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.

enter image description here enter image description here

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. enter image description here

Here is a little more enter image description here

  • 1
    $\begingroup$ Here's an approximation for the root: $$\begin{multline}z = 1.3179490483476020312550822896459296949554671129638*10^{-63} + \\ 6.8246178956306702512881753455550038029953841561527i\end{multline}$$ and $F(\Im(z)i) \approx 10^{-49}$. (ed ajf) $\endgroup$ – Alexander Vlasev Nov 21 '14 at 9:02
  • 1
    $\begingroup$ This paper says that $\zeta_1(s) = \pi^{s/2}\Gamma(s/2)\zeta(s)$ and not as you have $\pi^{-s/2}$. $\endgroup$ – Alexander Vlasev Nov 21 '14 at 9:52
  • 1
    $\begingroup$ An MO question has it as per the OP. mathoverflow.net/questions/7656/…. $\endgroup$ – daniel Nov 21 '14 at 11:31
  • 1
    $\begingroup$ @daniel. Thanks for the suggestion. I just deleted extra plots associated with $\pi^{s/2}$ and $G(z)$. $\endgroup$ – mike Nov 21 '14 at 12:12
  • 1
    $\begingroup$ @AleksVlasev I double checked the formula in the paper you linked. It reads $\zeta_1(s) = \Gamma(s/2)\zeta(s)/\pi^{s/2}$. So this formula is identical to what I had in the OP. $\endgroup$ – mike Nov 21 '14 at 12:27

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.

  • $\begingroup$ please see the 3D plot that I made using your $F(s)$. $\endgroup$ – mike Nov 22 '14 at 5:55
  • 1
    $\begingroup$ @mike: I have gone back and forth about this but your most recent plots of the function as given in the paper seem to confirm the assertion in the paper--that all the complex zeros lie on the line 1/2. The two exceptions seem to be real numbers, (a,0),(b,0). $\endgroup$ – daniel Nov 22 '14 at 9:34
  • 1
    $\begingroup$ To be nit-picky, all numbers in $\mathbb{C}$ are complex. I'm sure the language "non-real zeros" would have avoided the problem. $\endgroup$ – Alexander Vlasev Nov 22 '14 at 9:39
  • 2
    $\begingroup$ @AleksVlasev: Nit-picky but correct. I will edit to reflect this, thanks. $\endgroup$ – daniel Nov 22 '14 at 9:43

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. $s$.

enter image description here

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.

enter image description here enter image description here


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.