Timeline for Using D to find a symbolic derivative
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 12, 2012 at 23:15 | history | tweeted | twitter.com/#!/StackMma/status/201450670040158210 | ||
| May 12, 2012 at 10:09 | history | edited | Artes |
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| May 11, 2012 at 8:59 | answer | added | Artes | timeline score: 26 | |
| May 11, 2012 at 8:14 | answer | added | celtschk | timeline score: 14 | |
| May 11, 2012 at 7:59 | vote | accept | C. E.♦ | ||
| May 11, 2012 at 7:55 | answer | added | Szabolcs | timeline score: 15 | |
| May 11, 2012 at 7:48 | comment | added | C. E.♦ | Ah... yes, substituting values with expr /. pars is a lot better. I was looking for something like this, I assume it will solve my $-problem to so if you post this I will accept it as an answer. | |
| May 11, 2012 at 7:46 | comment | added | Szabolcs | If you go with the approach I suggested, please see this as well. | |
| May 11, 2012 at 7:44 | comment | added | Szabolcs |
In this particular example D[L[s, L0, L1, a], s] is sufficient because s has no value. Or do you mean don't want to substitute in the values for L0, etc. at all? In that case I suggest you don't give them values. Make a rule list instead pars = {a -> 0.04, L1 -> 1, L0 -> 1}, and do expr /. pars when you need to substitute them into expr.
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| May 11, 2012 at 7:29 | history | asked | C. E.♦ | CC BY-SA 3.0 |