超幾何系の級数の解き方備忘録

$$\sum^{\infty}_{n=0} \frac{x^{2n}}{\binom{2n}{n}} = \frac{4}{4-x^2} \left(1+\frac{x }{\sqrt{4-x^2}}\sin^{-1}\frac{x}{2} \right)$$

$$ \begin{align*} f(x)&=\sum^{\infty}_{n=0} \frac{x^{2n}}{\binom{2n}{n}} \\ \int^{x}_{0} f(t) dt &= \sum^{\infty}_{n=0} \frac{(n!)^2}{(2n)!(2n+1)} x^{2n+1} \\ &= \sum^{\infty}_{n=0} \frac{(n!)^2(2n+2)}{(2n+1)!(2n+2)} x^{2n+1} \\ &= 2 \sum^{\infty}_{n=0} \frac{(n!)^2(n+1)}{(2n+1)!(2n+2)} x^{2n+1} \\ &= 2 \sum^{\infty}_{n=0} \frac{n!(n+1)!}{(2(n+1))!} x^{2n+1} \\ x\int^{x}_{0} f(t) dt &= 2 \sum^{\infty}_{n=0} \frac{n!(n+1)!}{(2(n+1))!} x^{2n+2} \\ \frac{d}{dx}x\int^{x}_{0} f(t) dt &= 2 \sum^{\infty}_{n=0} \frac{n!(n+1)!(2n+2)}{(2(n+1))!} x^{2n+1} \\ &= 4 \sum^{\infty}_{n=0} \frac{((n+1)!)^2}{(2(n+1))!} x^{2n+1} \\ &= 4 \sum^{\infty}_{n=1} \frac{(n!)^2}{(2n)!} x^{2n-1} \\ &= 4 \sum^{\infty}_{n=0} \frac{(n!)^2}{(2n)!} x^{2n-1} - \frac{4}{x} \\ &= \frac{4}{x} f(x) - \frac{4}{x} \\ \int^{x}_{0} f(t) dt+xf(x)&= \frac{4}{x}(f(x)-1) \\ f(x)+f(x)+xf'(x)&= 4\frac{f'(x)}{x}-4\frac{f(x)-1}{x^2} \\ 2x^2f(x)+x^3f'(x)&= 4xf'(x)-4f(x)+4 \\ (x^3-4x)f'(x)+(2x^2+4)f(x)&=4 \end{align*} $$