Euler公式与广义连分数

作者: 茶水不太凉丶 | 来源:发表于2020-08-05 19:31 被阅读0次

Euler公式与广义连分数
连分数与历法
2018-04-26 开胃学习数学系列 - 欧拉法
欧拉公式
【数学建模算法】(11)图的应用：Euler图和Hamilton
连分数
学习笔记《Euler's formula》
从连分数的几何意义谈起
欧拉角与四元数
2018-01-06

闲来无事，搬运下我以前的文章。

Euler公式

所谓(无穷)连分数，就是指形如 $\large x_0+\dfrac{1}{x_1+\dfrac{1}{x_2+\dfrac{1}{x_3+\ddots}}}\\$ 的式子，其中 $x\in \mathbb{Z}$

我们来介绍下Euler连分数公式

$a_0+a_0a_1+a_0a_1a_2+\cdots +a_0a_1\cdots a_n=\dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2+\dfrac{\ddots}{\cdots \dfrac{a_{n-1}}{1+a_{n-1}-\dfrac{a_n}{1+a_n}}}}}}$

令 $s_i=a_i+a_ia_{i+1}+a_ia_{i+1}a_{i+2}+\cdots a_ia_{i+1}\cdots a_n$

那么

$\large \begin{align}s_{i+1}&=a_ia_{i+1}+a_ia_{i+1}a_{i+2}+\cdots +a_{i+1}\cdots a_n\notag \\s_0&=a_0+a_0a_1+a_0a_1a_2+\cdots +a_0a_1\cdots a_n\end{align}\\$

不难得到 $\large s_i=a_i\left( 1+a_{i+1}+a_{i+1}a_{i+2}+\cdots \right) =a_i\left( 1+s_{i+1} \right) \tag{2} \\$

由(1)知

$\large \begin{align}&\quad a_0+a_0a_1+a_0a_1a_2+\cdots +a_0a_1\cdots a_n\notag \\&=s_0\notag \\&=a_0\left( 1+s_1 \right) \notag \\&=\dfrac{a_0}{\dfrac{1}{1+s_1}}\notag \\&=\dfrac{a_0}{1-\dfrac{s_1}{1+s_1}}\end{align}\\$

由（2）又知

$\large \begin{align*}\dfrac{s_i}{1+s_{\text{i}}}&=\dfrac{a_i\left( 1+s_{i+1} \right)}{1+s_i} \\&=\dfrac{a_i}{\dfrac{1+s_i}{1+s_{i+1}}}\\&=\dfrac{a_i}{\dfrac{1+a_i\left( 1+s_{i+1} \right)}{1+s_{i+1}}}\\&=\dfrac{a_i}{\dfrac{\left( 1+s_{i+1} \right) +a_i\left( 1+s_{i+1} \right)}{1+s_{i+1}}-\dfrac{s_{i+1}}{1+s_{i+1}}} \\&=\boxed{\dfrac{a_i}{1+a_i-\dfrac{s_{i+1}}{1+s_{i+1}}}}\end{align*}\\$

用上式带入（3）继续化简

$\large \begin{align*}&\quad \dfrac{a_0}{1-\dfrac{s_1}{1+s_1}}\\&=\dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{s_2}{1+s_2}}}\\&=\dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2-\dfrac{s_3}{1+s_3}}}}\\&=\dfrac{a_0}{1-\dfrac{a_1}{1+a_1-\dfrac{a_2}{1+a_2+\dfrac{\ddots}{\cdots \dfrac{a_{n-1}}{1+a_{n-1}-\dfrac{a_n}{1+a_n}}}}}}\end{align*}\\$

应用

已知 $\large e^x=\displaystyle \sum_{n=0}^{\infty}{\dfrac{x^n}{n!}}=1+x+1\cdot x\cdot \dfrac{x}{2}+1\cdot x\cdot \dfrac{x}{2}\cdot \dfrac{x}{3}+\cdots \\$

代入上述公式可得

$\large \begin{align*}e^x&=\dfrac{1}{1-\dfrac{x}{1+x-\dfrac{\dfrac{1}{2}x}{1+\dfrac{1}{2}x-\dfrac{\dfrac{1}{3}x}{1+\dfrac{1}{3}x-\ddots}}}}\\&=\dfrac{1}{1-\dfrac{x}{1+x-\dfrac{x}{2+x-\dfrac{\dfrac{2}{3}x}{1+\dfrac{1}{3}x-\ddots}}}}\\&=\dfrac{1}{1-\dfrac{x}{1+x-\dfrac{x}{2+x-\dfrac{2x}{3+x-\dfrac{3x}{4+x-\ddots}}}}}\end{align*}\\$

