Timeline for Challenging problems concerning Jacobian elliptic functions with complex modulus
Current License: CC BY-SA 3.0
9 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 25, 2016 at 9:43 | vote | accept | Twi | ||
| Apr 25, 2016 at 9:42 | answer | added | Twi | timeline score: 2 | |
| Oct 13, 2015 at 20:58 | comment | added | Twi | to Sergei: I see now that the two points $k\neq\pm1$ would not cause any trouble: math.stackexchange.com/questions/234484/… | |
| Oct 13, 2015 at 13:35 | comment | added | Twi | to Sergei: Good idea, nevertheless I am not sure if Maximum modulus principle applies even if $|sn(Ku,k)|\leq1$ for all $|k|=1, k\neq\pm1$. Function $k\mapsto sn(K(k)u,k)$ has a branch cut in $(-\infty,-1]\cup[1,\infty)$ and hence it is not analytic in points $k=\pm1$. Maybe one would have to known something like $\limsup_{|k|\leq1,k\to\pm1}|sn(Ku,k)|\leq1$. | |
| Oct 12, 2015 at 9:15 | comment | added | Sergei | the problem estimate is true for real $0<k<1$ and for imaginary $ik$ with $0<k<1$. If it will be proved also for $|k|=1$ it will be true for all $k$ due to the max principle for the corner, as it will be true on its three parts of boundary. | |
| Oct 12, 2015 at 8:41 | comment | added | Twi | to Sergei: Currently, I do not have any proof even for the case $|k|=1$. With this restriction, the answer to the problem would be of interest to me, too. | |
| Oct 12, 2015 at 5:46 | comment | added | Sergei | can we prove it for $|k|=1$? | |
| Oct 11, 2015 at 9:30 | history | asked | Twi | CC BY-SA 3.0 |