一系列的无穷乘积

2019-10-04  本文已影响0人  洛玖言

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.

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