Editing Weierstrass elliptic function (section)

==Addition theorems==
Let <math>z,w\in\mathbb{C}</math>, so that <math>z,w,z+w,z-w\notin\Lambda </math>. Then one has:<ref name=":3">{{citation| surname1=Rolf Busam|title=Funktionentheorie 1|edition=4., korr. und erw. Aufl|publisher=Springer|publication-place=Berlin|at=p.&nbsp;286|isbn=978-3-540-32058-6|date=2006|language=German}}</ref>
<math display="block">\wp(z+w)=\frac14 \left[\frac{\wp'(z)-\wp'(w)}{\wp(z)-\wp(w)}\right]^2-\wp(z)-\wp(w).</math>

As well as the duplication formula:<ref name=":3" />
<math display="block">\wp(2z)=\frac14\left[\frac{\wp''(z)}{\wp'(z)}\right]^2-2\wp(z).</math>

==== Proofs ====
1. These formulas can come with a geometric interpretation. If one looks at the elliptic curve <math>C_{g_2,g_3}^{\mathbb{C}} </math> a line <math>\lambda= \{(x,y)\in\mathbb{C}^2:y=mx+q\}</math> intersects it in three points:<math>C_{g_2,g_3}^{\mathbb{C}} \cap \lambda=\{P,Q,R\} </math>. Since these points belong to the elliptic curve they can be labeled as <math>P=(\wp(u),\wp'(u)) \quad Q=(\wp(v),\wp'(v)) \quad</math> <math> R=(\wp(u+v),\wp'(u+v))</math> with <math>(u,v)\notin \Lambda </math>. From the formula of a secant line we have <math>m=\frac{y_P-y_Q}{x_P-x_Q}=\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)}</math> letting <math>C_{g_2,g_3}^{\mathbb{C}} = \lambda </math> we have the equation <math> (mx+q)^2=4x^3-g_2x-g_3</math> which becomes <math> 4x^3-m^2x^2-(2mq+g_2)x-g_3-q^2=0</math> using [[Vieta's formulas]] one obtains: 
<math> x_P+x_Q+x_R=\frac{m^2}4 </math> which provides the wanted formula 
<math>\wp(u+v)+\wp(u)+\wp(v)=\frac14 \left[ \frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)} \right]^2 </math>

2. A second proof from Akhiezer's book<ref>Akhiezer's book Elements of the theory of elliptic functions https://www.ams.org/books/mmono/079/mmono079-endmatter.pdf</ref> is the following:
 
if <math> f </math> is arbitrary elliptic function then: 
<math>f(u)=c\prod_{i=1}^n \frac{\sigma(u-a_i)}{\sigma(u-b_i)} \quad c \in \mathbb{C}</math> 

where <math> \sigma </math> is one of the [[Weierstrass functions ]] and <math> a_i , b_i</math> are the respective zeros and poles in the period parallelogram. We then let a function 
<math>k(u,v)=\wp(u)-\wp(v)</math> From the previous lemma we have: 
<math>k(u,v)= 
\wp(u)-\wp(v)=c\frac{\sigma(u+v)\sigma(u-v)}{\sigma(u)^2}
 </math>

From some calculations one can find that
<math>c=\frac1{\sigma(v)^2} \implies\wp(u)-\wp(v)=\frac{\sigma(u+v)\sigma(u-v)}{\sigma(u)^2\sigma(v)^2}</math> 

By definition the Weierstrass Zeta function: <math> \frac{d}{dz}\ln \sigma(z)=\zeta(z)</math> therefore we logarithmicly differentiate both sides obtaining:
<math>\frac{\wp'(u)}{\wp(u)-\wp(v)}=\zeta(u+v)-2\zeta(u)-\zeta(u-v)</math> 
Once again by definition <math> \zeta'(z)=-\wp(z)</math> thus by differentiating once more on both sides and rearranging the terms we obtain 

<math>-\wp(u+v)=-\wp(u)+\frac12 \frac{ \wp''(v)[\wp(u)-\wp(v) ]-\wp'(u)[\wp'(u)-\wp'(v)] }{ [\wp(u)-\wp(v) ]^2 } </math>  

Knowing that <math>\wp'' </math> has the following differential equation <math>2\wp''=12\wp^2-g_2</math> and rearranging the terms one gets the  wanted formula  
<math display="block">\wp(u+v)=\frac14 \left[\frac{\wp'(u)-\wp'(v)}{\wp(u)-\wp(v)}\right]^2-\wp(u)-\wp(v).</math>