Spring 2020 - Problem 8

Solution.

1. First, observe that $2^{2n}\p{n!}^2 \geq \p{2n}!$, so

   $$\abs{F\p{z}} \leq \sum_{n=0}^\infty \frac{\abs{z}^{2n}}{2^{2n}\p{n!}^2} \leq \sum_{n=0}^\infty \frac{\abs{z}^{2n}}{\p{2n}!} \leq \sum_{n=0}^\infty \frac{\abs{z}^n}{n!} = e^{\abs{z}}.$$

   Hence, the sum converges absolutely on compact sets, so $F$ is entire.

2. Notice that $F$ is even, so if $a$ is a root of $F$, then so is $-a$. Also, $F\p{0} = 1 \neq 0$. Since $F$ is entire of order at most $1$, we have the following Hadamard factorization, where $\set{-a_n, a_n}_n$ are the roots of $F$:

   $$F\p{z} = e^{az+b} \prod_{n=1}^\infty \p{1 - \frac{z^2}{a_n^2}},$$

   and the product converges locally uniformly. Next, notice that

   $$F\p{0} = 1 \implies e^b = 1.$$

   Since $F$ is even, we have

   $$1 = \frac{F\p{z}}{F\p{-z}} = \frac{e^{az}}{e^{-az}} = e^{2az},$$

   so $a = 0$. Hence,

   $$F\p{z} = \prod_{n=1}^\infty \p{1 - \frac{z^2}{a_n^2}},$$

   which was what we wanted to show.