1
$\begingroup$

I know next to nothing about analytic number theory, or the theory of the Riemann $\zeta$ function in particular, so the following might be too elementary to deserve more than derision; even so it seems it wouldn't hurt to ask where the following question has been considered and what the outcome was.

$\zeta : \mathbb{C} \rightarrow \mathbb{C}$ is a meromorphic function, and therefore, locally has a Laurent series expansion, and in any disc $D(c,r)$ centered at $c$ and of radius $r$ in $\mathbb{C}$ that avoids the poles, it has a Taylor series expansion, which can be truncated at $n$-th order to obtain a polynomial approximation, $p_{\zeta, D, n}$ of degree $n$.

Let's take such a disc in the critical strip. The question is: what are the zeros of $p_{\zeta, D, n}$ and how do they behave as $n \rightarrow \infty$? What happens to the asymptotic behavior as we move the disc around? Implicit in the question, of course, is curiosity about any light zeros of $p_{\zeta, D, n}$ might shed on zeros of $\zeta$.

$\endgroup$
2
$\begingroup$

Since $\zeta$ has a single pole, at $z=1$, the radius of convergence of the Taylor series at $c$ is $r=|c-1|$. Moreover, if $$\zeta(z)=\sum_0^\infty a_n(z-c)^n$$ is the Taylor expansion at $c$, then the limit in Hadamard's formula exists $\lim|a_n|^{1/n}=1/r$ (this is an easy exercise: if a function has a single pole on its circle of convergence and no other singularities in a slightly bigger disk then the limit exists). Now a general theorem of Jentzsch implies that the zeros of partial sums are: a) those which tend to the zeros of $\zeta$ in this disc, and b) additional zeros which are uniformly distributed near the circle $|z-c|=r$.

I don't think that this sheds any light on the zeros of $\zeta$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ While not directly related, another example where finite truncations shed no light on zeros of the zeta-function is related to a failed attempt at RH suggested by Turan (1948): the full series $\zeta(s) = \sum_{n \geq 1} 1/n^s$ has no zeros when ${\rm Re}(s) > 1$ and this suggests that the truncations $\zeta_N(s) = \sum_{n=1}^N 1/n^s$ should have "few" zeros there for large $N$. Turan showed that if $\zeta_N(s)$ for large $N$ has no zeros when ${\rm Re}(s) > 1$ then RH is a consequence, but Monach (1980) showed $\zeta_N(s)$ does have a zero with ${\rm Re}(s) > 1$ when $N \geq 31$. $\endgroup$ – KConrad May 5 '19 at 6:49

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.