Timeline for Expand in series an equation of two variables
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| when toggle format | what | by | license | comment | |
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| Jan 9, 2020 at 10:09 | vote | accept | ThunderBiggi | ||
| Jan 9, 2020 at 4:02 | answer | added | Hayashi Yoshiaki | timeline score: 1 | |
| Jan 7, 2020 at 15:34 | comment | added | ThunderBiggi | @HayashiYoshiaki This works well for now - I have used it for a few problems. If you write it as an answer I will accept it. | |
| Dec 31, 2019 at 21:49 | comment | added | Hayashi Yoshiaki |
@ThunderBiggi OK I got it. First you can try Subscript (Control key + _) to make tx and ty look more like a coefficient, then use Subscript[tx,0] -> Function[{x,y},x^2 (1-x^2) C1] . The reason of making the function of x and y even it is a function of only x is to convince Partial derivative function that its y derivative is 0. Another way to do similar thing is to use SetAttributes and Constant, but I don’t know whether it can make eg ‘a constant only respect to y’. Now I can’t use MA to test my suggestion. I hope you meet a good result.
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| Dec 31, 2019 at 11:16 | comment | added | ThunderBiggi |
@HayashiYoshiaki I want to replace tx[0, x], where the 0 is a dummy variable. At next order in the expansion it will be tx[1, x]. I don't see how making tx a function of x and y will help
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| Dec 29, 2019 at 16:48 | comment | added | Hayashi Yoshiaki |
I don't fully understand your question but how about using tx -> Function[{x,y}, x^2 (1 - x^2)^2 C1] instead of tx -> Function[x, x^2 (1 - x^2)^2 C1]?
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| Dec 29, 2019 at 15:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Nov 29, 2019 at 14:08 | comment | added | Alx | Please, see my attempt to answer. | |
| Nov 29, 2019 at 14:05 | answer | added | Alx | timeline score: 0 | |
| Nov 29, 2019 at 13:02 | comment | added | ThunderBiggi | @Alx I know that. This is just a simple example to indicate my question. My real problem includes 5 equations with 5 unknown variables which are very long and there this approach doesn't work. | |
| Nov 29, 2019 at 12:28 | comment | added | Alx |
You can simply write DSolve[Normal@Series[expr, {y, 0, 0}] == {0, 0}, {tx[0, x], sx[0, x]}, x] and get {{sx[0, x] -> C[1], tx[0, x] -> x^2 (1 - x^2)^2 C[2]}}.
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| Nov 29, 2019 at 10:27 | history | edited | ThunderBiggi | CC BY-SA 4.0 |
added 277 characters in body
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| Nov 29, 2019 at 10:12 | history | asked | ThunderBiggi | CC BY-SA 4.0 |