Taylor's results on zeros of the linear combination of Gamma-completed Riemann Zeta functions

Question

Let $s,z$ be two complex variables, $\zeta(s)$ be the Riemann $\zeta$-function. Let

\begin{equation} \zeta_1(s)=\pi^{-s/2}\Gamma\left(\frac{s}{2}\right)\zeta(s), \end{equation}

be the Gamma-completed Riemann zeta function, and let

\begin{equation} f(s)=\zeta_1(s+1/2)-\zeta_1(s-1/2) \end{equation}

It satisfies the functional relation $f(s)=-f(1-s)$

Let \begin{equation} F(z)=f(iz+1/2) \end{equation}

Taylor proved in 1945 (On the Riemann Zeta function) that all the zeros of $F(z)$ are real.

We plotted $Re(F(x+I y))=0$ and $Im(F(x+I y))=0$ in Mathematica 9.0.

We found out that at $x=0,y=\pm 6.8246$, $F(x+I y)=0$. Thus $F(z)$ seemed to have non-real zeros.

Has anyone notice this problem?

EDIT: Here is a snapshot of Taylor's formulas.

Here is a little more

Answer 1

The paper simply says that all the non-real zeros of $F(s)$ lie on the line $\sigma = 1/2.$ The latest images posted by OP seem to show just that. There are non-real zeros at $\sigma = 1/2$ and two zeros located symmetrically about that line at $t = 0.$ These last have imaginary component $t = 0$ so the assertion of the theorem doesn't address them.

---

I went back and looked at this again, not to beat a dead horse but because I am interested in the sort of trompe l'oeil that occurred here.

The reasoning behind the question about the paper's claim was:

Let σ+it=s=1/2+iz=1/2+ix−y. Here σ, t, x, y ∈ R. Thus σ = 1/2 − y, t = x. All zeros of f(s) lie on σ = 1/2 are equivalent to all zeros of f(s) lie on y = 0. Thus all zeros of F(z) = F(x+iy) are real. –

It's an inviting idea and seems to simplify things but it mis-states the claim of the paper, which is that "all complex zeros of F(s)...lie on the line $\sigma = 1/2.$"

This took a while to see because one automatically focuses on the particulars of the transformation: is the claim of equivalence correct? Well, it is. The transformation gives a valid new picture of the same information but then we misinterpret the appearance of complex zeros.

The transformation takes the original points (x,y) of $G(s)$ and assigns new coordinates $(\hat{x},\hat{y}):$

$\hat{x} = y;$

$\hat{y}=1/2 - x;$

So when two complex zeros are found in the new system at $(0,\pm 6.82)$ these correspond to points in the original system at about $(7.32,0),(6.52,0)$ as a quick check of the graph confirms. The sequence of complex zeros on the line $x = 1/2$ in the original system are mapped to a sequence of real zeros $(x_i,0)$ in the new system.

The images posted by @mike confirm this.

Answer 2

Here is a 3D plot of daniel's function $F(s)=F(\sigma+i t)=F(1/2+x+i y)$. The dots that pops out of the figure are the zeros because I am plotting $-\log|F(s)|$ vs. $s$.

It seems to me that these two zeros that are not on the $\sigma=1/2$ line are the only exceptions. See two figures below.