Average orders of multiplicative functions - MathOverflow

Average orders of multiplicative functions

2012-12-04T23:18:13Z

For a multiplicative function $f$ and $x>0$ let $$S_f(x)= \sum_{n \leq x} f(n).$$ Studying sums of this type is a favourite pastime of analytic number theorists. I'm trying to understand what kind of behaviour can occur for such sums. In particular, my question is the following.

Does there exist a multiplicative function $f$ and a constant $c_f>0$ such that $$S_f(x) \sim c_f\frac{x}{\log x},$$ as $x \to \infty$?

Here is some motivation for how I came across this problem. Analytic number theorists often study sums of the above type by studying the analytic properties of associated Diriclet series $$L(f,s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}.$$

Here, if $L(f,s)$ has a pole of order $r>0$ at $s=1$ and is well-behaved for $\text{Re}( s) >1$, then one can often show (using e.g. a Tauberian theorem such as Perron's formula) that we have an asymptotic formula $$S_f(x) \sim c_f x (\log x)^{r-1}.$$

More generally there is the Selberg-Delange method, here one works with complex powers $\zeta^z(s)$ of the Riemann zeta function. This method, when it works, will give an asymptotic formula of the shape $$S_f(x) \sim c_f x (\log x)^{z-1}.$$ In particular, it does not seem that one can obtain an asymptotic formula like the one I am seeking using this approach.

Note that one cannot use something like the prime number theorem to construct an example of the shape I am looking for, since $\pi(n)$ is not a multiplicative function!

Answer by Alexey Ustinov for Average orders of multiplicative functions

2012-12-05T00:49:28Z

You can find a many such functions in the book Sándor, J.; Mitrinović, D. S. & Crstici, B. Handbook of number theory. I Springer, 2006 (see for example ch. IV). Some asymptotic formulae have the form $\frac{x^2}{\log x}$ but you can easely transform them in the form $\frac{x}{\log x}$ using summation by parts.

Answer by Anthony Quas for Average orders of multiplicative functions

2012-12-07T01:50:17Z

I think an explicit multiplicative function that should do the job for you is this one: $f(2^n)=2^n/\big((n+1)\sqrt{\log(n+e)}\big)$, $f(3^n)=3^n/\big((n+1)\sqrt{\log(n+e)}\big)$, $f(p^n)=0$ for $n\ge 1$ and primes $p\ge 5$. The $+1$'s and $+e$'s are just there to make the formula make sense for $n=0$ also.

Here's an outline of why: It suffices to show that for all $\epsilon>0$, for all sufficiently large $N$ one has $$ \sum_{N < 2^k3^l\le (1+\epsilon) N} \frac{1}{(k+1)\sqrt{\log(k+e)}}\cdot \frac{1}{(l+1)\sqrt{\log(l+e)}}\sim c\epsilon /\log N $$ Taking logs, the summation range is essentially $\log N\le k\log 2+l\log 3\le \log N+\epsilon$. For large $N$, there are approximately $\epsilon\log N/(\log 2\log 3)$ pairs $(k,l)$ in the range. These are reasonably uniformly distributed in the band of $\mathbb R^2$, $x\log 2+y\log 3\in [\log N,\log N+\epsilon]$. Hence (with a bit of work making sure the ends of the integral don't dominate), the sum is close to $$ \frac{\epsilon}{\log 2}\int_{0}^{\log N/\log 3} \frac{1}{(y+1)\sqrt{\log (y+e)}}\frac{1}{(x+1)\sqrt{\log(x+e)}}\ dx, $$ where $y=(\log N-x\log 3)/\log 2$.

In the first half of the range, the integral is something like $$ \frac{\epsilon}{\log N\sqrt{\log\log N}} \int_0^{\log N/(2\log 3)} \frac{1}{(x+1)\sqrt{\log(x+e)}}\ dx $$ which is $\sim c\epsilon/\log N$. Similarly for the second half of the range.

If you prefer your multiplicative functions to be integer-valued, of course, you can just take the floor of everything.