*OH* ok I think I get it, so for my term
\frac{\partial}{\partial p_{j}} ln p_{i}
all of the terms would be zero apart from the once instance in which the j = i in the summation? And that justifies getting rid of the summation sign? Is that right?
Thanks!
Hannah
And just to clarify, I might be wrong, but thought that I needed to differentiate wrt
\frac{\partial}{\partial p_{i}}
on
-k \sum_{i} p_{i} ln p_{i}
all with subscript "i" because I hope to differentiate each projection (?) separately? Sorry I'm maybe not making myself clear :-)
Hey, thanks, but
in your example, if I were to multiply that out via the product rule
- k \sum^{r}_{j=1}[\frac{\partial p_{j}}{\partial p_{i}} ln p_{j} + p_{j} \frac{\partial}{\partial p_{i}} ln p_{j}]
Where
\frac{\partial p_{j}}{ \partial p_{i} } = \delta _{ij} and that would sum over...
Hello!
I'm getting confused when differentiating summations.
I understand that if you differentiate an expression and it gives a kroneker delta, that then sums over the appropriate summation and it disappears. But in my notes it has
\frac{\partial}{\partial p_{i}} [-k \sum_{i=1}^{r} p_{i} ln...