Is it true that all zeros of the Riemann Zeta Function are of order 1?

Let $h(z) = \frac{\zeta'(z)}{\zeta(z)}\frac{x^z}{z}$, where $x$ is a positive real number ($x > 1$, probably?) , and $\zeta$ is the Riemann zeta function. I'm computing the residues of $h$.

The only place I'm having problems computing the residue of $h$ are at the zeros of $\zeta$. I just need to know the order of the zeros of $\zeta$. It seems like for the trivial zeros (the negative even integers) the orders are 1.

But I don't know what to do about the non-trivial zeros. I'm probably confused about something. When I look in the derivation of the von Mangoldt formula shown here, it lists the residues of $h$, and from their computation, it seems that the orders of the zero at the non-trivial zeros are also 1.

I was under the impression that the order of the the zeros of $\zeta$ were not all known.

What am I missing?

EDIT: I found this. On page 2, it says that if $\rho$ is a zero of $\zeta$ contributes $\frac{x^\rho}{\rho}$ to the residue of $h$, counted with multiplicity. (I'm off by a negative sign because of the way I defined $h$). Is there a special way to compute the residue without knowing the order of $\rho$?

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    $\begingroup$ See mathoverflow.net/questions/59770/… . In summary, it is widely believed, but not proved, that the zeroes are simple. $\endgroup$ – David E Speyer Jul 18 '11 at 20:34
  • $\begingroup$ Then I don't understand how the residue of $h$ is computed at the zeros... $\endgroup$ – Braindead Jul 18 '11 at 20:40
  • $\begingroup$ If I remember correctly, It is still not known whether non-trivial zeros are all simple. But I think it's believed to be true, and all zeros computed so far were indeed simple. $\endgroup$ – Joel Cohen Jul 18 '11 at 20:42
  • $\begingroup$ So far as I remember, all the zeroes being simple can be shown to be equivalent to that certain hypothesis... $\endgroup$ – J. M. isn't a mathematician Jul 20 '11 at 12:17

Your Google Books link doesn't work for me; here's one that does.

I think there's nothing mysterious going on here, and no assumption about the multiplicities of the zeros of $\zeta$ is being made. The sum is written as a sum over the zeros, but it's implied that these contribute according to their multiplicity. The residue isn't computed in any special way (unless you count forming the residue of the logarithmic derivative as a special way); the residue from $\zeta'/\zeta$ is the multiplicity, and this gets multiplied by $x^\rho/\rho$.

  • $\begingroup$ Oh I see. I was just reading it incorrectly. Thank you! $\endgroup$ – Braindead Jul 18 '11 at 21:15

You might be interested in this recent arXiv paper: On simple zeros of the Riemann zeta-function

We show that at least 19/27 of the zeros of the Riemann zeta-function are simple, assuming the Riemann Hypothesis (RH). This was previously established by Conrey, Ghosh and Gonek [Proc. London Math. Soc. 76 (1998), 497--522] under the additional assumption of the Generalised Lindelöf Hypothesis (GLH). We are able to remove this hypothesis by careful use of the generalised Vaughan identity.


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