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Can anyone help me figure out how the identity below was obtained?

$ \frac{1}{\sqrt{(e_1-e_3)(e_2-e_3)}} = R \prod \limits_{n=1}^{\infty} \left(1 - \frac{1}{R^{4n}} \right)^{-4}\left(1 + \frac{1}{R^{2n}} \right)^{-4} = R\left(\sum_{n=1}^{\infty} \frac{1}{R^{n(n-1)}}\right)^{-4}. \tag{1}$

(Komatu, "A coefficient problem for functions univalent in an annulus", Kodai Math. Sem. Rep. 8(2), 1956, p. 49-70., theorem 2 on page 57.)

$(e_2-e_3)= (\frac{\pi}{\omega_1})^2 \theta_2^4 \quad \textrm{and} \quad (e_1-e_3)= (\frac{\pi}{\omega_1})^2 \theta_3^4 \tag{2}$

https://books.google.co.nz/books?id=zyxAb4ro-oMC&printsec=frontcover&source=gbs_ge_summary_r&cad=0#v=onepage&q&f=false page no 178 & 179( Elliptic functions by Armitage & Eberlein)

$\theta_2 = 2 q^\frac{1}{4}\prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n})^2 , \quad \theta_3 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1+q^{2n-1})^2 \tag{3}$ and

$\theta_4 = \prod \limits_{n=1}^{\infty} (1-q^{2n}) (1-q^{2n-1})^2 \tag{4}$ page no 104

As $(e_2 -e_3)(e_1-e_3)= (\frac{\pi}{\omega_1})^4 q \prod \limits_{n=1}^{\infty} (1-q^{2n})^8 (1+q^{2n})^8(1+q^{2n-1})^8 \tag{5}$

Where $ \sqrt{e_2 -e_3} \sqrt{e_1-e_3}= (\frac{\pi}{\omega})^2\theta_2^2\theta_3^2 \tag{6}$ by Abramowitz and Stegun

$\theta_2^2 \theta_3^2 = 4 q^\frac{1}{2} \prod \limits_{n=1}^{\infty} (1-q^{2n})^4 (1+q^{n})^4 = \Big( \frac{\theta_2(\sqrt{q })^2}{2} \Big)^2 = 4q^\frac{1}{2}\Big( \sum \limits_{n=1}^{\infty}q^{n(n-1)/2}\Big)^4 \tag{7}$

$q= 1/R^2 \tag{8}$

$\sqrt{e_2 -e_3} \sqrt{e_1-e_3} \!=\! (\frac{\pi}{\omega_1})^2 \! \frac{1}{R} \prod \limits_{n=1}^{\infty} (1\!-\!R^{-4n})^4 (1\!+\!R^{-2n})^4 \!=\! (\frac{\pi}{\omega_1})^2 \frac{1}{R}\Big(\sum \limits_{n=1}^{\infty}R^{-n(n-1)} \Big)^4 \tag{9}$

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    $\begingroup$ The identity makes no sense to anyone who does not have access to the book you have mentioned. Please define the terms. The second equation is just a formal identity, so I guess you want the first equation. You should then define what $e_1,e_2,e_3,R$ are. $\endgroup$
    – Venkataramana
    Commented Sep 1, 2017 at 0:26
  • $\begingroup$ where $e_{\lambda} = \wp(w_{\lambda}; w_1, w_3) (\lambda = 1,2,3)$, $ \wp$ denotes the Weierstrass $\wp$-function, $ 2w_1 = 2i\pi$, $2w_3 = -4\log (R)$, $ w_1+w_2+w_3 = 0$. $\endgroup$
    – Fareeda
    Commented Sep 1, 2017 at 0:34
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    $\begingroup$ Please add a link to any paper you are referencing whenever you can. I have done this for you here, but please add it yourself the next time. $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 1, 2017 at 2:23
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    $\begingroup$ The equation involving the first and third component looks a lot like a theta function identity in disguise; I don't have a computer or paper to work this out with in detail right now, but you might want to see this and this. $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 1, 2017 at 2:30
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    $\begingroup$ See also formulae 18.10.12 and 18.10.13 in Abaramowitz and Stegun. $\endgroup$
    – J. M. isn't a mathematician
    Commented Sep 1, 2017 at 2:40

