While going through the list of problems posed by Ramanujan in Journal of Indian Mathematical Society I came across this problem involving theta functions:

Prove that $$\frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2})=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}\sum_{n=1}^{\infty}e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and deduce the following:

  • ${\displaystyle \frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2}=\sqrt{5\sqrt{5}-10}\left(\frac{1}{2}+\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)} $
  • ${\displaystyle \sum_{n=1}^{\infty} e^{-\pi n^2}\left(\pi n^2-\frac{1}{4}\right)=\frac{1}{8}} $

The sums in above problem are clearly based on theta functions and we use a simplified notation here to define them. If $\tau$ is any complex number with positive imaginary part then we define $$\vartheta(\tau) =\sum_{n\in\mathbb {Z}} e^{\pi i\tau n^2}$$ and one of the key properties of theta function defined above is $$\vartheta(\tau) =(-i\tau) ^{-1/2}\vartheta(-1/\tau)$$ Ramanujan's first formula probably assumes that $x\in(0,1)$ and hence one can write $x=\cos t$ with $t\in(0,\pi/2)$ and we can consider the complex number $\tau=\sin t +i\cos t$ which clearly has positive imaginary part. The choice of $\tau$ in this manner is done because it gives us $$(-i\tau) ^{-1/2}=\cos(t/2) +i\sin(t/2)=\sqrt{\frac{1+x}{2}}+i\sqrt{\frac{1-x}{2}}$$ and $$-1/\tau=-\sin t+i\cos t=-\sqrt{1-x^2}+ix$$ Using this value of $\tau$ in the transformation formula for theta functions we get $$1+2A+2iB=\frac{\sqrt{1+x}+i\sqrt{1-x}}{\sqrt{2}}(1+2A-2iB)$$ where $$A=\sum_{n=1}^{\infty}e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2}),B=\sum_{n=1}^{\infty} e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and equating real parts we get $$1+2A=(1+2A)\sqrt {\frac{1+x}{2}}+2B\sqrt{\frac{1-x}{2}}$$ or $$\frac{1}{2}+A=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}B$$ In this manner the key formula of Ramanujan is established.

Out of the next two corollaries I was able to prove the second one easily by dividing the main formula by $\sqrt{1-x^2}$ and then taking limits as $x\to 1^{-}$. The first one dealing with $\sum e^{-5\pi n^2}$ was really looking difficult to obtain.

My question is

How to obtain the first corollary dealing with $\sum e^{-5\pi n^2}$ from the main formula of Ramanujan?

Since the formula appears to be using $x\in(0,1)$ I don't see a way to put $x=5$. Even if one does that both sides will contain the sums involving $\sum e^{-5\pi n^2}$ and it appears rather mysterious to obtain a link between $\sum e^{-\pi n^2}$ and $\sum e^{-5\pi n^2}$.

The link between these two sums can be obtained using a modular equation of degree 5, but the calculations involved are tedious (for this technique in action see this answer which evaluates $\sum_{n\in\mathbb {Z}} e^{-3\pi n^2}$). I was therefore hoping for some easier approach as indicated by Ramanujan. Maybe I am mising something obvious here.

  • 6
    $\begingroup$ Up to the question, this is impressive (as usual). Cheers and (+1). $\endgroup$ Jun 5, 2020 at 2:50
  • $\begingroup$ $$\sum_{n\in \mathbb{Z}} e^{-5\pi n^2}$$ can be done is a more condensed way, without using modular equation. One can even calculate $$\sum_{n\in \mathbb{Z}} e^{-N\pi n^2}$$ for $N\leq 100$ in a reasonable amount of time. $\endgroup$
    – pisco
    Jun 5, 2020 at 3:16
  • $\begingroup$ @reuns : there is some issue with your equations. If $x=\sqrt{5},i\sqrt {1-x^2}=2$ how come they add up upto $5$. Note that $$(x+i\sqrt {1-x^2})(x-i\sqrt{1-x^2})=1$$ so if own factor is $5$ the other has to be $1/5$. $\endgroup$
    – Paramanand Singh
    Jun 5, 2020 at 3:20
  • $\begingroup$ @pisco: Is it really possible by hand, pen, paper? I don't have access to any math software and I don't even know how to use them. Anyway if you can post your ideas as an alternative approach do give an answer. $\endgroup$
    – Paramanand Singh
    Jun 5, 2020 at 3:22
  • $\begingroup$ @AlapanDas: I have mentioned about that. Divide by $\sqrt{1-x^2}$ and take limits as $x\to 1^-$. $\endgroup$
    – Paramanand Singh
    Jun 5, 2020 at 3:24

