正接・双曲線正接の総和展開
正接・双曲線正接の総和展開
次が成り立つ。
次が成り立つ。
(1)
\begin{align*} \tan\pi z & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}} \end{align*}(2)
\begin{align*} \tan^{-1}\pi z & =\frac{1}{z\pi}-\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}-z^{2}}\\ & =-\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}-z^{2}} \end{align*}(3)
\begin{align*} \tanh\pi z & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}+z^{2}}\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}+z^{2}} \end{align*}(4)
\begin{align*} \tanh^{-1}\pi z & =\frac{1}{z\pi}+\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}+z^{2}}\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}+z^{2}} \end{align*}次の式は等号が成り立ちません。
\begin{align*} \tan\pi z & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}\\ & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(\left(k-\frac{1}{2}\right)-z\right)\left(\left(k-\frac{1}{2}\right)+z\right)}\\ & =\frac{1}{\pi}\sum_{k=1}^{\infty}\left(\frac{1}{\left(k-\frac{1}{2}\right)-z}-\frac{1}{\left(k-\frac{1}{2}\right)+z}\right)\\ & =-\frac{1}{\pi}\sum_{k=1}^{\infty}\left(\frac{1}{z-\left(k-\frac{1}{2}\right)}+\frac{1}{z+\left(k-\frac{1}{2}\right)}\right)\\ & \ne-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=1}^{\infty}\frac{1}{z+\left(k-\frac{1}{2}\right)}\right)\cmt{\because\text{絶対収束しないので順番を入れ替えれない}}\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{-1}\frac{1}{z+\left(-k-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z+\left(-\left(k-1\right)-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z+\left(-k+\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z-\left(k-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=\infty}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}\right) \end{align*}
\begin{align*} \tan\pi z & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}\\ & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(\left(k-\frac{1}{2}\right)-z\right)\left(\left(k-\frac{1}{2}\right)+z\right)}\\ & =\frac{1}{\pi}\sum_{k=1}^{\infty}\left(\frac{1}{\left(k-\frac{1}{2}\right)-z}-\frac{1}{\left(k-\frac{1}{2}\right)+z}\right)\\ & =-\frac{1}{\pi}\sum_{k=1}^{\infty}\left(\frac{1}{z-\left(k-\frac{1}{2}\right)}+\frac{1}{z+\left(k-\frac{1}{2}\right)}\right)\\ & \ne-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=1}^{\infty}\frac{1}{z+\left(k-\frac{1}{2}\right)}\right)\cmt{\because\text{絶対収束しないので順番を入れ替えれない}}\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{-1}\frac{1}{z+\left(-k-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z+\left(-\left(k-1\right)-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z+\left(-k+\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}+\sum_{k=-\infty}^{0}\frac{1}{z-\left(k-\frac{1}{2}\right)}\right)\\ & =-\frac{1}{\pi}\left(\sum_{k=\infty}^{\infty}\frac{1}{z-\left(k-\frac{1}{2}\right)}\right) \end{align*}
(1)
\(\cos\left(\pi z\right)\)の無限乗積展開より、\[ \cos\left(\pi z\right)=\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right) \] であるので、
\begin{align*} \tan\pi z & =-\frac{1}{\pi}\frac{d}{dz}\log\cos\left(\pi z\right)\\ & =-\frac{1}{\pi}\frac{d}{dz}\log\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)\\ & =-\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{d}{dz}\log\left(1-\frac{z^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)\cmt{\because\text{微分をするので定数の違いは無視できる}}\\ & =-\frac{1}{\pi}\sum_{k=1}^{\infty}\left(1-\frac{z^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)^{-1}\left(-\frac{2z}{\left(k-\frac{1}{2}\right)^{2}}\right)\\ & =-\frac{1}{\pi}\sum_{k=1}^{\infty}\left(\frac{\left(k-\frac{1}{2}\right)^{2}-z^{2}}{\left(k-\frac{1}{2}\right)^{2}}\right)^{-1}\left(-\frac{2z}{\left(k-\frac{1}{2}\right)^{2}}\right)\\ & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}\mrk *\\ & =\frac{z}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}+\sum_{k=-\infty}^{-1}\frac{1}{\left(-k-\frac{1}{2}\right)^{2}-z^{2}}\right)\\ & =\frac{z}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}+\sum_{k=-\infty}^{-1}\frac{1}{\left(k+\frac{1}{2}\right)^{2}-z^{2}}\right)\\ & =\frac{z}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}+\sum_{k=-\infty}^{0}\frac{1}{\left(k-1+\frac{1}{2}\right)^{2}-z^{2}}\right)\\ & =\frac{z}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}+\sum_{k=-\infty}^{0}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}\right)\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}} \end{align*} となり与式は成り立つ。
(2)
\(\sin\left(\pi z\right)\)の無限乗積展開より、\[ \sin\left(\pi z\right)=\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right) \] であるので、
\begin{align*} \tan^{-1}\pi z & =\frac{1}{\pi}\frac{d}{dz}\log\sin\left(\pi z\right)\\ & =\frac{1}{\pi}\frac{d}{dz}\log\left(\pi z\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)\right)\\ & =\frac{1}{\pi}\frac{d}{dz}\left(\log\pi z+\log\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)\right)\\ & =\frac{1}{\pi}\left(\frac{d}{dz}\log\pi z+\frac{d}{dz}\log\prod_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)\right)\\ & =\frac{1}{\pi}\left(\frac{d}{dz}\log\pi z+\sum_{k=1}^{\infty}\frac{d}{dz}\log\left(1-\frac{z^{2}}{k^{2}}\right)\right)\cmt{\because\text{微分をするので定数の違いは無視できる}}\\ & =\frac{1}{\pi}\left(\frac{\pi}{\pi z}+\sum_{k=1}^{\infty}\left(1-\frac{z^{2}}{k^{2}}\right)^{-1}\left(-\frac{2z}{k^{2}}\right)\right)\\ & =\frac{1}{\pi}\left(\frac{1}{z}-2z\sum_{k=1}^{\infty}\left(\frac{k^{2}-z^{2}}{k^{2}}\right)^{-1}\frac{1}{k^{2}}\right)\\ & =\frac{1}{z\pi}-\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}-z^{2}}\mrk *\\ & =-\frac{z}{\pi}\sum_{k=0}^{0}\frac{1}{k^{2}-z^{2}}-\frac{z}{\pi}\left(\sum_{k=1}^{\infty}\frac{1}{k^{2}-z^{2}}+\sum_{k=-\infty}^{-1}\frac{1}{k^{2}-z^{2}}\right)\\ & =-\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}-z^{2}} \end{align*} となり与式は成り立つ。
(3)
\begin{align*} \tanh\pi z & =-i\tan\pi zi\\ & =\left(-i\right)\frac{2zi}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-\left(zi\right)^{2}}\cmt{\because\tan\pi z=\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}}\\ & =\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}+z^{2}} \end{align*} \begin{align*} \tanh\pi z & =-i\tan\pi zi\\ & =-i\frac{zi}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-\left(zi\right)^{2}}\cmt{\because\tan\pi z=\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}-z^{2}}}\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{\left(k-\frac{1}{2}\right)^{2}+z^{2}} \end{align*}(4)
\begin{align*} \tanh^{-1}\pi z & =\frac{1}{-i}\tan^{-1}\pi zi\\ & =i\left(\frac{1}{zi\pi}-\frac{2zi}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}-\left(zi\right)^{2}}\right)\cmt{\because\tan^{-1}\pi z=\frac{1}{z\pi}-\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}-z^{2}}}\\ & =\frac{1}{z\pi}+\frac{2z}{\pi}\sum_{k=1}^{\infty}\frac{1}{k^{2}+z^{2}} \end{align*} \begin{align*} \tanh^{-1}\pi z & =\frac{1}{-i}\tan^{-1}\pi zi\\ & =i\left(-\frac{zi}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}-\left(zi\right)^{2}}\right)\cmt{\because\tan^{-1}\pi z=-\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}-z^{2}}}\\ & =\frac{z}{\pi}\sum_{k=-\infty}^{\infty}\frac{1}{k^{2}+z^{2}} \end{align*}
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三角関数を正接の半角、双曲線関数を双曲線正接の半角で表す。
\[
\sin z=\frac{2\tan\frac{z}{2}}{1+\tan^{2}\frac{z}{2}}
\]
逆三角関数と逆双曲線関数の級数表示
\[
\sin^{\bullet}x=\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\qquad,(|x|\leq1)
\]
逆三角関数の負角、余角、逆数
\[
\cos^{\bullet}x+\sin^{\bullet}x=\frac{\pi}{2}
\]
x tan(x)とx tanh(x)の積分
\[
\int z\tan^{\pm1}\left(z\right)dz=i^{\pm1}\left\{ \frac{1}{2}z^{2}-iz\Li_{1}\left(\mp e^{2iz}\right)+\frac{1}{2}\Li_{2}\left(\mp e^{2iz}\right)\right\} +C
\]