most recent 30 from http://mathoverflow.net 2013-05-25T05:39:29Z http://mathoverflow.net/feeds/question/76030 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>

http://mathoverflow.net/questions/76030/estimating-a-sum-of-gauss-sums/76061#76061 Felipe Voloch 2011-09-21T13:39:12Z 2011-09-21T13:39:12Z

<p>There should be a bunch of cancellation. Here is an idea. You need to relate your sums $\sum_a e(a/p)/ord_p a$ to the sums $\sum_{ord_p a | m} e(a/p)/m$. Now, if $mr = p-1$, </p> <p>$\sum_{ord_p a | m} e(a/p) = (1/r)\sum_{n=1}^{p-1} e(n^r/p) = O(p^{1/2})$</p> <p>by the Weil bound. This will deal with the elements of large order, I believe. There is work to do, but this should get you going.</p>

http://mathoverflow.net/questions/76030/estimating-a-sum-of-gauss-sums/76079#76079 Eric Naslund 2011-09-21T20:00:51Z 2011-09-21T20:57:36Z

<p>This is a comment rather than an answer, but it is too long. Let $g$ be some generator of the multiplicative group. Then </p> <p>$$\sum_{a=1}^{p-1}e\left(\frac{a}{p}\right)\frac{p-1}{ord_{p}a}=\sum_{k=1}^{p-1}e\left(\frac{g^{k}}{p}\right)\gcd,\left(p-1,k\right).$$ </p> <p>Rearranging yields $$\sum_{d|p-1} \phi(d) \sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right) $$ so that the entire sum is $$\sum_{p\leq x}\frac{1}{p-1}\sum_{d|p-1}\phi(d)\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right).$$ </p> <p>My hope in posting this is that there are existing bounds on sums of the form $\sum_{k\leq\frac{p-1}{d}}e\left(\frac{g^{dk}}{p}\right)$. It might be strange to deal with, as it is a sum over elements chosen for their multiplicative properties. Essentially, we would need a theorem regarding how these multiplicative elements are distributed among the residue classes, and that it cannot be "too far from uniform".</p>