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The problem:

Let $\tau(n)$ denote the Ramanujan $\tau$-function and $\sigma(n)$ be the sum of the positive divisors of $n$. Show that $$ (1-n)\tau(n) = 24\sum_{j=1}^{n-1} \sigma(j)\tau(n-j).$$

I'm afraid I don't even know how to start on this one.. My main problem is the $\tau$ function is defined as the coefficients of the $q$-series of the modular form $\Delta$, but that isn't (at least to my knowledge) very helpful in handling it. The properties that we have are all of a multiplicative nature (i.e., $\tau$ is multiplicative and $\tau(p^{a+2}) = \tau(p)\tau(p^{a+1}) - p^{11}\tau(p^a)$) which don't seem particularly well suited in a problem involving a sum.

Any hints, even just how to get started, would be greatly appreciated. Thanks!


As per Matt E's message, one can define, for any modular form $f$, an operator $\delta$ so that $\delta f = 12\theta f - 12E_2f$, where $\theta = q\frac{d}{dq}$. We wish to look at $\delta\Delta$. We have

\begin{align*} \delta\Delta &= 12\theta\Delta - 12E_2\Delta\newline &= 12\theta\left(\sum_{n=1}^\infty \tau(n)q^n\right) - 12E_2\Delta\newline &= 12q\left(\sum_{n=1}^\infty n\tau(n)q^{n-1}\right) - 12E_2\Delta\newline &= 12\left(\sum_{n=1}^\infty n\tau(n)q^n\right) - 12E_2\Delta. \end{align*} Now $\delta\Delta$ is a cusp form of weight 14, of which there are none that are non-zero. Thus we must have $$ 12\left(\sum_{n=1}^\infty n\tau(n)q^n\right) = 12E_2\Delta.$$ Thus we may simplify a little and plug in the a series representation of $E_2$ to find $$ \sum_{n=1}^\infty n\tau(n)q^n = (1 - 24\sum_{n=1}^\infty \sigma(n)q^n)\sum_{n=1}^\infty \tau(n)q^n.$$ Expanding the right hand side, and simplifying some more gives $$ \sum_{n=1}^\infty n\tau(n)q^n = \sum_{n=1}^\infty \tau(n)q^n - 24\sum_{n=1}^\infty \sigma(n)q^n\sum_{n=1}^\infty \tau(n)q^n,$$ so we have $$ \sum_{n=1}^\infty n\tau(n)q^n - \sum_{n=1}^\infty \tau(n)q^n = - 24\sum_{n=1}^\infty \sigma(n)q^n\sum_{n=1}^\infty \tau(n)q^n.$$ The end of the tunnel is starting to appear, as the left hand side looks quite nice at this point. We settle both sides into a single sum, so that we may match the coefficients on the left and right hand sides easily. We have $$ \sum_{n=1}^\infty \tau(n)(n - 1)q^n = - 24\sum_{n=1}^\infty \left(\sum_{j=1}^{n-1} \sigma(j)\tau(n-j)q^n\right),$$ so multiplying both sides by $-1$ and matching up the coefficients gives $$ \tau(n)(n - 1) = - 24 \left(\sum_{j=1}^{n-1} \sigma(j)\tau(n-j)\right)$$ for all $n\geq 1$, as desired.

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2 Answers 2

up vote 4 down vote accepted

Apply the operator $\Delta$ discussed in this question to the modular form $\Delta$ (sorry for the conflict of notation vis-a-vis $\Delta$!). The result is a modular form of weight $14$. What can you say about it?

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I believe I've come across some progress, which I've editted in. I'll continue to think about this, and thanks for the answer! I have not seen the $\Delta$ operator before, so definitely would not have thought of this. –  Alex Nov 22 '11 at 7:47
@Alex: Dear Alex, Try differentiating $\Delta$ in its series form (so that $\tau(n)$ appears directly in your answer) rather than in its product form. Regards, –  Matt E Nov 22 '11 at 7:53
Wow! That worked out so smoothly! Though it begs the question, what is the $\delta$ operator? Thanks again! –  Alex Nov 22 '11 at 8:29
@Alex: Dear Alex, It's part of the general theory of Rankin--Cohen brackets. I think it's easiest to understand from an automorphic point of view, rather than the more classical modular forms point of view. But that's the topic of another question ... ! Regards, –  Matt E Nov 22 '11 at 8:32

This result can be generalized : Let $f$ be the following product : $$f(x)=\prod_{n=1}^{+\infty}{\left(1-x^n\right)^{a_n}}=\sum_{n=0}^{+\infty}{p(n)x^n}$$ By taking the logrtihmic derivitive of $f$ we obtain :

$$\frac{f'(x)}{f(x)}=-\sum_{n=1}^{+\infty}{na_n\frac{x^{n-1}}{1-x^n}}=-\frac{1}{x}\sum_{n=1}^{+\infty}{na_n\frac{x^n}{1-x^n}}$$ with $$\sum_{n=1}^{+\infty}{na_n\frac{x^n}{1-x^n}}=\sum_{n=1}^{+\infty}{na_n\left(\sum_{m=1}^{+\infty}{x^{nm}}\right)}=\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}$$ Thus : $$f'(x)=-\frac{f(x)}{x}\left(\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}\right)$$ and so $$\sum_{n=0}^{+\infty}{np(n)x^{n}}=-\left(\sum_{n=0}^{+\infty}{p(n)x^n}\right)\left(\sum_{n=1}^{+\infty}{\left(\sum_{d|n}{da_d}\right)x^n}\right)$$

A Cauchy product gives then the recursion formula $$p(0)=1$$ $$p(n)=\frac{-1}{n}\sum_{k=1}^n{p(n-k)\left(\sum_{d|n}{da_d}\right)}$$

(sorry form my approximative english... again)

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