An application of Mobius Inversion in a paper of Shintani - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:02:58Z http://mathoverflow.net/feeds/question/109561 http://www.creativecommons.org/licenses/by-nc/2.5/rdf 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> http://mathoverflow.net/questions/109561/an-application-of-mobius-inversion-in-a-paper-of-shintani/109568#109568 Answer by Eric Naslund for An application of Mobius Inversion in a paper of Shintani Eric Naslund 2012-10-13T23:37:06Z 2012-10-14T16:47:13Z <p>Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals </p> <p>$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$</p> <p>where $*$ denotes Dirichlet convolution. We may rewrite the given asymptotic as </p> <p>$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$</p> <p>noticing that this is of the form</p> <p>$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$</p> <p>Mobius inversion tells us that </p> <p>$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$ </p> <p>where $\alpha^{-1}$ is the multiplicative inverse of $\alpha$ with respect to Dirichlet convolution. Applying Mobius inversion to our sum, we have that </p> <p>$$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$</p> <p>which equals </p> <p>$$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$ </p> <p>$$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$ </p> <p>Notice that the Dirichlet inverse to the function $g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.</p>