当 $x=1$ 时可以得到一个关于 $e$ 的广义连分数

$\large \boxed{e=\dfrac{1}{1-\dfrac{1}{2-\dfrac{1}{3-\dfrac{2}{4-\dfrac{3}{5-\dfrac{4}{6-\ddots}}}}}}}\\$

同理，

$\large \begin{align*}&\quad \sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\cdots \\&=x+\left( x \right) \left( \dfrac{-x^2}{2\cdot 3} \right) +\left( x \right) \left( \dfrac{-x^2}{2\cdot 3} \right) \left( \dfrac{-x^2}{4\cdot 5} \right) +\left( x \right) \left( \dfrac{-x^2}{2\cdot 3} \right) \left( \dfrac{-x^2}{4\cdot 5} \right) \left( \dfrac{-x^2}{6\cdot 7} \right) +\cdots \\&=\dfrac{x}{1-\dfrac{\dfrac{-x^2}{2\cdot 3}}{1+\dfrac{-x^2}{2\cdot 3}-\dfrac{\dfrac{-x^2}{4\cdot 5}}{1+\dfrac{-x^2}{4\cdot 5}-\ddots}}}\\&=\dfrac{x}{1+\dfrac{x^2}{2\cdot 3-x^2+\dfrac{2\cdot 3x^2}{4\cdot 5-x^2+\ddots}}}\end{align*}\\$

当 $x=1$ ,得到 $\large \boxed{\sin 1=\dfrac{1}{1+\dfrac{1}{2\cdot 3-1+\dfrac{2\cdot 3}{4\cdot 5-1+\dfrac{4\cdot 5}{6\cdot 7-1+\dfrac{5\cdot 7}{8\cdot 9-1+\ddots}}}}}}$

$\large\begin{align*}&\quad \cos x=\sum_{n=0}^{\infty}{\left( -1 \right) ^n\dfrac{x^{2n}}{\left( 2n \right) !}}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\cdots \\&=1+1\cdot \left( -\dfrac{x^2}{1\cdot 2} \right) +1\cdot \left( -\dfrac{x^2}{1\cdot 2} \right) \left( -\dfrac{x^2}{3\cdot 4} \right) +1\cdot \left( -\dfrac{x^2}{1\cdot 2} \right) \left( -\dfrac{x^2}{3\cdot 4} \right) \left( -\dfrac{x^2}{5\cdot 6} \right) +\cdots \\&=\dfrac{1}{1-\dfrac{-\dfrac{x^2}{1\cdot 2}}{1-\dfrac{x^2}{1\cdot 2}-\dfrac{-\dfrac{x^2}{3\cdot 4}}{1-\dfrac{x^2}{3\cdot 4}-\dfrac{-\dfrac{x^2}{5\cdot 6}}{1-\dfrac{x^2}{5\cdot 6}-\ddots}}}}\\&=\dfrac{1}{1+\dfrac{\dfrac{x^2}{1\cdot 2}}{1-\dfrac{x^2}{1\cdot 2}+\dfrac{\dfrac{x^2}{3\cdot 4}}{1-\dfrac{x^2}{3\cdot 4}+\dfrac{\dfrac{x^2}{5\cdot 6}}{1-\dfrac{x^2}{5\cdot 6}+\ddots}}}}\\&=\dfrac{1}{1+\dfrac{x^2}{1\cdot 2-x^2+\dfrac{1\cdot 2x^2}{3\cdot 4-x^2+\dfrac{3\cdot 4x^2}{5\cdot 6-x^2+\ddots}}}}\end{align*}\\$

令 $x=1$ 得

$\large \boxed{\cos 1=\dfrac{1}{1+\dfrac{1}{1\cdot 2-1+\dfrac{1\cdot 2}{3\cdot 4-1+\dfrac{3\cdot 4}{5\cdot 6-1+\dfrac{5\cdot 6}{7\cdot 8-1+\ddots}}}}}}\\$

