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Hello, Is it possible to reconstruct the Riemann zeta function given the precise location of its infinitely many zeroes? Thanks

edited Jul 16 at 10:18

asked Jul 16 at 10:07

Math-player
1,391●1●3●13

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I don’t understand the question. The Riemann zeta function is a well-defined unique object, so it is possible to construct it even without knowing any additional information about its zeros. – Emil Jeřábek Jul 16 at 10:23

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The question is badly phrased. I infer that you are asking whether there is a unique meromorphic function with a prescribed set of zeroes. That is obviously not true, since $x$ and $x^2$ have the same zeroes. Of course also $x$ and $\frac{x}{1-x}$ have the same zeroes. I suggest the following question: "Does there exists unique meromorphic function with prescribed poles (with degrees) and zeroes (with multiplicities)?" (And I suspect the answer to be negative.) – robot Jul 16 at 10:35

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You should probably have a look at the Weierstrass factorization theorem. – Marc Palm Jul 16 at 13:04

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It seems clear that the intended question was "If $f$ is a meromorphic function whose zeroes coincide with those of the $\zeta$ function, must $f=\zeta$ up to multiplication by a unit?" The example of $x$ and $x^2$ seems quite off topic. – Steven Landsburg Jul 16 at 16:55

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If $f$ and $g$ are meromorphic functions on a simply connected domain $U$ with the same zeros and poles (having the same multiplicities and orders), then $f/g = e^h$ for some function $h$ analytic on $U$. Conversely, you can multiply any meromorphic function on $U$ by $e^h$ and it will have the same zeros and poles. – Robert Israel Jul 16 at 18:22

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juan · Accepted Answer · 2012-07-16 10:29:07Z UTC

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This is well known (Riemann could have writen it)

$$\zeta(s)=\frac{1}{2}\frac{\pi^{s/2}}{(s-1)\Gamma(1+s/2)}\prod_{\Im\rho > 0}\Bigl\{ \Bigl(1-\frac{s}{\rho}\Bigr)\Bigl(1-\frac{s}{\overline{\rho}}\Bigr)\Bigr\}$$

Here $\rho$ runs through the non trivial zeros with positive imaginary part.

It is this what you call reconstruct?

answered Jul 16 at 10:29

juan
1,730●7●12

Thank you. That might be ok for me. – Math-player Jul 16 at 10:30

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Hadamard's refinement of Weierstrass factorization is relevant here: if/when we know the order of growth of zeta, and its zeros, we know it up to an exponentiated polynomial (as in R. Israel's comment), and we have a sharp constraint on the degree of that polynomial. The functional equation of zeta and Stirling for the Gamma function (and Phragmen-Lindelof) give the requisite estimate to know that zeta is of order 1 in the growth sense. – paul garrett Jul 16 at 18:43

Is it possible to reconstruct $\zeta$-function knowing its zeroes?

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