1 Answer 1

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From standard definitions of $e_1,e_2,e_3$ and $\theta_2$ in reference works such as Abramowitz and Stegun, (e.g., $\theta_2(0,x)=2x^{1/4}\sum_{n>0} x^{n(n-1)}$), it turns out that the following identity holds $$(e_1-e_3)(e_2-e_3)=(\pi/(4\omega_1))^4\theta_2(0,\sqrt{q})^8$$ which does agree with the stated identity. Let $q:=1/R^2$, then if we define $$Q_0:=\prod_{n>0}1-q^{2n},\;Q_1:=\prod_{n>0}1+q^{2n}, \;Q_2:=\prod_{n>0}1+q^{2n-1},\;Q_3:=\prod_{n>0}1-q^{2n-1},$$ then also $[e_1,e_2,e_3]=(\pi/\omega_1)^2Q_0^4/12\;[Q_2^8+Q_3^8,\;16qQ_1^8-Q_3^8,\;-16qQ_1^8-Q_2^8].$

There are many standard identities between theta functions and infinite products such as $Q_0,Q_1,Q_2,Q_3.$ For example DLMF equation 20.4.3. Also $1=Q_1Q_2Q_3$, and $Q_2^8=Q_3^8+16qQ_1^8.$ Note that $\theta_2(0,q)\theta_3(0,q)=\theta_2(0,\sqrt{q})^2/2=2q^{1/4}Q_0^2/Q_3^2$.

Note that there are varying conventions. Some use $\ \omega = 2\omega_1. \ $ For your equations you need to assume $\ \pi = \omega_1 \ $ to avoid $\pi$ factors in equation (1) for example.

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  • $\begingroup$ There is also some form of the Jacobi triple product involved in the middle step $\endgroup$
    – reuns
    Commented Sep 1, 2017 at 18:29
  • $\begingroup$ $(e_1-e_3)= (\frac{\pi}{2\omega_1})^2 \theta_3^4 $ by Abramowitz and I could not find standard identities between theta functions and infinite products, could you please explain me in details how $(e_1 - e_3)(e_2 -e_3)= \theta_2^8$. $\endgroup$
    – Fareeda
    Commented Sep 3, 2017 at 2:35
  • $\begingroup$ @Fareeda There are slight differences in reference sources. My sources use $(e_1-e_3)=(\pi/\omega_1)^2\theta_3^4$. The important things is to be consistent. The numbering of the roots $e_1,e_2,e_3$ is a convention. $\endgroup$
    – Somos
    Commented Sep 3, 2017 at 3:27
  • $\begingroup$ $ \theta_2^2(\sqrt{q })/2= 2 q^\frac{1}{4} \prod \limits_{n=1}^{\infty} (1-q^{n})^2 (1+q^{n})^4 $ how $ \theta_2^2(\sqrt{q })/2=2 q^\frac{1}{4}\prod \limits_{n=1}^{\infty} (1-q^{2n})^2/(1-q^{2n-1})^2$ $\endgroup$
    – Fareeda
    Commented Sep 3, 2017 at 3:27
  • $\begingroup$ Because $Q_0Q_3=\prod_{n>0} (1-q^n),$ and $Q_1Q_2=\prod_{n>0} (1+q^n)$. Thus, $(Q_0Q_3)(Q_1Q_2)^2=(Q_0/Q_3)(Q_1Q_2Q_3)^2$ and since $1=Q_1Q_2Q_3$ we get $Q_0/Q_3$. $\endgroup$
    – Somos
    Commented Sep 3, 2017 at 4:06

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