Asymptotic for Ramanujan's $\tau$-function

Question

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.

Answer 1

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.

4. 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

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

Answer 2

No, $\tau(n)$ fluctuates wildly, and it cannot be described in simpler terms. It is "irreducible arithmetic data", and we just love that. Same for its absolute value.

Answer 3

I would like to complement the other answers to mention a couple facts about the erratic nature of the signs of $\tau(n)$. One result is the result of Wilton saying that $$ \sum_{n \leq N} \frac{\tau(n)}{n^{11/2}} e(\alpha n) \ll N^{1/2+\varepsilon},$$ with uniformity in $\alpha \in \mathbb{R}$. This implies that the $\tau$ function changes sign on any arithmetic progression.

Another result is that $\tau(n)$ tends to have its sign pointing in the same direction as $\cos(2 \pi \sqrt{n})$. For this result, let $w$ be a fixed smooth compactly supported function. Then $$ \sum_{n=1}^{\infty} w(n/N) e(2 \sqrt{n}) \frac{\tau(n)}{n^{11/2}} \sim c N^{3/4}, $$ for some $c \neq 0$. As far as I know, this was first proved by Iwaniec, Luo, and Sarnak (Low lying zeros of families of $L$-functions.)

Both of the above results generalize to other holomorphic modular forms and Maass forms.

Answer 4

I am not too familiar with this stuff but it seems highly unlikely Ramanujan's tau function would have general asymptotics due to its arithmetic functional quality. Though you could probably find some limit inferior and limit superior which it oscillates around as the input increases since we get: $$\tau(n)=n^4\sigma(n)-24\sum_{i=1}^{n-1}i^2(35i^2-52in+18n^2)\sigma(i)\sigma(n-i)$$ Thus I see that RHS approximating some polynomial times some power of a logarithm times probably some insanely demented divisor-like function which doesn't asymptote to anything.