逆三角関数と逆双曲線関数の級数表示
逆三角関数の級数表示
(1)
\begin{align*} \sin^{\bullet}x & =\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\cnd{|x|\leq1}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1}\cnd{|x|\leq1} \end{align*}(2)
\begin{align*} \cos^{\bullet}x & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\cnd{|x|\leq1}\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1}\cnd{|x|\leq1} \end{align*}(3)
\[ \tan^{\bullet}x=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1}\cnd{|x|\leq1,x\ne\pm i} \](4)
\begin{align*} \sin^{-1,\bullet}x & =\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\cnd{1\leq|x|}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)}\cnd{1\leq|x|} \end{align*}(5)
\begin{align*} \cos^{-1,\bullet}x & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\cnd{1\leq|x|}\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)}\cnd{1\leq|x|} \end{align*}(6)
\[ \tan^{-1,\bullet}xdx=\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1}\cnd{|x|\leq1,x\ne\pm i} \](1)
\begin{align*} \sin^{\bullet}x & =\sin^{\bullet}0+\int_{0}^{x}\frac{d\sin^{\bullet}t}{dt}dt\\ & =\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}C\left(-\frac{1}{2},k\right)(-t^{2})^{k}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}}t^{2k}dt\\ & =\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\tag{*}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k\right)!}{4^{k}(2k+1)k!k!}x^{2k+1}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k\right)!}{(2k+1)\left(2k\right)!!\left(2k\right)!!}x^{2k+1}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1} \end{align*}(2)
\begin{align*} \cos^{\bullet}x & =\frac{\pi}{2}-\sin^{\bullet}x\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\tag{*}\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1} \end{align*}(2)-2
\begin{align*} \cos^{\bullet}x & =\cos^{\bullet}0+\int_{0}^{x}\frac{d\cos^{\bullet}t}{dt}dt\\ & =\cos^{\bullet}0+\int_{0}^{x}\frac{-1}{\sqrt{1-t^{2}}}dt\\ & =\frac{\pi}{2}-\int_{0}^{x}\sum_{k=0}^{\infty}C\left(-\frac{1}{2},k\right)(-t^{2})^{k}dt\\ & =\frac{\pi}{2}-\int_{0}^{x}\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}}t^{2k}dt\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1} \end{align*}(3)
\begin{align*} \tan^{\bullet}x & =\tan^{\bullet}0+\int_{0}^{x}\frac{d\tan^{\bullet}t}{dt}dt\\ & =\int_{0}^{x}\frac{1}{1+t^{2}}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}(-t^{2})^{k}dt\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1} \end{align*}(4)
\begin{align*} \sin^{-1,\bullet}x & =\sin^{\bullet,\bullet,-1,\bullet}x\\ & =\sin^{\bullet}(x^{-1})\\ & =\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\tag{*}\\ & =\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)} \end{align*}(5)
\begin{align*} \cos^{-1,\bullet}x & =\cos^{\bullet,\bullet,-1,\bullet}x\\ & =\cos^{\bullet}x^{-1}\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\tag{*}\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)} \end{align*}(6)
\begin{align*} \tan^{-1,\bullet}x & =\frac{\pi}{2}-\tan^{\bullet}x\\ & =\frac{\pi}{2}-\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1} \end{align*}逆双曲線関数の級数表示
(1)
\begin{align*} \sinh^{\bullet}x & =\sum_{k=0}^{\infty}\frac{(-1)^{k}C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\cnd{|x|<1}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1}\cnd{|x|<1} \end{align*}(2)
\begin{align*} \cosh^{\bullet}x & =\log(2x)-\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{-2k}\cnd{1<|x|}\\ & =\log(2x)-\sum_{k=1}^{\infty}\frac{\left(2k-1\right)!!}{2k\left(2k\right)!!}x^{-2k}\cnd{1<|x|} \end{align*}(3)
\[ \tanh^{\bullet}x=\sum_{k=0}^{\infty}\frac{1}{2k+1}x^{2k+1}\cnd{|x|<1} \](4)
