Timeline for Does there exist a Weierstrass/Hadamard factorization for $\chi(s)-1$?
Current License: CC BY-SA 4.0
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| S Apr 13, 2019 at 8:30 | history | suggested | Glorfindel | CC BY-SA 4.0 |
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference); for more info, see https://meta.mathoverflow.net/a/4058/70594
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| Apr 13, 2019 at 7:21 | review | Suggested edits | |||
| S Apr 13, 2019 at 8:30 | |||||
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 2, 2012 at 20:01 | comment | added | Agno | @Juan Exactly. This is what I found as well. What I am now looking for, is whether $\chi(s)-1$ can be factored into an infinite product of its zeros (i.c. the Gram points). If such a product exists, I want to remove those Gram points for which $\zeta(1/2+i g_n,2)<0$ and to see if that produces $f(t)$ (since this still includes the $\rho$'s at $y=2$). I searched the web on Weierstrass products for $\chi(s)$, but could not find any. I did experiment with the simpler Weierstrass factors of $sin(\pi z)$ and randomly removed the (integer) zeros. The sine wave gets chaotic pretty quickly this way. | |
| Sep 2, 2012 at 17:42 | comment | added | juan | @Agno The real points $t$ where $\chi(t)=1$ are the Gram points. At a Gram point $\zeta(1/2+i g_n,2)=\zeta(1/2-i g_n,2)$ your function $f(t)$ would be $0$ at $g_n$ unless $\zeta(1/2+i g_n,2)<0$. This is equivalent to $\zeta(1/2+i g_n)<1$. This happen for $g_3 = 31.71$, $g_8=48.71$, $g_{12}=60.35$, $g_{18}=76.17$, $g_{23}=88.38$, $g_{26}=95.41$ $\dots$ | |
| Sep 2, 2012 at 17:09 | comment | added | Agno | @Juan. I now suddenly see that I have put you on the wrong foot. $f(t)$ is a difference between the roots of the Hurwitz zetas $(a=2)$ and not of the regular ones $(a=1)$. The correct formula is listed in red in the picture, but in the text I forgot to put the roots in the formula. Have corrected it now. | |
| Sep 2, 2012 at 17:08 | history | edited | Agno | CC BY-SA 3.0 |
Add the square roots in f(t).
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| Sep 2, 2012 at 16:37 | comment | added | juan | @Agno Observe that $\zeta(s)=\chi(s)\zeta(1-s)$ gives $\chi(\frac12+it)=\zeta(\frac12+it)/\zeta(\frac12-it)$. Then your function $f(t)=|\sqrt{\zeta(\frac12-it)}-\sqrt{\zeta(\frac12+it)}|=|\sqrt{\zeta(\frac12-it)}|\cdot|1-\sqrt{\chi(\frac12+it)}|$. (If you take adequate roots). I think that the graph is not correct $f(t)$ vanish at the Gram point t=31.717979954764 or you are taking differents roots of $\zeta(\frac12+it)=\zeta(\frac12-it)=$ that at this point is a negative real number. At this point $\chi(t)=1$. | |
| Sep 2, 2012 at 12:29 | comment | added | Agno | @Juan. You are right and I have corrected it in my question. Note that I do not necessarily need the modulus, since I am predominantly interested in the overlapping zeros, however that still doesn't make $\chi(s)-1$ entire of course. Do you think it is possible to find such an infinite product (similar to the Hadamard product for $\zeta(s)$)? | |
| Sep 2, 2012 at 12:20 | history | edited | Agno | CC BY-SA 3.0 |
Corrected incorrect use of 's' and 't'. Changed 'entire' into 'meromorphic' per Juan's comment.
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| Sep 2, 2012 at 7:48 | comment | added | juan | $\chi(s)$ is a meromorphic function. I do not understand the assertion that $|\chi(s)-1|$ is entire. The modulus of a non constant meromorphic function is not entire. If you think only on the critical line $|\chi(0.5+it)-1|= 2 |\sin\vartheta(t)|$ has real zeros, and the modulus is not differentiable as a function of a real variable. | |
| Sep 1, 2012 at 18:36 | history | asked | Agno | CC BY-SA 3.0 |