$\large \begin{align*}\arctan x&=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}+\cdots \\&=x+x\left( \dfrac{-x^2}{3} \right) +x\left( \dfrac{-x^2}{3} \right) \left( \dfrac{-3x^2}{5} \right) +x\left( \dfrac{-x^2}{3} \right) \left( \dfrac{-3x^2}{5} \right) \left( \dfrac{-5x^2}{7} \right) +\cdots\\&=\dfrac{x}{1-\dfrac{\dfrac{-x^2}{3}}{1+\dfrac{-x^2}{3}-\dfrac{\dfrac{-3x^2}{5}}{1+\dfrac{-3x^2}{5}-\ddots}}}\\&=\dfrac{x}{1+\dfrac{x^2}{3-x^2+\dfrac{\left( 3x \right) ^2}{5-3x^2+\dfrac{\left( 5x \right) ^2}{7-5x^2+\ddots}}}}\end{align*}\\$

特别的，当 $x=1$ , $\arctan 1=\dfrac{\pi}{4}$

代入我们的公式得到

$\large\boxed{\pi =\dfrac{4}{1+\dfrac{1^2}{2+\dfrac{3^2}{2+\dfrac{5^2}{2+\dfrac{7^2}{2+\ddots}}}}}}\\$

$\large\begin{align*}&\quad \sinh x=\sum_{n=0}^{\infty}{\dfrac{x^{2n+1}}{\left( 2n+1 \right) !}}=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\cdots \\&=x+x\cdot \dfrac{x^2}{2\cdot 3}+x\cdot \dfrac{x^2}{2\cdot 3}\cdot \dfrac{x^2}{4\cdot 5}+x\cdot \dfrac{x^2}{2\cdot 3}\cdot \dfrac{x^2}{4\cdot 5}\cdot \dfrac{x^2}{6\cdot 7}+\cdots \\ &=\dfrac{x}{1-\dfrac{\dfrac{x^2}{2\cdot 3}}{1+\dfrac{x^2}{2\cdot 3}-\dfrac{\dfrac{x^2}{4\cdot 5}}{1+\dfrac{x^2}{4\cdot 5}-\ddots}}}\\ &=\dfrac{x}{1-\dfrac{x^2}{2\cdot 3+x^2-\dfrac{2\cdot 3x^2}{4\cdot 5+x^2-\dfrac{4\cdot 5x^2}{6\cdot 7+x^2-\ddots}}}} \end{align*}\\$

$\large \begin{align*}&\quad \cosh x=\sum_{n=0}^{\infty}{\dfrac{x^{2n}}{\left( 2n \right) !}}=1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\cdots \\&=1+1\cdot \dfrac{x^2}{1\cdot 2}+1\cdot \dfrac{x^2}{1\cdot 2}\cdot \dfrac{x^2}{3\cdot 4}+1\cdot \dfrac{x^2}{1\cdot 2}\cdot \dfrac{x^2}{3\cdot 4}\cdot \dfrac{x^2}{5\cdot 6}+\cdots \\&=\dfrac{1}{1-\dfrac{\dfrac{x^2}{1\cdot 2}}{1+\dfrac{x^2}{1\cdot 2}-\dfrac{\dfrac{x^2}{3\cdot 4}}{1+\dfrac{x^2}{3\cdot 4}-\ddots}}}\\&=\dfrac{1}{1-\dfrac{x^2}{1\cdot 2+x^2-\dfrac{1\cdot 2x^2}{3\cdot 4+x^2-\dfrac{3\cdot 4x^2}{5\cdot 6+x^2-\ddots}}}}\end{align*}\\$

上面2个结果也可以用双曲函数的性质得到

$\large \boxed {\begin{align*}\sin \left( ix\right)&=i\sinh x\\\cos\left( ix\right)&=\cosh x\\i^2&=-1\end{align*}}\\$

