# Weierstrass factorization theorem

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In mathematics, and particularly in the field of complex analysis, the Weierstrass factorization theorem asserts that entire functions can be represented by a product involving their zeroes. In addition, every sequence tending to infinity has an associated entire function with zeroes at precisely the points of that sequence. The theorem is named after Karl Weierstrass.

A second form of the theorem extends to meromorphic functions and allows one to consider a given meromorphic function as a product of three factors: terms depending on the function's poles and zeroes, and an associated non-zero holomorphic function.

## Motivation

The consequences of the fundamental theorem of algebra are twofold.[1] Firstly, any finite sequence ${\displaystyle \{c_{n}\}}$ in the complex plane has an associated polynomial ${\displaystyle p(z)}$ that has zeroes precisely at the points of that sequence, ${\displaystyle p(z)=\,\prod _{n}(z-c_{n}).}$

Secondly, any polynomial function ${\displaystyle p(z)}$ in the complex plane has a factorization ${\displaystyle \,p(z)=a\prod _{n}(z-c_{n}),}$ where a is a non-zero constant and cn are the zeroes of p.

The two forms of the Weierstrass factorization theorem can be thought of as extensions of the above to entire functions. The necessity of extra machinery is demonstrated when one considers the product ${\displaystyle \,\prod _{n}(z-c_{n})}$ if the sequence ${\displaystyle \{c_{n}\}}$ is not finite. It can never define an entire function, because the infinite product does not converge. Thus one cannot, in general, define an entire function from a sequence of prescribed zeroes or represent an entire function by its zeroes using the expressions yielded by the fundamental theorem of algebra.

A necessary condition for convergence of the infinite product in question is that each factor ${\displaystyle (z-c_{n})}$ must approach 1 as ${\displaystyle n\to \infty }$. So it stands to reason that one should seek a function that could be 0 at a prescribed point, yet remain near 1 when not at that point and furthermore introduce no more zeroes than those prescribed. Weierstrass' elementary factors have these properties and serve the same purpose as the factors ${\displaystyle (z-c_{n})}$ above.

## The elementary factors

These are also referred to as primary factors.[2]

For ${\displaystyle n\in \mathbb {N} }$, define the elementary factors:[3]

${\displaystyle E_{n}(z)={\begin{cases}(1-z)&{\text{if }}n=0,\\(1-z)\exp \left({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}\right)&{\text{otherwise}}.\end{cases}}}$

Their utility lies in the following lemma:[3]

Lemma (15.8, Rudin) for |z| ≤ 1, n ∈ No

${\displaystyle \vert 1-E_{n}(z)\vert \leq \vert z\vert ^{n+1}.}$

## The two forms of the theorem

### Existence of entire function with specified zeroes

Sometimes called the Weierstrass theorem.[4]

Let ${\displaystyle \{a_{n}\}}$ be a sequence of non-zero complex numbers such that ${\displaystyle |a_{n}|\to \infty }$. If ${\displaystyle \{p_{n}\}}$ is any sequence of integers such that for all ${\displaystyle r>0}$,

${\displaystyle \sum _{n=1}^{\infty }\left(r/|a_{n}|\right)^{1+p_{n}}<\infty ,}$

then the function

${\displaystyle f(z)=\prod _{n=1}^{\infty }E_{p_{n}}(z/a_{n})}$

is entire with zeros only at points ${\displaystyle a_{n}}$. If number ${\displaystyle z_{0}}$ occurs in sequence ${\displaystyle \{a_{n}\}}$ exactly m times, then function f has a zero at ${\displaystyle z=z_{0}}$ of multiplicity m.

### The Weierstrass factorization theorem

Sometimes called the Weierstrass product/factor theorem.[5]

Let ƒ be an entire function, and let ${\displaystyle \{a_{n}\}}$ be the non-zero zeros of ƒ repeated according to multiplicity; suppose also that ƒ has a zero at z = 0 of order m ≥ 0 (a zero of order m = 0 at z = 0 means ƒ(0) ≠ 0). Then there exists an entire function g and a sequence of integers ${\displaystyle \{p_{n}\}}$ such that

${\displaystyle f(z)=z^{m}e^{g(z)}\prod _{n=1}^{\infty }E_{p_{n}}\!\!\left({\frac {z}{a_{n}}}\right).}$[6]

### Hadamard factorization theorem

If ƒ is an entire function of finite order ρ then it admits a factorization

${\displaystyle f(z)=z^{m}e^{g(z)}\displaystyle \prod _{n=1}^{\infty }E_{p}(z/a_{n})}$

where g(z) is a polynomial of degree q, q ≤ ρ and p=[ρ] .[6]

## Notes

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