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 $1$.


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 $q \to 1^{-}$.

  • $\begingroup$ Thanks,your blog post is very informative.Thumbs up $\endgroup$ – Nicco Aug 9 '15 at 11:42
  • $\begingroup$ @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. $\endgroup$ – Tito Piezas III Aug 20 '15 at 15:11
  • $\begingroup$ @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. $\endgroup$ – Paramanand Singh Aug 21 '15 at 3:27

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 $\frac{\pi}{4}$

See wikipedia for more about pi

  • $\begingroup$ very nice proof for the limit +1 $\endgroup$ – Paramanand Singh Aug 9 '15 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.