# Zeros of polynomial approximations of the Riemann $\zeta$ function

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 $$\zeta$$.

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 $$\zeta$$.
• 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 $N \geq 31$. – KConrad May 5 '19 at 6:49