CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their properties; for example, trascedetal fuctios. This is the techique of Ifiite Series. Computer algorithms for determiig the value of a fuctio deped upo the usual arithmetic operatios; thus a exact determiatio ca oly be achieved for ratioal fuctios (quotiets of polyomials). If a fuctio is trascedetal, its values ca oly be approximated. For example, we kow that (9.) e x lim This expressio tells us that if for ay we do the calculatio described by the expressio o the right, that these umbers will, for large eough, be close to the true value of e x. Now, it turs out that this is a very iefficiet way to calculate e x, ad the expressio as a ifiite series (which we will discover later i this chapter) (9.2) e x x x x 2 2! x 3 3! x! is far better. Equatio 9.2 is take to mea: add up the umbers of the form x!, startig with 0. If we add up eough terms, we have a good approximatio to e x. Of course, it is importat to have estimates o how good this approximatio is, ad more geerally, to have ways of fidig these approximatig sums. That is what we study i this chapter, startig with the idea of covergece i the sese of good approximatio. Defiitio 9. A sequece is a list of umbers, deoted a, where a is the th term of the sequece. A sequece may be defied by a specific formula or a algorithm for determiig the members of the sequece successively, or recursively. 28
9. Covergece: Defiitio ad Examples 29 Example 9. The formulae a sequeces, respectively: (9.3) 2 3 ; ; b 2 ; c 3 2 0 defie the 3 4 5 2 3 The last sequece ca be defied recursively by: a 0 is give by the recursio a a. a ; 3 5 7 9 3, ad for 0 c 3 2 c 2. Similarly, the first The symbol! (read -factorial ) is used to deote the product of the first itegers. This also has the recursive defiitio: a 0, ad for 0, a a. (Note that we have defied 0! ). Of the sequeces described i equatio 9.2, the first ad the third clearly grow without boud, but the secod is bouded; i fact, if we rewrite the geeral term as (9.4) b we see that the sequece b approaches. We say that b coverges to, as i the followig defiitio. Defiitio 9.2 A sequece a a 2 a coverges to a limit L, writte (9.5) lim a L if, for every ε 0, there is a such that for all we have a L ε. This just says that we ca be sure that a is as close to L as we eed it to be, just by takig the idex large eough. We will rarely have to actually use this defiitio, relyig more o uderstadig what it says, ad kow facts about limits. For example: Propositio 9. If the geeral term a of a sequece ca be expressed as f for a cotiuous fuctio f ad if we kow that lim x f x L, the we ca coclude that lim x a L. As a applicatio, usig results from the precedig chapter, we have Propositio 9.2 a) lim b) lim p for p 0 p 0 for p 0 c) lim A if A 0 Let p ad q be polyomials. d) lim p q 0 if deg p deg q lim p q if deg p deg q e) If the polyomials p ad q have the same degree, the lim p q a b leadig coefficiets of p ad q. p f) lim e for ay polyomial p. p g) lim for ay polyomial of positive degree ad ay positive c. l c where a ad b are the
Chapter 9 Sequeces ad Series 30 These ca all be derived by replacig by x, ad usig limit theorems already discussed (such as l Hôpital s rule). Example 9.2 Example 9.3 2 lim 2 lim by e above 0 sice the umerator oscillates betwee - ad, ad the deomiator goes to ifiity. We should ot be perturbed by such oscillatio, so log as it remais bouded. For example we also have (9.6) lim si sice the term si remais bouded. The followig propositios state the geeral rule for hadlig such cases. Propositio 9.3 a) (Squeeze theorem) Give three sequeces a b c, if a b c for all adlim a lim c L the also lim b L b) If a b c, the sequece b is bouded, c 0 ad lim c Let s see why b) is true, usig a). Let M be the boud of the b. The (9.7) Mc b c Mc 0 0, the also lim a so a) applies ad the coclusio follows. I some cases where oe of the above rules apply, we have to retur to the defiitio of covergece. a Example 9.4 For ay a 0, lim! 0 To see why this is true, we thik of the sequece as recursively defied: the first term, a is a, ad each a is obtaied by multiplyig its predecessor by a. Now, evetually, that is, for large eough, a 2. Thus each term after that is less tha half its predecessor. This ow surely looks like a sequece covergig to zero. To be more precise, let N be the first iteger for which a N 2. The for ay k 0, (9.8) a N k N k! a N 2 k N! Now the sequece o the right is a fixed umber (a N N ) times a sequece ( 2 k ) which teds to zero. Thus our sequece coverges to zero, also by the squeeze theorem (propositio 9.3a). Note that i the above argumet, we oly had to show that the geeral term of our sequece is domiated by the geeral term of a sequece covergig to zero from some poit o. What happes to ay fiite collectio of terms of a sequece is ot relevat to the questio of covergece. We shall use the word evetually to mea from some poit o, or more precisely, for all greater tha some fixed iteger N. We restate propositio 9.3, usig the word evetually : 0.
