0!! 2! 3! EXAPLE 3 Fid the Taylor series for f x e x at a 2. SOLUTION We have f 2 e 2 ad so, puttig a 2 i the defiitio of a Taylor series (6), we get 0 f 2! x 2 Agai it ca be verified, as i Example, that the radius of covergece is R. As i Example 2 we ca verify that lim l R x 0, so 0 e 2 x 2! 3 e x 0 e 2 x 2! for all x We have two power series expasios for e x, the aclauri series i Equatio ad the Taylor series i Equatio 3. The first is better if we are iterested i values of x ear 0 ad the secod is better if x is ear 2. EXAPLE 4 Fid the aclauri series for si x ad prove that it represets si x for all x. SOLUTION We arrage our computatio i two colums as follows: f x si x f 0 0 f x cos x f 0 f x si x f 0 0 f x cos x f 0 Sice the derivatives repeat i a cycle of four, we ca write the aclauri series as follows: f 0 f 0! x x 3 f 4x si x f 40 0 x f 0 2! 3! x 5 5! x 7 x 2 f 0 3! x 3 7! 0 x 2 2!

67 740 CHAPTER INFINITE SEQUENCES AND SERIES N Figure 2 shows the graph of si x together with its Taylor (or aclauri) polyomials T x x T 3x x x 3 Notice that, as icreases, T x becomes a better approximatio to si x. y=si x y 3! T 5x x x 3 3! x 5 5! T 0 x T T Sice f x is si x or cos x, we kow that for all x. So we ca take i Taylor s Iequality: 4 By Equatio 0 the right side of this iequality approaches 0 as l, so R x l 0 by the Squeeze Theorem. It follows that R x l 0 as l, so si x is equal to the sum of its aclauri series by Theorem 8. We state the result of Example 4 for future referece. 5 R x si x x x 3 x 2 2! 0! x 3! x 5 5! x 7 f x x! 7! for all x FIGURE 2 EXAPLE 5 Fid the aclauri series for cos x. SOLUTION We could proceed directly as i Example 4 but it s easier to differetiate the aclauri series for si x give by Equatio 5: cos x d dx si x d x x 3 dx 3! x 5 5! x 7 7! 3x 2 3! 5x 4 5! 7x 6 7! x 2 2! x 4 4! x 6 6! N The aclauri series for e x, si x, ad cos x that we foud i Examples 2, 4, ad 5 were discovered, usig differet methods, by Newto. These equatios are remarkable because they say we kow everythig about each of these fuctios if we kow all its derivatives at the sigle umber 0. Sice the aclauri series for si x coverges for all x, Theorem 2 i Sectio.9 tells us that the differetiated series for cos x also coverges for all x. Thus 6 cos x x 2 x 2 2! 0 2! x 4 4! x 6 6! for all x EXAPLE 6 Fid the aclauri series for the fuctio f x x cos x. SOLUTION Istead of computig derivatives ad substitutig i Equatio 7, it s easier to multiply the series for cos x (Equatio 6) by x: x cos x x x 2 2! x 2 2! 0 0 EXAPLE 7 Represet f x si x as the sum of its Taylor series cetered at 3.

68 SECTION.0 TAYLOR AND ACLAURIN SERIES 74 N We have obtaied two differet series represetatios for si x, the aclauri series i Example 4 ad the Taylor series i Example 7. It is best to use the aclauri series for values of x ear 0 ad the Taylor series for x ear 3. Notice that the third Taylor polyomial T 3 i Figure 3 is a good approximatio to si x ear 3 but ot as good ear 0. Compare it with the third aclauri polyomial T 3 i Figure 2, where the opposite is true. y y=si x 0 π x 3 T FIGURE 3 SOLUTION Arragig our work i colums, we have ad this patter repeats idefiitely. Therefore the Taylor series at 3 is f 2 f x x x! 2! 3! f 3 s3 2 2! x The proof that this series represets si x for all x is very similar to that i Example 4. [Just replace x by x 3 i (4).] We ca write the series i sigma otatio if we separate the terms that cotai s3 : si x 0 f x si x f x cos x f x si x f x cos x f 3 3 s3 22! x s3 2 2! x 3 2 f f f f The power series that we obtaied by idirect methods i Examples 5 ad 6 ad i Sectio.9 are ideed the Taylor or aclauri series of the give fuctios because Theorem 5 asserts that, o matter how a power series represetatio f x c x a is obtaied, it is always true that c f a!. I other words, the coefficiets are uiquely determied s s x 2 3! 22!x EXAPLE 8 Fid the aclauri series for f x x k, where k is ay real umber. SOLUTION Arragig our work i colums, we have f x x k f 0 f x k x k f 0 k f x kk x k2 f 0 kk f x kk k 2 x k3 f 0 kk k 2.. f x kk k x k f 0 kk k Therefore the aclauri series of f x x k is 0 f 0! x 0 kk k! x

