2
$\begingroup$

Sometimes we meet with situation when we don't know asymptotics of a given but we know its average behaviour on $[1,n]$. For example, we can easily get that

$\sum_{i=1}^n \tau(n) = n \log (n) (1 + o(1))$

but we have only upper bound on $\tau(n)$ itself.

So my question is the next: is there some research (maybe for another NT fucntion) where authors found out minimal number of consecutive arguments (or its growth), such that average value of a given function on such arguments has clear asymptotics, i. e. a function $g(n)$ such that:

$\frac{1}{g(n)}\sum_{i=n+1}^{n+g(n)} \tau (i) = S(n) (1+o(1))$

for some function $S(n)$, which is a combination of polynomials, logarithms and exponents?

Edit: Thanks, Fedor and Gerry, I just try to summarize the best unconditional results on this moment. Really, this problem is closely related with a problem of error tern estimating for an average. From Huxley work "Exponential sums and lattice points III" (2003) it follows that $\Delta (x) << x^{131/416+\epsilon}$, where

$ \Delta (x) = \sum_{n\leq x} \tau (n) - x(\log x + 2\gamma -1)$,

so one can easily get using a method desribed by Fedor, that

$\frac{1}{h(x)}\sum_{x<n\leq x} \tau (n) = \log (x)(1 + o(1))$

for $h(x) = x^{\theta}$ with $\theta > 131/146$.

Also Shiu in his work "A Brun-Titchmarsh theorem for multiplicative functions" (1980) proved that

$\sum_{x<n\leq x + h(x)} \tau (n) << h(x) \log (x)$

for $x^\epsilon \leq h(x) \leq x$ with no additional restictions on positive $\epsilon$.

$\endgroup$
  • 3
    $\begingroup$ If you have reasobable bounds for the remainder in the asymptotic formula for the sum $T(n)$ (for $i$ from 1 to $n$), you may subtract $T(n+g(n))-T(n)$ and still get an asymptotics, if $g(n)$ is not too small. $\endgroup$ – Fedor Petrov Jun 6 '17 at 19:26
  • $\begingroup$ $O(1)$ means any bounded function. So it makes little sense to write $1+O(1)$, because it is just another way of writing $O(1)$. You probably meant $1+o(1)$ throghout. ($o(1)$ means any function tending to zero.) $\endgroup$ – GH from MO Jun 7 '17 at 15:46
6
$\begingroup$

It is known that $\sum_{x<n\le x+h(x)}\tau(n)$ is asymptotic to $h(x)\log x$ for $h(x)=x^{\theta}$ with $\theta>131/416$, and it is conjectured that the asymptotic holds for $h(x)\gg x^{\epsilon}$. See, e.g., Danilo Bazzanella, On the divisor function in short intervals, Arch Math 97 (2011) 453-458.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.