Editing Weierstrass factorization theorem (section)

==Weierstrass factorization theorem==
Let {{math|''ƒ''}} be an entire function, and let <math>\{a_n\}</math> be the non-zero zeros of {{math|''ƒ''}} repeated according to multiplicity; suppose also that {{math|''ƒ''}} has a zero at {{math|1=''z'' = 0}} of order {{math|''m'' ≥ 0}}.{{efn|A zero of order {{math|1=''m'' = 0}} at {{math|1=''z'' = 0}} is taken to mean {{math|''&fnof;''(0) ≠ 0}} — that is, <math>f</math> does not have a zero at <math>0</math>.}}
Then there exists an entire function {{math|''g''}} and a sequence of integers <math>\{p_n\}</math> such that

: <math>f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\!\!\left(\frac{z}{a_n}\right).</math><ref name="conway">{{citation|last=Conway|first=J. B.|title=Functions of One Complex Variable I, 2nd ed.|publisher=Springer|location=springer.com|year=1995|isbn=0-387-90328-3}}</ref>

The case given by the fundamental theorem of algebra is incorporated here. If the sequence <math>\{a_n\}</math> is finite then we can take <math>p_n = 0</math>, <math>m=0</math> and <math>e^{g(z)}=c</math> to obtain <math>\, f(z) = c\,{\displaystyle\prod}_n (z-a_n)</math>.

=== Examples of factorization ===
The trigonometric functions [[sine]] and [[cosine]] have the factorizations
<math display=block>\sin \pi z = \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)</math>
<math display=block>\cos \pi z = \prod_{q \in \mathbb{Z}, \, q \; \text{odd} } \left(1-\frac{2z}{q}\right)e^{2z/q} = \prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+\tfrac{1}{2}} \right)^2 \right) </math>
while the [[gamma function]] <math>\Gamma</math> has factorization 
<math display=block>\frac{1}{\Gamma (z)}=e^{\gamma z}z\prod_{n=1}^{\infty }\left ( 1+\frac{z}{n} \right )e^{-z/n},</math>
where <math>\gamma</math> is the [[Euler–Mascheroni constant]].{{citation needed|date=April 2019}} The cosine identity can be seen as special case of
<math display=block>\frac{1}{\Gamma(s-z)\Gamma(s+z)} = \frac{1}{\Gamma(s)^2}\prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+s} \right)^2 \right) </math>
for <math>s=\tfrac{1}{2}</math>.{{citation needed|date=April 2019}}