# Weierstrass factorization theorem

(Redirected from Weierstrass product)

In mathematics, and particularly in the field of complex analysis, the Weierstrass factorization theorem asserts that every entire function can be represented as a (possibly infinite) product involving its zeroes. The theorem may be viewed as an extension of the fundamental theorem of algebra, which asserts that every polynomial may be factored into linear factors, one for each root.

The theorem, which is named for Karl Weierstrass, is closely related to a second result that every sequence tending to infinity has an associated entire function with zeroes at precisely the points of that sequence.

A generalization of the theorem extends it to meromorphic functions and allows one to consider a given meromorphic function as a product of three factors: terms depending on the function's zeros and poles, and an associated non-zero holomorphic function.[citation needed]

## 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 for each z, the factors ${\displaystyle (z-c_{n})}$  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

Consider the functions of the form ${\displaystyle \exp(-{\tfrac {z^{n+1}}{n+1}})}$  for ${\displaystyle n\in \mathbb {N} }$ . At ${\displaystyle z=0}$ , they evaluate to ${\displaystyle 1}$  and have a flat slope at order up to ${\displaystyle n}$ . Right after ${\displaystyle z=1}$ , they sharply fall to some small positive value. In contrast, consider the function ${\displaystyle 1-z}$  which has no flat slope but, at ${\displaystyle z=1}$ , evaluates to exactly zero. Also note that for |z| < 1,

${\displaystyle (1-z)=\exp(\ln(1-z))=\exp \left(-{\tfrac {z^{1}}{1}}-{\tfrac {z^{2}}{2}}-{\tfrac {z^{3}}{3}}+\cdots \right)}$ .

Plot of ${\displaystyle E_{n}(x)}$  for n = 0,...,4 and x in the interval [-1,1].

The elementary factors ,[2] also referred to as primary factors ,[3] are functions that combine the properties of zero slope and zero value (see graphic):

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

For |z| < 1 and ${\displaystyle n>0}$ , one may express it as ${\displaystyle E_{n}(z)=\exp(-{\tfrac {z^{n+1}}{n+1}}\Sigma _{k=0}^{\infty }{\tfrac {z^{k}}{1+k/(n+1)}})}$  and one can read off how those properties are enforced.

The utility of the elementary factors En(z) lies in the following lemma:[2]

Lemma (15.8, Rudin) for |z| ≤ 1, ${\displaystyle n\in \mathbb {N} }$

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

## The two forms of the theorem

### Existence of entire function with specified zeroes

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 a number ${\displaystyle z_{0}}$  occurs in the sequence ${\displaystyle \{a_{n}\}}$  exactly m times, then function f has a zero at ${\displaystyle z=z_{0}}$  of multiplicity m.

• The sequence ${\displaystyle \{p_{n}\}}$  in the statement of the theorem always exists. For example, we could always take ${\displaystyle p_{n}=n}$  and have the convergence. Such a sequence is not unique: changing it at finite number of positions, or taking another sequence pnpn, will not break the convergence.
• The theorem generalizes to the following: sequences in open subsets (and hence regions) of the Riemann sphere have associated functions that are holomorphic in those subsets and have zeroes at the points of the sequence.[2]
• Also the case given by the fundamental theorem of algebra is incorporated here. If the sequence ${\displaystyle \{a_{n}\}}$  is finite then we can take ${\displaystyle p_{n}=0}$  and obtain: ${\displaystyle \,f(z)=c\,{\displaystyle \prod }_{n}(z-a_{n})}$ .

### The Weierstrass factorization theorem

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

${\displaystyle f(z)=z^{m}e^{g(z)}\prod _{n=1}^{\infty }E_{p_{n}}\!\!\left({\frac {z}{a_{n}}}\right).}$ [4]

#### Examples of factorization

The trigonometric functions sine and cosine have the factorizations

${\displaystyle \sin \pi z=\pi z\prod _{n\neq 0}\left(1-{\frac {z}{n}}\right)e^{z/n}=\pi z\prod _{n=1}^{\infty }\left(1-\left({\frac {z}{n}}\right)^{2}\right)}$

${\displaystyle \cos \pi z=\prod _{q\in \mathbb {Z} ,\,q\;{\text{odd}}}\left(1-{\frac {2z}{q}}\right)e^{2z/q}=\prod _{n=0}^{\infty }\left(1-\left({\frac {z}{n+{\tfrac {1}{2}}}}\right)^{2}\right)}$

while the gamma function ${\displaystyle \Gamma }$  has factorization
${\displaystyle {\frac {1}{\Gamma (z)}}=e^{\gamma z}z\prod _{n=1}^{\infty }\left(1+{\frac {z}{n}}\right)e^{-z/n},}$

${\displaystyle \gamma }$  is the Euler–Mascheroni constant.[citation needed] The cosine identity can be seen as special case of
${\displaystyle {\frac {1}{\Gamma (s-z)\Gamma (s+z)}}={\frac {1}{\Gamma (s)^{2}}}\prod _{n=0}^{\infty }\left(1-\left({\frac {z}{n+s}}\right)^{2}\right)}$

for ${\displaystyle s={\tfrac {1}{2}}}$ .[citation needed]

If ƒ is an entire function of finite order ρ and m is the order of the zero of ƒ at z=0, then it admits a factorization

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

where g(z) is a polynomial of degree q, qρ and p = [ρ] is the integer part of ρ.[4]