Help with differentiating summations

  • #1
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

[tex]\frac{\partial}{\partial p_{i}} [-k \sum_{i=1}^{r} p_{i} ln p_{i}][/tex]

Which I then differentiate to get


[tex]-k \sum_{i=1}^{r}[\frac{\partial p_{i}}{\partial p_{i}} ln p_{i} + p_{i} \frac{\partial}{\partial p_{i}} ln p_{i}][/tex]

Which I can get down to

[tex]-k \sum_{i=1}^{r}[ln p_{i} + 1][/tex]


But in the notes that summation sign has gone.....and I don't understand why. I know that if it had been...


[tex]\frac{\partial}{\partial p_{j}}[/tex] acting on it instead that the kroneker deltas would sum over the summation for me but it definitely has to be a subscript i.


Am I missing the point somewhere?


Thank you!


Hannah
 

Answers and Replies

  • #2
uart
Science Advisor
2,776
9
Hannah, you seem really confused about the "kroneker delta" thing. There are no delta functions involved here, the delta is being used as a partial derivative symbol.

Back to the problem of differentiating and as to why the summation "disappears". Consider rewriting it slightly as I have below. Note that I've used a different variable "j" for the dummy variably of summation.

[tex]\frac{\partial}{\partial p_{i}} [-k \sum_{j=1}^{r} p_{j} ln p_{j}][/tex]

In particular, think about what happens when [itex] j = i [/itex] and what happens when [itex] j \neq i [/itex].
 
  • #3
Hey, thanks, but


in your example, if I were to multiply that out via the product rule


[tex]- 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}][/tex]



Where

[tex] \frac{\partial p_{j}}{ \partial p_{i} } = \delta _{ij}[/tex] and that would sum over my indices, but would leave me with the second term being [tex]\frac{p_{j}}{p_{i}}[/tex] which doesn't correlate with my notes, it has

[tex]-k[ln p_{i } + 1][/tex] as the answer, and that would suggest to me that I should have [tex] \frac{p_i}{p_i}[/tex] as my second term?

???



Thanks

Hannah
 
  • #4
uart
Science Advisor
2,776
9
Hannah. [itex]p_i[/itex] is one variable in your system, [itex]p_j[/itex] is another different variable (for [itex]i \neq j [/itex]). What do you get when you take the partial derivative of a variable with respect to a different variable?
 
  • #5
And just to clarify, I might be wrong, but thought that I needed to differentiate wrt

[tex]\frac{\partial}{\partial p_{i}}[/tex]

on

[tex] -k \sum_{i} p_{i} ln p_{i}[/tex]

all with subscript "i" because I hope to differentiate each projection (?) separately? Sorry I'm maybe not making myself clear :-)
 
  • #6
uart
Science Advisor
2,776
9
^^^ No that's where you are going wrong. You need to find the PD for a particular value of the indexed variable. This is a common source of misunderstanding in this type of problem. The sum should run over a "dummy" variable.


BTW. See my previous post.
 
  • #7
Stephen Tashi
Science Advisor
7,541
1,452
but would leave me with the second term being [tex]\frac{p_{j}}{p_{i}}[/tex]
I don't think so. For example, what is [tex] x \frac{\partial}{\partial y} ln(x) [/tex] ?
 
  • #8
Mute
Homework Helper
1,388
10
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

[tex]\frac{\partial}{\partial p_{i}} [-k \sum_{i=1}^{r} p_{i} ln p_{i}][/tex]
You have an error in your notes. The expression as written above cannot be correct. The letter i is a dummy index that you are summing over. The index i must not be present after you complete the sum, so you cannot then differentiate with respect to p_i. You should be differentiating with respect to p_j (or the dummy index should be different than i if you are differentiating with respect to p_i).

Your problem with the calculation

[tex]
\frac{\partial}{\partial p_{i}} [-k \sum_{j=1}^{r} p_{j} ln p_{j}]
[/tex]

is that although you recognize that


[tex]
\frac{\partial p_{j}}{ \partial p_{i} } = \delta _{ij}
[/tex]

you seem to not have realized that for a differentiation such as

[tex]\frac{\partial f(p_j)}{\partial p_i}[/tex]

you must use the chain rule:

[tex]\frac{\partial f(p_j)}{\partial p_i} = \frac{\partial f(p_j)}{\partial p_j}\frac{\partial p_{j}}{ \partial p_{i} } = \frac{\partial f(p_j)}{\partial p_j} \delta_{ij} [/tex]

In particular,

[tex]\frac{\partial \ln(p_j)}{\partial p_i} = \frac{1}{p_j} \delta_{ij}[/tex]
 
  • #9
*OH* ok I think I get it, so for my term

[tex]\frac{\partial}{\partial p_{j}} ln p_{i}[/tex]

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
 
  • #10
Mute! I see! Woops! Thank you that makes sense :-)
 
  • #11
uart
Science Advisor
2,776
9
*OH* ok I think I get it, so for my term

[tex]\frac{\partial}{\partial p_{j}} ln p_{i}[/tex]

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
Yes you've got it. :) And that's all the discrete delta function [itex]\delta_{i,j}[/itex] means, it's one when i = j and zero when it's not. It's up to you whether or not you want to use that notation or to just consider the cases where i=j and where [itex]i \neq j[/itex] separately.
 

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