$\large \begin{align*}&\quad \ln \left( 1+x \right) =\sum_{n=1}^{\infty}{\dfrac{\left( -1 \right) ^{n-1}}{n}x^n}=x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-\dfrac{1}{4}x^4+\cdots \\&=x+x\left( -\dfrac{1}{2}x \right) +x\left( -\dfrac{1}{2}x \right) \left( -\dfrac{2}{3}x \right) +x\left( -\dfrac{1}{2}x \right) \left( -\dfrac{2}{3}x \right) +x\left( -\dfrac{1}{2}x \right) \left( -\dfrac{2}{3}x \right) \left( -\frac{3}{4}x \right) +\cdots \\&=\dfrac{x}{1+\dfrac{\dfrac{1}{2}x}{1-\dfrac{1}{2}x+\dfrac{\dfrac{2}{3}x}{1-\dfrac{2}{3}x+\ddots}}}\\&=\dfrac{x}{1+\dfrac{x}{2-x+\dfrac{2^2x}{3-2x+\dfrac{3^2x}{4-3x+\ddots}}}}\end{align*}\\$

当 $x=1$ 时可以得到

$\large\boxed{\ln 2=\dfrac{1}{1+\dfrac{1^2}{1+\dfrac{2^2}{1+\dfrac{3^2}{1+\dfrac{4^2}{1+\ddots}}}}}}\\$

$\large\begin{align*}&\quad \arcsin x=\sum_{n=0}^{\infty}{\dfrac{\left( 2n-1 \right) !!}{\left( 2n \right) !!}\dfrac{x^{2n+1}}{2n+1}}=x+\left( \dfrac{1}{2} \right) \dfrac{x^3}{3}+\left( \dfrac{1\cdot 3}{2\cdot 4} \right) \dfrac{x^5}{5}+\left( \dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right) \dfrac{x^7}{7}+\cdots \\&=x+x\left( \dfrac{x^2}{2\cdot 3} \right) +x\left( \dfrac{x^2}{2\cdot 3} \right) \left( \dfrac{\left( 3x \right) ^2}{4\cdot 5} \right) +x\left( \dfrac{x^2}{2\cdot 3} \right) \left( \dfrac{\left( 3x \right) ^2}{4\cdot 5} \right) \left( \dfrac{\left( 5x \right) ^2}{6\cdot 7} \right) +\cdots \\&=\dfrac{x}{1-\dfrac{\dfrac{x^2}{2\cdot 3}}{1+\dfrac{x^2}{2\cdot 3}-\dfrac{\dfrac{\left( 3x \right) ^2}{4\cdot 5}}{1+\dfrac{\left( 3x \right) ^2}{4\cdot 5}-\dfrac{\dfrac{\left( 5x \right) ^2}{6\cdot 7}}{1+\dfrac{\left( 5x \right) ^2}{6\cdot 7}-\ddots}}}}\\&=\dfrac{x}{1-\dfrac{x^2}{2\cdot 3+x^2-\dfrac{2\cdot 3\left( 3x \right) ^2}{4\cdot 5+\left( 3x \right) ^2-\dfrac{4\cdot 5\left( 5x^2 \right)}{6\cdot 7+\left( 5x \right) ^2-\ddots}}}}\end{align*}\\$

利用性质

$\large \boxed{\begin{align*}\arcsin ^{-1}\left( ix \right) &=i\sinh^{-1}x\\\arctan ^{-1}\left( ix \right) &=i\tanh ^{-1}x\\\end{align*}}\\$

容易得到

$\large \boxed{\tanh ^{-1}x=\dfrac{x}{1-\dfrac{x^2}{3+x^2-\dfrac{\left( 3x \right) ^2}{5+3x^2-\dfrac{\left( 5x \right) ^2}{7+5x^2+\dfrac{\left( 7x \right) ^2}{9+7x^2+\ddots}}}}}}\\$

$\large \boxed{\sinh^{-1}x=\dfrac{x}{1+\dfrac{x^2}{2\cdot 3-x^2+\dfrac{2\cdot 3\left( 3x \right) ^2}{4\cdot 5-\left( 3x \right) ^2+\dfrac{4\cdot 5\left( 5x \right) ^2}{6\cdot 7-\left( 5x \right) ^2+\ddots}}}}}\\$

其他

除了上述的12个等式，数学界还有一些非常“优美”的等式，大家可以一起欣赏下。但用我们的公式是得不出这些结果的，故不作证明.

