chapter 8 summary - 7/5 Summary of Chapter Eight Sequences...

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Unformatted text preview: 7/5 Summary of Chapter Eight Sequences A sequence {an} is an ordered list of numbers a1, a2, a3, - - v . The first number is al, the second is a2, and so on. The list is infinite in length, although some or all of the terms (numbers) in the sequence may be equal. A sequence {an} converges to a number L if, given any tolerance (positive number) 6, we can find some number a N in the list so that all terms beyond a N lie within 6 of L. In this case, we write limnnoo an 2 L, or an —> L as n —> 00‘ If a sequence does not converge to any number, we say that the sequence diverges. This can happen in several ways: the aims may approach 00 or ——00, or the a’ns may not approach anything at all. Example 1 lim (4 — 2 4, n—aoo 'n, hm —2n3+n—6 _ _g Twice 5n3 + 9 — 5’ {n2} diverges, and {sin(n)} diverges because the numbers sin(n) do not approach any numerical value as n increases ( although sin(n) is always between —1 and 1 If limnnoo an : L and limnnoo bn = K, then n—roo 77,400 lim (an+bn=L+K, lirn (an—bn)=L—K, lim can 2 cL for any number c, Tin—’00 lim (anbn) = LK, ’I'L-—>OO and lim (32> = 5 if each bn e 0 and K 7e 0. n—roo bn K We call {an} monotone increasing if each an < an“. A number U is an upper bound for {an} if each an 3 U. A monotone increasing sequence that has an upper bound must converge (although this result does not tell us what the limit is). We call {an} monotone decreasing if each an+1 < an. A number L is a lower bound for {an} if each an 2 L. A monotone decreasing sequence that has a lower bound. must converge (again, this does not tell us what the limit is). Example 2 Let a1 = x/3 and, for n = 1,2,3, ~«, let an+1 : i/3an. The first few terms of this sequence are 0.4 3(13 2 H 3V3\/;/§, and so on. It is not obvious whether this sequence converges, or what the limit might be. However, ask two questions. (I) Is the sequence bounded above? Notice that al s3 1.7321, a2 x 2.2795, a3 is 2.6151 and a3 % 2.8009. Based on this flimsy evidence, make an initial guess that each term of the sequence is less than 3. Now see if this is true. Certainly a1 < 3. Now proceed by induction. If, for any n, an < 3, then a3,“ 2 3a,, < 3(3) 2 9 implies that an+1 < 3 also. By induction, an < 3 for each positive integer n, so 3 is an upper bound for {an}. (2) Is the sequence increasing? Again, from our numerical evidence, it ap- pears to be. To see if this is true, notice that, for an < an+1 to be true, we must have a,L < m. This is true if a3, < 3am or an < 3, which we know to be true from (1)- Since {an} is a monotone increasing sequence that is bounded above, this sequence converges. , What does this sequence converge to? Let limnnDO an 2 L. Then limn_,oo an+1 = L also. Now take the limit as n -~> oo in the defining relationship an+1 = m. We get lim an+1 = 3( lim an). 71"“)00 '71—’00 L: V3L. Solve this to get L = 3. This sequence converges to 3. If you compute some of the ags, you can see the numbers increasing toward 3. For example, a9 ’5 2.9936. Then lnfimte Series A sequence is an infinitely long list of numbers. An infinite series (or just seies) is an infinite sum 2:021 ak 2: a1 + a2 + The numbers a1, a2, form a sequence, and the infinite series as the sum of all the numbers in this sequence. The nth partial sum of the series 22:1 a}, is the sum of its first it terms: n sn22ak=a1+a2+a3+-H+an- 19:1 3 Each 5” is a finite sum, hence is a number, so 81, 52, $3, is a sequence, the sequence of partial sums of the series 21:1 ak. As n is chosen larger, . . . ()0 . 