6
$\begingroup$

Hello

Let $p$ be a prime number. According to Davenport (Multiplicative Number Theory, page 137) Schur proved (Indeed he proved much more, but let consider the simplest case) $$ \max_{t}\left|\sum_{n\leq t}\left(\frac{n}{p}\right)\right|>\frac{\sqrt{p}}{2\pi}. $$ What can we say about $t$ which we obtain the maximum? In other words, can we find $t\gg p^{\frac{1}{2}+\varepsilon}$ such that $$ \left|\sum_{n\leq t}\left(\frac{n}{p}\right)\right|>\frac{\sqrt{p}}{2\pi}. $$

$\endgroup$
4
  • $\begingroup$ Are you sure you want to write $\gg$ (large $t$) rather than $\ll$? $\endgroup$
    – fedja
    Dec 14, 2011 at 0:37
  • 9
    $\begingroup$ Burgess proved the character sum is $O(p^{3/16} t^{1/2} \log p)$, so if it's $\gg p^{1/2}$ then $t \gg p^{5/8 - \epsilon}$. $\endgroup$ Dec 14, 2011 at 0:44
  • 1
    $\begingroup$ Fedja@: Yes. I am sure I want $\gg$. $\endgroup$
    – M.B
    Dec 14, 2011 at 2:23
  • $\begingroup$ Even if it were possible to have $t \ll p^{1/2 + \epsilon}$ then $p-t$ would work too and satisfy $t \gg p^{1/2 + \epsilon}$ by a wide margin... $\endgroup$ Dec 14, 2011 at 3:44

1 Answer 1

2
$\begingroup$

The character sum you ask about is $\gg \sqrt{p}$ for at least one value of $n$. I learned the following slick proof from a paper of Leo Goldmakher: We have

$$\tau(p) = \sum_{n \leq p} \bigg( \frac{n}{p} \bigg) e^{2 \pi i n/p}$$ and that is a Gauss sum with absolute value $\sqrt{p}$... now use partial summation.

In addition, there is an infinite family of characters for which the lower bound may be multiplied by an additional fractional power of $\log \log p$. A result like this was first proved by Paley. Please see the paper I linked to for proofs and references to earlier work.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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