69 742 CHAPTER INFINITE SEQUENCES AND SERIES This series is called the biomial series. If its th term is a, the a kk k k x! a! kk k x k k x x l x as l Thus, by the Ratio Test, the biomial series coverges if ad diverges if. x x The traditioal otatio for the coefficiets i the biomial series is ad these umbers are called the biomial coefficiets. The followig theorem states that x k is equal to the sum of its aclauri series. It is possible to prove this by showig that the remaider term R x approaches 0, but that turs out to be quite difficult. The proof outlied i Exercise 7 is much easier. 7 k kk k 2k! THE BINOIAL SERIES If k is ay real umber ad, the x k 0 x k kx kk x 2 2! x kk k 2 x 3 3! x Although the biomial series always coverges whe, the questio of whether or ot it coverges at the edpoits,, depeds o the value of k. It turs out that the series coverges at if k 0 ad at both edpoits if k 0. Notice that if k is a positive iteger ad k, the the expressio for ( k ) cotais a factor k k, so ( k ) 0 for k. This meas that the series termiates ad reduces to the ordiary Biomial Theorem whe k is a positive iteger. (See Referece Page.) V EXAPLE 9 Fid the aclauri series for the fuctio of covergece. f x s4 x ad its radius SOLUTION We write f x i a form where we ca use the biomial series: s4 x 4 4 x 2 x 4 2 x 42

70 SECTION.0 TAYLOR AND ACLAURIN SERIES 743 Usig the biomial series with k 2 ad with x replaced by x4, we have x 2 s4 x x x ( 2)( 3 2) x 2 2! 4 ( 2)( 3 2)( 5 2) x 3 3! 4 ( 2)( 3 2)( 5 2) ( 2 ) 2 8 x! 4 3 x 2!8 x x 3 x 2 3!8 3!8 x4 x 4 We kow from (7) that this series coverges whe, that is,, so the radius of covergece is R 4. We collect i the followig table, for future referece, some importat aclauri series that we have derived i this sectio ad the precedig oe. TABLE Importat aclauri Series ad Their Radii of Covergece x e x 0 x! x! x 2 si x x 2 2! 0 cos x x 2 2! 0 0 ta x x x x x 2 x 3 x x 3 x 2 x x 3 x k 0 x k kx 2! x 3 3! 3! x 5 5! x 7 2! x 4 4! x 6 3 x 5 5 x 7 kk x 2 2! 7! 6! 7 kk k 2 x 3 3! R R R R R R TEC odule.0/. eables you to see how successive Taylor polyomials approach the origial fuctio. Oe reaso that Taylor series are importat is that they eable us to itegrate fuctios that we could t previously hadle. I fact, i the itroductio to this chapter we metioed that Newto ofte itegrated fuctios by first expressig them as power series ad the itegratig the series term by term. The fuctio f x e x2 ca t be itegrated by techiques discussed so far because its atiderivative is ot a elemetary fuctio (see Sectio 7.5). I the followig example we use Newto s idea to itegrate this fuctio.

71 744 CHAPTER INFINITE SEQUENCES AND SERIES V EXAPLE 0 (a) Evaluate x e x2 dx as a ifiite series. (b) Evaluate x dx correct to withi a error of ex2 SOLUTION (a) First we fid the aclauri series for f x e x2. Although it s possible to use the direct method, let s fid it simply by replacig x with x 2 i the series for e x give i Table. Thus, for all values of x, e x2 0 x 2! x 2 0! x 2! x 4 2! x 6 3! Now we itegrate term by term: y e x2 dx y x 2 C x x 3! x 4 2! x 6 2 x 3!! dx 3! x 5 5 2! x 7 7 3! x 2 2! This series coverges for all x because the origial series for e x2 coverges for all x. (b) The Fudametal Theorem of Calculus gives y e x2 dx x x 3 0 3! x 5 5 2! x 7 7 3! x 9 9 4! 0 N We ca take C 0 i the atiderivative i part (a) The Alteratig Series Estimatio Theorem shows that the error ivolved i this approximatio is less tha 5! Aother use of Taylor series is illustrated i the ext example. The limit could be foud with l Hospital s Rule, but istead we use a series. e x x EXAPLE Evaluate lim. x l 0 x 2 SOLUTION Usig the aclauri series for e x, we have x x 2 e x x! 2! x 3 3! x lim lim x l 0 x 2 x l 0 x 2 N Some computer algebra systems compute limits i this way. lim x l 0 lim x l 0 because power series are cotiuous fuctios. x 2 2! x 3 x4 3! 4! x 2 2 x 3! x 2 4! x 3 5! 2