9. Covergece: Defiitio ad Examples 3 Propositio 9.4 a) (Squeeze theorem) Give three sequeces a b c, if evetually (9.9) a b c for all ad lim a the also (9.0) lim b b) Suppose that a bouded ad L lim c L b c evetually, that is, for all larger tha some N. If the sequece b is (9.) lim c the also (9.2) lim a p Example 9.5 For ay positive iteger p, lim 0! The idea here is that the umerator is a product of p terms, whereas the deomiator is a product of terms, so grows faster tha the umerator. To make this precise, write (9.3) p! 0 0 p p! Now, if is so large that p 2 2p will do), the the first factor is bouded by 2 p. Thus, for 2p, that is, evetually, (9.4) p! 2p p! Sice p! 0 ad, the result follows from the squeeze theorem. Fially, we ote that the limit of a sum is the sum of the limits: Propositio 9.5 If a b c, ad the sequeces b ad c coverge, the so does the sequece a, ad (9.5) lim a lim b lim c Series For may sequeces, i fact, the most importat oes, the geeral term is formed by addig somethig to its predecessor; that is, the sequece is formed by the recursio s s a, where a is from aother sequece. Such a sequece is called a series. Explicitly, the terms of the series are (9.6) a a a 2 a a 2 a 3 a a 2 a 3 a
Chapter 9 Sequeces ad Series 32 It is useful to use the summatio symbol: (9.7) a a 2 a 3 a k a k Defiitio 9.3 The series (9.8) a k is to be cosidered as the limit of the sequece (9.9) s a k If the limit L of the sequece s exists, the series is said to coverge, ad L is called its sum. If the limit does ot exist, the series diverges. The terms of the sequece s are called the partial sums of the series. Example 9.6 k 2 k Let s look at a few partial sums: (9.20) 2 3 4 7 8 5 6 We see that each term adds half the distace of its predecessor from, from which we guess that the partial sum: s 2. Let s ow show that to be true. As we have see it is true for the first four terms. If it is true for the th term, it is also true for the th term: (9.2) s s 2 2 2 2 2 2 Thus, our guess holds for the fifth, ad the the sixth, ad, by cotiued applicatio of equatio 9.2, ultimately, for all terms. So the result is easy to coclude: (9.22) k 2 k lim s lim 2 Now, remember that the idex is a way of relatig the partial sums of the series to the geeral term from which it is defied, so if we chage that relatio cosistetly, we do t chage the series. For example, (9.23) k a k a a k m 9a m 8 ad so forth. Each represetatio comes about by replacig the idex with a ew idex. For example, if we substitute for k, we get the first equality; if we substitute k for we get the secod equality, ad
9. Covergece: Defiitio ad Examples 33 if we replace k examples show. Example 9.7 For (9.24) Example 9.8 (9.25) by m 8, we get the last oe. It is ofte useful to make a chage of idex as the ext 2 k 2 2 k k 2 k 2 k 2 k 2 First, chage the idex by k m, ad the factor out 2 : k 2 k m 0 2 m 2 m 0 2 m 2 2 2 Propositio 9.6 (Geometric Series) : (9.26) x k x for x (9.27) x k diverges for x To show this, we obtai (by a clever little observatio) a formula for the partial sums (9.28) s Note that (9.29) s (9.30) s Equatig these expressios for s x k x x 2 x x x 2 x x x x 2 x, we obtai s x s x xs ad xs. Solvig this for s : (9.3) s so (9.32) x k lim s x k x x x lim x
Chapter 9 Sequeces ad Series 34 which equals x if x ad diverges if x. We look at the cases x separately. For x s, so the series diverges. For x, the sequece s is the sequece 0 0 0, so caot coverge to ay particular umber. Example 9.9 (9.33) k k Thus the partial sum s ca be calculated: (9.34) s 2 We first use the fact that k k 3 2 k k 4 3 (9.35) 2 2 3 3 4 4 (9.36) which coverges to as goes to ifiity. This is a example of a telescopig series. Fially, we observe that if a series coverges, its geeral term must go to zero. Be careful: there are may series whose geeral term goes to zero which do ot coverge. Propositio 9.7 If a k coverges, the lim a k To see this, let s a k t k 0 a k. The, sice these are both sequeces of the partial sums of the series, but idexed differetly, lim s lim t. Thus lim s t 0. But s t a. Fially, Propositio 9.5 gives us: Propositio 9.8 If a b c, ad the series b ad c coverge, the so does the series a, ad a b c 0 9.2. Tests for covergece Throughout this sectio, uless otherwise specified, we will be cosiderig series, all of whose terms are positive. For such a series, the sequece of partial sums is icreasig. If they remai bouded, the - just as i the assertio of theorem 8. for fuctios - the sequece of partial sums will coverge. Propositio 9.9 If a k 0 for all k, ad there is a M k 0 such that a M for all, the a k coverges.