5,, includes more terms of the 1nfin1te sum Zkzl ak. We say that the series 22:1 ak converges to a number L if the sequence of partial sums converge to L as n approaches infinity: DO 712 2 LL], : lim 3“: lim E ak. 711—900 71—900 k=1 k=l If the partial sums do not have a finite limit, then we say that the series 220:1 ak diverges. Example 3 (I) The harmonic series = 1 + % + % + i + - - - diverges. (2) Ifr is a given number, the series Zkzl r1“ = r + r2 + r3 + - ~- is called a geometric series. This series converges if —1 < r < 1, and diverges if r 2 1 or r g —1. If —1 < r < 1, this series converges to r/(l — r). For example, i3k—3+9+27I 81+... 3 § 8 —8 64 512'4096 ‘1_§-5, k=1 8 i<—§)k— §+25 125' 625 ‘3 _3 and 00 7 19 Z diverges. Ic=1 The series 22:0 rk : 1 + r + r2 + r3 + . : - is also called a geometric series, and converges (to 1/(1 — r)) exactly when ~«1 < r < 1. (3) The series 220:1 E1; is called a 13— series, and it converges when p > 1 and diverges when p S 1. When p : 1 this is the divergent harmonic series. For the 19— series, it is generally unclear what the series actually converges to for a given p > 1. Tests for Convergence or Divergence of Series The following tests help us determine whether a series converges or diverges. Limit Test - If lirnn_>00 an 75 0, then 2:021 an diverges. This test can never be used to show that a series converges. If ,limn_s00 a,L = O, the series may converge or it may diverge. For example 2:021(1/n) diverges and 220:1(1 /n2) converges, even though limnnoofl = limnnoofl/r?) = 0. Having limn_,00 an 2 0 does not settle the issue of convergence or divergence. Example 4 °° 5n+2 Z n+7 91:1 - - 5 2 _ diverges, because limnnoo 7:35— ... 5 # O. 00 Z cos(n) n=1 diverges, because lirnn_>00 cos(n) does not exist { and hence is not zero The integral test, comparison test, and ratio test are only for positive series :3le an, in Which each an > 0. Integral Test — Given 22:1 am suppose we can replace n by at in an to get a function f that is continuous and decreasing for a: Z 1. Then 20m and/1 f(a:)d:t both converge, or both diverge. Comparison Test — If 0 < an 3 19m then: 00 00 if E an diverges, so does E (3”, n=1 71.21 and 00 00 if E bn converges, so does E an. n=1 71:1 If all we know is that Z:le bn diverges, or 22le an converges, no conclusion can be drawn about the other series. Ratio Test — Suppose each an > O and . a 1 hm M” = L. n-aoo an Then 00 if 0 S L < 1, then 2 an converges, 71:1 and 00 if 1 < L g 00, then Zen diverges. 11:]. But, if L = 1, no conclusion can be drawn from this test. Example 5 Consider 230:2 Notice that lining“, m = 0, so this series may converge - we don”t know yet. Use the integral test.“ / 1 dleim 1 dm. 2 mln(m) ‘ r—ioo 2 acln(:c) i l For this integral, let u = 1n(x). Then clu : ids: and f 1 dQIZ/ldUIIHIUI =1n|1n(as)|. m1n(a:) it Then Iim T 1 dd“ lim T 1 also race 2 azln(m) — r—+oo 2 m1n(a:) lim [ln(ln(z))]§ = lim (ln(1n(r)) —1n(1n(2))) = 00. Therefore 2:022 m diverges. Example 6 Consider 220:1 Since —1 < Sin(n) < 1, then 0 < 1 + sin(n) < 2 and 1 + sin(n) < __2_ 10H 10’“ But 220:1 converges (2 times a convergent geometric series ), so by compar- ison, 2:021 converges. O< Example 7 Consider 220:1 Notice that 1 1 1 0<———- <———:—. 