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Unformatted text preview: MATH 138 Calculus 2 Fall 2010 Solutions # 10 [a _ .1 1 _.1 ,r. T 3 .r. ,3
l" .1 .r .1 .r . _ . _ ’_ .r .r J
i J HinJ': 2 [ — .l' . r] _! "“.bt'hlllJ.l~ _§?+Tﬁ?_ﬁ+'naﬂd : — —— + — — _
n H [2” l I)! I” ‘l, f 1
., _ .1
h'lllJ" _ ” 1 if" ii" ‘ . 5mm — J' _ . l i F J’ '_ _ l
T '— —i + 7'5 E l Thu“ l'ﬂh‘Ju— _ 9'3}. (‘ [1 * 120.1111!) * " (i' 1U?) _ 1_m” . 1ﬂ(1—.1;::“+ﬁm‘a‘r1‘°+'“) mus—e: '19}:on—[1+x+§lm3+§1z=+ﬁr4+§x5+ézﬁ+nq a; m.
r 2 _ If! is an odd ﬁmclion. then J'(—:r) = —f{r) =: ): {—!)"c,.m" = Z —c,,:r“. The coemcicms ofany pnwer series 1:: "“0
are uniquely delcmained (by Theorem 1 1.10.5). so (4)" c" = —c,..
”1'1 i5 even.thcn (—3)" = 1. 50 r?” = —c“ = 2r." = 0 => on = (1. Thus. all even coefficients are 0. that is‘ afl=c2=c4:n.=[)l at; w 2 N on '2“ no 1 n will I C‘cu an H I '
% ELI :’ I‘m)“: Elfin? :n§u%=.§uﬁm2 515.1,“le gum“ "‘ "~ WM as
_ [2") 011 n} (______2n 11)!
a I‘M byﬂm): Z fugfmar" Comparing cnefﬁciems for k— .— 21: we have f——(2n§!.— }— Frlz'l. => 2 (ﬂ)=1——t—! .
k—'[I ' 4; : Since f{"}{[l) = {n + ”L“ the Maclaurin series J10 07’” W is ;_ f_‘"lﬂw)" w (7: +1 ! ,, m .
Zu— = Z “I } .‘1‘ = 2 (11+ l}.T". Applying the Ratio Test with a" = (n + 1].r” gives us
!1'— ”=0  11:0
‘ finp 1 _ ‘ [11 + 2);::" H . n + '2 _.
"111.120 T‘ — ‘32:; W = ::: "151; n. + l = .r. » ] = lnrl. l‘or convergence. we must have :r. < L so the radius ofcmwergence h’ = 1. at Cl: {hﬂl ) (n+l lln'll) :7 {21‘1”} §il\Il—;;V_L'i :1— n‘a‘A 31—390 “1'2 W)
on n x no "—1 3 no
:(11) E: :_ _ z_ .e—~1= I [a __1 =
8 _ n} =* 3 1_:E:fﬂ => E :5: nl =5 I dz O‘FEE;
n=0 “=1 “=1 "—1 2n—1 z= Efl)"— => m3l=§(_l}n :2 mag1” "3
2n)! =1 =¢> z _ "21(4) (2")! ( ml (Twirl 1d 2 : —(1'.— a.) + a“, _ £03 ...
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