\begin{align*} \sinh^{-1,\bullet}x & =\sum_{k=0}^{\infty}\frac{(-1)^{k}C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\cnd{1<|x|}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)}\cnd{1<|x|} \end{align*}(5)
\begin{align*} \cosh^{-1,\bullet}x & =\log\frac{2}{x}-\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{2k}\cnd{0<x\leq1}\\ & =\log\frac{2}{x}-\sum_{k=1}^{\infty}\frac{\left(2k-1\right)!!}{2k\left(2k\right)!!}x^{2k}\cnd{0<x\leq1} \end{align*}(6)
\[ \tanh^{-1,\bullet}xdx=\sum_{k=0}^{\infty}\frac{1}{2k+1}x^{-(2k+1)}\cnd{1<|x|} \](1)
\begin{align*} \sinh^{\bullet}x & =\sinh^{\bullet}0+\int_{0}^{x}\frac{d\sinh^{\bullet}t}{dt}dt\\ & =\int_{0}^{x}\frac{1}{\sqrt{1+t^{2}}}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}C\left(-\frac{1}{2},k\right)(t^{2})^{k}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}\frac{(-1)^{k}C\left(2k,k\right)}{4^{k}}t^{2k}dt\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}C\left(2k,k\right)}{4^{k}(2k+1)}x^{2k+1}\tag{*}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{2k+1} \end{align*}(2)
\begin{align*} \cosh^{\bullet}x & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\frac{d\cosh^{\bullet}t}{dt}dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\frac{1}{\sqrt{t^{2}-1}}dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\frac{1}{t\sqrt{1-t^{-2}}}dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\sum_{k=0}^{\infty}C\left(-\frac{1}{2},k\right)t^{-1}(-t^{-2})^{k}dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\sum_{k=0}^{\infty}\frac{C\left(2k,k\right)}{4^{k}}t^{-(2k+1)}dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a+\int_{a}^{x}\left(t^{-1}+\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}}t^{-(2k+1)}\right)dt\right)\\ & =\lim_{a\rightarrow\infty}\left(\cosh^{\bullet}a-\log a\right)+\log x-\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{-2k}\\ & =\lim_{a\rightarrow\infty}\left(\log(a+\sqrt{a^{2}+1})-\log a\right)+\log x-\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{-2k}\\ & =\log(2x)-\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{-2k}\tag{*}\\ & =\log(2x)-\sum_{k=1}^{\infty}\frac{\left(2k\right)!}{4^{k}2kk!k!}x^{-2k}\\ & =\log(2x)-\sum_{k=1}^{\infty}\frac{\left(2k\right)!}{2k\left(2k\right)!!\left(2k\right)!!}x^{-2k}\\ & =\log(2x)-\sum_{k=1}^{\infty}\frac{\left(2k-1\right)!!}{2k\left(2k\right)!!}x^{-2k} \end{align*}(3)
\begin{align*} \tanh^{\bullet}x & =\tanh^{\bullet}0+\int_{0}^{x}\frac{d\tanh^{\bullet}t}{dt}dt\\ & =\int_{0}^{x}\frac{1}{1-t^{2}}dt\\ & =\int_{0}^{x}\sum_{k=0}^{\infty}(t^{2})^{k}dt\\ & =\sum_{k=0}^{\infty}\frac{1}{2k+1}x^{2k+1} \end{align*}(4)
\begin{align*} \sinh^{-1,\bullet}x & =\sinh^{\bullet,\bullet,-1,\bullet}x\\ & =\sinh^{\bullet}(x^{-1})\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}C\left(2k,k\right)}{4^{k}(2k+1)}x^{-(2k+1)}\tag{*}\\ & =\sum_{k=0}^{\infty}\frac{(-1)^{k}\left(2k-1\right)!!}{(2k+1)\left(2k\right)!!}x^{-(2k+1)} \end{align*}(5)
\begin{align*} \cosh^{-1,\bullet}x & =\cosh^{\bullet,\bullet,-1,\bullet}x\\ & =\cosh^{\bullet}x^{-1}\\ & =\log\frac{2}{x}-\sum_{k=1}^{\infty}\frac{C\left(2k,k\right)}{4^{k}2k}x^{2k}\tag{*}\\ & =\log\frac{2}{x}-\sum_{k=1}^{\infty}\frac{\left(2k-1\right)!!}{2k\left(2k\right)!!}x^{2k} \end{align*}(6)
\begin{align*} \tanh^{-1,\bullet}x & =\tanh^{\bullet,\bullet,-1,\bullet}x\\ & =\sum_{k=0}^{\infty}\frac{1}{2k+1}x^{-(2k+1)} \end{align*}ページ情報
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三角関数・双曲線関数の微分
\[
\left(\sin x\right)'=\cos x
\]
三角関数・双曲線関数の一次結合の逆数の積分
\[
\int\frac{1}{\alpha\sin z+\beta\cos z+\gamma}dz=-\frac{2}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}\tanh^{\bullet}\frac{\left(\gamma-\beta\right)\tan\frac{z}{2}+\alpha}{\sqrt{\alpha^{2}+\beta^{2}-\gamma^{2}}}+C
\]
三角関数と双曲線関数の２倍角と３倍角公式
\[
\sin2x=2\sin x\cos x
\]
三角関数(双曲線関数)の逆三角関数(逆双曲線関数)が恒等写像になる条件
\[
\sin^{\bullet}\sin z=?z
\]