$\large\dfrac{e^{\pi x}-2\sin \left( \pi x \right) -e^{-\pi x}}{e^{\pi x}+2\sin \left( \pi x \right) -e^{-\pi x}}=\dfrac{2x^2}{1+\dfrac{4x^4+1^4}{3+\dfrac{4x^4+2^4}{5+\dfrac{4x^4+3^4}{7+\dfrac{4x^4+4^4}{9+\ddots}}}}}\\$

$\large\varphi =\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\ddots}}}}}\\$

注： $\varphi$ 即为黄金分割率，是方程 $x^2+x-1=0$ 的正根，即 $\dfrac{-1+\sqrt{5}}2$

$\large\pi =3+\dfrac{1^3}{6+\dfrac{1^3+2^3}{6+\dfrac{1^3+2^3+3^3+4^3}{6+\dfrac{1^3+2^3+3^3+4^3+5^3+6^3}{6+\dfrac{1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3}{6+\ddots}}}}}\\$

$\large\dfrac{\sqrt{\pi}}{2}e^{z^2}\text{efr}\left( z \right) =\dfrac{z}{1-\dfrac{z^2}{\dfrac{3}{2}+\dfrac{z^2}{\dfrac{5}{2}-\dfrac{\frac{3}{2}z^2}{\dfrac{7}{2}+\dfrac{2z^2}{\dfrac{9}{2}-\dfrac{\frac{5}{2}z^2}{\dfrac{11}{2}+\dfrac{3z^2}{\dfrac{13}{2}-\dfrac{\frac{7}{2}z^2}{\dfrac{15}{2}+\ddots}}}}}}}}\\$

注： $\text{erf}\left( z\right)$ 是Gauss分布的积分，即误差函数. $\text{erf}\left( z\right) =\displaystyle \dfrac2{\sqrt{\pi}}\int_0^ze^{-t^2}\text{d}t$

$\largee=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3+\dfrac{3}{4+\dfrac{4}{5+\ddots}}}}}\\$

$\large\sqrt{\dfrac{1+\sqrt{5}}{2}+2}-\dfrac{1+\sqrt{5}}{2}=\dfrac{e^{-\dfrac{2\pi}{5}}}{1+\dfrac{e^{-2\pi}}{1+\dfrac{e^{-4\pi}}{1+\dfrac{e^{-6\pi}}{1+\dfrac{e^{-8\pi}}{1+\ddots}}}}}\\$

记 $P\left( n \right) =34n^3+51n^2+27n+5$

则 $\large\zeta \left( 3 \right) =\dfrac{6}{P\left( 0 \right) -\dfrac{1^6}{P\left( 1 \right) -\dfrac{2^6}{P\left( 2 \right) -\dfrac{3^6}{P\left( 3 \right) -\dfrac{4^6}{P\left( 4 \right) -\ddots}}}}}\\$

$\zeta \left( z\right)$ 即Riemann $\zeta$ 函数， $\zeta \left( s\right) =\displaystyle \sum_{n=1}^{\infty}\dfrac1{n^s}$ , $s\in \mathbb{C}$ 且 $\Re s>1$

Catalan常数: $\large G=\sum_{n=0}^{\infty}\frac{\left( -1\right) ^n}{\left( 2n+1\right) ^2}=-\int_0^{\frac{\pi}4}\ln \left( \tan x\right) \text{d}x\approx 0.915966\\$

$\largeG=1-\dfrac{1}{3^2+\dfrac{3^4}{5^2-3^2+\dfrac{5^4}{7^2-5^2+\dfrac{7^4}{9^2-7^2+\dfrac{9^4}{11^2-9^2+\ddots}}}}}\\$

$\large\tan x=\dfrac{1}{\dfrac{1}{x}-\dfrac{1}{\dfrac{3}{x}-\dfrac{1}{\dfrac{5}{x}-\dfrac{1}{\dfrac{7}{x}-\dfrac{1}{\dfrac{9}{x}-\ddots}}}}}\\$

$\large\dfrac{e^x-1}{e^x+1}=\dfrac{1}{\dfrac{2}{x}+\dfrac{1}{\dfrac{6}{x}+\dfrac{1}{\dfrac{10}{x}+\dfrac{1}{\dfrac{14}{x}+\dfrac{1}{\dfrac{18}{x}-\ddots}}}}}\\$

