Find any function $f(x)$ such that $f^{(n)}(0)=\dfrac{n!}{2^n}$.

Question

Find any function $f(x)$ such that $f^{(n)}(0)=\dfrac{n!}{2^n}$.

I tried $f(x)=e^{\dfrac{x}2}$, but then $f^{(n)}(0)=\dfrac1{2^n}$. Is there an easy way to find $f(x)$?

Answer 1

$$f(x)= \sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{2^n}.$$ Convergence? Sum?

Answer 2

Hint

Consider the Taylor expansion $$f(x)=\sum_{n=0}^p \frac{x^n}{n!}f^{(n)}(0)$$ It looks that you then have some geometric progression.

I am sure that you can take from here.

Answer 3

Check this: Every real sequence is the derivative sequence of some function and the links there.

This is a particular case of a classical (1) result of Borel, (2) undergrad real analysis exercise.