Forgot password?
 Create new account
Author: kuing

补录最近人教群几道题(包括一些未解决的)

[Copy link]

3147

Threads

8497

Posts

610K

Credits

Credits
66183
QQ

Show all posts

hbghlyj Posted at 2023-5-6 18:12:34
Last edited by hbghlyj at 2024-4-15 13:37:00wolframalpha.com/input?i=Table%5BSum%5BSin%5B … %5D%2C%7Bn%2C6%7D%5D
类似:math.stackexchange.com/questions/2670750/sum- … and-tan?noredirect=1
kuing 发表于 2021-8-19 09:33
粤B学生86鱼 2021/8/2 10:18:15
谁来看看这个恒等式,如何证明
\begin{equation}\label1\sum_{k=1}^{2 n-1} \frac{\sin \left(\frac{k^{2} \pi}{2 n}\right)}{\sin \left(\frac{k \pi}{2 n}\right)}=n\end{equation}
根据下面的恒等式
hbghlyj 发表于 2022-11-13 15:47
Variant of identity: ${\displaystyle \sum _{k=0}^{N-1}e^{ikx}=e^{i(N-1)x/2}{\frac {\sin(N\,x/2)}{\sin(x/2)}}}$

可以将\eqref{1}中的$\frac{\sin \left(\frac{k^{2} \pi}{2 n}\right)}{\sin \left(\frac{k \pi}{2 n}\right)}$写成级数, 得到二重级数
$$n=\sum _{j=0}^{2 n-1} \sum _{k=0}^{2 n-1-j} \exp \left(\frac{i \pi  }{2 n}(j+k) (j-k+1)\right)$$
证明见此帖

手机版Mobile version|Leisure Math Forum

2025-4-21 01:24 GMT+8

Powered by Discuz!

× Quick Reply To Top Return to the list