Question about zeros of partial sums of Taylor series of Riemann xi-function

Question

Consider the entire function $ \psi(z) $ whose Taylor series about 1/2 "looking like" the Riemann xi-function $ \xi(z)=\sum_{n=0}^{\infty}a_{2n}(z-1/2)^{2n} $, but with different coefficients $a\psi_{2n}$. It is intuitively clear that one can choose these coefficients so that all partial sums of the Taylor series of this function would have zeros only with the real part of 1/2 (for example, one can choose $a\psi_{2n} \sim exp(-n^2)$). Is it also intuitively clear that all the zeros of the function itself $ \psi(z) $ will also lie on the straight line $x=1/2$? Or it is not obvious? I would like to clarify this question.

Answer 1

Let $f_N(z) = \sum_{n=0}^N c_n z^n$ with $c_n $ real, if its $N$ zeros $a_{N,1},\ldots,a_{N,N}$ are real and simple there is $r_{N+1} \le 1$ such that for any $c_{N+1}\in [- r_{N+1},r_{N+1}]$, $f_{N+1}(z) = c_{N+1} z^{N+1}+f_N(z)$ has sign changes and $f_{N+1}'(z)$ stays of constant sign near $a_{N,j}$, so that $a_{N+1,j} \approx a_{N,j}$ is real and simple.

The $N+1$-th zero of $f_{N+1}$ is simple and it can't be complex since complex zeros come by pair.

Letting $c_0 = 1$ and $c_{N+1} = \min(r_{N+1}, e^{-N^2})$ yields an entire function $f(z) = \sum_{n=0}^\infty c_n z^n$ whose zeros are all real and whose every partial sums have only real zeros.

It is not obvious to check if $(e^{-n^2})$ decreases fast enough for $\sum_{n=0}^\infty e^{-n^2} z^n$ to be in that class.