72 SECTION.0 TAYLOR AND ACLAURIN POLYNOIALS 745 ULTIPLICATION AND DIVISION OF POWER SERIES If power series are added or subtracted, they behave like polyomials (Theorem.2.8 shows this). I fact, as the followig example illustrates, they ca also be multiplied ad divided like polyomials. We fid oly the first few terms because the calculatios for the later terms become tedious ad the iitial terms are the most importat oes. EXAPLE 2 Fid the first three ozero terms i the aclauri series for (a) e x si x ad (b) ta x. SOLUTION (a) Usig the aclauri series for e x ad si x i Table, we have e x si x x! x 2 2! x 3 3! x x 3 3! We multiply these expressios, collectig like terms just as for polyomials: x x 2 x 2 x x x 2 6 x 3 6 x 3 6 x 4 2 x 3 6 x 4 6 x 3 x 2 3 x 3 Thus e x si x x x 2 3 x 3 (b) Usig the aclauri series i Table, we have ta x si x cos x We use a procedure like log divisio: x x 3 3! x 5 5! x 2 2! x 4 4! x 3 x x 5 2 x 2 24 x 4 )x 6 x 3 20x 5 x 2 x 3 24 x 5 Thus ta x x 3 x x 5 3 x 3 30 x 5 3 x 3 6 x x 5 Although we have ot attempted to justify the formal maipulatios used i Example 2, they are legitimate. There is a theorem which states that if both f x c x ad tx b coverge for x x R ad the series are multiplied as if they were polyomials, the the resultig series also coverges for x R ad represets f xtx. For divisio we require b 0 0; the resultig series coverges for sufficietly small. x

73 746 CHAPTER INFINITE SEQUENCES AND SERIES.0 EXERCISES. If f x 0 b x 5 for all x, write a formula for. 2. The graph of f is show. b 8 7. f x cos x, a 8. f x si x, 9. f x sx, a f x x 2, a 2 a (a) Explai why the series y 0 x.6 0.8x 0.4x 2 0.x 3 is ot the Taylor series of f cetered at. (b) Explai why the series x 2.5x x 2 3 is ot the Taylor series of f cetered at If f 0! for 0,, 2,..., fid the aclauri series for f ad its radius of covergece. 4. Fid the Taylor series for f cetered at 4 if f 4! 3 What is the radius of covergece of the Taylor series? 5 2 Fid the aclauri series for f x usig the defiitio of a aclauri series. [Assume that f has a power series expasio. Do ot show that R x l 0.] Also fid the associated radius of covergece. 5. f x x2 6. f x l x 7. f x si x 8. f x cos 3x 9. f x e 5x 0. f x xe x. f x sih x 2. f x cosh x 3 20 Fid the Taylor series for f x cetered at the give value of a. [Assume that f has a power series expasio. Do ot show that R x l 0.] 3. f x x 4 3x 2, a 4. f x x x 3, a 2 5. f x e x, a 3 6. f x x, a 3 f 2. Prove that the series obtaied i Exercise 7 represets si x for all x. 22. Prove that the series obtaied i Exercise 8 represets si x for all x. 23. Prove that the series obtaied i Exercise represets sih x for all x. 24. Prove that the series obtaied i Exercise 2 represets cosh x for all x Use the biomial series to expad the fuctio as a power series. State the radius of covergece. 25. s x x Use a aclauri series i Table to obtai the aclauri series for the give fuctio. 29. f x si x 30. f x cosx2 3. f x e x e 2x 32. f x e x 2e x 33. f x x cos( 2 x 2 ) 34. f x x 2 ta x x x 2 f x 36. f x s4 x 2 s2 x 37. f x si 2 x Hit: Use si 2 x 2 cos 2x.] x si x if x f x x 3 6 if x 0 [ ; Fid the aclauri series of f (by ay method) ad its radius of covergece. Graph f ad its first few Taylor polyomials o the same scree. What do you otice about the relatioship betwee these polyomials ad f? 28. x 4 x f x cosx f x ex cos x 4. f x xe x 42. f x l x Use the aclauri series for e x to calculate e 0.2 correct to five decimal places.

74 SECTION.0 TAYLOR AND ACLAURIN SERIES Use the aclauri series for si x to compute si 3 correct to five decimal places. 6. y x 62. si x y e x l x 45. (a) Use the biomial series to expad s x 2. (b) Use part (a) to fid the aclauri series for si x Fid the sum of the series. 46. (a) Expad s 4 x as a power series. (b) Use part (a) to estimate s 4. correct to three decimal places Evaluate the idefiite itegral as a ifiite series. 47. y x cosx 3 dx 48. y e x x cos x 49. y dx 50. y arctax 2 dx x dx ! x 4 0! ! 27 3! l 2 8 4! l 22 2! l 23 3! ! 3 5! 5 54 Use series to approximate the defiite itegral to withi the idicated accuracy. 5. x cosx 3 dx (three decimal places) 52. ta x 3 six 3 dx (five decimal places) y 0 y y 0.4 s x 4 dx 0 y 0.5 x 2 e x 2 dx Use series to evaluate the limit. x ta x 55. lim 56. x l0 x 3 ( error ) ( error 0.00) lim x l0 cos x x e x 69. Prove Taylor s Iequality for, that is, prove that if for, the 2 f x x a d R2x 6 x a (a) Show that the fuctio defied by f x 2 ex 0 is ot equal to its aclauri series. ; (b) Graph the fuctio i part (a) ad commet o its behavior ear the origi. 7. Use the followig steps to prove (7). (a) Let. Differetiate this series to show that tx 0 ( k )x for x a d if x 0 if x si x x 6 x 3 lim x l 0 x 5 tx ktx x x 58. Use the series i Example 2(b) to evaluate We foud this limit i Example 4 i Sectio 4.4 usig l Hospital s Rule three times. Which method do you prefer? Use multiplicatio or divisio of power series to fid the first three ozero terms i the aclauri series for the fuctio. 59. y e x 2 cos x ta x x lim x l 0 x y sec x (b) Let hx xk tx ad show that hx 0. (c) Deduce that tx x k. 72. I Exercise 53 i Sectio 0.2 it was show that the legth of the ellipse x a si, y b cos, where a b 0, is L 4a y 0 2 s e 2 si 2 where e sa 2 b 2 a is the eccetricity of the ellipse. Expad the itegrad as a biomial series ad use the result of Exercise 46 i Sectio 7. to express L as a series i powers of the eccetricity up to the term i e 6. d

