Timeline for When is $\Re(\zeta(s)) - \Im(\zeta(s)) = 0 $ with $\Re(\zeta(s))\neq 0$ and $\Im(\zeta(s))\neq 0$? [closed]
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26 events
| when toggle format | what | by | license | comment | |
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| Jan 21 at 19:53 | comment | added | Steven Clark | @free_lancer Your formulas for $\Re(\zeta(s))$ and $\Im(\zeta(s))$ are actually formulas for $\Re(\eta(s))$ and $\Im(\eta(s))$ since they don't account for the $\frac{1}{1-2^{1-s}} $ factor in $$\zeta(s)=\frac{1}{1-2^{1-s}} \eta(s)=\frac{1}{1-2^{1-s}} \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}\,,\quad\Re(s)>0.$$ | |
| Jan 21 at 14:08 | history | closed |
Stopple GH from MO Daniele Tampieri Wojowu Sam Hopkins |
Not suitable for this site | |
| Jan 21 at 3:35 | answer | added | მამუკა ჯიბლაძე | timeline score: 4 | |
| Jan 21 at 3:22 | answer | added | Steven Clark | timeline score: 1 | |
| Jan 21 at 3:13 | comment | added | MrPie | first I think we need the correct form of $\Re(\zeta)$ and $\Im(\zeta)$ as many people disagreed with my formula involving the cosine and sine for some reason. | |
| Jan 21 at 3:08 | comment | added | MrPie | well I was thinking we can find points when the real part is zero and then the imaginary part separately like a system of equations because they both need to be zero for us to call it a root. It seems that it has something to do with the trig functions somehow if they magically can only be zero together in the center. Its possible that both the series share many values but never the value zero unless its at the center. | |
| Jan 21 at 3:03 | comment | added | Conrad | not sure what you say as of course the fact that $\zeta$ has zeroes only on the critical line is RH and it's still a conjecture that may or may not be true; note that $\eta$ has zeroes on the line $\Re s =1$ so it's tricky to deduce stuff from the partial sums since those are similar | |
| Jan 21 at 3:01 | comment | added | MrPie | they are equal infinitely often but not zero somehow unless we in the middle of the plane wierd. On the line $\alpha = \dfrac{1}{2}$ there are points when the real part is zero but not the imaginary part and vise versa there is points when the imaginary part is zero but not the real part. They dont need to be both zero at the same time. It only needs to happen to be a root | |
| Jan 21 at 2:58 | comment | added | MrPie | this kinda says that both the real and imaginary parts are zero only at the half line | |
| Jan 21 at 2:53 | comment | added | MrPie | but possible never zero unless its the strip in the center. | |
| Jan 21 at 2:51 | comment | added | Conrad | they are equal very often on any vertical line | |
| Jan 21 at 2:49 | comment | added | MrPie | @Conrad if they are only equal on a vertical line to prove Riemann statement we only need to show they cant be two lines they are both on. Meaning there is only one vertical line that corresponds to each time the real part and imaginary are equal. So if one zero is on the line then they all must be on the line. But i dont see how each time they are equal we get a new vertical line yet. | |
| Jan 21 at 2:36 | comment | added | Conrad | See comment at the other post but essentially the real and imaginary part are equal very often since the argument is a real continuous function with jumps at zeta zeroes that oscillates a lot so passes often through $\pi/4+2k\pi$ or $3\pi/4+2k\pi$ I would look up Selberg limit theorem and its simpler analogs on $1/2<\sigma<1$ - there is a book by Laurincikas called Limit Theorems for the RZ function which is post Titchmarsh and discusses these things - a lot is known about the distribution of the values of zeta on vertical lines in the critical strip and that book gives a good account | |
| Jan 21 at 2:08 | comment | added | MrPie | @Stopple it seems there are times when the real part is zero but not the imaginary part and similair the imaginary part can be zero without the real part. They both are zero maybe at the value $\alpha = \dfrac{1}{2}$ | |
| Jan 21 at 2:04 | comment | added | MrPie | sorry i mess up title a bit | |
| Jan 21 at 2:03 | history | edited | MrPie | CC BY-SA 4.0 |
edited title
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| Jan 21 at 1:51 | comment | added | MrPie | he bounds the number of times $\zeta(s) = a$ with some function called $K$ and $a\neq 0$ | |
| Jan 21 at 1:45 | comment | added | MrPie | okay thanks while spending my next 30 minutes please tell me my intuition is not purely baloney | |
| Jan 21 at 1:42 | comment | added | Stopple | Read the whole chapter, including the notes at the end. @Conrad already alluded to this material in his comment on your previous question. | |
| Jan 21 at 1:40 | review | Close votes | |||
| Jan 21 at 14:10 | |||||
| Jan 21 at 1:39 | comment | added | MrPie | @Stopple that chapter start with the modulus. I want to find when real and imaginary part are equal. | |
| Jan 21 at 1:29 | comment | added | MrPie | @Stopple so there are multiple vertical lines or no? | |
| Jan 21 at 1:26 | comment | added | MrPie | @Stopple never read the book but I was interested because it would clarify my previous question. When i asked it the zeta function repeated itself with respect to its imaginary part | |
| Jan 21 at 1:25 | comment | added | MrPie | negative ding because i point out the obvious | |
| Jan 21 at 1:24 | comment | added | Stopple | The answers to the kinds of question you are asking can be found in Chapter XI Distribution of Values of $\zeta(s)$ of Titchmarsh's Theory of the Riemann Zeta Function. PS not Reimann. | |
| Jan 21 at 1:14 | history | asked | MrPie | CC BY-SA 4.0 |