超幾何系の級数の解き方備忘録

$$\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{(\genfrac{}{}{0ex}{}{2n}{n})}=\frac{4}{4-x^2}(1+\frac{x}{\sqrt{4-x^2}}{\sin }^{-1}\frac{x}{2})$$

$$\begin{array}{rl}f(x)& =\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{(\genfrac{}{}{0ex}{}{2n}{n})}\\ {\int }_0^{x}f(t)dt& =\sum _{n=0}^{\mathrm{\infty }}\frac{(n!{)}^2}{(2n)!(2n+1)}{x}^{2n+1}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{(n!{)}^2(2n+2)}{(2n+1)!(2n+2)}{x}^{2n+1}\\ & =2\sum _{n=0}^{\mathrm{\infty }}\frac{(n!{)}^2(n+1)}{(2n+1)!(2n+2)}{x}^{2n+1}\\ & =2\sum _{n=0}^{\mathrm{\infty }}\frac{n!(n+1)!}{(2(n+1))!}{x}^{2n+1}\\ x{\int }_0^{x}f(t)dt& =2\sum _{n=0}^{\mathrm{\infty }}\frac{n!(n+1)!}{(2(n+1))!}{x}^{2n+2}\\ \frac{d}{dx}x{\int }_0^{x}f(t)dt& =2\sum _{n=0}^{\mathrm{\infty }}\frac{n!(n+1)!(2n+2)}{(2(n+1))!}{x}^{2n+1}\\ & =4\sum _{n=0}^{\mathrm{\infty }}\frac{((n+1)!{)}^2}{(2(n+1))!}{x}^{2n+1}\\ & =4\sum _{n=1}^{\mathrm{\infty }}\frac{(n!{)}^2}{(2n)!}{x}^{2n-1}\\ & =4\sum _{n=0}^{\mathrm{\infty }}\frac{(n!{)}^2}{(2n)!}{x}^{2n-1}-\frac{4}{x}\\ & =\frac{4}{x}f(x)-\frac{4}{x}\\ {\int }_0^{x}f(t)dt+xf(x)& =\frac{4}{x}(f(x)-1)\\ f(x)+f(x)+x{f}^{\prime }(x)& =4\frac{f^{\prime }(x)}{x}-4\frac{f(x)-1}{x^2}\\ 2x^{2}f(x)+x^3{f}^{\prime }(x)& =4x{f}^{\prime }(x)-4f(x)+4\\ (x^3-4x)f^{\prime }(x)+(2x^2+4)f(x)& =4\end{array}$$