While studying Ramanujan's tau function, I observed that the function satisfies a beautiful identity that I had not seen previously in the literature.

Let $\tau(n)$ be Ramanujan's tau function, such that $$q\prod_{n=1}^\infty (1-q^n)^{24} = \sum_{n=1}^\infty \tau(n)\,q^n,$$ then $$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}}$$

I believe it is well known due to the popularity of the function. Can anyone prove the identity? The identity implies the congruence property $$\tau(2)\equiv0\pmod{8}$$ Thus proving the identity is akin to proving the congruence property.

  • $\begingroup$ let me know if I edited wrong! :) $\endgroup$ – Ant Jul 16 '15 at 20:08
  • $\begingroup$ You edited nicely. $\endgroup$ – Nicco Jul 16 '15 at 20:13
  • $\begingroup$ if you are referring to en.wikipedia.org/wiki/Ramanujan_tau_function#Values the values at integers are integers, and tabulated for small $n.$ $\endgroup$ – Will Jagy Jul 16 '15 at 20:22

This answer is meant to demonstrate some ways of calculating $\tau(n)$, ranging from elegant via straight to weird.

Elegant: Ramanujan's $\tau(n)$ is a multiplicative arithmetic function; so $\tau(mn) = \tau(m)\,\tau(n)$ whenever $\gcd(m,n)=1$. However, its multiplicativity properties are also known for non-coprime $m,n$: $$\begin{align} \tau(m)\,\tau(n) &= \sum_{d\mid\gcd(m,n)} d^{11}\tau\!\left(\frac{mn}{d^2}\right) \\ \tau(mn) &= \sum_{d\mid\gcd(m,n)} \mu(d)\,d^{11}\tau\!\left(\frac{m}{d}\right)\tau\!\left(\frac{n}{d}\right) \end{align}$$ where $\mu$ is the Möbius function. Since $\tau$ is not the zero function, its multiplicativity implies $\tau(1)=1$, which I will implicitly use from now on. From the special multiplicativity properties above we get $$\begin{align} \tau(4) &= \tau(2\cdot 2) = \tau(2)^2 - 2^{11} \\ \tau(6) &= \tau(2)\,\tau(3) \\ \tau(8) &= \tau(2\cdot 4) = \tau(2)\,\tau(4) - 2^{11}\tau(2) = \tau(2)^3 - 2^{12}\tau(2) \end{align}$$ Plugging in and assuming $\tau(2)\neq0$, which will be justified later, we find that your claim is equivalent to $$7\,\tau(2)^2 = 16\,\tau(3) \tag{*}$$

Straight: So let us work out $\tau(2),\tau(3)$ now. Denoting $$\begin{align} \operatorname{D}(q) &= q\prod_{n=1}^\infty(1-q^n)^{24} = \sum_{n=1}^\infty\tau(n)\,q^n && \text{(modular discriminant)} \\ \operatorname{P}(q) &= 1 - 24\sum_{n=1}^\infty \sigma(n)\,q^n && \text{(quasi-modular Eisensein series)} \\ \text{where}\quad \sigma(n) &= \sum_{d\mid n}d && \text{(divisor sum)} \end{align}$$ we have $$q\frac{\mathrm{d}\operatorname{D}(q)}{\mathrm{d}q} = \operatorname{P}(q) \operatorname{D}(q)$$ which after expansion into $q$-series and comparing coefficients yields the recurrence relation $$(n-1)\,\tau(n) = -24\sum_{k=1}^{n-1}\sigma(k)\,\tau(n-k)$$ In particular, $$\begin{align} \tau(2) &= -24 \\ \tau(3) &= -12\left(3 + \tau(2)\right) = 2^2 3^2 7 = 252 \end{align}$$ and that fulfills $(*)$.

Weird: In fact, $(*)$ is a direct consequence of another general recurrence which I had worked out years ago just for fun, without finding any use for it.

Namely, $\operatorname{P}(q)$ fulfills a differential equation of order 3 and Chazy type III: $$2\operatorname{P}'''\!{} - 2\operatorname{P}\operatorname{P}''\!{} + 3\operatorname{P}'^2 = 0 \quad\text{where}\quad (\,)' = q\frac{\mathrm{d}(\,)}{\mathrm{d}q}$$ Plugging in $\operatorname{P} = \frac{\operatorname{D}'}{\operatorname{D}}$ gives a fourth-order differential equation for $\operatorname{D}$: $$2\operatorname{D}^3\operatorname{D}''''\!{} - 10\operatorname{D}^2\operatorname{D}'\operatorname{D}'''\!{} - 3\operatorname{D}^2\operatorname{D}''^2\!{} + 24\operatorname{D}\operatorname{D}'^2\operatorname{D}''\!{} - 13\operatorname{D}'^4 = 0$$ Plugging in the $q$-series for $\operatorname{D}$ then results in a nonlinear recurrence relation for $\tau(n)$ that does not use other arithmetic functions: $$\begin{align} 2(n-2)(n-1)^3\tau(n) &= \sum_{\substack{a,b,c,d=1,\ldots,n-1\\a+b+c+d=n+3}} f(a,b,c,d)\,\tau(a)\,\tau(b)\,\tau(c)\,\tau(d) \\ \text{with}\quad f(a,b,c,d) &= -2 a^4 + 10 a^3 b + 3 a^2 b^2 - 24 a^2 b c + 13 a b c d \end{align}$$ Surely you want to quit, but we are now ready to apply this for $n=3$. Then $(a,b,c,d)$ runs through the six distinct permutations of $(1,1,2,2)$, and we get $$16\,\tau(3) = (15+34+16-14-80+36)\,\tau(2)^2 = 7\,\tau(2)^2$$ which confirms $(*)$ without even needing an explicit value for $\tau(2)$.

As mentioned before, I do not believe such relations to be of great interest; but if you are in a playful mood, working out such stuff might be fun.

By the way, one can work out an order-$3$ differential equation for $\operatorname{D}$ (which unfortunately is even more complex) and try to work out a recurrence relation for $\tau(n)$ from that. Interested?

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The identity follows immediately from the fact that 1728 can be expressed in terms of ramanujan tau function in two different ways,namely




After equating the two identities and using the multiplicative property of the function, $\tau(mn)=\tau(m)\tau(n)$ for $\gcd(m,n)=1$, Then we formally obtain the identity after some simple algebraic manipulation.

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  • $\begingroup$ Quite witty and observant! $\endgroup$ – T.A.Tarbox Mar 17 '17 at 3:19

Looking at the link,

$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}} $

is $-24 =8\frac{84480-2\ (-6048)}{-1472-15\cdot 2048} =-24 $

so it checks.

How did you find it?

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  • $\begingroup$ I found it by simple observation $\endgroup$ – Nicco Jul 16 '15 at 20:41

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