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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.

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    $\begingroup$ For fixed $z$ we have that $|z/a_n| < 1$ for large enough $n$ as $|a_n|\to\infty$. Since the term inside $\exp(\ldots)$ is the $n$'th order Taylor series of $-\log(1-z)$ we will have $E_{p_n}(z/a_n) \sim 1$ as $n\to\infty$. Of course to prove convergence you need to know how fast it approaches $1$ which is discussed here (and you need to place some restrictions on $a_n$ and $p_n$, see this $\endgroup$ – Winther Jan 15 '17 at 16:34
  • $\begingroup$ @Winther Thank you for the reply! The key insight for me was that $|z/a_n|<1$ for large enough $n$. $\endgroup$ – Konstantin Jan 15 '17 at 16:44
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    $\begingroup$ btw a small correction to the comment above: I'm assuming the case that $p_n \to \infty$ as $n\to\infty$. In most common examples the set of $p_n$ values are finite like for $\sin(z)$ where $p_n = 1$ for all $n$, but also in this case it's not hard to see that $E_{p_n}(z/a_n) \to 1$ for the same reason that $|z/a_n|\to 0$. $\endgroup$ – Winther Jan 15 '17 at 16:52
  • $\begingroup$ @Winther Very useful comment! Maybe you can write a reply to the question so we may close it as solved? $\endgroup$ – Konstantin Jan 15 '17 at 16:58
  • $\begingroup$ Useful link: math.uiuc.edu/~r-ash/CV/CV6.pdf $\endgroup$ – Konstantin Jan 15 '17 at 18:15
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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.

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  • $\begingroup$ Thank you very much for the amazing explanation! Could you please explain how you get $\left|E_{p_n}\left(\frac{z}{a_n}\right)\right| \leq 1 + \left|1-E_{p_n}\left(\frac{z}{a_n}\right)\right|$ $\endgroup$ – Konstantin Jan 15 '17 at 17:52
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    $\begingroup$ @Konstantin Using the triangle inequality $|x| = |1 + x-1| \leq |1| + |x-1| = |1| + |1-x|$ $\endgroup$ – Winther Jan 15 '17 at 17:54

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