# A problem posed by Ramanujan involving $\sum e^{-5\pi n^2}$

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.

• Up to the question, this is impressive (as usual). Cheers and (+1). Jun 5, 2020 at 2:50
• $$\sum_{n\in \mathbb{Z}} e^{-5\pi n^2}$$ 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. Jun 5, 2020 at 3:16
• @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 $1/5$. Jun 5, 2020 at 3:20
• @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. Jun 5, 2020 at 3:22
• @AlapanDas: I have mentioned about that. Divide by $\sqrt{1-x^2}$ and take limits as $x\to 1^-$. Jun 5, 2020 at 3:24

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.