【微分積分】5-4-3 整級数展開｜問題集

\(\displaystyle =\log(1+x)-\log(1-x)\)
\(\displaystyle =x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots+\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots\right)\)
\(\displaystyle =2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)\)
\(\displaystyle =2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\)

\(\displaystyle =\frac{1}{(3x-1)(x-1)}\)
\(\displaystyle =\frac{1}{2}\left(-\frac{3}{3x-1}+\frac{1}{x-1}\right)\)
\(\displaystyle =\frac{1}{2}\left(\frac{3}{1-3x}-\frac{1}{1-x}\right)\)
\(\displaystyle =\frac{1}{2}\left(3\sum_{n=0}^{\infty}(3x)^n-\sum_{n=0}^{\infty}x^n\right)\)
\(\displaystyle =\sum_{n=0}^{\infty}\left(\frac{3^{n+1}}{2}-\frac{1}{2}\right)x^n\)

一般二項展開の公式より、\(|t|<1\)のとき
\(\displaystyle \frac{1}{\sqrt{1-t^2}}=\sum_{n=0}^{\infty}\begin{pmatrix}-\frac{1}{2} \\ n \end{pmatrix}(-t^2)^n\)
\(\displaystyle \frac{1}{\sqrt{1-t^2}}=\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}t^{2n}\)
両辺を積分すると、
\(\displaystyle \int_0^x\frac{1}{\sqrt{1-t^2}}dt=\int_0^x\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!}t^{2n}dt\)
\(\displaystyle [\sin^{-1}t]_0^x=\sum_{n=0}^{\infty}\int_0^x\frac{(2n-1)!!}{(2n)!!}t^{2n}dt\)
\(\displaystyle \sin^{-1}x=\sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}x^{2n+1}\)