75 748 CHAPTER INFINITE SEQUENCES AND SERIES LABORATORY PROJECT CAS AN ELUSIVE LIIT This project deals with the fuctio f x sita x tasi x arcsiarcta x arctaarcsi x. Use your computer algebra system to evaluate fx for x, 0., 0.0, 0.00, ad Does it appear that f has a limit as x l 0? 2. Use the CAS to graph f ear x 0. Does it appear that f has a limit as x l 0? 3. Try to evaluate lim x l 0 fx with l Hospital s Rule, usig the CAS to fid derivatives of the umerator ad deomiator. What do you discover? How may applicatios of l Hospital s Rule are required? 4. Evaluate lim x l 0 fx by usig the CAS to fid sufficietly may terms i the Taylor series of the umerator ad deomiator. (Use the commad taylor i aple or Series i athematica.) 5. Use the limit commad o your CAS to fid lim x l 0 fx directly. (ost computer algebra systems use the method of Problem 4 to compute limits.) 6. I view of the aswers to Problems 4 ad 5, how do you explai the results of Problems ad 2? WRITING PROJECT HOW NEWTON DISCOVERED THE BINOIAL SERIES The Biomial Theorem, which gives the expasio of a b k, was kow to Chiese mathematicias may ceturies before the time of Newto for the case where the expoet k is a positive iteger. I 665, whe he was 22, Newto was the first to discover the ifiite series expasio of a b k whe k is a fractioal expoet (positive or egative). He did t publish his discovery, but he stated it ad gave examples of how to use it i a letter (ow called the epistola prior) dated Jue 3, 676, that he set to Hery Oldeburg, secretary of the Royal Society of Lodo, to trasmit to Leibiz. Whe Leibiz replied, he asked how Newto had discovered the biomial series. Newto wrote a secod letter, the epistola posterior of October 24, 676, i which he explaied i great detail how he arrived at his discovery by a very idirect route. He was ivestigatig the areas uder the curves y x 2 2 from 0 to x for 0,,2,3,4,... These are easy to calculate if is eve. By observig patters ad iterpolatig, Newto was able to guess the aswers for odd values of. The he realized he could get the same aswers by expressig x 2 2 as a ifiite series. Write a report o Newto s discovery of the biomial series. Start by givig the statemet of the biomial series i Newto s otatio (see the epistola prior o page 285 of [4] or page 402 of [2]). Explai why Newto s versio is equivalet to Theorem 7 o page 742. The read Newto s epistola posterior (page 287 i [4] or page 404 i [2]) ad explai the patters that Newto discovered i the areas uder the curves y x 2 2. Show how he was able to guess the areas uder the remaiig curves ad how he verified his aswers. Fially, explai how these discoveries led to the biomial series. The books by Edwards [] ad Katz [3] cotai commetaries o Newto s letters.. C. H. Edwards, The Historical Developmet of the Calculus (New York: Spriger-Verlag, 979), pp Joh Fauvel ad Jeremy Gray, eds., The History of athematics: A Reader (Lodo: acilla Press, 987). 3. Victor Katz, A History of athematics: A Itroductio (New York: HarperCollis, 993), pp D. J. Struik, ed., A Sourcebook i athematics, (Priceto, NJ: Priceto Uiversity Press, 969).

