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Looking into the discussion in this post, I was naturally led to consider the following general identity

Given the two well known jacobi theta functions, namely $\theta_2(q)=\sum_{n=-\infty}^\infty q^{(n+1/2)^2}$ and $\theta_3(q)=\sum_{n=-\infty}^\infty q^{n^2}$ where $q=e^{2 \pi i\tau}$, $|q|\lt1$

It is then conjectured that the following identity is true

$\frac{\cfrac{2\,q^{\frac{1}{2}}}{1-q^2+\cfrac{q^2(1-q^2)^2}{1-q^6+\cfrac{q^4(1-q^4)^2}{1-q^{10}+\cfrac{q^6(1-q^6)^2}{1-q^{14}+\ddots}}}}}{\cfrac{1}{1-q+\cfrac{q(1+q)^2}{1-q^3+\cfrac{q^2(1+q^2)^2}{1-q^5+\cfrac{q^3(1+q^3)^2}{1-q^7+\ddots}}}}}=\theta_2(q^2)\,\theta_3(q^2)$

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  • $\begingroup$ If your power has more than one digit, then you have to put it in parenthesis like q^{10}, so the other digit will not fall off. $\endgroup$ Sep 9, 2015 at 14:52
  • $\begingroup$ Your LaTex preview is not working, right? $\endgroup$ Sep 9, 2015 at 15:07
  • $\begingroup$ If you continue having editing issues, try to format your question in the sandbox before repeatedly editing it here. $\endgroup$
    – davidlowryduda
    Sep 9, 2015 at 15:30
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    $\begingroup$ So this implies $A(q^2) = A(q)\,B(q)$. Functional equations of that form are often given in the OEIS, typically by Michael Somos. $\endgroup$
    – ccorn
    Sep 9, 2015 at 22:38
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    $\begingroup$ Another consequence: $A(q)/B(q) = \theta_3^2(q^2)$ $\endgroup$
    – ccorn
    Sep 9, 2015 at 22:53

1 Answer 1

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For the above continued fraction we have (see paper [1] Proposition 2.9) $$ \frac{2q^{1/2}}{1-q^2+}\frac{q^2(1-q^2)^2}{1-q^6+}\frac{q^4(1-q^4)^2}{1-q^{10}+}\frac{q^6(1-q^6)^2}{1-q^{14}+}\ldots=2q^{1/2}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}} $$ and ([1] relations (13),(27)) $$ \frac{k'_r}{1-k_r}=-1+\frac{2}{1-u_0(q^{1/2},q)} $$ where $$ \frac{u_0(q^{1/2},q)}{2q^{1/2}}=\frac{1}{1-q+}\frac{q(1+q)^2}{1-q^3+}\frac{q^2(1+q^2)^2}{1-q^5+}\frac{q^3(1+q^3)^2}{1-q^7+}\ldots $$ Combining the above relations we get $$ CF(q)=4q\frac{k'_r-k_r+1}{k'_r+k_r-1}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}}=\frac{4q}{\sqrt{k_{4r}}}\sum^{\infty}_{n=0}\frac{q^{2n}}{1+q^{4n+2}} $$ Hence we only have to show that $$ \frac{4q^{1/2}}{\sqrt{k_r}}\sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n+1}}=\theta_2(q)\theta_3(q)=\sqrt{\frac{2Kk_r}{\pi}}\sqrt{\frac{2K}{\pi}} $$ or equivalently $$ \frac{2q^{1/2}}{k_r}\sum^{\infty}_{n=0}\frac{q^n}{1+q^{2n+1}}=\frac{K}{\pi} $$ But the Jacobi elliptic function $cn(u)$ has expansion (see [2] pg.55): $$ cn(u)=\frac{2\pi}{Kk_r}\sum^{\infty}_{n=0}\frac{q^{n+1/2}\cos((2n+1)z)}{1+q^{2n+1}}\textrm{, where }u=\frac{2Kz}{\pi} $$ and $cn(0)=1$ (see [2],pg.17), hence $$ 1=\frac{2\pi}{Kk_r}\sum^{\infty}_{n=0}\frac{q^{n+1/2}}{1+q^{2n+1}} $$ By this last note the conjecture follows.

References

[1] N.D. Bagis and M.L. Glasser "Evaluations of a Continued Fraction of Ramanujan". Rend. Sem. Mat. Univ. Padova, Vol. 133, (2015)

[2] J.M. Armitage and W.F. Eberlein "Elliptic Functions". Cambridge University Press, (2006)

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