$\largex^{\frac{1}{x}}=1+\dfrac{2\left( x-1 \right)}{x^2+1-\dfrac{\left( x^2-1 \right) \left( x-1 \right) ^2}{3x\left( x+1 \right) -\dfrac{\left( 4x^2-1 \right) \left( x-1 \right) ^2}{5x\left( x+1 \right) -\dfrac{\left( 9x^2-1 \right) \left( x-1 \right) ^2}{7x\left( x+1 \right) -\dfrac{\left( 16x^2-1 \right) \left( x-1 \right) ^2}{9x\left( x+1 \right) -\ddots}}}}}\\$

记

$P\left( n \right) =5n^2+2,Q\left( n \right) =55n^6+165n^5+242n^4+209n^3+172n^2+50n+12$

则

$\large\pi \cot\text{h}\pi =1+\dfrac{28}{Q\left( 0 \right) +\dfrac{1^2\left( 1^2+1 \right) ^2\left( 1^2+4 \right) P\left( 0 \right) P\left( 2 \right)}{Q\left( 1 \right) +\dfrac{2^2\left( 2^2+1 \right) ^2\left( 2^2+4 \right) P\left( 1 \right) P\left( 3 \right)}{Q\left( 2 \right) +\dfrac{\ddots}{Q\left( n-1 \right) +\dfrac{n^2\left( n^2+1 \right) ^2\left( n^2+4 \right) P\left( n-1 \right) P\left( n+1 \right)}{Q\left( n \right) +\ddots}}}}}\\$

但同时不能用上述公式展开（甚至有的根本不能用连分数展开！）的函数还有很多，例如下面这些，

$\arcsin x=\sum_{n=0}^{\infty}{\frac{\left( 2n \right) !}{4^n\left( n! \right) ^2\left( 2n+1 \right)}x^{2n+1}}=x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{112}x^7+\cdots x\in \left( -1,1 \right)$

$\tan x=\sum_{n=1}^{\infty}{\frac{B_{2n}\left( -4 \right) ^n\left( 1-4^n \right)}{\left( 2n \right) !}x^{2n-1}}=x+\frac{1}{3}x^3+\frac{2}{15}x^5+\frac{17}{315}x^7+\cdots x\in \left( -1,1 \right)$

$\sec x=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^nE_{2n}x^{2n}}{\left( 2n \right) !}}=1+\frac{1}{2}x^2+\frac{5}{54}x^4+\frac{61}{720}x^6+\cdots x\in \left( -\frac{\pi}{2},\frac{\pi}{2} \right)$

$\csc x=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^{n+1}2\left( 2^{2n-1}-1 \right) B_{2n}}{\left( 2n \right) !}x^{2n-1}}=\frac{1}{x}+\frac{1}{6}x+\frac{7}{360}x^3+\frac{31}{15120}x^5+\cdots x\in \left( 0,\pi \right)$

$\cot x=\sum_{n=0}^{\infty}{\frac{\left( -1 \right) ^n2^{2n}B_{2n}}{\left( 2n \right) !}}x^{2n-1}=\frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3-\frac{2}{945}x^5-\cdots x\in \left( 0,\pi \right)$

$\tanh x=\sum_{n=1}^{\infty}{\frac{2^{2n}\left( 2^{2n}-1 \right) B_{2n}x^{2n-1}}{\left( 2n \right) !}=x-\frac{1}{3}x^3+\frac{2}{15}x-\frac{17}{315}x^7+\cdots \left| x \right|<\frac{\pi}{2}}$

$\mathrm{arsinh} x=\sum_{n=0}^{\infty}{\left( \frac{\left( -1 \right) ^n\left( 2n \right) !}{2^{2n}\left( n! \right) ^2} \right) \frac{x^{2n+1}}{2n+1}}=x-\frac{1}{6}x^3+\frac{3}{40}x^5-\frac{5}{112}x^7+\cdots \left| x \right|<1$

$\arccos x=\ln \left( 2x \right) -\sum_{n=1}^{\infty}{\left( \frac{\left( -1 \right) ^n\left( 2n \right) !}{2^{2n}\left( n! \right) ^2} \right) \frac{x^{-2n}}{2n}}=\ln \left( 2x \right) -\left( \frac{1}{4}x^{-2}+\frac{3}{32}x^{-4}-\frac{15}{288}x^{-6}+\cdots \right) ,\left| x \right|>1$

$B_{n}$ , $E_{n}$ 分别为Bernoulli级数和Euler级数,关于特殊函数，可以参考北大物理系图书《特殊函数概论》，作者为王竹溪，郭敦仁。

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