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 $\prod_{k=1}^{\infty}f_{k}(z)$ 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 $G\subset{\mathbb{C}}$ be a domain, let $\{a_{k}\}$ be a sequence of points in $G$ with no accumulation points  in $G$, and let $\{n_{k}\}$ 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

 $\displaystyle E_{0}(z)$ $\displaystyle:=1-z,$ $\displaystyle E_{p}(z)$ $\displaystyle:=(1-z)e^{z+\frac{1}{2}z^{2}+\cdots+\frac{1}{p}z^{p}}\qquad\text{% for p\geq 1},$

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

Theorem (Weierstrass Factorization).

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

 $f(z)=z^{m}e^{g(z)}\prod_{k=1}^{\infty}E_{p_{k}}\left(\frac{z}{a_{k}}\right).$

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 $\sin(\pi z)$, which has zeros at all the integers. Applying the Weierstrass factorization theorem directly we get that

 $\sin(\pi z)=ze^{g(z)}\prod_{k=-\infty,k\not=0}^{\infty}\left(1-\frac{z}{k}% \right)e^{z/k},$

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

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

This is an example where we could choose the $p_{k}=1$ 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 .

References

• 1 John B. Conway. . Springer-Verlag, New York, New York, 1978.
• 2 Theodore B. Gamelin. . Springer-Verlag, New York, New York, 2001.
Title Weierstrass factorization theorem WeierstrassFactorizationTheorem 2013-03-22 14:19:31 2013-03-22 14:19:31 jirka (4157) jirka (4157) 8 jirka (4157) Theorem msc 30C15 Weierstrass product theorem MittagLefflersTheorem elementary factor