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Unformatted text preview: 9.5 POWER SERIES AND INTERVAL OF CONVERGENCE ! if r > 1, then ! an diverges 8.5 Power Series 9-4h94 an could converge or diverge. if r = 1, then √ s test works since limn→∞ n an = r tells us that the ses comparable to a geometric series with ratio r.) Use this o determine the behavior of the series. 94. ∞ # $ " 2 n 9-4h95 n n=1 95. 521 ∞ # $ " 5n + 1 n n=1 3 n2 Announcements: " •  Webassign due Friday 108. (−1) cos(2 πn) is an alternating series. ! The series (−1) /n converges by the alternating •  Written " HW due Sunday series test. ! 109. The series (−1) 2 converges. •  Quiz next Monday The series 1/(n + 1) converges by the ratio test. ngthen Your Understanding ∞ oblems ??, explain what is wrong with the statement. 9-4h108 2n 2 The series ! 2 1/n . ∞ 9-4h109 2 ! n n=0 n n n=0 1/n3/2 converges by comparison9-4h110 with 110. If ! a converges, then !(−1)n a converges. n n 9-4h111 111. If an alternating series converges by the alternating series test, then the error in using the first n terms of the series to approximate the entire series is less in magnitude than !∞ that converges but |a | dithe first term omitted. n n=1 oblems ??, give an example of: A series verges. !∞ n=1 an 9-4h112 112. If an alternating series converges, then the error in using the first n terms of the series to approximate the entire series is less in magnitude than the first term omitted. 9-4h113 113. If 9-4h114 114. ! To find the sum of the alternating harmonic series (−1)n−1 /n to within 0.01 of the true value, we can sum the first 100 terms. An alternating series that does not converge. A series ! an such that |an+1 | lim = 3. n→∞ |an | ! |an | converges, then ! (−1)n |an | converges. ! de if the statements in Problems ?? are true or false.9-4h115 Give 115. If an is absolutely convergent, then it is convergent. xplanation for your answer. ! 9-4h116 116. If an is conditionally convergent, then it is absolutely If the terms sn of a sequence alternate in sign, then the convergent. ! sequence converges. 9-4h117 117. If a > 0.5 b > 0 for all n and bn diverges, then n ! !n If 0 ≤ a ≤ b for all n and a converges, then n n n a diverges. n ! bn converges. 9-4h118 118. Which test will help you determine if the series converges ! ! If 0 ≤ anSection ≤ bn for n and bn or diverges? 8.5 all Power Series an diverges, then 2010 Kiryl Tsishchanka ∞ The goal of this section is to show how series like below " are used diverges. 1 to approximate Power series: ! x sinx, cosx,.. functions such If bn ≤ an ≤ 0 for all n and as ben ,converges, thenPower Series k3 + 1 ! k=1 an converges. ! ! (a) Integral test A power series is a series of the form and x is a variable, then a series of the form DEFINITION: If an converges, then |aIfn |cconverges. 0 , c1 , c2 , . . . are constants(b) Comparison test ! If 5 (c) Ratio test ∞ an converges, then lim |an+1 |/|an! | ̸= 1. n→∞ c n xn = c 0 + c 1 x + c 2 x2 + . . . + c n x n + . . . n=0 POWER SERIES AND INTERVAL OF CONVERGENCE is called a power series in x and a series of ! then form In Section saw that the axcalled converges for −1 < xof<the 1and diverges where ?? x∞iswe a variable andgeometric c ’s are series constants the coefficients series. ! section studies then convergence of more general series constructed from powers. otherwise. This n 2 n c (x − x ) = c + c (x − x ) + c (x − x ) + . . . + c (x − x ) + . .cos . x, x n 0 0 1 0 2 0 n 0 Chapter shows how such power series are used to approximate functions such as e , sin x, More??generally: n=0 and ln x. is called a power series in x − x0 . A power series about x = a is a sum of constants times powers of (x − a): EXAMPLES: ∞ ∞Section 8.5 Power Series " ! 2 n C01++Cx1 (x Cn (x − a)n . 2 (x 1. xn = +− x2a) ++x3C+ . .− . a) + · · · + Cn (x − a) + · · · = 2010 Kiryl Tsishchanka n=0 Power Series n=0 ∞ ! xn x2 x 3 2. DEFINITION: = 1 + x + If+c , c+, c. . ,. . . . are constants and x is a variable, then a series of the form n! 2! 03! 