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Does there exist a (meromorphic) elementary function $f(z)$ that is zero at all the positive integers $z = 1, 2, 3, \ldots$ and only at those points?

Edit: an elementary function can be written as a finite composition of constants, rational functions, exponentials and logarithms.

Obviously a function with those zeros can be constructed using the gamma function or a Weierstrass product, but the question is whether there is an elementary function.

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  • $\begingroup$ sin(pi*x) maybe? $\endgroup$
    – faridrb
    Commented May 8, 2010 at 18:45
  • $\begingroup$ oh sorry, misread the question. $\endgroup$
    – faridrb
    Commented May 8, 2010 at 18:48

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$1/\Gamma(1-z)$.

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  • $\begingroup$ The gamma function is not elementary. $\endgroup$
    – Fredrik Johansson
    Commented May 8, 2010 at 18:46
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    $\begingroup$ This is, however, the "most elementary" function with this property, for some reasonable notions of elementary (at least, assuming you want simple zeroes). For instance, any other function with the same zeroes has larger asymptotic growth; this is a version of the Hadamard Factorization Theorem. More precisely, if $f(z)$ has simple zeroes at the positive integers, then there is a polynomial $g(z)$ so that $f(z) = e^{g(z)}/\Gamma(1-z)$. So if you don't think the gamma function is elementary, the answer to your original question is "no". $\endgroup$
    – Dylan Thurston
    Commented May 9, 2010 at 1:50
  • $\begingroup$ Sorry, I just reread the question and saw "meromorphic", which changes things. $\endgroup$
    – Dylan Thurston
    Commented May 9, 2010 at 1:51
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I don´t completely understand what you mean by elementary, but you can look at http://en.wikipedia.org/wiki/Weierstrass_factorization_theorem. Sorry if the question had not to do with this.

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  • $\begingroup$ To spell this out further: all the OP has left unspecified in the question is the degrees of vanishing. So to rephrase: is there any sequence d_1,d_2,... of strictly positive integers s.t. the function Weierstrass gives one is elementary? The OP has declared that having all d_i = 1 doesn't pass muster. $\endgroup$
    – Allen Knutson
    Commented May 9, 2010 at 1:29
  • $\begingroup$ @Allen: We're also allowed to have poles. $\endgroup$
    – Dylan Thurston
    Commented May 9, 2010 at 14:42

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