一系列的无穷乘积

作者: 洛玖言 | 来源:发表于2019-10-04 09:06 被阅读0次

第一类

利用 $\sin x$ 的无穷乘积展开式

$\displaystyle\sin x=x\left(1-\dfrac{x^2}{\pi^2}\right)\left(1-\dfrac{x^2}{2^2\pi^2}\right)\cdots=x\prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2\pi^2}\right)$

Q1

$\lim_{n\to\infty}\prod_{k=1}^n\left(1+\dfrac{1}{k^2}\right)$

Sol:
$\displaystyle\dfrac{\sin x}{x}=\prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2\pi^2}\right)$
当 $x=i\pi$ 时，
$\displaystyle\dfrac{\sin i\pi}{i\pi}=\prod_{n=1}^{\infty}\left(1+\dfrac{1}{n^2}\right)$

又 $\sin z = \dfrac{e^{iz}-e^{-iz}}{2i}$

$\therefore \displaystyle\prod_{n=1}^{\infty}\left(1+\dfrac{1}{n^2}\right)=\dfrac{e^{\pi}-e^{-\pi}}{2\pi}$

下面几题同理

Q2

$\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(2n)^2}\right)$

Sol:
$\displaystyle\sin\dfrac{i\pi}{2}=\dfrac{i\pi}{2}\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(2n)^2}\right)=\dfrac{e^{-\frac\pi2}-e^{\frac\pi2}}{2i}$

$\therefore\displaystyle\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(2n)^2}\right)=\dfrac{e^{\frac\pi2}-e^{-\frac\pi2}}{\pi}$

Q3

$\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^2}\right)$

Sol:
$\displaystyle\sin\dfrac{\pi}{2}=\dfrac{\pi}{2}\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^2}\right)=1$

$\therefore\displaystyle\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^2}\right)=\dfrac{2}{\pi}$

Q4

$\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^4}\right)$

Sol:
$\displaystyle\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^4}\right)=\prod_{n=1}^{\infty}\left(1+\dfrac{1}{(2n)^2}\right)\cdot\prod_{n=1}^{\infty}\left(1-\dfrac{1}{(2n)^2}\right)=\dfrac{2(e^{\frac{\pi}{2}}-e^{-\frac\pi2})}{\pi^2}$

Q5

$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^2}\right)$

Sol:
$\begin{aligned} &\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^2}\right)=\lim_{x\to\pi}\dfrac{\displaystyle x\prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2\pi^2}\right)}{x\left(x-\dfrac{x^2}{\pi^2}\right)}\\ =&\lim_{x\to\pi}\dfrac{\sin x}{\pi-x}\cdot\dfrac{\pi^2}{(\pi+x)x}\\ =&\dfrac12 \end{aligned}$

Sol2:
$\begin{aligned} &\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^2}\right)\\ =&\lim_{n\to\infty}\left(1-\dfrac12\right)\left(1+\dfrac12\right)\cdots\left(1-\dfrac1n\right)\left(1+\dfrac1n\right)\\ =&\lim_{n\to\infty}\dfrac12\cdot\dfrac32\cdot\dfrac23\cdot\dfrac43\cdots\dfrac{n-1}{n}\cdot\dfrac{n+1}{n}\\ =&\dfrac12 \end{aligned}$

Q6

$\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^4}\right)$

Sol:
$\begin{aligned} &\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^4}\right)\\ =&\dfrac12\cdot\prod_{n=1}^{\infty}\left(1+\dfrac{1}{n^2}\right)\cdot\prod_{n=2}^{\infty}\left(1-\dfrac{1}{n^2}\right)\\ =&\dfrac{e^{\pi}-e^{-\pi}}{8\pi} \end{aligned}$

如果我之后懂了怎么证明
$\sin x=x\prod_{n=1}^{\infty}\left(1-\dfrac{x^2}{n^2\pi^2}\right)$
再补上.

一系列的无穷乘积

第一类

Q1

Q2

Q3

Q4

Q5

Q6

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