9.2 Tests for covergece 35 Because of this propositio, for a series with positive terms, the statemets a k coverges, a k diverges, are usually writte simply as (9.37) a k coverges Here is a importat applicatio of this propositio: a k diverges 0 Propositio 9.0 (Compariso Test). Give two sequeces a k b k with a a) If b k the a k b) If a k the b k k b k. The As for (a), the sequece of partial sums s a k is bouded by 0 b k, so coverges by Propositio 9.9. I the secod case, sice the sequece of partial sums a k has o boud, either does the sequece of partial sums of b k. It is importat to observe that it is ot ecessary that the iequalities i the hypothesis of propositio 9.0 hold for all k, oly that they evetually hold. That is because the issue of covergece series is determied by the ed of the series, ad ot affected by ay fiite umber of terms. Example 9.0 r k r if 0 r Sice r k r k r, (9.38) so the compariso test applies. Example 9. k r k if r r k r r k Now, here the trouble is that the umerator grows without boud - but it does t grow as fast as a power. So, what we do is borrow somethig from the deomiator to compesate for the umerator. We ote that evetually k k r 2 ; i fact, this is true as soo as k 2lk lr (which evetually happes, sice k lk ). The for all k larger tha this umber (9.39) k r k Sice r, we also have r, ad so the series k r k r k r k (9.40) r k coverges, ad thus, by compariso, our origial series coverges. Example 9.2 (9.4) Now, 2 2
Chapter 9 Sequeces ad Series 36 so our series is domiated by a telescopig series which coverges (see example 9.9. above). A very useful applicatio of the compariso test is the followig. Propositio 9. (The Itegral Test). Suppose that f is a oegative, oicreasig fuctio defied o a iterval M. Suppose the a is a sequece such that for M, a a) If f x dx the a M b) If M f x dx the a Let (9.42) b f x dx f. The The, sice the fuctio is oicreasig, f b f a ; that is a b a compariso theorem. For example, if f x dx, the b coverges, so by compariso a coverges. Example 9.3 (The harmoic series). We apply the itegral test usig the fuctio f x (9.43) as we saw i chapter 8, the result follows. dx x x. Sice Now, use the also If we apply example 7 of chapter 8 to series via the itegral test we have a result which is very useful for comparisos: Propositio 9.2 Let p be a positive umber. a) p if p p b) p if This follows from the same result for the itegral of x p. Example 9.4 2 The fuctio f x l p x lx p is decreasig. We itegrate usig the substitutio u lx: (9.44) A 2 dx la du x lx p l2 u p We kow (agai from example 7, chapter 8) that this coverges if p, ad otherwise diverges. Thus, by the itegral test, (9.45) 2 l if p p
9.2 Tests for covergece 37 ad otherwise diverges. Fially, we eed a tool to test for covergece whe we caot realize the geeral term of the series i the form f for some fuctio f. For example, if the expressio for a ivolves the factorial, we proceed to the followig. Propositio 9.3 (Ratio Test). Give the series a, cosider (9.46) lim a a if the limit exists. If L, the series coverges; if L, the series diverges. For the case L draw o coclusio. Suppose that L. The there is a umber r with L r such that evetually a there is a iteger N such that a a r for all N. We coclude (9.47) a N a N r a N 2 a N r a N r2 a N 3 a N 2 r a N r3 L, we ca a r. That is, ad so forth. Thus, we have, for all k, a N k a N r k, so by compariso with the geometric series, our series coverges. If o the other had, L, there is a umber r, L r, such that evetually a a r. Followig the same argumet but with the iequalities reversed, we coclude that for all k, a N k a N r k, so we have divergece by compariso with the geometric series. We ca coclude othig if L. This is the case for the all the series of the type p, ad as we have see, for some p we get covergece, ad divergece for other p. Example 9.5 a! We try the ratio test. (9.48) a a a!! a a 0 as, so the ratio test gives us covergece. Example 9.6 (9.49) Try the ratio test: 2 3 3 2 a a 3 3 3 2 2 3 3 3 2 3 so we have covergece. Example 9.7 (9.50) r Here the ratio test gives a a r
