Talk:Weierstrass factorization theorem

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Typos

Hi! there are typos: it should be as follows

Then there exists an entire function that has (only) zeroes at every point of {zi}; in particular, P is such a function1:

   P(z)=\prod_{i=1}^\infty E_{p_i}\left(\frac{z}{z_i}\right).

I doubt that this statement is correct:

 *
         The theorem may be generalized  ..... respectively;

then: f(z)=\frac{\prod_i(z-z_i)}{\prod_j(z-p_j)}.

Why should any of these products converge?

cu , F

Clean-up needed

Article says:

Holomorphic functions can be factored: 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…

What is $\Omega$ ? Simply connected? Are the zeroes in $\Omega$ ? (where else!) Are they simple zeroes or multiple? I really doubt that the statement, has it stands now, is correct. Everything would be fine if $\Omega =\mathbf {C}$ and if the zeroes allow multiplicity, but I don't know what the editor really meant. --Bdmy (talk) 18:37, 10 April 2009 (UTC)[reply]

Cosine formula

Is the cosine formula correct? I only ask because it goes to zero when z=0. Psalm 119:105 (talk) 11:12, 20 August 2009 (UTC)[reply]

You're right, there was a mistake in the formula. I removed the \pi z factors. - Greg

Examples of factorization

In the right hand of both examples there are exponentials lacking.

It would be nice to give an example when the entire function does not have any zero (some exponential). —Preceding unsigned comment added by 163.10.1.186 (talk) 17:41, 21 October 2009 (UTC)[reply]

These factorisations are correct. The intermediate expression is the Weierstrass factorisation, which can be obtained for example by expressing sin and cos using gamma functions. The RHS can be found in, for example, Gradshteyn/Ryzhik 1.431. 138.38.106.191 (talk) 11:35, 22 January 2013 (UTC)[reply]

First sentence of the Motivation

The fact that every finite sequence of complex numbers is the set of zeros of some polynomial function is true, but not a consequence of the Fundamental of Algebra. Proof is that that claim is true for real numbers, and actually for any field. — Preceding unsigned comment added by 175.223.31.17 (talk) 10:10, 14 June 2019 (UTC)[reply]

Recognition of brilliance

There is something striking about the formal statement of the theorem. It is the entirety of the subsection "The Weierstrass factorization theorem," which ought to be important to this article. I have zero complaints about nothing regarding none of the null characteristics of this non-explanation. Nothing isn't empty in the vacuousness of its zeroness:

Let $ƒ$ be an entire function, and let $\{a_{n}\}$ be the non-zero zeros of $ƒ$ repeated according to multiplicity; suppose also that $ƒ$ has a zero at $z = 0$ of order $m \geq 0$ (a zero of order $m = 0$ at $z = 0$ is taken to mean $ƒ (0) \neq 0$ —that is, $f$ does not have a zero at $0$ )...

OK, but in seriousness, surely there is a less confusing way to describe this. While I actually find the paragraph about zeroes somewhat enchanting, it clearly does not help with understanding the theorem. Eebster the Great (talk) 05:11, 4 June 2023 (UTC)[reply]