How to obtain the function knowing its higher derivatives at $0$

Question

does some one knows how to obtain $f(x)$ knowing that in x=0 they have the following value

$f^{n}(0)= \frac{1}{n-s}$ if $ n=1,3,5,\cdots$ and $f^{n}(0)=0$ otherwise

Answer 1

Using the derivatives to fill in a Taylor series expansion... $$ f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots \\ \frac{x}{1! (1-s)} + \frac{x^3}{3! (3 - s)} + \frac{x^5}{5! (5 - s)} + \cdots \\ \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)} $$

This would be notoriously difficult to manually evaluate ot recognize as the expansion of a known function. Luckily, Wolfram Alpha gives us: $$ \sum_{n=1}^\infty \frac{x^{2n-1}}{(2n-1)!(2n-1-s)}= \frac{_1F_2(\frac{1}{2}-\frac{s}{2};\frac{3}{2}, \frac{3}{2}-\frac{s}{2}; \frac{x^2}{4})x}{s-1} $$