Weierstrass factorization theorem: Difference between revisions

Revision as of 17:36, 8 April 2023

Revision as of 17:37, 8 April 2023

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== Hadamard factorization theorem ==

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>.

If {{math|''ƒ''}} is an entire function of finite [[Entire function|order]] {{math|''ρ''}} and {{math|''m''}} is the order of the zero of {{math|''ƒ''}} at {{math|1=''z'' = 0}}, then it admits a factorization

: <math>f(z) = z^m e^{g(z)} \displaystyle\prod_{n=1}^\infty E_{p}\!\!\left(\frac{z}{a_n}\right)</math>

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>.

where {{math|''g''(''z'')}} is a polynomial of degree {{math|''q''}}, {{math|''q'' ≤ ''ρ''}} and {{math|1=''p'' = [''ρ'']}} is the integer part of {{math|''ρ''}}.<ref name="conway" />

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

==See also==