Forgot password?
 Create new account
View 111|Reply 0

$∑c_nk^{-n}=0∀k∈ℕ$ 求证 $c_n$ 全为零

[Copy link]

3151

Threads

8498

Posts

610K

Credits

Credits
66208
QQ

Show all posts

hbghlyj Posted at 2022-11-27 08:01:04 |Read mode
H. Priestley Complex Analysis Exercise 15.7
Let $\{c_n\}$ be a sequence of complex numbers such that $\sum\abs{c_n}$ converges and $\sum_{n=0}^\infty c_nk^{-n}=0$ for $k=1,2,3,\dots$. Prove that $c_n=0$ for all $n$.
想法:
因为 $\sum\abs{c_n}$ converges, 可以定义 $D(0,1)$ 上的 全纯函数 $f(z)=\sum_{n=0}^\infty c_nz^n$.
$k^{-1}$是$f(z)$的根, 0是a limit point, 由15.7 Identity Theorem, $f(z)=0$, 所以$c_n$全为零.
这样就可以了吗

手机版Mobile version|Leisure Math Forum

2025-4-21 19:03 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list