{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw-5-sec-12.9-solns - (3 = 332 =Â<1l—lx/3:3 that is...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: @(3) = 332 =§<1l—lx/3) :3: that is, when |w1 < 3, 50R 2 3andI =(—3 1 1 1 6'f(m)=x+10=1_0(1_—W)= when ‘ 10 fine/37H. co 2 n The geometric series 2: [—(E) ] converges when — 3 n=0 R=3andI= (—3,3). @113): 2:52: 1 ‘ 47.71757) = “0:, W] = 3:: Ha)" :< >— = we": SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES U 999 )) or, equivalently, 2 23—— 111+ —-—x” The series converges when l— 3 n=03 W18 ‘<1, ; °° :1: '1 . °° n 1 10 71;) (—1—0-) or, equ1valently, EJ—l) 10““ :c”. The series converges 1| < 1, that is, when 1x| < 10, so R = 10 and I = (—10,10). 2 <1 4: II ——|<1 4: |m|2<9 4e [z|<3,so Z (—2$2)" or, equivalently, Z (—1)"2":r2"+1. The series converges when n=0 1 |—2z2|<1 : |m2|<% r» |x|<—,soR=——an 1 1—m 1+2: l—x 9- f($)= =(1+a:)( =0 The series converges when |z| < 1, so R: 1 and I: (—11). 1+m_—(1—m)+2:_1+2( A second approach: f (11:) = _ 1—1: 1—17 f \r _ ‘”‘W(‘%%) 00 00 OO 00 00 >=(1+z) :2" “11:2"+Em"+1=1+Zx"+2m”=1+22x". 11:0 11:1 11:1 11:1 1—9: )=—1+2zx =1+221”.x n=0 A third approach: f(x):1+:=(1+:L‘)(1ix>=(1+m)(1+$+$2+x3+...) :(1+x+32+$3+"’)+($+$2+$3+$4+-~)=1+2m+2m2+2x3+-~=1+2 ix". x2 m2 123 n (,0 303"” . 10.f(:1:)= 3__$3 =;3“m§=a—3 $:n:0<:— 3): —n§0 a3n+3.The seriesconvergeswhen{ma/a3! <1 (i; '13} < |:3| 4i) Irrl < |a|,soR= ]a[ and] = (~—|a| , |a|). 11. f(:c) = ——3—— = 3 xz—z—2 (x—2)(:c+1):a:—2 x+1 :1: = —1 to get B = —1. Thus _;__1_1__1_( = [-$ 6)" - W] A B + :> 3=A(x+1)+B(x—2).Letx=2togetA=land Trim) ‘ 17125 = ‘% 2:3 (3)" EH)” as" = f; [(—1)"+1 — 27:1]75" 7120 We represented f as the sum of two geometric series; the first converges for a: 6 (—2, 2) and the second converges for (—1, 1). Thus, the sum converges for x E (—1, 1) = _\ ”m 1000 U CHAPTER 12 INFINITE SEQUENCES AND SERIES S C Q I? ‘ q 3+2 z+2 A B 1. = V :\= ~ = _ : 2f(a:) 2m2_z_1 (2x+1)($_1) 2z+1+z—1 => a:+2 A(:c 1)+B(2x+1).Letz ltoget 3=3B => B=1andx= —§toget§=—§A é, A=—1.Thus, $+2 _ *1 1 _ 1 1 _ oo 71. 9° 11, 2x2—z—1_2x+1+$—1_ 1(1_(_2$))“1(1_3)— Z(*2x) —Zm n=0 71:0 00 = — 2 [(—2)" + 1] .73" 71:0 We represented f as the sum of two geometric series; the first converges for a: e (—-;— , %) and the second converges for (—1,1). Thus, the sum converges fora: E (— 1 1 = I. @)f(z)= 1 =i(‘1 d w n n ' (1+$)2 dz 1+$> — _E [n§()(—l) x J [from Exercrse3] I = i(—1)"+1mc”“1 [from Theorem 2(i)] = n—l fla)f(x)— 1 =-li[ 1 ]_ 1d[°° ‘ — (1 + $)3 2 d2: (1 + x)2 _§E n§0(*1)"(n + 1)x"] [from part (a)] It =—% 3.:(~1)"(n+1)nac"‘1 = g §(—1)"(n+2)(n+1)znwithR= 1_ u 3 n=1 n=0 I g 2 w I ; 09 f0”) = (1:303 = 2:2 - m = 9* é >=ZO(—1)"(n + 2)(n + Um" [from part on =§ i (—1)"<n + 2)<n + 1W2 n=0 To write the ower series with x" rather than en“), we will decrease each occurrence of n in the term b 2 and increase P y the initial value of the summation variable by 2. This gives us % E (—1)"(n)(n — Dz” with R = 1. n=2 14 (a) 1 —- ‘1 — §(—1)”x"[ eometric series withR— 1] so ' 1+x_1—(—$)_n=o g ~ ’ I . dg; 00 n 71 oo nxn+1 oo (_1)n—1$n I f(z)—ln(1+x)—f1+$—f[’§0(—1)dex—C+n§0(—l) n+1—nz=:1 n I [C = 0sincef(0) =ln1 = 0], withR =1 co _ 11—111 00 _ n—l n+1 co _ nn (b)f(m)=a:ln(1+z)=a:[z( 1)” at] [bypart(a)] = Z( 1) x Z( 1) x n=1 n=1 n n=2 ”‘1 co _ ”-1 2 n 00 _ n—l 2n (c)f(m)=ln(z2+1)= z( 1) ”(93) [bypart(a)] = z( I)” 9” withR: 11:1 11:1 ’ dz): 1 d9: 1 °° m" 1 °° mm” °° 2:" 15 (as)~ln(5—:L')—— 5—:v—*g 1—x/5__§/[n;0(g)Idx_ _§nz=: 5"(n+l)_C—n;n5” Puttingx = 0, we getC = ln 5. The series converges for [av/5| < 1 er lzl < 5, so R = 5. “mm We.» - - A SECTION 12.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES C] 1001 1 °° . . °° 16. We know that = 2 (2x)". Differentiating, we get ——2-——2 = i 2"nz"'1 = Z 2"“(11 + 1):c", so 1 _ 21' 11:0 (1 “‘ 2113) 11:1 11:0 m ””2 $2 2 32°°2+1( °° +2 °° 2 :1:=——=—~———=— " n+1m": 2"n+lz" or 2’“ —1 ", (1-2)2 2 (1—223)2 23:20 ) "2:30 ( ) "Z32 (” )m withR=% 1 1 1 °° :1: '1 °° 1 n :1: V @T§_W=§n§o(§) =1§02n+1$ for|§‘<1 4: |m|<2.Now 1 _d 1 _d°°1n_°"nn_1__°°n+1n (35.2)2 ' dz(2—x) ‘ dm(,§02n+lm)_,§12n+lm T1230 2M2” 'S" 3 _ z _3°°n+1n__°°n+1 n+3 °°n—2n _ _ f(m)— (m—2)2 —a: ":0 2n+2z —"§0W1 mugs 271—15” forlzl<2.Thus,R~2andI—(—2,2). 18F E 17 °° NW“ Th . rom xampe ,g(m)—arctanx—’f:_:0(—) 2n+1‘ us, 00 n (3/3)2n+1 00 n 1 2n+1 (E f(ac) arc an(a:/3) "a; 1) 2”“ 7;; 1) 32n+1(2n+1)w for|3|<1 4: |x|<3,soR 3 _ z _m 1 _x°° $2n_$°° n12n °° n1 2"“ ®)—mz+16—16(1—(—I2/16))_16,;0(_16) —16,§0_1) 16"” “E5 1) 16"+1x ‘ The series converges when |—a:2/16| < 1 4: x2 < 16 4:) |x| < 4, so R = 4. The paltial sumsare 31 = %, 9 3 5 7 32 = 51 — 1%, 33 : 32 + %, 34 = 53 — 1—2:, 35 = 34 + lit—65’ . . . . Notethat $1 correspondstothefirsttenn ofthe infinite sum, regardless‘of the value of the summation variable and the value of the exponent. co 2 n. 00 2n+1 20. f(z) = ln(:c2 +4) 2 f’(m) = 26—22% = 242(1tT—1fim) = 923'? (—24-) = Z(—1)":2n+1 , __ n=0 (,0 n m211+1 d3; C 0° 1 n z2n+2 1 4 0° 1 n z2n+2 W)- "ax-D — +§o<-> ——(—-)- n +254 (WW [f(0) = ln4, so C = ln4]. The series converges when |—m2/4| < 1 c) m2 < 4 4: |m| < 2, so R = 2. If SECTION 12.9 REPRESENTATIONS 0F FUNCTIONS AS POWER SERIES El 1003 f(-"c)= 20% 1)"+12 ”1+ 1, ,respectively. Both series converge by the Alternating Series Test. The partial sums are 81=T,82=81— 3 5 ,.... As 71 increases, 3" (x) approximates f better on the interval of convergence, which is [— é , é]. 