Weierstrass factorization theorem: Difference between revisions

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Hadamard factorization theorem: copy pasted a superior statement from Entire_function#Genus
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Entire functions of finite [[Entire function|order]] have [[Jacques Hadamard|Hadamard]]'s canonical representation<ref name="conway" />:<math display="block">f(z)=z^me^{P(z)}\prod_{n=1}^\infty\left(1-\frac{z}{z_n}\right)\exp\left(\frac{z}{z_n}+\cdots+\frac{1}{p} \left(\frac{z}{z_n}\right)^p\right),</math>where <math>z_k</math> are those [[Zero of a function|roots]] of <math>f</math> that are not zero (<math>z_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}{|z_n|^{p+1}}</math>converges. The non-negative integer <math>g=\max\{p,q\}</math> is called the genus of the entire function <math>f</math>.
 
 
We always have <math>p, q \leq floor(\rho)</math>, but we can make a more precise statement: 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>.