Weierstrass factorization theorem

It is clear that any finite set ${\displaystyle \{c_{n}\}}$ of points in the complex plane has an associated polynomial ${\textstyle p(z)=\prod _{n}(z-c_{n})}$ whose zeroes are precisely at the points of that set. The converse is a consequences of the fundamental theorem of algebra: any polynomial function ${\displaystyle p(z)}$ in the complex plane has a factorization ${\textstyle p(z)=a\prod _{n}(z-c_{n}),}$ where a is a non-zero constant and ${\displaystyle \{c_{n}\}}$ is the set of zeroes of ${\displaystyle p(z)}$.[1]

The two forms of the Weierstrass factorization theorem can be thought of as extensions of the above to entire functions. The necessity of additional terms in the product is demonstrated when one considers ${\textstyle \prod _{n}(z-c_{n})}$ where 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.

Consider the functions of the form ${\textstyle \exp \left(-{\tfrac {z^{n+1}}{n+1}}\right)}$ 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 ${\textstyle E_{n}(z)=\exp \left(-{\tfrac {z^{n+1}}{n+1}}\sum _{k=0}^{\infty }{\tfrac {z^{k}}{1+k/(n+1)}}\right)}$ and one can read off how those properties are enforced.

The utility of the elementary factors ${\textstyle E_{n}(z)}$ lies in the following lemma:[2]

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

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

Define the Hadamard canonical factors

$${\displaystyle E_{n}(z):=(1-z)\prod _{k=1}^{n}e^{z^{k}/k}}$$

Entire functions of finite order have Hadamard's canonical representation[4]:

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

where ${\displaystyle a_{k}}$ are those roots of ${\displaystyle f}$ that are not zero (${\displaystyle a_{k}\neq 0}$), ${\displaystyle m}$ is the order of the zero of ${\displaystyle f}$ at ${\displaystyle z=0}$ (the case ${\displaystyle m=0}$ being taken to mean ${\displaystyle f(0)\neq 0}$), ${\displaystyle P}$ a polynomial (whose degree we shall call ${\displaystyle q}$), and ${\displaystyle p}$ is the smallest non-negative integer such that the series

$${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{|a_{n}|^{p+1}}}}$$

converges. The non-negative integer ${\displaystyle g=\max\{p,q\}}$ is called the genus of the entire function ${\displaystyle f}$. In this notation,

$${\displaystyle g\leq \rho \leq g+1}$$

In other words: If the order ${\displaystyle \rho }$ is not an integer, then ${\displaystyle g=[\rho ]}$ is the integer part of ${\displaystyle \rho }$. If the order is a positive integer, then there are two possibilities: ${\displaystyle g=\rho -1}$ or ${\displaystyle g=\rho }$.

For example, ${\displaystyle \sin }$, ${\displaystyle \cos }$ and ${\displaystyle \exp }$ are entire functions of genus ${\displaystyle g=\rho =1}$.

• Mittag-Leffler's theorem
• Wallis product, which can be derived from this theorem applied to the sine function
• Blaschke product

• "Weierstrass theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]

• Visualization of the Weierstrass factorization of the sine function due to Euler at the Wayback Machine (archived 30 November 2018)