Timeline for When is $\Re(\zeta(s)) - \Im(\zeta(s)) = 0 $ with $\Re(\zeta(s))\neq 0$ and $\Im(\zeta(s))\neq 0$?
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21 at 20:29 | comment | added | Steven Clark | @free_lancer I don't understand what you're looking for at this point, but I added a region plot that perhaps provides some additional insight. | |
| Jan 21 at 20:28 | history | edited | Steven Clark | CC BY-SA 4.0 |
Added region plot.
|
| Jan 21 at 18:26 | comment | added | MrPie | We can examine cases like when is $\Im(\zeta(s))>0$ and $\Re(\zeta(s))<0$ because this will never produce a root. So we can ask when they share the same sign. | |
| Jan 21 at 18:24 | comment | added | MrPie | which would imply the distance between them is not zero. | |
| Jan 21 at 18:22 | comment | added | MrPie | actually its more then just the distance. We need to find when they are equal and then at these points if they are positive or negative. If they both are zero only at this middle point we should be able to show they are positive or negative at other points. | |
| Jan 21 at 18:15 | comment | added | MrPie | hmmmm maybe we should then look for when the distance $\Re(\zeta)-\Im(\zeta)>0$ and $\Re(\zeta)-\Im(\zeta)<0 $. The distance seems so close to each other maybe its just coicendence they cant both be zero unless the real part is in the middle. The distance between the real part and imaginary part seems similar to the function itself. We need analytic formula for $\Re(\zeta)-\Im(\zeta)$ | |
| Jan 21 at 18:08 | comment | added | Steven Clark | @free_lancer hey are also both zero at the negative even integers which correspond to the trivial zeta zeros. | |
| Jan 21 at 17:58 | comment | added | MrPie | They are really close to one another | |
| Jan 21 at 17:57 | comment | added | MrPie | I see so there infintly many points they are both zero, but they seem to be both zero only at the line $1/2$ | |
| Jan 21 at 15:44 | comment | added | Steven Clark | The blue curve is the list of all points $(\alpha, i \beta)$ in the complex plane where $\Re(\zeta(\alpha+i \beta))=\Im(\zeta(\alpha+i \beta))$. This corresponds to zero at the negative even integers and sometimes, but not always, when $\alpha=\frac{1}{2}$. Note the blue contour lines are continuous and so there are an uncountably infinite number of points on any segment of any of the blue contour lines where $\Re(\zeta(\alpha+i \beta))=\Im(\zeta(\alpha+i \beta))$. | |
| Jan 21 at 15:35 | history | edited | Steven Clark | CC BY-SA 4.0 |
Made a a clarification and a correction.
|
| Jan 21 at 4:23 | comment | added | MrPie | meaning the roots of real and imaginary part are at the same location then | |
| Jan 21 at 4:20 | comment | added | MrPie | so the plots indicate they are both zero only at the value $1/2$ | |
| Jan 21 at 4:18 | comment | added | Steven Clark | The blue curve in the plots above corresponds to $\Re(\zeta(\alpha+i \beta))=\Im(\zeta(\alpha+i \beta))$, and this value is only zero at the zeta zeros. This same contour was also plotted in the other answer but over a larger range of $\alpha$ values. | |
| Jan 21 at 4:08 | comment | added | MrPie | if you can please I would like to see one when they dont have a root but are equal. Cause me and Conrad where just talking how real and imaginary can be equal infinity often. So I dont see how they can only cross at this point. They should cross more times. | |
| Jan 21 at 4:05 | comment | added | Steven Clark | I can do a density plot or 3D plot of $\Re(\zeta(s))-\Im(\zeta(s))$, but this is not a contour. The three contours listed and plotted in the answer above only intersect at $\Re(\zeta(s))=\Im(\zeta(s))=0$ which are the roots of $\zeta(s)$. | |
| Jan 21 at 4:02 | comment | added | MrPie | the plot show they only intersect at the roots of $\zeta$ | |
| Jan 21 at 3:58 | comment | added | MrPie | can you do contour plot of $\Re(\zeta) - \Im(\zeta)$ and plot roots in red like you did above | |
| Jan 21 at 3:56 | comment | added | MrPie | yes but it doesnt show when $\Im(\zeta) = 0$ or $\Re(\zeta) = 0$. It only show when they are both zero | |
| Jan 21 at 3:54 | comment | added | Steven Clark | I'm not sure what you're asking. Do you understand the three contours listed in the answer above corresponding to the three curves? The orange and green curves show where the real and imaginary parts are zero, and the blue curve shows where they're equal to each other. | |
| Jan 21 at 3:48 | comment | added | MrPie | Your plot still indicate they cross at those points like we suspect. | |
| Jan 21 at 3:46 | comment | added | MrPie | I want to find zeroes of both real and imaginary part separately. | |
| Jan 21 at 3:45 | comment | added | MrPie | yes but what about the zeroes of both of the series separately. It doesnt seem they both need to be zero at the same time | |
| Jan 21 at 3:43 | comment | added | Steven Clark | I'm not sure I understand your second and third comments, but the horizontal axis is the real part $\alpha$ and the vertical axis is the imaginary part $i \beta$. I combined a contour plot of the three curves specified in the answer above with a list plot of the zeta zeros which are shown as the red discrete dots. All of the zeta zeros appear on the vertical line corresponding to $\alpha=\frac{1}{2}$. | |
| Jan 21 at 3:35 | comment | added | MrPie | did you pick $\alpha =1/2$? | |
| Jan 21 at 3:35 | comment | added | Steven Clark | @free_lancer Of course since the nontrivial zeta zeros are the only points in the critical strip which satisfy all three conditions. | |
| Jan 21 at 3:34 | comment | added | MrPie | these plots look like they indicate they only equal each other at the middle of the plane not even the roots. Or am I reading them wrong? | |
| Jan 21 at 3:30 | comment | added | MrPie | And they all magically intersect at these red points. | |
| Jan 21 at 3:22 | history | answered | Steven Clark | CC BY-SA 4.0 |