双曲線関数と三角関数の級数展開
双曲線関数の級数展開
(1)
\[ \sinh x=\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!} \](2)
\[ \cosh x=\sum_{k=0}^{\infty}\frac{x^{2k}}{(2k)!} \](3)
\[ \tanh x=\sum_{k=1}^{\infty}\frac{2^{2k}\left(2^{2k}-1\right)B_{2k}}{(2k)!}x{}^{2k-1} \](4)
\[ \sinh^{-1}x=\sum_{k=0}^{\infty}\frac{2(1-2^{2k-1})B_{2k}}{(2k)!}x{}^{2k-1} \](5)
\[ \cosh^{-1}x=\sum_{k=0}^{\infty}\left(\frac{E_{2k}}{(2k)!}x^{2k}\right) \](6)
\[ \tanh^{-1}x=\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}x{}^{2k-1} \](1)
\begin{align*} \sinh x & =\frac{e^{x}-e^{-x}}{2}\\ & =\sum_{k=0}^{\infty}\frac{x^{k}-(-x)^{k}}{2k!}\\ & =\sum_{k=0}^{\infty}\left(\frac{x^{2k+1}-(-x)^{2k+1}}{2(2k+1)!}+\frac{x^{2k}-(-x)^{2k}}{2(2k)!}\right)\\ & =\sum_{k=0}^{\infty}\left(\frac{x^{2k+1}+x{}^{2k+1}}{2(2k+1)!}+\frac{x^{2k}-x{}^{2k}}{2(2k)!}\right)\\ & =\sum_{k=0}^{\infty}\frac{x^{2k+1}}{(2k+1)!} \end{align*}(2)
\begin{align*} \cosh x & =\frac{e^{x}+e^{-x}}{2}\\ & =\sum_{k=0}^{\infty}\frac{x^{k}+(-x)^{k}}{2k!}\\ & =\sum_{k=0}^{\infty}\left(\frac{x^{2k+1}+(-x)^{2k+1}}{2(2k+1)!}+\frac{x^{2k}+(-x)^{2k}}{2(2k)!}\right)\\ & =\sum_{k=0}^{\infty}\left(\frac{x^{2k+1}-x{}^{2k+1}}{2(2k+1)!}+\frac{x^{2k}+x{}^{2k}}{2(2k)!}\right)\\ & =\sum_{k=0}^{\infty}\frac{x^{2k}}{(2k)!} \end{align*}(3)
\begin{align*} \tanh x & =\frac{\sinh x}{\cosh x}\\ & =\frac{\sinh^{2}x+\cosh^{2}x-\cosh^{2}x}{\cosh x\sinh x}\\ & =2\frac{\cosh2x}{\sinh2x}-\frac{\cosh x}{\sinh x}\\ & =2\tanh^{-1}2x-\tanh^{-1}x\\ & =2\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}(2x){}^{2k-1}-\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}x{}^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{2^{2k}\left(2^{2k}-1\right)B_{2k}}{(2k)!}x{}^{2k-1}\\ & =\sum_{k=1}^{\infty}\frac{2^{2k}\left(2^{2k}-1\right)B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}(4)
\begin{align*} \sinh^{-1}x & =\frac{1}{\sinh x}\\ & =\frac{2\cosh^{2}\frac{x}{2}-\cosh x}{\sinh x}\\ & =\frac{2\cosh^{2}\frac{x}{2}}{2\sinh\frac{x}{2}\cosh\frac{x}{2}}-\frac{\cosh x}{\sinh x}\\ & =\tanh^{-1}\frac{x}{2}-\tanh^{-1}x\\ & =\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}\left(\frac{x}{2}\right){}^{2k-1}-\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}x{}^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{2(1-2^{2k-1})B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}(5)
\begin{align*} \cosh^{-1}x & =\sum_{k=0}^{\infty}\frac{E_{k}}{k!}x^{k}\\ & =\sum_{k=0}^{\infty}\left(\frac{E_{2k+1}}{(2k+1)!}x^{2k+1}+\frac{E_{2k}}{(2k)!}x^{2k}\right)\\ & =\sum_{k=0}^{\infty}\left(\frac{E_{2k}}{(2k)!}x^{2k}\right) \end{align*}(6)