Chapter 9 Sequeces ad Series 38 so we coclude that the series coverges if r, ad diverges if r. This may seem to be a simplificatio of propositio 9.6, but i fact it is a fraud. The argumet is circular, for we have used the covergece of the geometric series to derive the ratio test. We observe that we did t really eed to kow that the limit of a a exists, oly that evetually these ratios are either less tha some umber less tha to coclude covergece, or greater tha some umber greater tha, for divergece. 9.3. Absolute covergece There are ew difficulties whe we have to cosider series of egative as well as positive terms. For example, although the harmoic series diverges, if we alterately chage sigs, the series ow coverges. Example 9.8 The series 2 3 4 coverges. To see this, we look at the sequeces of eve partial sums ad odd partial sums separately. Sice (9.5) s 2 s 2 2 2 the sequece of eve partial sums is icreasig. Similarly, (9.52) s 2 s 2 2 s 2 2 2 2 3 s 2 tells us that the sequece of odd partial sums is decreasig. Now (9.53) s 2 s 2 2 s 2 that is, the odd partial sums are all greater tha all the eve partial sums. So both sequeces are bouded, ad thus coverge. But, they coverge to the same limit, as we see by takig the limit i equatio 9.53: (9.54) lim s 2 lim s 2 lim 2 lim s 2 sice 2 0. Sice they both coverge to the same limit, the full sequece also coverges, ad to the same limit. This argumet actually geeralizes to ay alteratig series, a series whose terms alterate i sig. Propositio 9.4 If a is a decreasig sequece, ad lim a 0 the the series a coverges. Defiitio 9.4 Give a sequece a, we say the series a coverges absolutely if, for the series formed of the absolute values a, we have covergece: a. Propositio 9.5 If a series coverges absolutely, it coverges. That is, (9.55) If a the a coverges
9.4 Power Series 39 To see that, let s be the th partial sum of the sequece, p the sum of all the positive terms makig up s, ad q the sum of the absolute values of all the egative terms. The (9.56) s p q Both sequeces p ad q are icreasig, ad bouded by a, so coverge, to, say p, q respectively. The (9.57) a lim s lim p lim q p q Because of this peculiarity of sequeces of terms with alteratig sigs, we shall be most iterested i absolute covergece. We ca use the tests of sectio 9.3 (applied to the series of absolute values), to test for absolute covergece. Example 9.9 x coverges for x This is because the sum of the absolute values is just the geometric series. Example 9.20 2 x coverges for x Here we use the ratio test for the absolute values; a (9.58) a 2 x 2 x Thus, we get covergece for x of absolute value less tha. 2 x x 9.4. Power Series Defiitio 9.5 the power series. A power series is a series of the form a x c The poit c is called the ceter of A power series defies a fuctio o the set of poits for which it coverges by (9.59) f x a x c The series provides a effective way of approximately evaluatig the fuctio f ; our goal i these last sectios is to show that most fuctios do have a power series represetatio. We ca use the ratio test to determie the questio of covergece. We take the ratio of successive terms of (4): a (9.60) x c a a x c a x c L x c
Chapter 9 Sequeces ad Series 40 if the limit L lim a a exists. I this case the series coverges absolutely for x c L, ad diverges for x c L. Thus, the domais of covergece ad divergece of the series are separated by the circle, cetered at c of radius L. It ca be show that, i geeral, there is a circle separatig these domais, eve if the limit of the ratio of successive coefficiets does t exist. 0 R Propositio 9.6 Give the power series represetatio f x a x c there is a umber R such that we get absolute covergece for all x x c R, ad divergece for all x x c R. a R is called the radius of covergece of the power series. We have this value of R: lim a R if the limit exists. The first example of a power series represetatio is that of the geometric series: Example 9.2 Example 9.22 x x successive coefficiets is k x coverges for for x has the radius of covergece R. k (9.6) k k as. Example 9.23 (9.62) so R (9.63) x! has radius of covergece R x for ay umber k. We use the ratio test. The ratio of!! Usig the ratio test:, ad the series coverges for all x. O the other had, the ratio test shows us that the series!x has radius of covergece R 0, so coverges oly for x 0. Newto thought of power series as geeralized polyomials - that is, as polyomials, oly loger. This is justified, because we ca operate with power series just as we operate with polyomials: we ca add, multiply, ad substitute i them by doig so term by term. Example 9.24 (9.64) x x x x x x x for R For x x x 2 x 3 0 x x 2 x 3 x 4 Example 9.25 x 2 x 2 x 2 x 2 for x