76 SECTION. APPLICATIONS OF TAYLOR POLYNOIALS 749. APPLICATIONS OF TAYLOR POLYNOIALS I this sectio we explore two types of applicatios of Taylor polyomials. First we look at how they are used to approximate fuctios computer scietists like them because polyomials are the simplest of fuctios. The we ivestigate how physicists ad egieers use them i such fields as relativity, optics, blackbody radiatio, electric dipoles, the velocity of water waves, ad buildig highways across a desert. APPROXIATING FUNCTIONS BY POLYNOIALS Suppose that f x is equal to the sum of its Taylor series at a: f x 0 f a! x a I Sectio.0 we itroduced the otatio T x for the th partial sum of this series ad called it the th-degree Taylor polyomial of f at a. Thus T x i0 f ia i! x a i f a f a! x a f a 2! x a 2 f a! x a FIGURE T 2x T 4x T 6x T 8x T 0x e x y x 0.2 y= y=t (x) (0, ) y=t (x) y=t (x) 0 x x Sice f is the sum of its Taylor series, we kow that T x l f x as l ad so T ca be used as a approximatio to f : f xt x. Notice that the first-degree Taylor polyomial T x f a f ax a is the same as the liearizatio of f at a that we discussed i Sectio 3.0. Notice also that T ad its derivative have the same values at a that f ad f have. I geeral, it ca be show that the derivatives of T at a agree with those of f up to ad icludig derivatives of order (see Exercise 38). To illustrate these ideas let s take aother look at the graphs of y e x ad its first few Taylor polyomials, as show i Figure. The graph of T is the taget lie to y e x at 0, ; this taget lie is the best liear approximatio to e x ear 0,. The graph of T is the parabola y x x 2 2 2, ad the graph of T 3 is the cubic curve y x x 2 2 x 3 6, which is a closer fit to the expoetial curve y e x tha T 2. The ext Taylor polyomial T 4 would be a eve better approximatio, ad so o. The values i the table give a umerical demostratio of the covergece of the Taylor polyomials T to the fuctio y e x x. We see that whe x 0.2 the covergece is very rapid, but whe x 3 it is somewhat slower. I fact, the farther x is from 0, the more slowly T coverges to e x x. Whe usig a Taylor polyomial T to approximate a fuctio f, we have to ask the questios: How good a approximatio is it? How large should we take to be i order to achieve a desired accuracy? To aswer these questios we eed to look at the absolute value of the remaider: R x f x T x

77 750 CHAPTER INFINITE SEQUENCES AND SERIES There are three possible methods for estimatig the size of the error:. If a graphig device is available, we ca use it to graph ad thereby estimate the error. 2. If the series happes to be a alteratig series, we ca use the Alteratig Series Estimatio Theorem. 3. I all cases we ca use Taylor s Iequality (Theorem.0.9), which says that if,the V EXAPLE (a) Approximate the fuctio f x s 3 x by a Taylor polyomial of degree 2 at a 8. (b) How accurate is this approximatio whe 7 x 9? SOLUTION (a) f x R x! x a f x s 3 x x 3 f 8 2 R x Thus the secod-degree Taylor polyomial is The desired approximatio is (b) The Taylor series is ot alteratig whe x 8, so we ca t use the Alteratig Series Estimatio Theorem i this example. But we ca use Taylor s Iequality with 2 ad a 8: f x f x 3 x 23 f 8 2 f x 2 9 x 53 f 8 44 f x 0 27 x 83 T 2 x f 8 f 8! 2 2x 8 288x 8 2 s 3 x T 2 x 2 2x 8 288x 8 2 where. Because x 7, we have x ad so f x x Therefore we ca take Also 7 x 9, so ad. The Taylor s Iequality gives x 8 x 8 R 2x R2x ! x 8 f 8 2! 3! x x Thus, if 7 x 9, the approximatio i part (a) is accurate to withi

78 SECTION. APPLICATIONS OF TAYLOR POLYNOIALS T y=#œ x 0 FIGURE Let s use a graphig device to check the calculatio i Example. Figure 2 shows that the graphs of y s 3 x ad y T 2 x are very close to each other whe x is ear 8. Figure 3 shows the graph of computed from the expressio We see from the graph that R 2x R 2x s 3 x T 2 x R 2x whe 7 x 9. Thus the error estimate from graphical methods is slightly better tha the error estimate from Taylor s Iequality i this case. y= R (x) FIGURE 3 V EXAPLE 2 (a) What is the maximum error possible i usig the approximatio whe 0.3 x 0.3? Use this approximatio to fid si 2 correct to six decimal places. (b) For what values of x is this approximatio accurate to withi ? SOLUTION (a) Notice that the aclauri series si x x x 3 3! x 5 5! si x x x 3 is alteratig for all ozero values of x, ad the successive terms decrease i size because x, so we ca use the Alteratig Series Estimatio Theorem. The error i approximatig si x by the first three terms of its aclauri series is at most x 7 7! x If 0.3 x 0.3, the 0.3, so the error is smaller tha x To fid si 2 we first covert to radia measure. si 2 si 2 80 si ! 5 5 5! ! x 5 5! x Thus, correct to six decimal places, si (b) The error will be smaller tha if x !

