Ramanujan’s $\tau$-Function – Asymptotic Analysis

analytic-number-theoryca.classical-analysis-and-odeselliptic functionsnt.number-theorysequences-and-series

The Ramanujan's $\tau$-function is defined by
$$q\prod_{n=1}^\infty (1-q^n)^{24}=\sum_{n=1}^\infty \tau (n)q^n$$
where $|q|\lt 1$.

Is there a known asymptotic formula for $\tau (n)$ or $|\tau (n)|$, i.e. $\tau (n)\sim f(n)$ where $f$ is some "simple" function?

Context

The partition function $p$, defined by
$$\prod_{n=1}^\infty \frac{1}{1-q^n}=\sum_{n=0}^\infty p(n)q^n$$
has a known asymptotic formula, namely
$$p (n)\sim\frac{1}{4n\sqrt{3}}\exp\left(\pi\sqrt{\frac{2n}{3}}\right);$$
a proof of this uses elliptic/modular function theory.
It was quite shocking for me to observe that no asymptotic formula for $\tau (n)$ appears in the OEIS, but it is maybe hiding somewhere. The changes of signs of $\tau$ appear quite chaotic; still, one would wish to find an asymptotic formula for $|\tau (n)|$. We know that
$$|\tau (n)|=O(n^{\frac{11}{2}+\epsilon})$$
and
$$|\tau (p)|\le 2p^{\frac{11}{2}}$$
if $p$ is prime.
According to Zagier,

The proof of these formulas, if written out from scratch, has been estimated at 2000 pages.

and in his book Manin cites this as a probable record for the ratio: "length of proof:length of statement" in the whole of mathematics.

Best Answer

While an asymptotic for $|\tau(n)|$ does not exist, there are many results that help us to nail down the order of $|\tau(n)|$. First, let us write $$\tau(n)=n^{\frac{11}{2}}f(n).$$ Also, let $d(n)$ denote the number of divisors of $n$. By Deligne's bound and the classical bound for the maximal order of $d(n)$, there exists a constant $c_1>0$ such that

$$|f(n)|\leq d(n)\leq \exp\Big(c_1 \frac{\log n}{\log\log n}\Big)\ll_{\varepsilon}n^{\varepsilon}.$$

  1. There exists a density one subset of the positive integers such that $$|f(n)|\leq (\log n)^{-\frac{1}{2}+o(1)}.$$ This was proved by Luca, Radziwill, and Shparlinski.

  2. The constant $-\frac{1}{2}$ in the log exponent in the preceding item cannot be improved on a density one subset of the positive integers. This follows from the central limit theorem of Luca, Radziwill, and Shparlinski. In particular, if $v\in\mathbb{R}$, then $$\displaystyle\lim_{x\to\infty}\frac{\#\left\{n\leq x\colon \tau(n)\neq 0,~\displaystyle\frac{\log|f(n)|+\frac{1}{2}\log\log n}{\sqrt{(\frac{1}{2}+\frac{\pi^2}{12})\log\log n}}\geq v\right\}}{\#\{n\leq x\colon \tau(n)\neq 0\}}=\frac{1}{\sqrt{2\pi}}\int_v^{\infty}e^{-\frac{t^2}{2}}dt.$$

  3. M. Ram Murty proved that as a consequence of the (now proved) Sato-Tate conjecture, there exists a constant $c_2\in(0,c_1)$ such that $$f(n)=\Omega_{\pm}\Big(\exp\Big(c_2\frac{\log n}{\log\log n}\Big)\Big).$$ In light of the remarks above, we see that this is sharp up to the quality of $c_2$.

Items 1 and 3 together show that $|\tau(n)|$ has no asymptotic.

  1. For $\tau(p)=f(p)p^{11/2}$ along the primes, we have $|f(p)|\leq 2$ (Deligne), and the (now proven) Sato-Tate conjecture implies that if $I\subseteq[-1,1]$ is a fixed subinterval, then

$$\lim_{x\to\infty}\frac{\#\{p\leq x\colon f(p)/2\in I\}}{\#\{p\leq x\}}=\frac{2}{\pi}\int_{I}\sqrt{1-t^2}dt.$$

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