9
$\begingroup$

Is there a "naturally occurring" function $f$ which is meromorphic in the complex plane such that the poles of $f$ on the real axis are precisely at the primes? I say "naturally occurring" since we can always cook up a function with the right poles of the right order, but I would hope that it comes from number theory.

Alternatively, is there a naturally arising meromorphic function whose poles appear at all primes and powers of primes?

$\endgroup$
  • $\begingroup$ What you mean is «meromorphic function with poles at the primes», not «meromorphic at the primes». $\endgroup$ – Mariano Suárez-Álvarez Feb 14 '13 at 17:14
  • $\begingroup$ no there isnt... $\endgroup$ – user58512 Feb 14 '13 at 17:38
  • $\begingroup$ Fixed the title $\endgroup$ – mck Feb 14 '13 at 17:53
  • $\begingroup$ Can you construct a function with poles at the integers? $\endgroup$ – Thom Tyrrell Feb 14 '13 at 18:25
  • 2
    $\begingroup$ Yes, the Weierstrass $\mathfrak{p}$ function for the integer lattice will have poles at exactly the integers. The poles will be order 2, but at least for now I won't put any restrictions on that. $\endgroup$ – mck Feb 14 '13 at 18:29
5
$\begingroup$

Let $P(s) = \sum_{p} p^{-s}$ be the Prime zeta function.

Claim: the series expansion $$f(z) = 2 \sum_{n=1}^\infty P(2n)z^{2n-1}$$ defines a meromorphic function whose only poles are simple poles of residue $1$ at the primes and their negatives.

Let $F(z) = \prod_ p (1-z^2/p^2)^{-1}$. The product converges uniformly on compact subsets of $\mathbf C - P$, where $P$ is the set of all primes and their negatives. Therefore it is holomorphic there. It has simple poles at the points of $P$. (Note the special value $F(1) = \zeta(2) = \pi^2/6$).

Using the series expansion for $\log(1-x)$, we have the series expansion for $\log F$, $$\log F(z) = \sum_{n=1}^\infty \frac{P(2n)}{n}z^{2n}.$$

Taking the derivative we get $f(z)$. Since the poles of $F$ are simple and $F$ is never $0$ on $\mathbf C - P$ (on account of the convergence of the product), $f = F'/F$ has simple poles with residue $1$ at the points of $P$, and no other poles.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ And this function is "naturally occuring"? $\endgroup$ – mrf Sep 27 '13 at 6:23
  • $\begingroup$ @mrf Up to a unit of the ring of holomorphic functions on $\mathbf C$, it is unique! :) $\endgroup$ – Bruno Joyal Sep 27 '13 at 6:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.