\begin{align*} \tanh^{-1}x & =\frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\\ & =1+\frac{1}{x}\frac{2x}{e^{2x}-1}\\ & =1+\frac{1}{x}\sum_{k=0}^{\infty}\frac{B_{k}}{k!}(2x)^{k}\\ & =1+2B_{1}+\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}x{}^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}三角関数の級数展開
(1)
\[ \sin x=\sum_{k=0}^{\infty}\frac{(-1)^{k}x{}^{2k+1}}{(2k+1)!} \](2)
\[ \cos x=\sum_{k=0}^{\infty}\frac{(-1)^{k}x{}^{2k}}{(2k)!} \](3)
\[ \tan x=\sum_{k=1}^{\infty}\frac{(-1)^{k}2^{2k}\left(1-2^{2k}\right)B_{2k}}{(2k)!}x{}^{2k-1} \](4)
\[ \sin^{-1}x=\sum_{k=0}^{\infty}\frac{(-1)^{k}2(1-2^{2k-1})B_{2k}}{(2k)!}x{}^{2k-1} \](5)
\[ \cos^{-1}x=\sum_{k=0}^{\infty}\left(\frac{(-1)^{k}E_{2k}}{(2k)!}x{}^{2k}\right) \](6)
\[ \tan^{-1}x=\sum_{k=0}^{\infty}\frac{(-1)^{k}2^{2k}B_{2k}}{(2k)!}x{}^{2k-1} \](1)
\begin{align*} \sin x & =-i\sinh(ix)\\ & =-i\sum_{k=0}^{\infty}\frac{(ix)^{2k+1}}{(2k+1)!}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}x{}^{2k+1}}{(2k+1)!} \end{align*}(2)
\begin{align*} \cos x & =\cosh(ix)\\ & =\sum_{k=0}^{\infty}\frac{(ix)^{2k}}{(2k)!}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}x{}^{2k}}{(2k)!} \end{align*}(3)
\begin{align*} \tan x & =-i\tanh ix\\ & =-i\sum_{k=1}^{\infty}\frac{2^{2k}\left(2^{2k}-1\right)B_{2k}}{(2k)!}(ix){}^{2k-1}\\ & =\sum_{k=1}^{\infty}\frac{(-1)^{k}2^{2k}\left(1-2^{2k}\right)B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}(4)
\begin{align*} \sin^{-1}x & =i\sinh^{-1}(ix)\\ & =i\sum_{k=0}^{\infty}\frac{2(1-2^{2k-1})B_{2k}}{(2k)!}(ix){}^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}2(1-2^{2k-1})B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}(5)
\begin{align*} \cos^{-1}x & =\cosh^{-1}(ix)\\ & =\sum_{k=0}^{\infty}\left(\frac{E_{2k}}{(2k)!}(ix)^{2k}\right)\\ & =\sum_{k=0}^{\infty}\left(\frac{(-1)^{k}E_{2k}}{(2k)!}x{}^{2k}\right) \end{align*}(6)
\begin{align*} \tan^{-1}x & =-i\tanh^{-1}(ix)\\ & =i\sum_{k=0}^{\infty}\frac{2^{2k}B_{2k}}{(2k)!}(ix){}^{2k-1}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}2^{2k}B_{2k}}{(2k)!}x{}^{2k-1} \end{align*}ページ情報
| タイトル | 双曲線関数と三角関数の級数展開 |
| URL | https://www.nomuramath.com/p3gr5c7x/ |
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オイラーの公式の応用
\[
\cos z\pm i\sin z=e^{\pm iz}
\]
逆三角関数と逆双曲線関数の積分
\[
\int\sin^{\bullet}xdx=x\sin^{\bullet}x+\sqrt{1-x^{2}}
\]
1±itan(z)など
\[
1\pm i\tan z=\frac{1}{\cos\left(2\Re z\right)+\cosh\left(2\Im z\right)}\left(e^{\pm2i\Re z}+e^{\mp2\Im z}\right)
\]
ピタゴラスの基本三角関数公式
\[
\cos^{2}x+\sin^{2}x=1
\]