Has the Weierstass transform been used to give Hermite series representations of the Riemann zeta function? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T23:01:10Z http://mathoverflow.net/feeds/question/16456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16456/has-the-weierstass-transform-been-used-to-give-hermite-series-representations-of Has the Weierstass transform been used to give Hermite series representations of the Riemann zeta function? Craig Calcaterra 2010-02-25T21:59:59Z 2010-03-02T04:59:43Z <p>The inverse of the <a href="http://en.wikipedia.org/wiki/Weierstrass_transform" rel="nofollow" title="Weierstrass transform">Weierstrass transform</a> expands a function as a series of Hermite polynomials $H_{n}$. There are several ways to invert the Weierstrass transform which led me to the following expansions:</p> <ol> <li><p>Along the critical line </p> <blockquote> <blockquote> <p>$\zeta\left(\tfrac{1}{2}+ix\right)=\overset{\infty}{\underset{m=0}{{\textstyle \sum}}}\left[\overset{\infty}{\underset{n=1}{{\textstyle \sum}}}\dfrac{1}{n^{1/2+\ln n/4}}\left(\frac{-i\ln n}{2}\right)^{m}\right]\frac{H_{m}\left(x\right)}{m!}\text{.}$ </p> </blockquote> </blockquote></li> <li><p>Hasse's representation yields </p> <blockquote> <blockquote> <p>$\zeta\left(z\right)\left(1-2^{1-z}\right)=\overset{\infty}{\underset{m=0}{{\textstyle \sum}}}\left[\overset{\infty}{\underset{n=0}{{\textstyle \sum}}}\overset{n}{\underset{k=0}{{\textstyle \sum}}}\frac{\left(-1\right)^{k}\left(\ln\left(k+1\right)\right)^{m}}{2^{n+1}}\binom{n}{k}e^{\left[\ln\left(k+1\right)\right]^{2}/4}\right]\frac{\left(-1\right)^{m}H_{m}\left(z\right)}{2^{m}m!}\text{.}$ </p> </blockquote> </blockquote></li> <li><p>The Laurent series yields </p> <blockquote> <blockquote> <p>$\zeta\left(z+1\right)-\frac{1}{z}=\overset{\infty}{\underset{n=0}{{\textstyle \sum}}}\tfrac{1}{n!2^{n}}\left[\underset{k=0}{\overset{\infty}{{\textstyle \sum}}}\frac{\gamma_{n+2k}}{2^{2k+1}k!}\right]H_{n}\left(z\right)$ </p> <blockquote> <p>where the $\gamma_{n}$ are the Stieltjes constants.</p> </blockquote> </blockquote> </blockquote></li> </ol> <p>Et cetera. The calculations were all formal, and I've mostly ignored convergence, assuming this territory is well-trod. Has this been explored, or is there some flaw in this approach?</p> <p>Also, is there some better way to find a Hermite expansion of $\zeta$?</p>