1/n3+1 Mn?) 713/2 But 2:021 7131/2 converges (19— series with p = 3/2 > 1). Therefore the given series converges. - Exam 1e 8 Consider °°_ 1 . Here we can write p n__1 2n+3 0<—1—-< 1 6n 2n+3' forn: 1,2,3,---. This inequality is true because 2n+3 < 6n for positive integers n. But :3le 6i” diverges {1/6 times the divergent harmonic series ), so the given series diverges. Example 9 Consider 220:1 Here n! = l e 2 - - n, the product of the n! integers from 1 through n. We cannot use the integral test here, because as! is not defined if a: is not a positive integer. Take the ratio of successive terms: 10“+1 an+1 (n+1)! _ 10”+1 n! 10 > 0 an ” 12'!” _ 10n (n+ 1)! — n+ 1 as'n —> oo.By the ratio test, this series converges. Example 10 Consider 220:1 Take the ratio of successive terms: 2-...+1 an+1 (n+1)4 ” 4 2M1 2 1 4 2 z 2 : ) an 2" n + 1 2” 1 + 31- n4 as n —> 00. Since this limit is greater than 1, this series diverges. Estimating the Sum of a Positive Series Suppose we know that 22:, an converges to some number s, and that each an is a positive number. We want to estimate the sum of this series, to within whatever tolerance or error is given. The key is to compare 5 with the partial sum 5”, since we can make 3,, as close as we want to s by choosing n large enough. Let R41 = s — 3n =(a1+a2+-~+an+an+1+~')—(a1+a2+~-~+an) =0n+1+an+2+”'- Assume that we can replace n by at in an to get a continuous function f that decreases as a: ——> 00. By an area argument like that used to derive the integral test, we get 00 Rm 3/ But, for a positive series, we also have 0 < s - 8,, = Rn. Then 00 0<s—snS/ 71, If we choose ‘n so that f(ac)d:c g c, then we will have 0 < s —— sn 5 6, so 5,, will be within 6 of 5. Example 11 The series 22:, converges (pm series with p = 3 Suppose we want to approximate the sum 3 of this series to within 1/100. Let f(.’E) = 1/333. The strategy is to choose n so that /00 idm < 1 n m3 ‘“ 100' To do this, compute 00 1 7" 1 ’I" / ——3d:r : lim / 23—3da: = lim [——x‘2] n x T—>OO n T—iOO 2 71 . < 1 1 > 1 1 = hm + = <‘ - T—)O<) 2T2 2712 For this to be true, we need or n > 50. The smallest positive integer satisfying this inequality is n = 8. Thus 5 is approximated, to within 1/100, by 8 1 5822—158— k=1 2 =1 .1. + + l + 343 512 = 1.1952, to four decimal places. Alternating Series A series is alternating if the terms alternate in sign. Such a series is often written 00 Z(—1)k+1ak = 01 — a2 +043 — a4 + ' ", k=l with each ak positive. Alternating Series Test — 2:1(—1)k+1ak converges if the positive terms ak decrease to zero as k —> oo. . oo _ lc+1l 1 ' _ Example 12 Zk=1( 1) k converges, because k decreases to zero as k in creases If an alternating series 22:1(—1)’“+1ak converges to s, then the difference between 8 and the nth partial sum of the series is less than an+1z s — snl < an“. If the alls decrease to zero as n increases, this enables us to approximate the sum of the series by a partial sum to whatever degree of accuracy we want. Example 13 Approximate the sum of 2:021(~1)”+1—1— to within 1/100. n2 This alternating series converges because l/n2 decreases to zero as n in— creases. Let s be the sum of the series. Then 1 IS — Snl < an+1 : To get sn within 1/100 of 5, choose n so that 1 1 ___-_.___ <___. (n+ 1)2 M 100 This will hold if (n + 1)2 2 100, or n + 1 Z 10, so we can choose n = 9. The sum of the series is approximated by 9 1 59 2 Sam“? = 0.82796 k:=l I with an error of no more than 1 / 100. Absolute and Conditional Convergence In general, a series 220:1 ck may be neither positive nor alternating. None of the above tests would apply to such a series. We say that 220:1 ck converges absolutely if 22:1 |ckl converges. It is pos— sible for 22:1 ck to converge, but 2:021 [ckl to diverge. In this case we say that 2:11 ck converges conditionally. i Example 14 22:1(—1)k+1% converges