User mixedmath - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:42:44Z http://mathoverflow.net/feeds/user/14508 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125250/recovering-sum-n-leq-x-an-from-sum-n-leq-x-ane-n-x Recovering $\sum_{n \leq x} a(n)$ from $\sum_{n \leq x} a(n)e^{-n/x}$ mixedmath 2013-03-22T08:35:35Z 2013-03-25T18:11:15Z <p>In the theory of automorphic forms and multiple Dirichlet series, we often take <a href="http://en.wikipedia.org/wiki/Mellin_transform" rel="nofollow">inverse Mellin transforms</a> of Dirichlet series to come up with Tauberian theorems, like the <a href="http://en.wikipedia.org/wiki/Wiener%E2%80%93Ikehara_theorem" rel="nofollow">Ikehara Tauberian method</a>. In particular, from</p> <p>$\displaystyle \sum_n \frac{a(n)}{n^s}$, we might expect to use Mellin to get asymptotics for $\displaystyle \sum_{n \leq x} a(n)$.</p> <p>Sometimes, the series doesn't behave well enough to integrate. A common way around that seems to be to multiply by the gamma function to get better decay. But this has the side effect of 'smoothing' the sum to get asymptotics for sums that look like $\displaystyle \sum_{n \leq x} a(n) e^{-n/x}$.</p> <p>For a while, I just accepted that sometimes we need to accept smoothed sums. But I began to wonder how hard it would be to recover the regular, un-smoothed sum. So I'm prompted to ask:</p> <blockquote> <p>Suppose we can calculate $F(x) := \displaystyle \sum_{n \leq x} a(n) e^{-n/x}$ for any given $x > 1$. For a fixed $N$, can one recover $\displaystyle \sigma_N := \sum_{n \leq N} a(n)$ from $F(x)$? In particular, it is ever possible to recover $\sigma_N$ by calculating $F(x_i)$ for a suitably chosen, but <em>finite</em> set of $x_i \in \mathbb{R}$?</p> </blockquote> http://mathoverflow.net/questions/109561/an-application-of-mobius-inversion-in-a-paper-of-shintani An application of Mobius Inversion in a paper of Shintani mixedmath 2012-10-13T21:56:20Z 2012-10-14T16:47:13Z <p>I've been reading about Shintani zeta functions and in particular with respect to finding the density of cubic discriminants as in the theorem of Davenport-Heilbronn. In Shintani's paper "On zeta-functions associated with the vector space of quadratic forms" [Tokyo Univ. J. Fac. Sci Sect. 1A Math 1975], in the proof of Theorem 4, Shintani writes</p> <blockquote> <p>Hence, we have <em>(by earlier results in this paper and in a previous paper)</em>: $$\sum_{nk^4 \leq x} h_r(n) = 2^{-1}\zeta(2)\zeta(4)x + O(x^{2/3 + \epsilon}) \qquad (x \to +\infty, \forall \epsilon > 0)$$ An application of the Mobius inversion formula now yields that<br> $$\sum_{n \leq x} h_r(n) = 2^{-1}\zeta(2)x + O(x^{2/3 + \epsilon})$$</p> </blockquote> <p>When I think of <a href="http://en.wikipedia.org/wiki/M%C3%B6bius_inversion_formula" rel="nofollow">Mobius Inversion</a>, I think of two things: \begin{align*} g(n) &amp;= \sum_{d \mid n} f(n/d) \iff f(n) = \sum_{d \mid n} \mu(n)g(n/d) \quad \text{or}\\ g(x) &amp;= \sum_{n \leq x}f(n/x) \iff f(x) = \sum_{n \leq x} \mu(n)g(n/x) \end{align*}</p> <p>But I don't see how I can use these here. Unfortunately, I also know that there are many things that might be called Mobius Inversion. This is one of those steps that taunts me. Qualitatively, we remove the fourth-power condition and end up losing a factor of $\zeta(4)$, and that feels very reasonable.</p> <p>Further, in playing with it for a while, I re-stumbled upon the fact that $\displaystyle \dfrac{1}{\zeta(s)} = \sum_n \dfrac{\mu(n)}{n^s}$ (easy, but which I did not originally remember) and thus that $\displaystyle \dfrac{1}{\zeta(4)} = \sum \dfrac{\mu(n)}{n^4}$.</p> <p>To make my question explicit - how can we arrive at the second equation from the first?</p>