Lower bound for character sums - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T18:34:48Z http://mathoverflow.net/feeds/question/83378 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83378/lower-bound-for-character-sums Lower bound for character sums M.B 2011-12-14T00:00:09Z 2011-12-14T13:15:50Z <p>Hello</p> <p>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 $$\left|\sum_{n\leq t}\left(\frac{n}{p}\right)\right|>\frac{\sqrt{p}}{2\pi}.$$</p> http://mathoverflow.net/questions/83378/lower-bound-for-character-sums/83422#83422 Answer by Frank Thorne for Lower bound for character sums Frank Thorne 2011-12-14T13:15:50Z 2011-12-14T13:15:50Z <p>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 href="http://arxiv.org/pdf/0911.5547v2" rel="nofollow">a paper of Leo Goldmakher</a>: We have</p> <p>$$\tau(p) = \sum_{n \leq p} \bigg( \frac{n}{p} \bigg) e^{2 \pi i n/p}$$ and that is a Gauss sum with absolute value $\sqrt{p}$... now use partial summation.</p> <p>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.</p>