Order and type of an entire function $f$ such that the numbers $f^{(n)}(0)$ are integers.

Question

Let $f$ be an entire function with order $p=1$ and such that the numbers $f^{(n)}(0)$ are integers. Then show that the type $\sigma$ is at least $1$.

I appreciate any suggestions.

Answer 1

If its order $\rho$ is $1$, then $f$ can not be a polynomial and there are infinitely many $n$'s for which $f^{(n)}(0)\neq 0$. If the latter are assumed to be integers, we get a subsequence $n_k$ such that $$ |f^{(n_k)}(0)|\geq 1 \qquad \forall k $$

Now denote $M(r)$ the maximum of $|f|$ on the disk of radius $r$ centered at $0$. It follows from Cauchy's integral formula that $$ |f^{(n)}(0)|\leq \frac{n! M(r)}{2\pi r^n} \qquad \forall r>0\quad \forall n $$ In particular, we see that for $r=n=n_k$, we have $$ M(n_k)\geq \frac{2\pi n_k^{n_k}}{n_k!}\quad\Rightarrow \quad \frac{\ln M(n_k)}{n_k}\geq \frac{\ln (2\pi)}{n_k}+ \frac{1}{n_k}\ln \left( \frac{n_k^{n_k}}{n_k!}\right)\qquad \forall k $$ By Stirling's approximation, the limit of the rhs is $1$. Therefore, it follows readily from the definition of the type $\sigma$ of an order $\rho=1$ entire function that $$ \sigma =\limsup_{r\rightarrow +\infty}\frac{\ln M(r)}{r^\rho}\geq \limsup_{k\rightarrow +\infty}\frac{\ln M(n_k)}{n_k}\geq \lim_{k\rightarrow +\infty} \frac{\ln (2\pi)}{n_k}+ \frac{1}{n_k}\ln \left( \frac{n_k^{n_k}}{n_k!}\right)=1 $$

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Remains of a huge overkill...

Preliminary facts As noted by Stephen Montgomery-Smith, recall that the MacLaurin series expansion of an entire function $f(z)=\sum a_nz^n$ is given by $a_n=\frac{f^{(n)}(0)}{n!}$.

The order $\rho $ of $f$ satisfies (Theorem 9.4 p.257) $$ \rho =\limsup_{n\rightarrow +\infty}\frac{n\ln n}{-\ln |a_n|} $$

and its type $\sigma $ obeys (Theorem 9.5 p.259) $$ \sigma =\frac{1}{e\rho } \limsup_{n\rightarrow +\infty}n|a_n|^\frac{\rho}{n} $$

It follows that $$ (\rho \sigma)^\frac{1}{\rho}=e^{1-\frac{1}{\rho}}\limsup_{n\rightarrow +\infty}\frac{|f^{(n)}(0)|}{n^{1-\frac{1}{\rho}}} $$ where the derivatives could actually be taken at any $z_0$.

Solution If $\rho=1$, then $f$ can not be a polynomial and there are infinitely many $n$'s for which $f^{(n)}(0)\neq 0$. If the latter are assumed to be integers, we get a subsequence $n_k$ such that $$ |f^{(n_k)}(0)|\geq 1 \qquad \forall k $$ Plugging this in the second formula above and using Stirling's formula yields $$ \sigma \geq \frac{1}{e} \lim_{k\rightarrow +\infty} n_k \cdot (n_k!)^{-\frac{1}{n_k}}=\frac{1}{e} \lim_{k\rightarrow +\infty} n_k\cdot \left(\left(\frac{n_k}{e}\right)^{n_k}\sqrt{2\pi n_k}\right)^{-\frac{1}{n_k}}=1. $$ Or, plugging this in the third formula above gives $\sigma \geq 1$ directly. $\Box$