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In mathematics, the Weierstrass factorization theorem in complex analysis, named after Karl Weierstrass, 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. ## Motivation

### The elementary factors

## The two forms of the theorem

### Sequences define holomorphic functions

Sometimes called the Weierstrass theorem### Holomorphic functions can be factored

Sometimes called the Weierstrass Product/Factor/Factorization theorem. Sometimes called the Hadamard Factorization theorem; for example c.f. Boas.$f(z)=e^\{g(z)\}\; prod\_\{i=1\}^infty\; E\_\{p\_i\}left(frac\{z\}\{z\_i\}right)$
## References

## See also

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

The consequences of the fundamental theorem of algebra are twofold. Firstly, any finite sequence,$\{c\_n\}$, in the complex plane has an associated polynomial that has zeroes precisely at the points of that sequence:

- $,prod\_n\; (z-c\_n).$

Secondly, any polynomial function in the complex plane, $p(z)$, has a factorization

- $,p(z)=aprod\_n(z-c\_n),$

where a is a non-zero constant and c_{n} 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 whether the product

- $,prod\_n\; (z-c\_n)$

defines an entire function if the sequence, $\{c\_n\}$, is not finite. The answer is never, because the now-infinite product will 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: each factor, $(z-c\_n)$, must approach 1 as $ntoinfty$. 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. Enter the genius of Weierstrass' elementary factors. These factors serve the same purpose as the factors, $(z-c\_n)$, above.

These are also referred to as primary factors.

For $n\; in\; mathbb\{N\}$, define the elementary factors:

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

Their utility lies in the following lemma:

Lemma (15.8, Rudin) for |z| ≤ 1, n ∈ N_{o}

- $vert\; 1\; -\; E\_n(z)\; vert\; leq\; vert\; z\; vert^\{n+1\}.$

If $lbrace\; z\_i\; rbrace\_i\; subset\; mathbb\{C\}-\{0\}$ is a sequence such that:

- $vert\; z\_i\; vert\; rightarrow\; infty$ as $i\; rightarrow\; infty$
- there is a sequence, $lbrace\; p\_i\; rbrace\_i\; subset\; mathbb\{N\}\_o$, such that for all r > 0, $sum\_\{i\}\; left(frac\{r\}\{vert\; z\_i\; vert\}right)^\{1+p\_i\}\; <\; infty.$

Then there exists an entire function that has (only) zeroes at every point of $lbrace\; z\_i\; rbrace$; in particular, P is such a function:

- $P(z)=prod\_\{i=1\}^infty\; E\_\{p\_i\}left(frac\{z\}\{z\_i\}right).$

- The theorem generalizes to: 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.
- Note also that the case given by the fundamental theorem of algebra is incorporated here. If the sequence, $\{\; z\_i\; \}$ is finite then setting $p\_i\; =\; 0$ suffices for convergence in condition 2, and we obtain: $,\; P(z)\; =\; prod\_n\; (z-z\_n)$.

If f is a function holomorphic in a region, $Omega$, with zeroes at every point of $lbrace\; z\_i\; rbrace\_i\; subset\; mathbb\{C\}-\{0\}$ then there exists an entire function g, and a sequence $lbrace\; p\_i\; rbrace\_i\; subset\; mathbb\{R\}\_o^+$ such that:

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Last updated on Wednesday October 01, 2008 at 19:51:26 PDT (GMT -0700)

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Last updated on Wednesday October 01, 2008 at 19:51:26 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

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