most recent 30 from http://mathoverflow.net 2013-05-19T20:30:22Z http://mathoverflow.net/feeds/user/17977 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76030/estimating-a-sum-of-gauss-sums Chris 2011-09-21T08:09:06Z 2011-09-21T20:57:36Z

<p>Hey guys, I'm concerned with bounding the following sum of gauss sums from above $$\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)},$$ where $p$ runs through the primes $\leq x$, $\chi$ runs through the multiplicative characters modulo $p$ and $e\left(\frac{a}{p}\right)=\exp\left(\frac{2\pi ia}{p}\right)$. By using orthogonality relations of characters one gets $$\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}=(p-1)\sum_{a=1}^{p-1}~{e\left(\frac{a}{p}\right)\frac{p-1}{ord_pa}},$$ where $ord_pa$ denotes the multiplicative order of $a$ modulo $p$. The right side can be bounded trivially by $$(p-1)\sum_{a=1}^{p-1}~{\frac{p-1}{ord_pa}}=(p-1)^2\sum_{d\mid p-1}{\frac{\varphi(d)}{d}},$$ $\varphi(d)$ denoting Euler's totient function. Using $\varphi(n)\leq n$ one gets the estimate $$\left|\sum_{p\leq x}~{\frac{1}{(p-1)^2}}\sum_{m=1}^{p-1}~\sum_{\chi~(p)}~\sum_{a=1}^{p-1}{~\chi^m(a)e\left(\frac{a}{p}\right)}\right|\leq\sum_{p\leq x}{\tau(p-1)},$$ where $\tau(n)$ is the number of divisors of $n$. The latter sum can be shown to be asymptotically equivalent to a positive constant times $x$. I would like to know if there is a way to show that the sum is $o(x)$.</p>