Let $q = e^{2\pi i\tau}$, if $$\psi(q^2)=\sum_{n=0}^{\infty} q^{n(n+1)}$$ is one of ramanujan theta functions,is it possible to evaluate the limit $$\lim_{q\rightarrow 1} (1-q){\psi^2(q^2)}$$ In fact I'm interested in understanding the behaviour of the theta function around its natural boundary $1$.
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1This (rather nontrivial) behavior can be obtained by Jacobi imaginary transformation, see this answer. – Start wearing purple Aug 08 '15 at 16:40
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@ L.G Thanks a lot,that link really helped me – Nicco Aug 08 '15 at 16:46
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1A related question. – Lucian May 19 '16 at 20:16
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Clearly we know that $$\vartheta_{2}(q) = 2q^{1/4}\psi(q^{2})$$ so we have $$\psi^{2}(q^{2})(1 - q) = \frac{q^{-1/2}}{4}\vartheta_{2}^{2}(q)(1 - q) \to \frac{1}{4}\cdot \pi = \frac{\pi}{4}$$ from this answer to your previous question. For the sake of clarity, I have considered only real variable $q$ so that from equation $q = e^{2\pi i \tau}$ the variable $\tau$ is imaginary. And the limit above is taken when $q \to 1^{-}$.
Paramanand Singh
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@Paramanand: It seems Nicco is doing very interesting work on q-cfracs. See his two beautiful cfracs that I've placed side-by-side for easy comparison in this post. – Tito Piezas III Aug 20 '15 at 15:11
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@TitoPiezasIII: Yeah I have seen his posts on MSE and I am fascinated by his work. Someone should convince him to publish all the material (even if it is non-rigorous or perhaps not fully developed) somewhere (maybe on a blog) and give links on MSE and mathoverflow. – Paramanand Singh Aug 21 '15 at 03:27
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I used the following q-continued fraction, see Ramanujan theta function and its continued fraction ,which to my surprise is a q-analogoue of Gauss's well known continued fraction for $\pi$.Given
$$\psi^2(q^2) = \cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^2(1-q^2)^2}{1-q^5+\cfrac{q^3(1-q^3)^2}{1-q^7+\ddots}}}}$$
By multiplying both sides by $(1-q)$ and letting $q\rightarrow1$, then after equivalence transformation we have
$$\lim_{q\rightarrow 1} {(1-q)}{\psi^2(q^2)} = \lim_{q\rightarrow1} {\cfrac{1}{1+\cfrac{\cfrac{q(1-q)}{1-q^2}}{\cfrac{1-q^3}{1-q^2}+\cfrac{\cfrac{q^2(1-q^2)}{1-q^3}}{\cfrac{1-q^5}{1-q^3}+\cfrac{\cfrac {q^3(1-q^3)}{1-q^4}}{\cfrac{1-q^7}{1-q^4}+\ddots}}}}}$$
Which finally becomes
$$\lim_{q\rightarrow1} {(1-q)}{\psi^2(q^2)} = \cfrac{1}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9 + \ddots}}}}}$$
Which is the well known continued fraction for $\frac{\pi}{4}$