conditionally. This alternating series converges, bat the series of absolute values of its terms is 22:1 %, the divergent harmonic series. Absolute Convergence Test — If 22:1 ickf converges, then 22:1 ck converges also. Thus, absolute convergence implies convergence (but convergence does not imply absolute convergence). The value of this fact is that we can sometimes show that a series 22:, C}, converges by applying tests for positive series to 22:1 ickl. If 22:1 |ckl con— verges, so does 21:1 ck. However, if 22:1 Ick| diverges, then we do not know about 220:1 ck, which may converge conditionally. Example 15 Consider 22:1 This series is not positive, nor is it al— ternating, since sin(k) is positive sometimes, and negative sometimes, but the terms do not alternate in sign as k: increases. Notice that 1 sin(k) S k3/2‘ 0< k3,, sinUc! sitika leg/2., ’ converges. Therefore 2:021 ICE/2 . oo 1 g Loo Since 216:1 Egg—.2 converges, then Zkzl converges also. This series is absolutely convergent, therefore convergent. Summary of Convergence Tests 00 For any series E an, lim an 75 0 implies divergence TL-—)OO 71:1 integral test For positive series — comparison test ratio test 00 alternating series :(—1)"+lbn - convergence if the b’ns decrease to 0 71:1 absolute convergence implies convergence Power Series A series of the form 22:0 Cn(.’r — a)” = Co + 01(115 * a) + 02(35 '" C02 + ' " is called a power series, with center a, and coefficients c0, c1, (:2, Given the center and the coefficients, we ask: for what values of so does this series converge? One can show that there are three possibilities: (1) The series may converge only for r : a. This happens with the series 220:0 nix”, in which a = O and on = n! If as % 0, this series diverges. (2) The series may converge for all as. This happens with Zoo ix” n=0 n! (3) The series may converge for all a: in some interval (a — r, a + r), centered at a, and diverge for as > a + r and for a: < a — r. In case (1), the series is uninteresting. In case (2), the series converges for all m, and we say that it has infinite radius of convergence. In case (3), we say that the series has radius of convergence r. In case (3), we call (a — r, a + r) the open interval of convergence of the power series. We can often determine which case applies to a particular series by using the ratio test to check for absolute convergence. Example 16 Consider 2:020 The ratio of the absolute value of succes— sive terms is n! 1 m lml = lml ‘“’ 0 n + 1 as n ——> 00, for all x. This power series converges absolutely for all x, and therefore also converges for all m. Example 17 Consider Zoo 1 (a: + 2)”. The center of this series is —2. 11:0 n3“ T ake the ratio of the absolute value of successive terms: £m+21u+l (n+1)3“+" _ n 3" (:13 + 2)“+1 — n + 1 3n+1 (a: + 2)” n 1 1 =—-—|m+2|«+§|m+2| asn—>oo. From the ratio test, the series converges absolutely if this limit is less than 1. Therefore, the series converges if élm+2|<1, or |m+2| <3. This is the interval ——5 < :17 < 1, of radius 3, centered at —2. The series converges (absolutely) for —5 < at < 1, and diverges ifs: < —5 or ifs: > 1. This series has radius of convergence 3. Differentiation and Integration of Power Series Suppose 220:0 cn(x — a)" converges either on the entire real line (that is, for all m), or for a: in some open interval (a — r, a + r). For such as, the power series defines a function of as, and we can write f = 2:10 cn(:c —~ a)”. This function can be integrated by integrating the power series term by term, as long as we remain within an interval over which the series converges. Thus, as an indefinite integral, f(x)dx=ch (x~a)”do:=ch 1 (m—a)”+1+C’ nzo n+1 n=0 1 2 1 _ c :c a ‘ —c a: a + 0( ) I 2 1( ) 3 Further, within an interval over which the series converges, we can differen- tiate a power series term by term: C2(:B (1)3 I ~~+C'. mac) '= 2 d1; (one — or) n=0 : ZmnC’E — a)n—1 = C1 + 262(17 - a) + 3C3(m — a)2 + . 