1 2 n=0 3. ∞ ! (−1)n n=0 ∞ x2n x2 x4 !x6 n 2 n x . .= =1− + − cn+ . c0 + c1 x + c2 x + . . . + cn x + . . . (2n)! 2! 4! 6! n=0 ∞ ! (x − 1)n power series (x − 1)2 (x − 1)3 of the form 4. is called a = 1 + (x − 1) + in x and + a series + ... n! 2! 3! ∞ n=0 5. ∞ ! n (x (−1) ! n 2 n 2 − x0 ) + 3c2 (x − x0 ) + . . . + cn (x − x0 ) + . . . 0+ 1 (x + 2)n cn (x − x0 ) = c(x +c2) (x + 2) − + ... n=0 = 1 − (x + 2) + n! 2! 3! Some examples: is called a power series in x − x . n=0 THEOREM: For any power series EXAMPLES: ∞ ∞ ! 0 cn xn exactly one of the following is true: n=0 (a) The ! series nconverges only2for x3= 0. 1. x = 1 + x + x + x + ... (b) The n=0 series converges absolutely (and hence converges) for all real values of x. ∞ 3 (c) The ! seriesxnconverges absolutely hence converges) for all x in some finite open interval x2 x(and = 1 +ifxx+< −R+or x + . . At either of the values x = −R or x = R, the series (−R,2. R), andn! diverges > .R. 2! 3! n=0 may converge absolutely, converge conditionally, or diverge, depending on the particular series. ∞ ! x∞ x x ! n x 3. (−1) 1− + − )n , + . . . one of the following is true: THEOREM: For a(2n)! power=series exactly 2! cn (x4!− x06! n=0 2n 2 4 6 n=0 ∞ (a) The series converges ! (x − 1)n only for x = x0 . (x − 1)2 (x − 1)3 = 1 + (x − 1) + + + ... n! 2! (b) The series converges absolutely (and hence converges) 3! for all real values of x. n=0 4. ∞ converges absolutely n (c) The series (and hence converges) in3 some finite open interval ! (x + 2)2 for(xall+x2) n (x + 2) −x(x +R2)or+x > x0 + R. − + . .values . (x0 −5. R, x0 + (−1) R), and diverges = if x1 < At either of the x = x0 − R or 0− n! 2! 3! x = x0 + n=0 R, the series may converge absolutely, converge conditionally, or diverge, depending on the particular series. ∞ ! cn xn exactly one of the following is true: THEOREM: For any power series n=01 (a) The series converges only for x = 0. (b) The series converges absolutely (and hence converges) for all real values of x. the series alternates and is positive for n = 2, we multiply by (−1)n . For n = 2, we divide by 4, for n = 3 we divide by 9, and in general, we divide by n2 . One way to write this series is ∞ " (−1)n (x − 2)2n . n2 n=2 Practice: Chapter Nine SEQUENCES AND SERIES mple 2 ion ! ∞ We think of a as a constant. For" anyxnfixed x, the power series Cn (x − a)n is a series of Determine whether the power series converges or diverges for n numbers like those considered in Section2??. To investigate the convergence of a power series, we n=0 consider the partial sums, which in this case are the polynomials Sn (x) = C0 +C1 (x−a)+C2 (x− (a)2 x = −1 (b) x = 3 a) + · · · + Cn (x − a)n . As before, we consider the sequence8 (a) Substituting x = −1, we S have 0 (x), S1 (x), S2 (x), . . . , Sn (x), . . . . $n ∞ ∞ ∞ # " " " xn (−1)n 1 = = . − 2n 2n 2 For a fixed value of x, if this sequence of partial sums converges to a limit S, that is, if n=0 n=0 n=0 522 Chapter Nine SEQUENCES AND SERIES lim Sn (x) = S, then we say that the power series converges to S for this value of x. n→∞ This is a geometric series with ratio −1/2, so the to 1/(1 − (− 12 )) = 2/3. ! series converges We think of a as a constant. For any fixed x, the power series Cn (x − a)n is a series of (b) Substituting x = 3, we have numbers like those considered in Section ??. To investigate the convergence of a power series, we = C0 ∞ +C1# (x−a)+C consider the partial sums, which in this case are ∞ the polynomials $n 2 (x− ∞Sn (x) A power series may converge for some n values n x and "the " "not 8 3for others. xsequence 3of a)2 + · · · + Cn (x − a)n . As before, we consider = = . n n 2 S0 (x), S1 (x), S2 (x), 2. . . , Sn (x), . .2. . ! n=0 n=0 n=0 mple 1 Find an expression for the general term of the series and use it to write the series using notation: This is a geometric series withofratio greater than 1, so it diverges. For a fixed value of x, if this sequence partial sums is, if 4 6 converges to8 a limit S, that 10 (x − 2) (x − 2) (x − 2) lim Sn (x) = S, then