Given $g(z)$ an holomorphic function in the whole $\mathbb{C}$. Is there a way to find $f(z)$ such as:
$$f^{(2k)}(z_0) = g^{(2k)}( z_0 ) \frac{(2k)!}{ (k!)^2}$$
where $f^{(2k)}(z_0)$ denotes the 2kth derivative of $f(z_0)$ for a fixed $z_0$ ?
This question is particularly interesting when $k \to \infty$.
A suboptimal solution could be just finding $f(z)$ such as:
$$ |f^{(2k)}(z_0)| \ge |g^{(2k)}(z_0)| \frac{(2k)!}{ (k!)^2}, \forall k>1$$
With the Stirling formula you get $$ \binom{2k}{k}=\sqrt{\frac{2k+\theta_{2k}}{2\pi(k+\theta_k)^2}}2^{2n}\simeq\frac{4^k}{\sqrt{\pi k}}<4^k,~~ θ_j\in (0,1) $$ so that a simple re-scaling of the argument will give an example for the inequality case with $$ f(z)=g(z_0+2(z−z_0))=g(2z−z_0)\implies |f^{(2k)}(z_0)|=2^{2k}|g^{(2k)}(z_0)|>\binom{2k}{k}|g^{(2k)}(z_0)|. $$