79 752 CHAPTER INFINITE SEQUENCES AND SERIES Solvig this iequality for x, we get TEC odule.0/. graphically shows the remaiders i Taylor polyomial approximatios * _ FIGURE 4 y= Rß(x) y= _ 0 FIGURE 5 y= Rß(x) So the give approximatio is accurate to withi whe. What if we use Taylor s Iequality to solve Example 2? Sice have ad so f 7x x f 7x cos x, we So we get the same estimates as with the Alteratig Series Estimatio Theorem. What about graphical methods? Figure 4 shows the graph of ad we see from it that whe 0.3. This is the same estimate that we obtaied i Example 2. For part (b) we wat, so we graph both y R 6x R 6x ad y i Figure 5. By placig the cursor o the right itersectio poit we fid that the iequality is satisfied whe x Agai this is the same estimate that we obtaied i the solutio to Example 2. If we had bee asked to approximate si 72 istead of si 2 i Example 2, it would have bee wise to use the Taylor polyomials at a 3 (istead of a 0) because they are better approximatios to si x for values of x close to 3. Notice that 72 is close to 60 (or 3 radias) ad the derivatives of si x are easy to compute at 3. Figure 6 shows the graphs of the aclauri polyomial approximatios T x x T 3 x x x 3 T 5 x x x 3 R 6x si x (x 6 x 3 20 x 5 ) R 6x x 3! x 5 5! or R 6x 7! x 7 to the sie curve. You ca see that as icreases, T x is a good approximatio to si x o a larger ad larger iterval. y x T 7 x x x 3 T 3! 3! x 5 5! x 7 7! T x x FIGURE 6 T T y=si x Oe use of the type of calculatio doe i Examples ad 2 occurs i calculators ad computers. For istace, whe you press the si or e x key o your calculator, or whe a computer programmer uses a subroutie for a trigoometric or expoetial or Bessel fuctio, i may machies a polyomial approximatio is calculated. The polyomial is ofte a Taylor polyomial that has bee modified so that the error is spread more evely throughout a iterval. APPLICATIONS TO PHYSICS Taylor polyomials are also used frequetly i physics. I order to gai isight ito a equatio, a physicist ofte simplifies a fuctio by cosiderig oly the first two or three terms i its Taylor series. I other words, the physicist uses a Taylor polyomial as a

80 SECTION. APPLICATIONS OF TAYLOR POLYNOIALS 753 approximatio to the fuctio. Taylor s Iequality ca the be used to gauge the accuracy of the approximatio. The followig example shows oe way i which this idea is used i special relativity. N The upper curve i Figure 7 is the graph of the expressio for the kietic eergy K of a object with velocity v i special relativity. The lower curve shows the fuctio used for K i classical Newtoia physics. Whe v is much smaller tha the speed of light, the curves are practically idetical. K 0 FIGURE 7 K=mc@-m c@ K= 2 c V EXAPLE 3 I Eistei s theory of special relativity the mass of a object movig with velocity v is where m 0 is the mass of the object whe at rest ad c is the speed of light. The kietic eergy of the object is the differece betwee its total eergy ad its eergy at rest: (a) Show that whe v is very small compared with c, this expressio for K agrees with classical Newtoia physics: K 2m 0 v 2. (b) Use Taylor s Iequality to estimate the differece i these expressios for K whe 00 ms. v SOLUTION (a) Usig the expressios give for K ad m, we get With x v 2 c 2, the aclauri series for x2 is most easily computed as a biomial series with k 2. (Notice that because v c.) Therefore we have ad If v is much smaller tha c, the all terms after the first are very small whe compared with the first term. If we omit them, we get (b) If x v 2 c 2, f x m 0 c 2 x2, ad is a umber such that, the we ca use Taylor s Iequality to write f x x x2 2 x ( 2 )( 3 2 ) 2! 2 x 3 8 x x 3 K m 0 c 2 2 m 0 c 2 2 K mc 2 m 0 c 2 m 0 c 2 v2 v 2 c m s v 2 c 2 K mc 2 m 0 c 2 v 2 c v 4 c K m 0 c c Rx m 0 c 2 s v 2 c 2 m 0 c 2 x 2 ( 2)( 3 2)( 5 2) 3! v 4 c 5 v v 6 c 6 v 2 c 2 2m 0 v 2 2! x 2 We have f x 3 4m 0 c 2 x52 ad we are give that 00 ms, so m 0 v f x 3m 0 c 2 4 v 2 c m 0 c c 2 52 c 6 x 3