3 t 1 00 8 oo 8 +1 t 00 t8n+2 1 :1: . = - = n : t n = . ' < 23 > 1 _ t3 t 1 _ t8 tngou ) ”go => / 1 _ t8 dt 0 + "£30 8n + 2 The senes for 1 _ t8 converges when |t8| < 1 4:} |t| < 1, so R = 1 for that series and also the series for t/ (1 — t8). By Theorem 2, the series for t /1—t8 dtalsohasR—l. oo 71 _ co — 24. ByExample6,ln(1—t) = — Z % forltl < 1,so-ln(1—tfl = — 2t "=1 — ln(1—t) _ _°°fl /Tdt_c :4 By Theorem 2, R = 1. co 2n+1 —1 _ __ ”:1: ~ = i 25. By Example 7, tan a: _ n;0( 1) 271+ 1 With R 1, so 3 5 7 3 5 7 0° 2n+1 _1 x :c x z x a: n+1 x _ = _ __ ___ :___ __...: -1 d zt‘mmxc” 3+5 7+) 3 5+7 n§1()2n+1a“ x—tan 1x n 11:2"2 __Z(_1) +12 z+tan 12; +1 z2n—1 00 "+1 $2n—1 __3_ — n ———_= — .8 Th 2,R=1. fa: x3 xdx— 0+ "2(— 1) (2n+1)(2n— 1) C+n§1( 1) 4112 _1 y eorem nz( 2)2n+1 0° 11 $4n+3 . = -1 ——— thR=1. @yExample7,/tan12(x)da:= ":20(—1)"—2—n+1 da: C+n§0( ) (2n+1)(4n+3) w1 . 27 1 =—1§—:0(——5"=x) §(—1)"x5" : 1 + x5 1 — (—x5)— n=0 1 dx= Z(—1)"x5"dx= 0+ 2(1)" ””5”“ Thus 1+I5 "=0 ”:0 57l+1 , 0.2 6 11 0~2 6 11 I=/o 1+1x5 dx: [z-—%+Efi- —---]0 =0.2—(0%) + (0121) —---.Theseriesisalternating,soifweuse the first two terms, the error is at most (0.2)11/11 z 1.9 x 10—9. So I z 0.2 — (0.2)6/6 z 0.199989 to six decimal places. 1004 [:1 CHAPTER 12 INFINITE SEQUENCES AND SERIES < 28. from Example 6, we know 1n(1 — x) = — Z: 33—, so n21 n ("34)"—=.:(—1>"+1“—4 => n=1n 1n(1 +9.4) =1n[1 _ (—x4)] = 1:. 4n z4n+l /1n(1 +564) d2) = :1(_1)n+1 (tn—1&1): C + Z}- 1)n+1 m. Thus, 0.4 5 9 13 17 0.4 5 9 13 17 0.4) (0.4) (0.4) (0.4) I: 11 4d=£__"’_ L_$_ :( _ _ [0 n( +06) 98 I + 0 5 18 + 39 68 5 18 39 68 +"" The series is alternating, so if we use the first three terms, the error is at most (0.4)”/ 68 z 2.5 x 10—9. So I~ (0. 4)5/ 5 (0.4)9/ 18 + (0.9) 13/ 39 z 0.002034 to six decimal places. 29. We substitute 39: for z in Example 7, and find that n (2m):n—:l 32n+1 x2n+2 00 3271+). d3: Z(—1)"—2nj_—1—dx=0+2(“1)" n=0 x2n+3 /zarctan(3:1:)dm = $20 (— 1) 01 35 57 79 0.1 3:1: 3x 3x 32 So /0 warctan(3x)dm=[T—.-§—3.5+5'7 7.9+...] 103 5 X 105 +35 X 107 63 X 109 2187 m Ru“ 3.5 X 10—8. SO The series is alternating, so if we use three terms, the error is at most /0.1 marctan(3a:) da‘v N i — ———9:—+ L43— ~ 0 000983 to six decimal" laces 0 ‘ “”103 5x105 +35x107~' p ' 0.3 2 0.3 00 n 4n+3 0-3 00 n 4n+3 III _ 2 )7; $4" *1) {l} _ (—1) 3 3°” /0 1 + 2:4 dz _/0 E g0(1)d$=,;oI4n + 3 I0 _ ":0 (4n + 3)104"+3 23x103—7x107+11x1011 _"' . . . . . 