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Let $a_{m}$ be the imaginary part of the nontrivial roots of the Riemann zeta function $\zeta(s)$. Suppose we have their generating function $u(x)=\sum_{m=1}^{\infty} a_{m}x^{m}=14.134725\ldots{}x^{1}+21.022040\ldots{}x^{2}+\ldots$ What information about the nontrivial roots of $\zeta(s)$ are encoded by the roots and poles of $u(x)$?

Besides the theorem of Hadamard mentioned in twf:190 and standard generatingfunctionological techniques, what can this function tell us?

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  • $\begingroup$ I changed the code used for the links. The changes worked in the preview, but the links only work for me while the math is being rendered. I'll start a thread on meta. $\endgroup$ Apr 23, 2010 at 17:44

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I would be willing to wager vast sums of money that this function has the unit circle as a boundary of essential singularities. As $x\to 1$ it diverges like $u(x) \sim c (1-x)^{-2}\log{(1-x)}$.

Edit: Hi Ben, let me say a bit more. Write $N(t)$ for the number of zeros of the Riemann zeta function in the strip $0 < Re(s) < 1$ with imaginary part between $0$ and $t$. According to Riemann we have $N(t)=\frac{t}{2\pi}\log{(\frac{t}{2\pi e})}+S(t)$ where $S(t)$ satisfies $S=O(\log{t})$. This function $S(t)$ captures the fine variation in the distribution of the zeros of the zeta function. It is quite random: Selberg proved unconditionally, in a triumph of analytic number theory, that $(\log{\log{t}})^{-\frac{1}{2}}S(t)$ has the distribution of a Gaussian random variable with mean zero and variance one!

Now, what if you try to invert this? Write $\gamma_n$ for the imaginary part of the $n$th zero, ordered by height. After inverting the above functions you'll get something like $\gamma_n=f(n)+r(n)$ where $f$ is essentially the valute of $t$ such that of $n=\frac{t}{2\pi}\log{(\frac{t}{2\pi e})}$, and $r$ is some chaotically varying error term. Splitting $u$ as $u_f(x) + u_r(x)$ accordingly, I am sure that $u_f(x)$ has a nice analytic continuation, but for all intents and purposes $u_r(x)$ looks like a power series with "random" coeffecients. Such things generally do not have good analytic continuations

That being said, this is not a proof of my claim, and I would be delighted to see one.

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  • $\begingroup$ So, why would I be a sucker to take the other end of that bet? $\endgroup$
    – Ben Webster
    Apr 24, 2010 at 20:16
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In keeping with David Hansen's wager, I'd also be willing to bet that this is related to Hugh Montgomery's pair correlation conjecture http://en.wikipedia.org/wiki/Montgomery%27s_pair_correlation_conjecture

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  • $\begingroup$ Unless one has to look at the mean-square of $u(x)$, I am not sure exactly how the distribution of pairs of zeros comes into play. $\endgroup$ Aug 2, 2010 at 22:35
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My answer not exactly refers to the generating function you state above, but rather a kind of Dirichlet series which is built up from the zeta function. It is called the acid zeta function. I seem to remember to have seen even a monograph on this topic but I could not come up with a title by a quick google search.

The only thing I found is this recent preprint of Jining Gao: http://arxiv.org/abs/1003.3392 Maybe this is of interest for you too.

EDIT: I recently came across the following book. http://books.google.de/books?id=op0RejW-pvoC&printsec=frontcover&dq=Zeta+Functions+over+Zeros+of+Zeta+Functions&source=bl&ots=ytkjMifsmd&sig=kXXU9CFMvU5Cvtkc4vZr_9610v0&hl=de&ei=e11MTPz9KYuWOKa2zJUD&sa=X&oi=book_result&ct=result&resnum=2&ved=0CCQQ6AEwAQ#v=onepage&q&f=false Maybe that ist of interest

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