Timeline for Directional derivative of SiegelTheta
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2019 at 6:40 | history | edited | J. M.'s missing motivation♦ |
edited tags
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| Mar 23, 2013 at 17:07 | comment | added | J. M.'s missing motivation♦ | @ssch, the algorithm for it is slow, you see... | |
| Feb 15, 2013 at 13:54 | vote | accept | Emil Bostrom | ||
| Feb 14, 2013 at 18:59 | vote | accept | Emil Bostrom | ||
| Feb 14, 2013 at 18:59 | |||||
| Feb 14, 2013 at 18:58 | history | edited | Emil Bostrom | CC BY-SA 3.0 |
Greater specification of the problem
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| Feb 14, 2013 at 18:50 | answer | added | Dr. belisarius | timeline score: 3 | |
| Feb 14, 2013 at 17:44 | history | edited | Emil Bostrom | CC BY-SA 3.0 |
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| Feb 14, 2013 at 16:53 | comment | added | ssch | Wow, this was a really tricky function to do anything useful with, seems it evaluates really slowly at some points, doing an interpolation of it now I'll give it another 15minutes :p | |
| Feb 14, 2013 at 14:59 | history | edited | Dr. belisarius | CC BY-SA 3.0 |
added 5 characters in body
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| Feb 14, 2013 at 14:29 | history | edited | Emil Bostrom | CC BY-SA 3.0 |
added 132 characters in body
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| Feb 14, 2013 at 14:20 | comment | added | Emil Bostrom | The function is defined for all $\Omega$ with positive definite imaginary part, just as the error message says. In my case I'm only interested in the region $(x,y,z)\in[0,1]\times [0,1-x]\times [0,1]$. Also I'm only interested in the case where the second argument is $s=(0,0)$, and the reason I introduced a non-zero value for $s$ is only so it will allow me to take the derivative. | |
| Feb 14, 2013 at 12:45 | comment | added | ssch |
What's the domain for Ω? When I try things like SiegelTheta[Ω[1, 3, 3], {1, 0, 1}] I just get the error SiegelTheta::invmat: ... must be a symmetric matrix with a positive definite imaginary part
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| Feb 14, 2013 at 10:52 | comment | added | Emil Bostrom | I've used the expression $\Omega$ = {{(1/z - 1)^(1/2)*I/Pi*(1 - y), (1/z - 1)^(1/2)* I/Pi*(-1 + x + y)}, {(1/z - 1)^(1/2)* I/Pi*(-1 + x + y), (1/z - 1)^(1/2)*I/Pi*(1 - x)}} | |
| Feb 14, 2013 at 10:26 | review | First posts | |||
| Feb 14, 2013 at 13:10 | |||||
| Feb 14, 2013 at 10:22 | comment | added | ssch | Can you post the Mathematica expression for Ω | |
| Feb 14, 2013 at 10:08 | history | asked | Emil Bostrom | CC BY-SA 3.0 |