How to find a function from an infinite sequence of derivatives at $x=0$

Question

I need an odd function $f(x)$ which converges to $\pm \infty$ at $\pm a$ for some positive $a$. At $x=0$, the even derivatives must be $0$, and the odd derivatives must be factorials : $f(0)=0$, $f'(0)=0!$, $f''(0)=0$, $f'''(0)=2!$, … $f^{(2n)}(0)=0$, $f^{(2n+1)}(0)=(2n)!$

Does such a function exist?

Answer 1

Assuming the function is analytic (can be expressed as a power series), which places some requirements on the size of the derivatives over the domain of definition, we have $$ \begin{align} f(x) &=\sum_{k=0}^\infty\frac{(2k)!}{(2k+1)!}x^{2k+1}\\ &=\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}\\ &=\frac12\log\left(\frac{1+x}{1-x}\right) \end{align} $$

Answer 2

By taylors expansion:

$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=0+0!x+0+(2!/3!)x^3+0+(4!/5!)x^5+...$

$=x+(2!/3!)x^3+(4!/5!)x^5+...=\sum_{n=0}^\infty\frac{1}{2n+1}x^{2n+1}\to\pm\infty,$ at $\pm 1$.

You might need some more analysis, but $\sum_{n=1}^\infty\frac{1}{n}=\infty$