11:1 Notice that this series for f’ begins with n = 1, because the derivative of the constant term co in the power series for f is zero. The radius of convergence of the series for f’ is the same as that for f Example 18 From the geometric series, we know that 1 —.’E 00 anzl+m+$2+m3+x4+-~. n=0 M) = 1 for —l < m < 1. Then /f(w)da:=/ 1 dmz—lnil—m]+0 1 —:c _. _ +1 M _ 2 _ 3 —Z/x”dm—Zn+1mn —x+2m +39: + 71:0 71:0 And 1 00 d 00 I _ _ _ —1 a 2 3 f(30)- m —Zfix”—ann —1+2ac+3x +490 +-~-7 n—O n=1 for ~1 < m < 1. r Example 19 Put r = -—a:2 into the geometric series 1—; = 2:0:0 r” to get 1 OO 00 1M2 =Z<—wr =Z<-1>“w2”=1—w2+m4~w6+-~7 n=0 n=0 for —1 < m < 1. With = 1/(1 + :52), we have 29: 0° / = _ 2 __1 'n, 2 2n—1 f (w) —————(1 + W n; > < mm for —1<:I:<1. 10 i g i i i g i g Taylor and Maclaurin Series Expansions of Func— tions Suppose f 2 22:0 cn(m — a)”. How are the coefficients on related to 1%)? First notice that f(a) 2 c0, because (cc—a)” = O at x : a for n = 1, 2, 3, - ~ -. Now differentiate the series term by term: f'(93) = imam — a)””1 2 c1 + 2c2(x — a)2 + 303(33 , a)3 + I H‘ 7121 If we put CE 2 a, all terms involving x — a on the right vanish, and we are left with f’(a) : 01. Differentiate again: W) = 2m — 1>cn<m — a)“ n22 = 262 +6C3($ — a) + 1204(x —— 502 + . ... Again putting a: = a, we get f”(a) : 2C2, so 1 62 = if"(a). Differentiate a third time: f”’(x) : :nm — 1)(n — 2)cn(ac w (an—3 n=3 = 6C3 + 2404(95 — a) + 60c5(x —— 502 + . ... Then f”’(a) = 603, so ._ 1 //I _ 1 III C3 — 6f (6‘) — if (CL)- Continuing in this way, we find that in general 1 _— __ (n) an — n! f (a), where f (n) (a) is the nth derivative of f (x), evaluated at a: = a, and f(0)(a) = f (a) as a notational convenience. This means that the power series, which we have called f (ac), can be written in terms of and its derivatives at a as me) = Z imam — a)“. n=0 The power series on the right is the Taylor series of f about a. When a = 0, this Taylor series is often called a Maclaurin series. 11 i i Example 20 Let f = e“. We will write the Maclaurin series for f (so a = 0).The derivatives of em are all e“, so f(”)(x) : em and f(")(0) = e0 = 1. The Taylor coefficients of em about 0 are 1 1 __ (n) __._ Cn‘nlf (0)—n!‘ The Maclaurin series for f(a:) is 00 1 n_ 1 2 1 3 1 4 20—7551: _1+x+—2—!$ +333: +133 +--l-. n: It is easy to check that this series converges for all real numbers as. This raises the following question. Suppose f and a are given. If f can be differentiated as many times as we like at a, then we can write the Taylor series of f about a: Z %f‘"’(a)(w ~ a)”. n=0 . But how do we know that this series converges to f in some open interval about a? The answer is provided by the following. Suppose we are dealing with a: in some interval a - d 5 cc 3 a + d, and we can find a number M that bounds all derivatives f(”) for :0 in this interval: .f(”)(x)lSMfora~dS$Sa+dandn=1,2;3,”‘- Then the Tavlor series of f converges to f at least for a — d < :c < a + d: f(:1:) 22%f(n)(a)(a:—a)n for a—d<x<a+d. n=0 ' Example 21 Let f(ac) 2 cos(:13) and a : 0. We will expand f(ac) in a Taylor series about 0. First, the nth derivative of cos(w) must be either cos(9:), sin(a:), — cos(a:), or —- COS(.’13), depending on n. In any case, lf‘”)(rc)l s 1 on any interval. Therefore the Maclaurin series for cos(x) about 0 converges to cos(m) for all :10. All we have to do is write this series. Compute N3?) = — 5111(90), f”($) = — 005013), f”’(m) : sin(m) and f(4)($) : cos(:c). 