we(x say − that2) the power − series converges + to S for this−value of x. + ···. n→∞ 4 9 16 25 A power series may converge for some values of x and not for others. merical and View Convergence ion The Graphical series is about x = 2ofand the odd terms are zero. We use (x − 2)2n and begin with n = 2. Since Example 1 Solution !n . For n = 2, we divide by 4, for the alternates is term positive for nand =use 2, itwe multiply byusing (−1) Findseries an expression for the and general of the series to write the series notation: Consider the series 2 n = 3 we divide by 9, and in general, we 4 6 8divide by10n . One way to write this series is (x − 2) (x 2− 2) (x − 2)3 (x − 2) 4 (x − 1)n (x − 1) (x − 1) (x −− 1) + − + ···. 9+ 16∞ − 25 +2n · · · + (−1)n−1 + ···. (x − 1) − 4 n 2 3 " (−1) 4(x − 2) n . 2 and begin with n = 2. Since The series is about x = 2 and the odd terms are zero. We use (x − 2)n2n To investigate the convergence of thisn=2 series, we look at the sequence of partial sums graphed in the series alternates and is positive for n = 2, we multiply by (−1)n . For n = 2, we divide by 4, for n= 8 3 we divide by 9, and in general, we divide by n2 . One way to write this series is Here we call the first term in the sequence S (x) rather than S (x) so that the last term of S (x) is C (x − a) . Practice: 0 mple 2 Example 2 Determine whether the power series ∞ (a) x =whether −1 the power series " x Determine n=0 ion Solution 1 n n n ∞ " (−1)n (x − 2)2n . n2 n=2 n 2n converges or diverges(b) for (a) x = −1 (a) Substituting x = −1, we have (a) Substituting x = −1, we have ∞ " xn converges or diverges for 2n n=0 ∞ " (b) x=3 x=3 ∞ " n n x (−1) $n ∞ ∞ ∞ # n= " " " xn (−1) 1 n = = − 2n . 2 n n 2 2 2 n=0 n=0 n=0 n=0 n=0 = $n ∞ # " 1 . − 2 n=0 This a geometric series with ratio with −1/2, ratio so the series converges 1/(1 − converges (− 12 )) = 2/3.to 1/(1 − (− 1 )) = 2/3. Thisis is a geometric series −1/2, so thetoseries 2 (b) Substituting x = 3, we have (b) Substituting x = 3, we have ∞ ∞ ∞ # $n " " " 3 xn 3n ∞ ∞ .n ∞ # $n = " = n " 3 2n 2nx n=0 " 2 3 n=0 n=0 = = . n n 2 2 2 This is a geometric series with ratio greater than 1, so it diverges. n=0 n=0 n=0 This is a geometric series with ratio greater than 1, so it diverges. Numerical and Graphical View of Convergence Consider the series (x − 1) (x − 1) (x − 1) − + · · · + (−1) (x − 1) − View+of Convergence merical and Graphical 2 3 4 2 2 − 1)n + ···. n To investigate convergence of this series, we look at the sequence of partial sums graphed in Consider thetheseries 8 Here 3 4 n−1 (x n we call the first term in the sequence 2S0 (x) rather than S13(x) so that the last term 4 of Sn (x) is Cn (x − a) . (x − 1)n (x − 1) (x − 1) (x − 1) + − + · · · + (−1)n−1 + ···. 2 3 4 n To investigate the convergence of this series, we look at the sequence of partial sums graphed in (x − 1) − term in the sequence S (x) rather than S (x) so that the last term of S (x) is C (x − a) . Convergence of power series: ! ! 8 Here we call the first Chapter Nine SEQUENCES AND SERIES 0 1 For fixed the power Cn (x a)nfixed is ax,series of series Cn (x − a)n is a series of 22 any Chapter Ninex, SEQUENCES AND SERIES We think of a series as a constant. For−any the power Section ??. To investigate theconsidered convergence of a power we the convergence ! numbers like those in Section ??. To series, investigate of a power series, we n think of a sums, as a constant. fixed x, polynomials the power series CnC(x+C − a)(x−a)+C is a series(x− of the partial which this case are the Sn (x) = = Cin0For +Cany this case areconsider theWe polynomials Sn (x) 0 1 2 1 (x−a)+C 2 (x− numbers like those considered in Section ??. To investigate the8 convergence of a power series, we 2 n 8 + · · · + C (x − a) . As before, we consider the sequence a) n e, we consider the sequence consider the partial sums, which in this case are the polynomials Sn (x) = C0 +C1 (x−a)+C2 (x− 8 S (x), S (x), S (x), . . . , S (x), .... + · · · + Cn (x − a) . As before, we1 consider 2 the sequence n ), S1 (x), S2a)(x), . . . , Sn (x), . . . . 