Editing Weierstrass factorization theorem (section)

== Hadamard factorization theorem ==
{{Main page|Hadamard factorization theorem}}

A special case of the Weierstraß factorization theorem occurs for entire functions of finite [[Entire function|order]]. In this case the <math>p_n</math> can be taken independent of <math>n</math> and the function <math>g(z)</math> is a polynomial. Thus  <math display="block">f(z)=z^me^{P(z)}\prod_{k=1}^\infty E_p(z/a_k)</math>where <math>a_k</math> are those [[Zero of a function|roots]] of <math>f</math> that are not zero (<math>a_k \neq 0</math>), <math>m</math> is the order of the zero of <math>f</math> at <math>z = 0</math> (the case <math>m = 0</math> being taken to mean <math>f(0) \neq 0</math>), <math>P</math> a polynomial (whose degree we shall call <math>q</math>), and <math>p</math> is the smallest non-negative integer such that the series<math display="block">\sum_{n=1}^\infty\frac{1}{|a_n|^{p+1}}</math>converges. This is called [[Jacques Hadamard|Hadamard]]'s canonical representation.<ref name="conway" /> The non-negative integer <math>g=\max\{p,q\}</math> is called the genus of the entire function <math>f</math>. The order <math>\rho</math> of <math>f</math> satisfies <math display="block">g \leq \rho \leq g + 1</math>
In other words: If the order <math>\rho</math> is not an integer, then <math>g = [ \rho ]</math> is the integer part of <math>\rho</math>. If the order is a positive integer, then there are two possibilities: <math>g = \rho-1</math> or <math>g = \rho </math>.

For example, <math>\sin</math>, <math>\cos</math> and <math>\exp</math> are entire functions of genus <math>g = \rho = 1</math>.