三道趣题

wayne

4# creasson
能帮我下载这篇文章吗

wayne

我构造了一个数列。
a[1]=0; (a[n+1] - a[n] )/(1+a[n+1]a[n]) =1/(n^2+1)
不过，这个数列也麻烦。

wayne

西西的第一题结果不太对。
在creasson的帮助下，算出来结果是：
$ArcTan(\frac{\tan (\frac{\sqrt{2-\sqrt{2}} \pi }{2^{3//4}})-(\sqrt{2}-1) \tanh (\frac{\sqrt{2+\sqrt{2}} \pi }{2^{3//4}})}{\tanh (\frac{\sqrt{2+\sqrt{2}} \pi }{2^{3//4}})+(\sqrt{2}-1) \tan (\frac{\sqrt{2-\sqrt{2}} \pi }{2^{3//4}})})$

wayne

一般的：

yinhow

葡萄糖

wayne 发表于 2012-5-3 20:38
西西的第一题结果不太对。
在creasson的帮助下，算出来结果是：
$ArcTan(\frac{\tan (\frac{\sqrt{2-\sqrt{2}} \pi }{2^{3//4}})-(\sqrt{2}-1) \tanh (\frac{\sqrt{2+\sqrt{2}} \pi }{2^{3//4}})}{\tanh (\frac{\sqrt{2+\sqrt{2}} \pi }{2^{3//4}})+(\sqrt{2}-1) \tan (\frac{\sqrt{2-\sqrt{2}} \pi }{2^{3//4}})})$
...

\begin{align*}
&\arctan\left(\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)-\left(\sqrt{2}-1\right)\tanh\left(\frac{\sqrt{2+\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}
{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}{\sqrt[3]{8}}\pi\right)+\left(\sqrt{2}-1\right)\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}\right)\\
&\arctan\left(\frac{\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}
{\sqrt[3]{8}}\pi\right)}-\left(\sqrt{2}-1\right)}
{1+\left(\sqrt{2}-1\right)\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}
{\sqrt[3]{8}}\pi\right)}}\right)\\
&\arctan\left(\frac{\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}
{\sqrt[3]{8}}\pi\right)}-\tan\left(\frac{\pi}{8}\right)}
{1+\tan\left(\frac{\pi}{8}\right)\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}
{\sqrt[3]{8}}\pi\right)}}\right)\\
&\arctan\left(\frac{\tan\left(\frac{\sqrt{2-\sqrt{2}}}{\sqrt[3]{8}}\pi\right)}{\tanh\left(\frac{\sqrt{2+\sqrt{2}}}
{\sqrt[3]{8}}\pi\right)}\right)-\frac{\pi}{8}\\
&\arctan\left(\frac{\tan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\pi\right)}{\tanh\left(\sqrt{\frac{\sqrt{2}+1}{2}}\pi\right)}\right)-\frac{\pi}{8}\\
\end{align*}

\[\sum\limits_{n=1}^{\infty}\arctan\frac{1}{n^2+1}=\arctan\left(\frac{\tan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\pi\right)}{\tanh\left(\sqrt{\frac{\sqrt{2}+1}{2}}\pi\right)}\right)-\frac{\pi}{8}
\]
下面这个反正切函数无穷级数和刊载于AMM E3375
\[\sum\limits_{n=1}^{\infty}\arctan\frac{1}{n^2}=\arctan\left(\frac{\tan\left(\frac{\pi}{\sqrt2}\right)}{\tanh\left(\frac{\pi}{\sqrt2}\right)}\right)+\frac{3\pi}{4}
\]
【注】：
\[\arctan\left(\frac{x-\tan\left(\frac{\pi}{4}\right)}{1+\tan\left(\frac{\pi}{4}\right)x}\right)=\arctan\left(\frac{x-1}{1+x}\right)=\arctan\left(x\right)-\frac{\pi}{4}-\frac{\pi}{2}\operatorname{sign}\left(x+1\right)+\frac{\pi}{2}
\]
\[\arctan\left(\frac{\frac{\tan\left(\frac{\pi}{\sqrt{2}}\right)}{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}-\tan\left(\frac{\pi}{4}\right)}{1+\tan\left(\frac{\pi}{4}\right)\frac{\tan\left(\frac{\pi}{\sqrt{2}}\right)}{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}}\right)
=\arctan\left(\frac{\tan\left(\frac{\pi}{\sqrt{2}}\right)-\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)+\tanh\left(\frac{\pi}{\sqrt{2}}\right)}\right)
=\arctan\left(\frac{\tan\left(\frac{\pi}{\sqrt{2}}\right)}{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}\right)+\frac{3\pi}{4}
\]