Timeline for How to differentiate formally?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 8, 2015 at 1:35 | comment | added | Charlie Parker | @Jens sorry for taking this discussion further, I am genuinely curious about this and there's something that still "bothers me". In particular, why didn't mathematica return $1 - n + n^2 - n^3 + n^4 - n^ 5 $ that is also zero...and you might ask me, why did I choose that weird equation and my answer is...Its random. But is mathematica expression random? Or was there a reason it chose $1 - n$ and not (any) other expression that involves n and is zero (in this special case n=1). Do you know what I mean? I wish I knew. If I did I wouldn't have extended this discussion :p | |
| Aug 7, 2015 at 23:01 | comment | added | Jens | @Pinocchio Of course it looks weird, but nonetheless it's a true statement because the derivative is taken with respect to a variable that (for n=1) doesn't occur in the sum. Therefore its value would be 1-n==0. What makes it a little confusing is that the output is not in the most simplified form because n still appears even though it can only have one value in that last line. Still, it's correct. | |
| Aug 7, 2015 at 22:22 | comment | added | Charlie Parker | @Jens if n=1 then $\sum_i i x_i = 1 x_1$ and n-1 = 0. Isn't that still sort of a bizarre ting to do if we are taking the derivative with respect to x[2]? Sorry, I am just trying to understand why the output is the way it is. Currently, it still doesn't quite make sense to me. | |
| Aug 7, 2015 at 22:06 | comment | added | Jens | @Pinocchio No, the only other case in that sum is n=1. | |
| Aug 7, 2015 at 20:10 | comment | added | Charlie Parker | @Jens I see what true means now (weird, the programmer could have just said otherwise in a print statement...). But, how is the derivative wrt x[2] be 1 - n for n < 1? for n<1 the sum doesn't even exist...so wouldn't technically it be the derivative of a non-existent sum (so zero), so the derivative of zero is zero? Or am I completely off? | |
| Aug 7, 2015 at 20:05 | comment | added | Jens | @Pinocchio True is the entry that humans would call "otherwise" - it's the default alternative if none of the above holds. | |
| Aug 7, 2015 at 19:58 | comment | added | Charlie Parker | sorry if this a super newbie question, but why does it say $1-n$ for...true? :/ What does that condition even mean? | |
| Jun 17, 2015 at 7:44 | comment | added | Joel Bosveld | It might be worth mentioning that when doing this with more than one variable (say $x$ and $y$), it is necessary to have them Sorted in the NonConstants rule for the UpValue of the variables, otherwise it won't find a match when trying to evaluate D (and leave it unevaluated). | |
| Dec 22, 2012 at 17:03 | comment | added | Wizard | This is really helpful. Thank you very much. | |
| Dec 22, 2012 at 17:02 | vote | accept | Wizard | ||
| Dec 15, 2012 at 22:27 | history | edited | Jens | CC BY-SA 3.0 | More symbolic examples |
| Dec 15, 2012 at 22:03 | history | answered | Jens | CC BY-SA 3.0 |