4
$\begingroup$

The inverse of the Weierstrass transform 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:

  1. Along the critical line

    $\zeta\left(\frac{1}{2}+ix\right)=\sum \limits^{\infty}_{m=0}\left[\sum \limits^{\infty}_{n=1}\dfrac{1}{n^{1/2+\ln (n)/4}}\Bigl(\dfrac{-i\ln n}{2}\Bigr)^{m}\right]\dfrac{H_{m}\left(x\right)}{m!}\text{.}$

  2. Hasse's representation yields

    $\zeta\left(z\right)\left(1-2^{1-z}\right)=\sum \limits^{\infty}_{m=0}\left[\sum \limits^{\infty}_{n=0} \sum \limits^{n}_{k=0}\dfrac{\left(-1\right)^{k} \ln^m\left(k+1\right) }{2^{n+1}}\binom{n}{k}e^{\ln^2(k+1) /4}\right]\dfrac{\left(-1\right)^{m}H_{m}\left(z\right)}{2^{m}m!}\text{.}$

  3. The Laurent series yields

    $\zeta\left(z+1\right)-\frac{1}{z}=\sum \limits^{\infty}_{n=0}\dfrac{1}{2^nn!}\left[\sum \limits^{\infty}_{k=0}\dfrac{\gamma_{n+2k}}{2^{2k+1}k!}\right]H_{n}\left(z\right)$

    where the $\gamma_{n}$ are the Stieltjes constants.

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?

Also, is there some better way to find a Hermite expansion of $\zeta$?

$\endgroup$
  • 3
    $\begingroup$ What exactly is your question? $\endgroup$ – Qiaochu Yuan Feb 25 '10 at 22:30
  • $\begingroup$ I think he means to ask if these are actually correct, and what other similar expansions (in terms of Hermite polynomials) are there which are connected to $\zeta$. $\endgroup$ – Jacques Carette Feb 26 '10 at 0:35
  • $\begingroup$ Newly edited for clarification. $\endgroup$ – Craig Calcaterra Feb 26 '10 at 6:06
  • 1
    $\begingroup$ What don't you write the formal manipulations that lead to the 1st formula ? $\endgroup$ – reuns Oct 25 '17 at 2:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.