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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES D 79<br />

3<br />

11. f (x) =<br />

2 2 x - x-<br />

3 A B<br />

..,....------,-,-,---"7 = --+ -- => 3 = A(x + 1) + B(x - 2). Let x = 2 to get A = 1 and<br />

(x- 2)(x + 1) x - 2 x + 1 ·<br />

x = - 1 to get B = - 1. Thus<br />

3 __ 1<br />

x2- x - 2 - X- 2 X+ 1 - 2 1- (x/2) 1 - (- x) 2 n=O 2 n=O<br />

_ _ _ 1_=~( 1 )- 1 =-~ E (=)" _ E(-x)n<br />

= f: [-~ (1:)" -1(-1)"']xn = E [(- 1t+l- n~l]x"<br />

n=O 2 2 n=O 2<br />

We represent<strong>ed</strong> f as the sum of two geometric series; the first converges for x E (-2, 2) and the second converges for ( -1, ~) .<br />

Thus, the sum converges for x E ( - 1, 1) =I.<br />

13. (a) f(x ) = 1 = ·-d (-- -1 ) = , --d [ L..... ~ (- 1)n x "]<br />

(1+x) 2 dx 1+x dx n=O<br />

[from Exercise 3]<br />

~ ~<br />

= L (-1)"+ 1 nx"- 1 [from Theorem 2(i)] = L (-1)"(n + l)x" with R = ~ .<br />

n=l<br />

n=O<br />

In the last step, note that we decreas<strong>ed</strong> the initial value of the summation variable n by 1, and then increas<strong>ed</strong> each<br />

occurrence ofn in the term by 1 [also note that (-1)"+ 2 = (- 1)" ].<br />

(b)f(x)=<br />

1<br />

1<br />

=-1:j_[ ] =-1:j_[E(-1)"(n+ 1)x"] [frompart(a)]<br />

(1+x) 3 2dx (1+x) 2 2dx n=O<br />

~ ~<br />

= -~ L (-1)"(n + 1)nxn- 1 = ~ L (- 1)n(n + 2)(n + 1)x" with R = 1.<br />

n=l<br />

n=O<br />

x 2 1 1 ~<br />

(c)f(x) = ( 1<br />

+x) 3 =x 2 · (l+x) 3 = x 2 · 2nJ;<br />

0<br />

{- 1)''(n+2)(n+1)x" [frompart(b)]<br />

= 1: E (-1)"{n+2)(n+ l)x"+~<br />

2 n=O<br />

To write the power series with x " rather ~an ·x"+ 2 , we will decrease each occurrence of n in the term by 2 and increase<br />

1 . CX><br />

the initial value of the summation variable by 2. This gives us 2 .. (-1)"(n)(n- 1)x,. with R = 1.<br />

~2<br />

15. f(x) = ln(5 - x) = -~~ = _1: ~~ = -1:1 [f: (::\"] dx = C- ~ E xn+l = ·C..:... f: x"<br />

5-x 5 1 - x/5 5 n=O 5} 5n=o5"(n+ 1) n=tn5n<br />

Putting x = 0, we get C = ln 5. The series converges for Jx/51 < 1 Jx J < 5, so R = 5.<br />

1 1<br />

00<br />

17. We know that - 1 4<br />

= ( 4<br />

) = L ( -4x)". Differentiating, we get<br />

+ X 1- - X n=O<br />

f(x) = {1+x4x)2 = · -4x.<br />

for J-4xJ < 1 Jxl < ;}, so R = ;} .<br />

-4 2 =-X f;{- 4)n+1 (n+1)xn = f;(-1)n4n(n+1)xn+l<br />

(1 + 4x) 4 n=O n=O<br />

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