Function whose $n^{th}$ derivative at $x=0$ is $(n!)^2$

Question

There exists a function in math where $$f^{(n)}(0)=(n!)^2$$ They say there's an analytical formula for this function but I do not know how to find it. I am investigating functions who cannot be represented with Taylor Polynomials and this is one of the examples I came about. Any help is appreciated! Thanks!

Answer 1

You can obtain the function via Borel summation: $$ \sum\limits_{n = 0}^\infty {n!z^n } = \sum\limits_{n = 0}^\infty {\int_0^\infty {e^{ - t} t^n dt} z^n } = \int_0^\infty {e^{ - t} \sum\limits_{n = 0}^\infty {(zt)^n } dt} = \int_0^\infty {\frac{{e^{ - t} }}{{1 - zt}}dt} . $$ Along the path of integration, $\Re t$ must tend to $+\infty$ and it must omit the singularity at $t=1/z$. Note that $$ \left[ {\frac{{d^n }}{{dz^n }}\int_0^\infty {\frac{{e^{ - t} }}{{1 - zt}}dt} } \right]_{z = 0} = \left[ {n!\int_0^\infty {\frac{{e^{ - t} t^n }}{{(1 - zt)^{n + 1} }}dt} } \right]_{z = 0} = n!\int_0^\infty {e^{ - t} t^n dt} = (n!)^2 . $$