9.4 Power Series 4 To see the first, we ote that x 2 is obtaied from x by substitutig x 2 for x. Thus, the power series represetatio is obtaied i the same way. I the secod, we have substituted x 2 for x. Example 9.26 Fid a power series expasio for 5 2x cetered at the origi. What is its radius of covergece? To solve a problem like this, we have to relate the fuctio to aother fuctio, whose power series we kow. I this case that would be x. Now 5 2x 5 2 5 x, so our fuctio is obtaied from x by first replacig x by 2 5 x, ad the dividig by 5. We follow the same istructios with the power series. Start with (9.65) Replace x by 2 5 x: (9.66) Divide by 5 ad clea up: (9.67) x 2 5 x 5 2x 5 x 2 5 x 2 5 x 2 x 5 We ca calculate the radius of covergece usig propositio 9.6, or we ca reaso as follows; sice the series we started with coverges for x, our fial series coverges for 2 5 x, or x 5 2. Fially, we ca also itegrate ad differetiate power series term by term: Propositio 9.7 Suppose that f x (9.68) x (9.69) f x a x has radius of covergece R. The 0 f t dt ad both have the same radius of covergece, R. a x a x Example 9.27 arctax 2 x2 We kow that the derivative of the arc taget is x 2. Now, i example 9.25, we have already foud the power series represetatio of that fuctio, so we obtai the power series represetatio of arctax by itegratig term by term. x Example 9.28 e x! fid (9.70) f x for all x Let f x x! 0 x! The, differetiatig term by term, we x! x!
Chapter 9 Sequeces ad Series 42 where the last equatio is obtaied by replacig the idex by. Thus f x differetial equatio, y y, defiig the expoetial fuctio. Sice f 0 fuctio. f x, so satisfies the also, it is the expoetial Example 9.29 e x 2 x2! for all x Just replace x i example 9.28 by x 2. 9.5. Taylor Series Fially we tackle the questio: how do we fid the power series represetatio of a give fuctio? Recallig that the purpose of the power series is to have a effective way to approximate the values of a fuctio by polyomials, we tur to that questio: what is the best way to so approximate a fuctio? We start with a fuctio f that has derivatives of all orders defied i a iterval about the origi. To begi with, we recall the defiitio of the derivative i this cotext: (9.7) lim x 0 f x f 0 x If we rewrite this as f 0 (9.72) lim x 0 f x f 0 f 0 x x we see that the liear fuctio y f 0 f 0 x approximates f x to first order: f 0 f 0 x is closer to f x tha x is to zero, ad by a order of magitude. We ow ask, ca we fid a quadratic polyomial which approximates f to secod order? Let y a bx cx 2 be such a polyomial. The we wat f x a bx cx 2 (9.73) lim x 0 x 2 0 We calculate this limit usig l Hôpital s rule. First of all, for l Hôpital s rule to apply, we have to have a f 0. The f x f 0 bx cx 2 (9.74) lim l H f x b 2cx x 0 x 2 lim x 0 2x We ca apply l Hôpital s rule agai, if we have b f 0 : f x f 0 2cx (9.75) lim l H f x 2c lim x 0 2x x 0 2 if c f 0 2. We coclude that the polyomial (9.76) f 0 f 0 x approximates f to secod order: this is closer to f x tha x is to 0 by two orders of magitude. Furthermore, it is the uique quadratic polyomial to do so. f 0 2 x 2 0 0
