Timeline for How to increase performance of this code for plotting a contour plot?
Current License: CC BY-SA 4.0
15 events
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| Jul 16, 2018 at 7:35 | vote | accept | diffusiondiver11 | ||
| Jul 16, 2018 at 6:44 | history | edited | Coolwater | CC BY-SA 4.0 |
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| Jul 16, 2018 at 6:12 | answer | added | Akku14 | timeline score: 6 | |
| Jul 15, 2018 at 20:32 | history | edited | diffusiondiver11 | CC BY-SA 4.0 |
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| Jul 15, 2018 at 20:28 | comment | added | diffusiondiver11 | No. I got different outputs for both values. The code really works but is very inefficient. I have added the other output to the description of the problem. pot is not independent of $z$. Look the SquareRoot term has $(z-l)^2$ term. Infact pot is a function of $x$, $y$ and $z$. $pot(x,y,z)$ | |
| Jul 15, 2018 at 20:25 | history | edited | diffusiondiver11 | CC BY-SA 4.0 |
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| Jul 15, 2018 at 20:25 | comment | added | David G. Stork |
The reason you got the same picture "for different values of $z$" is that pot is independent of z, so of course you're merely plotting the same function.
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| Jul 15, 2018 at 20:22 | comment | added | diffusiondiver11 | It did was evaluated for both z=0 and z=0.5 . But it took really long time to to that. 10-15 minutes as I remember. The picture is the output for the same code I gave above in description for z=0. You can run the code if you want, it will give the same output. | |
| Jul 15, 2018 at 20:19 | comment | added | David G. Stork |
pot has no definition of the variables within it, so of course it will never be evaluated for z = 0.5 (or indeed any other value).
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| Jul 15, 2018 at 19:57 | comment | added | diffusiondiver11 | @MariuszIwaniuk it sure may help. Thanks! | |
| Jul 15, 2018 at 19:55 | comment | added | diffusiondiver11 | But even Vector3D plots are taking such a long time that I could not output them. I did the same with other 3 dimensional Integrals but they took at most a few seconds. This one's not. | |
| Jul 15, 2018 at 19:46 | comment | added | Henrik Schumacher | The problem with these integrals is that they are singular and three-dimensional. That makes it quite expensive. Unfortunately, one also cannot exploit symmetry (e.g. by polar coordinates) in an obvious way... | |
| Jul 15, 2018 at 19:21 | comment | added | Mariusz Iwaniuk |
Maybe this helps: mathematica.stackexchange.com/questions/173253/…. Try user Henrik Schumacher answer. You must only to modify the code.
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| Jul 15, 2018 at 17:48 | history | edited | diffusiondiver11 | CC BY-SA 4.0 |
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| Jul 15, 2018 at 17:28 | history | asked | diffusiondiver11 | CC BY-SA 4.0 |