12 g i a a Then f’(O) = 0, f”(0) = —1, f”’(O) = 0 and Ma» = 1. Every derivative of cos(ac) of odd order is zero at at = O, and every derivative of even order is either 1 or —1 at :c : 0. The pattern is: f<2“+1><0) = o, f<2"><0> = (~1)”~ The power series is 00 1 n n 00 (—1)” n where in the series on the right we used the fact that all the odd order derivatives in the series on the left are zero, and kept only the even order terms. Therefore, for all ac, 1 4 1 6 _oo(~—1)n2n_ 1 2 «so»: Wm ”17;; WC -6735 +.... 71:0 Example 2‘2 Suppose we want the Maclaurin series for cos(3m). We do not have to start from the beginning. Just substitute 3st for a: in the series just derived: °° (—1)” 2n 2n (2n)!3 IE . n=0 Example 23 Suppose we want the Maclaurin series for sin(x). Again, we do not have to start from the beginning. We have the Maclaurin series for cos(:c), so just difi‘erentiate: ~ _ d __d°°(‘1)" 2n 5111(93) - dm (303(93) — dm 7;) (2n)! 3: _. _ 00 (—1)” 211—1 _ “2:; (2n)! (2n):r . Now 2n/(2n)! = 1/(2n —— 1)!, so - 00 "llnH 271-1 1 3 1 5 1 7 sm($)::(—2;L—_—1—)—!m :m-‘im 71:11 This is correct, but this series is more traditionally written in the equivalent form ' for all m. 13 Some Applications of Power Series Expansions Often power series expansions are used to approximate such quantities as integrals. Example 24 Appromimate fol/2 cos(ac3)dsc to within 1/ 100000. We cannot evaluate this integral by the Fundamental Theorem. To approwi- mate it, first‘ewpand 003(333) in a power series about 0. To do this, replace a by m3 in the Maclaurin series for cos(.w) to get 00 “1 n 2n 00 _1 n n cos(m3) = Z ((273)! ($3) :6 ((271))! x6 . 71:0 11:0 Then 1/2 05 3 * DO (—1)” 1/2906” a: /0 c (whim—E (2n)!/0 d ~—— i til? [6.1+ 0 Now 1 #1180980 — 0000000846 < 100000. Therefore, with an error of less than 1/100000, the sum of this alternating series is % — £76, or 447/896 (about 0.49888 ). Example 25 Approzrimate [01/ 4 arctan(2x)dcc, with an error of no more than 0.001. One strategy is to expand arctan(293) in a Maclaurin series, then integrate term by term. We do not need to derive this series from scratch. Begin with 00 1 TL 1 + :32 Z 2(4)an > n=0 which we derived previously (from a geometric series). Recognizing that this is the derivative of arctan(x), write 1 f 1 + m2 dac — arctan(:1:) + C' : T;(—1)”/w2ndm 14 : i $2n+1. 2 +1 n=0 To evaluate the constant C, put :6 = O to get arctan(0)+C : 0. But arctan(0) : 0, so 0 = 0, and we have °° (4)” 1 1 1 arctan(:n) : 2: 2n + 1x2n+1 : x — 33:3 + 5335‘— ?x7 + . .. 71:0 Replacing 90 by 2m, we have arctan(2x) «_- Z 22n+1$2n+1 n=0 2n + 1 Now 1 1/4 00 (_1)n 1/4 / arctan(2m)dm = 22n+1/ mgnflda: 0 2n + 1 0 n=0 _ i0: (_1)n22n+1 ‘ n20 (2n + 1)(2n + 2)42n+2 — 2 23 + 25 27 + ‘ H (2X42) (3)(4)(44) (5)(6)(46) (7)(8)(48) ’ Now 23 (3)(4)44 = 0.0020 - a > 0.001, but 25 Therefore, the sum of the first two terms of this alternating series approximate the sum of this alternating series with an error less than 0.001, so the approxi- mation we want is 2 ————-————23 — 0 059896 (2X42) ’ <3><4><44> ‘ ' ’ 1/4 / arctan(2m)da: z 0 to five decimal places. 15 ééiis._._32aixiifisafifigéy, gbasicééxpéc£2?a? ...
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chapter 8 summary - 7/5 Summary of Chapter Eight Sequences...

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