81 754 CHAPTER INFINITE SEQUENCES AND SERIES Thus, with c ms, v R x 2 3m 0 c c m c 4 0 So whe 00 ms, the magitude of the error i usig the Newtoia expressio for kietic eergy is at most m 0. Aother applicatio to physics occurs i optics. Figure 8 is adapted from Optics, 4th ed., by Eugee Hecht (Sa Fracisco: Addiso-Wesley, 2002), page 53. It depicts a wave from the poit source S meetig a spherical iterface of radius R cetered at C. The ray SA is refracted toward P. r i A FIGURE 8 Refractio at a spherical iterface S s o L o h R V t s i L i C P Courtesy of Eugee Hecht Usig Fermat s priciple that light travels so as to miimize the time take, Hecht derives the equatio 2 o i 2s i R i s o o where ad 2 are idexes of refractio ad o, i, s o, ad s i are the distaces idicated i Figure 8. By the Law of Cosies, applied to triagles ACS ad ACP, we have N Here we use the idetity cos cos 2 o sr 2 s o R 2 2Rs o R cos i sr 2 s i R 2 2Rs i R cos Because Equatio is cumbersome to work with, Gauss, i 84, simplified it by usig the liear approximatio cos for small values of. (This amouts to usig the Taylor polyomial of degree.) The Equatio becomes the followig simpler equatio [as you are asked to show i Exercise 34(a)]: s o s i R The resultig optical theory is kow as Gaussia optics, or first-order optics, ad has become the basic theoretical tool used to desig leses. A more accurate theory is obtaied by approximatig cos by its Taylor polyomial of degree 3 (which is the same as the Taylor polyomial of degree 2). This takes ito accout rays for which is ot so small, that is, rays that strike the surface at greater distaces h above the axis. I Exercise 34(b) you are asked to use this approximatio to derive the

82 SECTION. APPLICATIONS OF TAYLOR POLYNOIALS 755 more accurate equatio h s o s i R 2s o 2 s o R 2 2s i R 2 s i The resultig optical theory is kow as third-order optics. Other applicatios of Taylor polyomials to physics ad egieerig are explored i Exercises 32, 33, 35, 36, ad 37 ad i the Applied Project o page EXERCISES ;. (a) Fid the Taylor polyomials up to degree 6 for f x cos x cetered at a 0. Graph f ad these polyomials o a commo scree. (b) Evaluate f ad these polyomials at x 4, 2, ad. (c) Commet o how the Taylor polyomials coverge to f x. ; 2. (a) Fid the Taylor polyomials up to degree 3 for f x x cetered at a. Graph f ad these polyomials o a commo scree. (b) Evaluate f ad these polyomials at x 0.9 ad.3. (c) Commet o how the Taylor polyomials coverge to f x. ; 3 0 Fid the Taylor polyomial T x for the fuctio f at the umber a. Graph f ad o the same scree. T 3 3. f x x, a 2 4. f x x e x, a 0 5. f x cos x, a 2 6. f x e x si x, a 0 7. f x arcsi x, a 0 8. f x l x, x a 9. f x xe 2x, a 0 0. f x ta x, a CAS 2 Use a computer algebra system to fid the Taylor polyomials T cetered at a for 2, 3, 4, 5. The graph these polyomials ad f o the same scree.. f x cot x, a 4 2. f x s 3 x 2, a (a) Approximate f by a Taylor polyomial with degree at the umber a. (b) Use Taylor s Iequality to estimate the accuracy of the approximatio f xt x whe x lies i the give iterval. ; (c) Check your result i part (b) by graphig. 3. f x sx, a 4, 2, 4 x f x x 2, a, 2, 0.9 x. 5. f x x 23, a, 3, 0.8 x.2 6. f x si x, a 6, 4, 7. f x sec x, a 0, 2, 8. f x l 2x, a, 3, 9. f x e x 2, a 0, 3, 20. f x x l x, a, 3, 2. f x x si x, a 0, 4, 22. f x sih 2x, a 0, 5, 23. Use the iformatio from Exercise 5 to estimate cos 80 correct to five decimal places. 24. Use the iformatio from Exercise 6 to estimate si 38 correct to five decimal places. 25. Use Taylor s Iequality to determie the umber of terms of the aclauri series for e x that should be used to estimate to withi How may terms of the aclauri series for l x do you eed to use to estimate l.4 to withi 0.00? ; Use the Alteratig Series Estimatio Theorem or Taylor s Iequality to estimate the rage of values of x for which the give approximatio is accurate to withi the stated error. Check your aswer graphically si x x x 3 6 cos x x 2 arcta x x x 3 ( error 0.0) 2 x ( error 0.05) 3 x x x x x.5 x x ( error 0.005) Rx 0.5 x.5 e 0.