0 2 n S0 (x), S1 (x), S2 (x), . . . , Sn (x), . . . . For a fixed value of x, if this sequence of partial sums converges to a limit S, that is, if sequence of partial to athat limit S, that is, if lim Snsums (x) =converges S, then we say the power series converges to S for this value of x. n→∞ For a fixed value of x, if this sequence of partial sums converges to a limit S, that is, if hat the power series to S for this value of x. lim Sconverges n (x) = S, then we say that the power series converges to S for this value of x. n→∞ A power series may converge for some values of x and not for others. mayothers. converge for some values of x and not for others. for some values Aofpower x andseries not for ! Find an expression for the general term of the series and use it to write the series using notation: ! 4 6 8 10 ! Example 1 Find an expression for(x the−general term it to−write notation: 2) (x −of 2)the series (x −and 2) use (x 2) the series using − + ···. term of the series and use it to write the −series using+ notation: 4 2)4 (x −9 2)6 (x − 162)8 (x −25 (x − 2)10 − + − + ···. (x − 2)6 (x − 2)8 (x − 2)104 9 16 25 +The series is − +the · · ·odd . terms olution about x = 2 and are zero. We use (x − 2)2n and begin with n = 2. Since 9 16 25 the series alternates and is positive for n = 2, we multiply by (−1)n2n . For n = 2, we divide by 4, for Solution The series is about x = 2 and the odd terms are zero. We use (x − 2) and begin with n = 2. Since n = 3 we divide by 9, and in general, we divide by n2 . One way ton write this series is the series alternates and is2npositive for n = 2, we multiply by (−1) . For n = 2, we divide by 4, for dd terms arenzero. We use (x − 2) and begin with n = 2. Since 2 ∞ divide n 2n way to write this series is " One = 3 we divide by 9, and in general, we (−1)by(xn−. 2) for n = 2, we multiply by (−1)n . For n = 2, we divide by 4, for. ∞ n n2 2n " 2 (−1) (x − 2) n=2 ral, we divide by n . One way to write this series is . n2 n=2 xample 1 ∞ " (−1)n (x − 2)2n 2 . ∞ n n n n=0 n=0 n=0 Figure ??. The graph suggests that the partial sums converge for x in 2). In Exam!the interval n(0, 1 is 9.5 POWER SERIES INTERVAL OF Weand think of we a asshow aseries constant. any fixedsox,the the power series C1/(1 a) CONVERGENCE series ofof 523 ples This ?? thatwith theFor series converges for 0 < x ≤ AND 2. to This is − called thea interval is a ??, geometric ratio −1/2, series converges (− n (x 2 )) = 2/3. numbers likeofthose convergence this series. (b) Substituting x =considered 3, we havein Section ??. To investigate the convergence of a power series, we Figure ??. The graph suggests that the partial sums converge for x in the interval (0, In Exam(x) = C0 quite +C1 (x−a)+C consider partial sums, which in this case are polynomials $nnconverge At x the =1 .4, which is inside the" interval, the series appears to rapidly: 2). 2 (x− ∞ ∞the ∞ # S n " " 2 ?? and ??, we show xconsider 3n sequence ples thewe series converges for 0 83< x . ≤ 2. This is called the interval of + · · · + Cn (x − a)n . As that before, the a) = = . . 2nS7 (1.4) =20.33653 . . . S5 (1.4) = 0.33698. 2n convergence of this series. n=0 S0 (x), S1 (x), Sn=0 (x), . . . n=0 , Sn (x), . . . . At x = 1.4, which Sis6 (1 inside interval, appears to converge quite rapidly: .4) =the 0.33630 . .2.the series S (1.4) = 0.33645 ... This is a geometric series with ratio greater than 1,8 so it diverges. Table ?? shows the results of using x = 1.9 andSx (1 =.4) 2.3=in0.33653 the power series. For x = 1.9, (1 .4) = 0.33698. . . . . . S 5 this sequence of partial sums 7 Forisa inside fixed value of x, ifof convergesthe to aseries limitconverges, S, that is, though if which the interval convergence but close to an endpoint, S6which (1 .4)that = 0.33630 . series . . interval S8 (1of .4)convergence, =to0.33645 . the . .value lim Sn (x)For = S, then we say powerthe converges S for this of x. rather slowly. x = 2.3, is the outside series diverges: the n→∞ Numerical and ofmore Convergence largerGraphical the value ofView n, the wildly the series contribution of theFor twentyTable ?? shows the results of using x = oscillates. 