Weierstrass Factorization Theorem allows representing an entire function $f$ (can be considered as an infinite polynomial) as a product involving zeros $\{a_n\}$ of $f$:
$$
f(z)=z^m e^{g(z)}\prod_{n=1}^{\infty} E_{p_n}(\frac{z}{a_n})
$$
In the formula above we substitute a simple expression $\prod_{n}(z-a_n)$ that would be used if the product was finite (fundamental theorem of algebra) by the product of elementary factors:
$$
E_n(z)=\left\{
\begin{array}{ll}
(1-z), n=0\\
(1-z)\exp(\frac{z}{1}+\frac{z^2}{2}+…+\frac{z^n}{n}), x\neq0\\
\end{array}
\right.
$$
These elementary factors should ensure that the product converges (terms become close to 1) and that the zeros are at $\{a_n\}$.
Question: Could you please explain how the term $\exp(\frac{z}{1}+\frac{z^2}{2}+…+\frac{z^n}{n})$ helps the product to converge? I'm especially confused about the case $z>1$ when $E_n(z)$ seems to grow very fast instead of being close to 1.
Best Answer
I will here show a short proof that the Weierstrass product converges given some conditions on $a_n$ and $p_n$.
We start by showing that $|1-E_n(z)| \leq |z|^{n+1}$ for all $|z|\leq 1$ where $E_n(z) = (1-z)e^{z + \frac{z^2}{2} \ldots + \frac{z^n}{n}}$. Take $g_n(z) = 1 - E_n(z)$ then
$$\frac{dg_n(z)}{dz} = -\left[-1 + (1-z)(1 + z + \ldots + z^{n-1})\right]e^{z + \frac{z^2}{2} \ldots + \frac{z^n}{n}}\\= z^n e^{z + \frac{z^2}{2} \ldots + \frac{z^n}{n}}\tag{1}$$
This shows that $g_n'(z)$ has a zero of order $n$ at $z=0$ and since $g_n(0) = 0$ it follows that $g_n(z)$ has a zero of order $n+1$. Taylor expanding the holomorphic (for $|z|\leq 1$) function $\frac{g_n(z)}{z^{n+1}}$ about $z=0$ gives us
$$\left|\frac{g_n(z)}{z^{n+1}}\right| = \left|\sum_{k=0}^\infty b_k z^k\right| \leq \sum_{k=0}^\infty b_k = g_n(1) = 1 \implies |1-E_n(z)| \leq z^{n+1}$$ where we have used that the coefficients in the Taylor expansion must be real and positive which follows from looking at $(1)$ and noting that the same property holds for $e^z$.
Using the result above we have that
$$\left|E_{p_n}\left(\frac{z}{a_n}\right)\right| \leq 1 + \left|1-E_{p_n}\left(\frac{z}{a_n}\right)\right| \leq 1 + \left|\frac{z}{a_n}\right|^{p_n+1}$$
as long as $|z| \leq |a_n|$. Since $\lim_{n\to\infty} |a_n| = \infty$ this holds for all sufficiently large $n$.
To complete the proof we use that $\prod (1+c_n)$ converges iff $\sum c_n$ converges when $c_n$ is a positive series and it follows that $\prod E_{p_n}\left(\frac{z}{a_n}\right)$ converges absolutely iff $\sum \left|\frac{z}{a_n}\right|^{p_n+1}$ converges which is a condition we must assume.