83 756 CHAPTER INFINITE SEQUENCES AND SERIES 30. Suppose you kow that ad the Taylor series of f cetered at 4 coverges to f x for all x i the iterval of covergece. Show that the fifthdegree Taylor polyomial approximates f 5 with error less tha A car is movig with speed 20 ms ad acceleratio 2 ms at a give istat. Usig a secod-degree Taylor polyomial, estimate how far the car moves i the ext secod. Would it be reasoable to use this polyomial to estimate the distace traveled durig the ext miute? 32. The resistivity of a coductig wire is the reciprocal of the coductivity ad is measured i uits of ohm-meters ( -m). The resistivity of a give metal depeds o the temperature accordig to the equatio where t is the temperature i C. There are tables that list the values of (called the temperature coefficiet) ad (the resistivity at 20C) for various metals. Except at very low temperatures, the resistivity varies almost liearly with temperature ad so it is commo to approximate the expressio for t by its first- or secod-degree Taylor polyomial at t 20. (a) Fid expressios for these liear ad quadratic approximatios. ; (b) For copper, the tables give C ad -m. Graph the resistivity of copper ad the liear ad quadratic approximatios for 250C t 000C. ; (c) For what values of t does the liear approximatio agree with the expoetial expressio to withi oe percet? By expadig this expressio for E as a series i powers of dd, show that E is approximately proportioal to D 3 whe P is far away from the dipole. P f 4! 3 t E q D 2 D 20e t A electric dipole cosists of two electric charges of equal magitude ad opposite sig. If the charges are q ad q ad are located at a distace d from each other, the the electric field E at the poit P i the figure is q D d (a) Derive Equatio 3 for Gaussia optics from Equatio by approximatig cos i Equatio 2 by its first-degree Taylor polyomial. (b) Show that if cos is replaced by its third-degree Taylor polyomial i Equatio 2, the Equatio becomes q 20 d _q Equatio 4 for third-order optics. [Hit: Use the first two terms i the biomial series for o ad i. Also, use.] si 35. If a water wave with legth L moves with velocity v across a body of water with depth d, as i the figure, the (a) If the water is deep, show that v stl2. (b) If the water is shallow, use the aclauri series for tah to show that v std. (Thus i shallow water the velocity of a wave teds to be idepedet of the legth of the wave.) (c) Use the Alteratig Series Estimatio Theorem to show that if L 0d, the the estimate v 2 td is accurate to withi 0.04tL. d 36. The period of a pedulum with legth L that makes a maximum agle with the vertical is 0 T 4 L y 2 t 0 where k si( 2 0) ad t is the acceleratio due to gravity. (I Exercise 40 i Sectio 7.7 we approximated this itegral usig Simpso s Rule.) (a) Expad the itegrad as a biomial series ad use the result of Exercise 46 i Sectio 7. to show that T 2 L 2 t 2 k k k 6 2 If 0 is ot too large, the approximatio T 2sLt, obtaied by usig oly the first term i the series, is ofte used. A better approximatio is obtaied by usig two terms: (b) Notice that all the terms i the series after the first oe have coefficiets that are at most 4. Use this fact to compare this series with a geometric series ad show that 2 L t v 2 tl 2 tah 2d L dx s k 2 si 2 x T 2 L t ( 4 k 2 ) ( 4 k 2 ) T 2 L t 4 3k 2 4 4k 2 (c) Use the iequalities i part (b) to estimate the period of a pedulum with L meter ad 0 0. How does it compare with the estimate T 2sLt? What if 0 42? L

84 APPLIED PROJECT RADIATION FRO THE STARS If a surveyor measures differeces i elevatio whe makig plas for a highway across a desert, correctios must be made for the curvature of the earth. (a) If R is the radius of the earth ad L is the legth of the highway, show that the correctio is C R seclr R (b) Use a Taylor polyomial to show that C L 2 4 5L 2R 24R 3 (c) Compare the correctios give by the formulas i parts (a) ad (b) for a highway that is 00 km log. (Take the radius of the earth to be 6370 km.) R L R C 38. Show that T ad f have the same derivatives at a up to order. 39. I Sectio 4.8 we cosidered Newto s method for approximatig a root r of the equatio f x 0, ad from a iitial approximatio x we obtaied successive approximatios x 2,,..., where x 3 x x f x f x Use Taylor s Iequality with, a x, ad x r to show that if f x exists o a iterval I cotaiig r, x, ad x, ad, for all x I, the f x f x K x r 2K x r 2 [This meas that if x is accurate to d decimal places, the x is accurate to about 2d decimal places. ore precisely, if the error at stage is at most 0 m, the the error at stage is at most 2K0 2m.] Luke Dodd, Photo Researchers, Ic. APPLIED PROJECT RADIATION FRO THE STARS Ay object emits radiatio whe heated. A blackbody is a system that absorbs all the radiatio that falls o it. For istace, a matte black surface or a large cavity with a small hole i its wall (like a blastfurace) is a blackbody ad emits blackbody radiatio. Eve the radiatio from the su is close to beig blackbody radiatio. Proposed i the late 9th cetury, the Rayleigh-Jeas Law expresses the eergy desity of blackbody radiatio of wavelegth as where is measured i meters, T is the temperature i kelvis (K), ad k is Boltzma s costat. The Rayleigh-Jeas Law agrees with experimetal measuremets for log wavelegths but disagrees drastically for short wavelegths. [The law predicts that f l as but experimets have show that f l 0.] This fact is kow as the ultraviolet catastrophe. I 900 ax Plack foud a better model (kow ow as Plack s Law) for blackbody radiatio: f 8hc5 e hckt where is measured i meters, T is the temperature (i kelvis), ad. Use l Hospital s Rule to show that h Plack s costat Js c speed of light ms k Boltzma s costat JK lim f 8kT 4 f 0 ad lim f 0 l 0 l l 0 for Plack s Law. So this law models blackbody radiatio better tha the Rayleigh-Jeas Law for short wavelegths.