311 . . The series 15 alternating, so If we use only two terms, the error 15 at most W m 0.000 000 16. So, to Six decrmal 1 0'3 “2 dm~ —33 —37~ 0008969. paces’/o 1+x4 ~3x103_7><107~' TI. 00 31. Using the result of Example 6, 1n(1 — z) = — Z 271-, with a: = —0.1, we have n=1 0.01 0.001 0.0001 0.00001 ln1.1=ln[1—(— 0.1)]:0.1——2—+T— 4 + 5 —-~~.Theseriesisalternating,soifweuseonly . . . 1 the first four terms, the error is at most 0 02001 = 0.000002. So In 1.1~ ~ 0.1 — g + 9—252 — O 020 z 0.09531. (—1)"z2" , _ °° (—1)"2n:c2"’1 . 32. f (x) :0 —(2—n—)—!—-— => f._(z) — g1 —-(—2—n_)T— [the first term disappears], so N 0° (_1)n(2n)(2n __ 1)z2n—-2 (_1)nx2(n—1) co (_1)n+1x2n ‘ _ a: = —— — ——- = —— substitutin n + 1 for n f ( ) .‘é. (2n)! 3:. [2(n — 1)]! E. On)! I g I 1%?" H =’>:‘L(2)nT= —f(a:) => f(m)+f(w)= n—O ..=o W37) 1006 I] CHAPTER 12 INFINITE SEQUENCESAND SERIES <6 e i ' Z. 97 (as. ‘9)f(w)=z :01—7 :1 f(w)= i”; 2:411:11" iii-=flx) (b) By Theorem 10.4.2, the only solution to the differential equation df(:1;) /d$ = f (x) is f (9:) = K ex, but f (O) = 1, soK: 1andf(:z:) :61. Or: We could solve the equation 11 f (z) / dz = f (m) as a separable differential equation. sin nx °° sin ms . d sin nan cos my 36. l 2 l S— 2 , so 2 2 converges by the Companson Test. — 2 = , so when a: = 2km n n ":1 n da: 71 n W 00 m y1 [Is an integer], Z f,’,(z) = E w = Z i, which diverges [harmonic series]. [1' (at) = —— sin me, so 11:1 11:1 n=1 2 f" ’(m): ’— 2 sin nx, which converges only if sin nm 2 0, or z 2 kn [k an integer]. 71:1 $71. an+1 $n+1 n2 . TL 2 37. If an = ,then by the Ratio Test, lim — lim ——- - — = lac] 11m = |z| < 1 for n2 "—roo an n—>oo (n + 1)2 a)" "-900 TL + 1 co 11 co 1 convergence, so R = 1. When :5 = :I:1, Z -n-2-— —Z 715 which Is a convergent p—series (p— —— 2 > 1), so the interval of 71:1 11:1 convergence for f is [—1, 1]. By Theorem 2, the radii of convergence of f ' and f ” are both 1, so we need only check the . ’ oo '1 oo 11—1 00 n . endpoints. f (x) = Z a; => f’(:c) = Z nacnz = Z , and this series diverges for a: = 1 (harmonic series) 11:1 11.: - 00 11—1 and converges for a: = —-1 (Alternating Series Test), so the interval of convergence is [—1, 1). f”(m) = Z n diverges 11:1 at both 1 and— 1 (Test for Divergence) since lim fl: 1 91E 0, so its interval of convergence is (—1, 1). °° n_1 °° d n d °° d 1 1 . = —— = —- n = — —— = 1 1 1- ” (a)1;1nx "220 dav:E dz ”2:303: dm 1 —:11_(1——_-1_a:_)2(_)=(1—_-_:13_)2 l$l < (b) (i) zlnxn = (I? Elnxn-l = 1: [(1 _ 102] [frompart (3)] = (1 fm)2 for IIL'] < 1. (101111111: -in(i). 7:: 571:7:1719)" = (If—6.27 =2. (c) (1) “am—1):; =12 )3 n(n—1)z"~ = 1:234; [gnaw-1] ”2525}; —:z: (1336):, = (12f:)3 forlxl <1 (ii)Putz=—'m(i) :"222" =§2M HG)" (1295):)3 = 2 2 °° TL ...
View Full Document

{[ snackBarMessage ]}