1 Answer 1


Finally I have managed to prove the identity in question. It appears that the main formula (proved in question) as well as its first corollary are both derived from the transformation formula for theta functions rather than being derived from each other. This means that the issue at hand is not really the corollary of the main result as I was expecting.

Let us write $$a=\vartheta(i), b=\vartheta(5i),c=\vartheta(i/5)\tag{1}$$ and we have to prove that $$a=b\sqrt {5\sqrt{5}-10}\tag{2}$$ This is done in two steps and the first one out of these two is obvious. Putting $\tau=5i$ in the transformation formula for theta functions (see the question) we get $$c=b\sqrt{5}\tag{3}$$ In order to prove $(2)$ we need another relation between $a, b$ and $c$ and use it together with $(3)$.

This is the second step where we put $\tau=i+2$ so that $$(-i\tau) ^{-1/2}=\frac{\sqrt{1+2i}}{\sqrt{5}}=\sqrt{\frac{\sqrt{5}+1}{10}}+i\sqrt{\frac{\sqrt{5}-1}{10}}=p+iq\text{ (say)} \tag{4}$$ and $$-\frac{1}{\tau}=\frac{i-2}{5}$$ Using these values in the transformation formula for theta function (and also noting that $\vartheta(\tau +2)=\vartheta(\tau)$) we get $$a=(p+iq)\left\{1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\left(\cos\frac{2\pi n^2}{5}-i\sin\frac{2\pi n^2}{5}\right)\right\}$$ Note that the left hand side is purely real and hence equating real parts we get $$a=p\left(1 +2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)+2q\sum_{n=1}^{\infty}e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}$$ and equating imaginary parts we get $$ 2p\sum_{n=1}^{\infty} e^{-\pi n^2/5}\sin\frac{2\pi n^2}{5}=q\left(1+2\sum_{n=1}^{\infty} e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ Combining these equations we have $$a=\frac{p^2+q^2}{p}\left(1+2\sum_{n=1}^{\infty}e^{-\pi n^2/5}\cos\frac{2\pi n^2}{5}\right)$$ And now we have the magic happening here. If $5\mid n$ then the cosine term equals $1$ otherwise it equals $\cos(2\pi/5)$. We can thus rewrite the above equation as $$a=\frac{p^2+q^2}{p}\left(1+2\cos\frac{2\pi}{5}\sum_{n>0,5\nmid n} e^{-\pi n^2/5}+2\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)$$ and this can be further rewritten as $$a=\frac{p^2+q^2}{p}\left\{1+2\cos\frac{2\pi}{5}\sum_{n=1}^{\infty} e^{-\pi n^2/5}+2\left(1-\cos\frac{2\pi}{5}\right)\sum_{n=1}^{\infty}e^{-5\pi n^2}\right\}$$ Finally this means that \begin{align} a&=\frac{p^2+q^2}{p}\left(1+(c-1)\cos\frac{2\pi}{5}+2(b-1)\sin^2\frac{\pi}{5}\right)\notag\\ &=b\cdot\frac{p^2+q^2}{p}\left(\sqrt{5}\cos\frac{2\pi}{5}+1-\cos\frac{2\pi}{5}\right)\text{ (using (3))}\notag\\ &=\frac{b} {p\sqrt{5}}\left(1+\frac{(\sqrt{5}-1)^2}{4}\right)\text{ (using (4))}\notag\\ &=\frac{b}{p}\cdot\frac{\sqrt{5}-1}{2}\notag\\ &=b\sqrt{\frac{5(\sqrt{5}-1)(3-\sqrt{5})}{4}}\notag\\ &=b\sqrt{5(\sqrt{5}-2)}\notag \end{align} I think this is almost what Ramanujan had in his mind when he posed the problem.


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