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Weierstraß $sigma$ function identity

asymptotic behavior of the real part of the Riemann zeta function for $0<sigma<1$Definition Weierstrass $zeta$-function unclearInverse Elliptic function questionIs right this application of Hadamard three-lines theorem for $ fraczeta(s)s- fracdzeta(s)dsigma$?“Direct” derivation of exponential form of the Riemann zeta function.Weierstrass zeta, sigma function pseudo-peridocity identityCalculating the lattice of the tori of a non-singular projective cubic curve$Theta$ function in terms of Weierstraß $sigma$ function?Zeroes of some degree of two elliptic functionsEisensteinseries identity

1

$begingroup$

Let $Lambda$ be a lattice and $sigma_Lambda_tau(z):= sigma(z)= prod_win Lambdasetminus 0 left(1 - fraczwright)expleft(fraczw+fracz^22w^2right) $ the Weierstraß sigma function. Furthermore let $eta_1$ be a quasiperiod of the Weierstraß $zeta$ function.

For $q = exp(2 pi i tau)$ and $u = exp(2 pi i z)$ the following identity holds
beginalign
sigma_Lambda_tau(z) = frac12 pi i expleft(fraceta_1 z^22right)left(u^frac12- u^-frac12right)prod_n=1^infty frac(1- q^nu)(1 -q^n u^-1)(1-q^n)^2.
endalign
I dont really know on how to prove it so i would be glad if anyone could help here. Thanks in advance.

complex-analysis elliptic-functions

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asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think this may work : fix $q$, log differentiate in $z$, compare the poles and the behavior at $z=0$ or use $frac1e^2i pi z-1 =frac12 +lim_N to inftysum_n=-N^N frac1z-n$ to make $zeta_tau(z)$ appear. Then go back to $sigma$ and compare the behavior at $z = 0$ (where $prod_n=1^infty frac(1- q^nu)(1 -q^n u^-1)(1-q^n)^2=1$)
  $endgroup$
  – reuns
  yesterday
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $Lambda$ be a lattice and $sigma_Lambda_tau(z):= sigma(z)= prod_win Lambdasetminus 0 left(1 - fraczwright)expleft(fraczw+fracz^22w^2right) $ the Weierstraß sigma function. Furthermore let $eta_1$ be a quasiperiod of the Weierstraß $zeta$ function.

For $q = exp(2 pi i tau)$ and $u = exp(2 pi i z)$ the following identity holds
beginalign
sigma_Lambda_tau(z) = frac12 pi i expleft(fraceta_1 z^22right)left(u^frac12- u^-frac12right)prod_n=1^infty frac(1- q^nu)(1 -q^n u^-1)(1-q^n)^2.
endalign
I dont really know on how to prove it so i would be glad if anyone could help here. Thanks in advance.

complex-analysis elliptic-functions

share|cite|improve this question

asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think this may work : fix $q$, log differentiate in $z$, compare the poles and the behavior at $z=0$ or use $frac1e^2i pi z-1 =frac12 +lim_N to inftysum_n=-N^N frac1z-n$ to make $zeta_tau(z)$ appear. Then go back to $sigma$ and compare the behavior at $z = 0$ (where $prod_n=1^infty frac(1- q^nu)(1 -q^n u^-1)(1-q^n)^2=1$)
  $endgroup$
  – reuns
  yesterday
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $Lambda$ be a lattice and $sigma_Lambda_tau(z):= sigma(z)= prod_win Lambdasetminus 0 left(1 - fraczwright)expleft(fraczw+fracz^22w^2right) $ the Weierstraß sigma function. Furthermore let $eta_1$ be a quasiperiod of the Weierstraß $zeta$ function.

For $q = exp(2 pi i tau)$ and $u = exp(2 pi i z)$ the following identity holds
beginalign
sigma_Lambda_tau(z) = frac12 pi i expleft(fraceta_1 z^22right)left(u^frac12- u^-frac12right)prod_n=1^infty frac(1- q^nu)(1 -q^n u^-1)(1-q^n)^2.
endalign
I dont really know on how to prove it so i would be glad if anyone could help here. Thanks in advance.

complex-analysis elliptic-functions

share|cite|improve this question

asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

EnzousEnzous

111

New contributor

Enzous is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

EnzousEnzous

111