9.5 Taylor Series 43 We ca repeat this procedure as may times as we care to, cocludig Propositio 9.8 The polyomial which approximates f ear 0 to th order is (9.77) f 0 f 0 x f 0 2 x 2 f 0! Of course we ca make the same argumet at ay poit, ot just the origi. To summarize: Defiitio 9.6 Suppose that f is a fuctio with derivatives at all orders defied i a iterval about the poit c. The Taylor polyomial of degree of f, cetered at c is (9.78) T c f x f k c k! x c k Propositio 9.9 The Taylor polyomial T c f is the polyomial of degree which approximates f ear c to th order. So, we ca compute effective approximatios to the values of f x ear c by these Taylor polyomials; but the questio is, how effective is this? More precisely, what is the error? We use this estimate: Propositio 9.20 Suppose that f is differetiable to order i the iterval c a c a cetered at the poit c. The the error i approximatig f i this iterval by its Taylor polyomial of degree, T c f is bouded by (9.79) M! x c where M is a boud of the values of f over the iterval c a c a. To be precise, we have the iequality (9.80) f x Tc M f x! x c I the ext chapter we will show how the error estimate is obtaied, ad see how to work with it. What we wat ow is to cocetrate o the represetatio by series. Defiitio 9.7 Let f be a fuctio which is differetiable to all orders i a eighborhood of the poit c. The Taylor series for f cetered at c is (9.8) T c f x f c x c! If c is the origi, this series is called the Maclauri series for f. Propositio 9.2 Suppose that f is a fuctio which has derivatives of all orders i the iterval c a c a, Let M be a boud for the th derivative of f i the iterval. If the sequece (9.82) M! x c 0
Chapter 9 Sequeces ad Series 44 coverges to zero for all x i the iterval, the f is give by its Taylor series: (9.83) f x i c c a. As a example, e x has the Maclauri series f c! x c (9.84) e x! x We have already show by other meas. We ca verify this usig propositio 9.2 as well, sice the th derivative of e x is still e x, ad the value at x 0 is. By a parallel calculatio we obtai the power series represetatio of e x cetered at ay poit: Example 9.30 For c ay poit, the fuctio e x has the Taylor series represetatio cetered at c: (9.85) e x e c! x c We do have to verify that the remaiders coverge to zero; that is the terms 9.82 coverge to zero. Sice e x is a icreasig fuctio, its maximum i the iterval a c a c is at x a c, so we ca take M e a c. The, for the expoetial fuctio we have M (9.86) lim! x c by example 9.24. It is useful to make the followig observatio e a c x c lim! Propositio 9.22 Suppose that f has a power series represetatio: (9.87) f x The, this is its Taylor series. More precisely: a x c (9.88) a f c! 0 This is easy to see; if we differetiate 9.87 k times we obtai: (9.89) f k x Now, let x c ad obtai f k c k k a x c k k!a k, for all terms but the first have the factor x c. So, if we have foud a power series represetative of a fuctio, the that is automatically the Taylors series for the fuctio.
9.5 Taylor Series 45 Example 9.3 Fid the Maclauri series for the fuctio f x x 5x 2 x 3. Sice a polyomial is already expressed as a sum of powers of x, that expressio is a power series, ad thus the Maclauri series for the polyomial. Example 9.32 Fid the Taylor series cetered at c for the fuctio f x have to fid the values of the derivatives of f at c : (9.90) f 4 x 5x 2 x 3. We (9.9) f x 0x 3x 2 so f 6 (9.92) f x 0 6x so f (9.93) f x 6 so f 6 4 ad all higher derivatives are zero. Thus the Taylor series is (9.94) f x 4 6 x 4 2! x 2 6 3! x 3 4 6 x 2 x 2 x 3 Now, we ca fid the Maclauri series for may fuctios, so log as we kow how to differetiate them. Followig is a list of the most importat Maclauri series. Propositio 9.23 a) x b) e x! x x x c) cosx d) six 2! x2 2! x2 e) arctax 2 x2 We have already see how to get (a), (b) ad (d). For the trigoometric fuctios, we proceed as follows. First, the cosie: (9.95) f 0 (9.96) f x six so f (9.97) f x cosx so f 0
Chapter 9 Sequeces ad Series 46 (9.98) f x (9.99) f iv x Thus, up to four terms we have six so f cosx so f iv 0 (9.00) cosx x 2 2! x 4 4! But, ow, sice we have retured to cosx, the cycle 0 0 coclude that (9.0) cosx x 2 2! x 4 x 6 4! 6! x 8 8! repeats itself agai ad agai. We which ca be rewritte as iii of propositio 9.23 above. Oe fial Taylor series is worth otig: sice the itegral of x is l x, we ca fid the Taylor series cetered at for lx as follows: (9.02) t t (9.03) l t Now, make the substitutio x t, so t x: x (9.04) lx t x