Weierstrass factorization theorem

There are several different statements of this theorem, but in essence this theorem will allow us to prescribe zeros and their orders of a holomorphic function. It also allows us to factor any holomorphic function into a product of zeros and a non-zero holomorphic function. We will need to know here how an infinite product converges. It can then be shown that if ${\prod }_{k=1}^{\mathrm{\infty }}{f}_k(z)$ converges uniformly and absolutely (http://planetmath.org/AbsoluteConvergenceOfInfiniteProduct) on compact subsets, then it converges to a holomorphic function given that all the $f_k(z)$ are holomorphic. This is what we will by the infinite product in what follows.

Note that once we can prescribe zeros of a function then we can also prescribe the poles as well and get a meromorphic function just by dividing two holomorphic functions $f/h$ where $f$ will contribute zeros, and $h$ will make poles at the points where $h(z)=0$. So let’s start with the existence statement.

Theorem (Weierstrass Product).

Let $G\mathrm{\subset }\mathrm{C}$ be a domain, let $\mathrm{\{}{a}_k\mathrm{\}}$ be a sequence of points in $G$ with no accumulation points in $G$, and let $\mathrm{\{}{n}_k\mathrm{\}}$ be any sequence of non-zero integers (positive or negative). Then there exists a function $f$ meromorphic in $G$ whose poles and zeros are exactly at the points $a_k$ and the order of the pole or zero at $a_k$ is $n_k$ (a positive order stands for zero, negative stands for pole).

Next let’s look at a more specific statement with more . For one let’s start looking at the whole complex plane and further let’s forget about poles for now to make the following formulas simpler.

Definition.

We call

$E_0(z)$ $:=1-z,$

$E_p(z)$ $:=(1-z)e^{z+\frac{1}{2}{z}^2+\mathrm{\cdots }+\frac{1}{p}{z}^p}\mathit{\hspace{1em}\hspace{1em}}\text{for }p\ge 1,$

an elementary factor.

Now note that for some $a\in ℂ\backslash \{0\}$, $E_p(z/a)$ has a zero (zero of order 1) at $a$.

Theorem (Weierstrass Factorization).

Suppose $f$ be an entire function and let $\mathrm{\{}{a}_k\mathrm{\}}$ be the zeros of $f$ such that $a_k\mathrm{\ne }\mathrm{0}$ (the non-zero zeros of $f$). Let $m$ be the order of the zero of $f$ at $z\mathrm{=}\mathrm{0}$ ($m\mathrm{=}\mathrm{0}$ if $f$ does not have a zero at $z\mathrm{=}\mathrm{0}$). Then there exists an entire function $g$ and a sequence of non-negative integers $\mathrm{\{}{p}_k\mathrm{\}}$ such that

$$f(z)=z^m{e}^{g(z)}\prod _{k=1}^{\mathrm{\infty }}{E}_{p_k}\left(\frac{z}{a_k}\right).$$

Note that we can always choose $p_k=k-1$ and the product above will converge as needed, but we may be able to choose better $p_k$ for specific functions.

Example.

As an example we can try to factorize the function $\sin \mathit{}\mathrm{(}\pi \mathit{}z\mathrm{)}$, which has zeros at all the integers. Applying the Weierstrass factorization theorem directly we get that

$$\sin (\pi z)=z{e}^{g(z)}\prod _{k=-\mathrm{\infty },k\ne 0}^{\mathrm{\infty }}\left(1-\frac{z}{k}\right)e^{z/k},$$

where $g\mathit{}\mathrm{(}z\mathrm{)}$ is some holomorphic function. It turns out that $e^{g\mathit{}\mathrm{(}z\mathrm{)}}\mathrm{=}\pi$, and rearranging the product we get

$$\sin (\pi z)=z\pi \prod _{k=1}^{\mathrm{\infty }}\left(1-\frac{z^2}{k^2}\right).$$

This is an example where we could choose the $p_k\mathrm{=}\mathrm{1}$ for all $k$ and thus we could then get rid of the ugly parts of the infinite product. For calculations in this example see Conway [1].

Title	Weierstrass factorization theorem
Canonical name	WeierstrassFactorizationTheorem
Date of creation	2013-03-22 14:19:31
Last modified on	2013-03-22 14:19:31
Owner	jirka (4157)
Last modified by	jirka (4157)
Numerical id	8
Author	jirka (4157)
Entry type	Theorem
Classification	msc 30C15
Synonym	Weierstrass product theorem
Related topic	MittagLefflersTheorem
Defines	elementary factor