Answers & Hints for Selected Exercises 693
- Assume t -1-2nn is fixed. By Lemma 8.5.4(b), £::cos kt= sin(n+ 21 ~~):;;sint/^2
k=l
:::; 12 si ; t/21 ·
- (a) When t = nn, convergence is absolute since every term is 0. Suppose
n n
t -1-mr. Then I sin ktl 2 sin^2 kt = ~(1 - cos 2kt), so .L t I sin ktl 2 ~ .L t -
n n n k=l k=l
~ .L t cos k(2t). Recall that .L t = +oo and ,L t cos k(2t) converges by
k=l k=l k=l
n
Example 8 .5.7(b). .". ,L ti sinktl diverges.
k=l
- Suppose {bk} is monotone increasing. Then bk ~ b =sup{ bk}, so {b - bk}
is a monotone decreasing sequence converging to 0. Thus, by Dirichlet's test,
,Lak(b-bk) converges. Then ,Lakbk = ,Lakb-,Lak(b-bk) converges. The
proof is similar if {bk} is monotone decreasing.
- (a) If ,L lak I converges, then by Ex. 4, ,L( ak)^2 converges. (b) {l/ k}
- {l/k} is square summable. Apply Thm. 8.5.16.
- Both { Jlik} and {l/k} are square summable. Apply Thm. 8.5.16.
EXERCISE SET 8.6
- Suppose .L ak(x1 -c)k does not converge absolutely, and lx2-cl > lx 1 -cl-If
,L ak(x 2 -c)k converges, then by Thm. 8.6.2, ,L ak(x 1 -c)k converges absolutely,
a contradiction.
- By Thm. 8 .2.16, ,L ak(x 1 - c)k converges absolutely when
lim tflakllx - elk< 1; that is, when Ix-cl lim ~ < 1, and diverges when
k-+oo k-+oo
Jx - cJ lim ~ > 1. Let R = lim ~· Let p = 1/ R (0 if R = oo, and oo
k-+oo k-+oo
if R = 0). Then the series converges absolutely when Jx - cl < p and diverges
when Ix - cl > p. Therefore, the radius of convergence is p.
- (a) (-2, 0), 1
(i) (2, 4), 1
(c) JR., oo
(k) [-3, -1], 1
(e) (1, 5), 2 (g) (-~, -j], ~
(m) [-1, 1), 1
- For (a) and (b), apply Thm. 8.1.12. For (c), observe that by Thm. 8.4.3,
f(x)g(x ) = f (..!£ ai(x - c)ibk-i(x - c)k-i) = f (..!£ aibk-i) (x-c)k.
k=O i=O k=O i=O
00
- Suppose f(x) = ,L ak(x - c)k on the interval Jx - cJ < p and c -1-0. Then
k=O 00
f(x) = g(u) where g(u) = .L akuk and u = x - c. By Part (a), d~g(u) =
k=O
00 00
,L kakuk-l. By the chain rule, f'(x) = d~g(u)u'(x) = ,L kak(x - c)k-^1.
k=O k=O