1.9 and x In=fact, 2.3 the in the power series. x = 1.9, fifth term isinside about 28;interval the contribution of the hundredth about −2,500,000,000. Figure though ?? which is the the of convergence but close term to anisendpoint, the series converges, Consider series A power series may converge for some values of x and not for others. shows interval convergence and is theoutside partial the sums. ratherthe slowly. Forofx(x = 2.3, which interval of convergence, the series diverges: the n (x − 1)3 (x − 1)4 − 1)2 n−1 (x − 1) + wildly the − series + · · Partial · + (−1) · . of the twenty− 1) of − n, the larger the(xvalue more oscillates. In fact, S Table 9.1 sumsthe for contribution series+ in· ·! 2 14 (x) S5 (x)3 4 n Interval for the xample 1 Find an expression term of of thethe series and use to write using ✲ thegeneral fifth✛ term isofabout 28; contribution hundredth term isx about −2,500,000,000. Figure ?? Example ??it with =the 1.9series inside intervalnotation: convergence To investigate the convergence of this series, we look at the sequence of partial sums in 9.5sums. POWER SERIES AND INTERVAL graphed OF CONVERGENCE 523 shows the interval of(x convergence 6 partial − 2)4 (xand − 2)the (xof−convergence 2)8 (x − and 2)10 x = 2.3 outside 8 Here we call the first term in the sequence n − + than S1 (x)−so that the last term + · ·of· .Sn (x) is Cn (x − a) . S0 (x) rather (x) S Table 9.1 Partial sums for series in 4 14 9 16 25 S5 (x) Interval n S (1 .9) n S (2 .3) n n ✛ graph ✲ Figure ??. The that the partial sums converge the interval of suggests Example ?? with for x =x 1.9ininside interval (0, 2). In Examconvergence 2.3 2We use 0.495 0.455 2n x terms are olution The series about x = 2that and the odd zero.of (x0− 2) with nis = called 2. Since the interval of convergence and =2. 2.3This outside ples ?? and ??, iswe show the series converges for < xand2x≤begin 1 1.9 3 n 5 0.69207 1.21589 the series alternates and is positive for n sec2tab1 = 2, we multiply by (−1) . For 5n = 2, we divide by 4, for convergence of this series. 2n Sn (1 way to.9) write 8thisn series isSn (2.3) n = 3 we divide by 9, and in general, we divide by n8 . One0.61802 0.28817 Numerical and Graphical view of convergence: At x = 1.4, which is inside the series appears to converge quite rapidly: 2.3the interval, 0.495 0.455 11 2 1.71710 n 11 2 2n0.65473 x∞ 1 fig9h11 1.9 " (−1) (x − 2) S8 (x) 3 . 0.69207 14 5 −0.70701 1.21589 14 5 0.63440 S7 (1.4) = 0.33653 . . . 8 0.61802 8 0.28817 sec2tab1 n2 .4) = 0.33698. .. n=2 x =S 25S(1 11 (x) S6for(1 .4) = Figure 9.11: Partial sums series in 0.33630 . . . S8 (x) Example ?? converge for 0 < x < 2 .4) = 11S8 (1 0.65473 14 0.33645 . . 1.71710 . 11 14 −0.70701 ∞ Table ?? shows thex = results of using x =POWER 1.9SERIES and AND x INTERVAL = 2.3OFinCONVERGENCE the power series. For x = 1.9, " xn 9.5 2 S11of(x) Notice whether that the the interval convergence, 0<x≤ is centered on x = 1. Since the523 Determine power series converges or2,diverges for n isextends inside the interval of convergence but close to an endpoint, the seriesinterval converges, though 2 one unit on sums eitherfor side, weinsay the radius of convergence of this series is 1. n=0 Figure 9.11 : Partial series Figure ??. Thexgraph suggests that theispartial sums the converge for x inoftheconvergence, interval (0, 2). In the Examrather slowly. For = 2.3, interval series diverges: the (a) x= Example ????, converge for that 0which < the x <series 2 outside ples ?? and−1 we show converges(b) for 0x<= x3 ≤ 2. This is called the interval of tervals Convergence largerofthe value ofofn, more wildly the series oscillates. In fact, the contribution of the twentyconvergence thisthe series. Notice that interval of convergence, 0 < x ≤ 2, is centered on x = 1. Since ...
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