Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.10 TAYLOR AND MACLAURIN SERIES ¤ 491
19.
n f (n) (x) f (n) (9)
0 x −1/2 1 3
1 − 1 2 x−3/2 − 1 · 1
2 3 3
3
2
4 x−5/2 − 1 · − 3
2 2 · 1
3 5
3 − 15 8 x−7/2 − 1 · − 3
2 2 · −
5
2 · 1
3 7
.
.
.
lim
n→∞ an+1
a n
1
√ = 1 x 3 − 1
3 (x − 9) 2
(x − 9) +
2 · 33 2 2 · 3 5 2!
− 3 · 5 (x − 9) 3
+ ···
2 3 · 3 7 3!
= 1 3 + ∞
n=1
1 · 3 · 5 · ··· ·(2n − 1)[2(n +1)− 1] |x − 9|
n+1
= lim
·
n→∞
2 n+1 · 3 [2(n+1)+1] · (n +1)!
(2n +1)|x − 9|
= lim
= 1 |x − 9| < 1
n→∞ 2 · 3 2 (n +1) 9
(−1) n 1 · 3 · 5 · ··· ·(2n − 1)
2 n · 3 2n+1 · n!
(x − 9) n .
2 n · 3 2n+1 · n!
1 · 3 · 5 ·····(2n − 1) |x − 9| n
for convergence, so |x − 9| < 9 and R =9.
21. If f(x) =sinπx,thenf (n+1) (x) =±π n+1 sin πx or ±π n+1
cos πx. In each case, f (n+1) (x) ≤ π n+1 ,sobyFormula9
with a =0and M = π n+1 , |R n (x)| ≤
πn+1
(n +1)! |x|n+1 = |πx|n+1
(n +1)! . Thus, |R n(x)| → 0 as n →∞by Equation 10.
So lim Rn(x) =0and, by Theorem 8, the series in Exercise 7 represents sin πx for all x.
n→∞
23. If f(x) =sinhx, thenforalln, f (n+1) (x) =coshx or sinh x. Since|sinh x| < |cosh x| =coshx for all x,wehave
f (n+1) (x) ≤ cosh x for all n. Ifd is any positive number and |x| ≤ d, then
Formula 9 with a =0and M =coshd,wehave|R n(x)| ≤
f (n+1) (x) ≤ cosh x ≤ cosh d,soby
cosh d
(n +1)! |x|n+1 . It follows that |R n(x)| → 0 as n →∞for
|x| ≤ d (by Equation 10). But d was an arbitrary positive number. So by Theorem 8, the series represents sinh x for all x.
25. The general binomial series in (17) is
(1 + x) k
= ∞ k
x n =1+kx +
n
n=0
(1 + x) 1/2
= ∞ 1
2
n
n=0
k(k − 1)
x 2 +
2!
x n =1+ 1
2
k(k − 1)(k − 2)
x 3 + ···.
3!
1
x +
2 −
1
2
x 2 +
2!
1
2 −
1
2
3!
=1+ x 2 − x2
2 2 · 2! + 1 · 3 · x3
− 1 · 3 · 5 · x4
+ ···
2 3 · 3! 2 4 · 4!
=1+ x 2 + ∞
n=2
(−1) n−1 1 · 3 · 5 ·····(2n − 3)x n
2 n · n!
−
3
2
x 3 + ···
for |x| < 1, soR =1.