Difference between revisions of "Weierstrass factorization theorem"

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__NOTOC__
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==Theorem==
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(Weierstrass factorization theorem) Let $f \colon \mathbb{C} \rightarrow \mathbb{C}$ be [[entire]] and let $\{a_n\}$ denote the set of zeros of $f$ repeated according to multiplicity. Suppose that $f$ has a zero at $z=0$ of [[order of zero|order]] $m \geq 0$ (if $0$ is a zero of order $0$ means that $f(0) \neq 0$). Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that
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$$f(z) = z^m e^{g(z)} \displaystyle\prod_{k=1}^{\infty} E_{p_n} \left( \dfrac{z}{a_k} \right),$$
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where $E_{p_n}$ denotes a [[Weierstrass elementary factors|Weierstrass elementary factor]].
  
=Examples of Weierstrass factorizations=
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==Proof==
{{:Weierstrass factorization of sine}}
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==See also==
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[[Weierstrass factorization of sine]]<br />
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[[Weierstrass factorization of cosine]]<br />
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[[Weierstrass factorization of sinh]]<br />
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[[Weierstrass factorization of cosh]]<br />
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[[Gamma function Weierstrass product]]<br />
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==References==
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[[Category:Theorem]]
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[[Category:Unproven]]

Latest revision as of 19:12, 26 November 2016

Theorem

(Weierstrass factorization theorem) Let $f \colon \mathbb{C} \rightarrow \mathbb{C}$ be entire and let $\{a_n\}$ denote the set of zeros of $f$ repeated according to multiplicity. Suppose that $f$ has a zero at $z=0$ of order $m \geq 0$ (if $0$ is a zero of order $0$ means that $f(0) \neq 0$). Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that $$f(z) = z^m e^{g(z)} \displaystyle\prod_{k=1}^{\infty} E_{p_n} \left( \dfrac{z}{a_k} \right),$$ where $E_{p_n}$ denotes a Weierstrass elementary factor.

Proof

See also

Weierstrass factorization of sine
Weierstrass factorization of cosine
Weierstrass factorization of sinh
Weierstrass factorization of cosh
Gamma function Weierstrass product

References