Weierstrass Factorization Theorem

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Theorem

Let $f$ be an entire function.

Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.

Let the sequence $\sequence {a_n}$ consist of the nonzero zeroes of $f$, repeated according to multiplicity.


First Form

Let $\sequence {p_n}$ be a sequence of non-negative integers for which the series:

$\ds \sum_{n \mathop = 1}^\infty \size {\dfrac r {a_n} }^{1 + p_n}$

converges for every $r \in \R_{> 0}$.


Then there exists an entire function $g$ such that:

$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$

where:

$E_{p_n}$ are Weierstrass's elementary factors
the continued product converges locally uniformly absolutely on $\C$.


Second Form

There exists a sequence $\sequence {p_n}$ of non-negative integers and an entire function $g$ such that:

$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$

where:

$E_{p_n}$ are Weierstrass's elementary factors
the continued product converges locally uniformly absolutely on $\C$.


Proof

From Weierstrass Product Theorem, the function:

$\ds \map h z = z^m \prod_{n \mathop = 1}^\infty \map {E_{p_n} } {\frac z {a_n} }$

defines an entire function that has the same zeroes as $f$ counting multiplicity.

Thus $f / h$ is both an entire function and non-vanishing.

As $f / h$ is both holomorphic and nowhere zero there exists a holomorphic function $g$ such that:

$e^g = f / h$

Therefore:

$f = e^g h$

as desired.

$\blacksquare$


Also known as

Some sources give this as the Weierstrass factor theorem or the Weierstrass infinite product theorem.


Also see


Source of Name

This entry was named for Karl Weierstrass.


Historical Note

The Weierstrass Factorization Theorem was proved by Karl Weierstrass during his work investigating the properties of entire functions.

This theorem finally provided a satisfactory context for Euler Formula for Sine Function.


Sources