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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

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ An MO question has it as per the OP. mathoverflow.net/questions/7656/…. $\endgroup$ – daniel Nov 21 '14 at 11:31
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    $\begingroup$ @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
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    $\begingroup$ @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
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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.

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  • $\begingroup$ please see the 3D plot that I made using your $F(s)$. $\endgroup$ – mike Nov 22 '14 at 5:55
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @AleksVlasev: Nit-picky but correct. I will edit to reflect this, thanks. $\endgroup$ – daniel Nov 22 '14 at 9:43
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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

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