Help with differentiating summations

[Radiohannah]

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

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

Which I then differentiate to get$$-k \sum_{i=1}^{r}[\frac{\partial p_{i}}{\partial p_{i}} ln p_{i} + p_{i} \frac{\partial}{\partial p_{i}} ln p_{i}]$$

Which I can get down to

$$-k \sum_{i=1}^{r}[ln p_{i} + 1]$$But in the notes that summation sign has gone...and I don't understand why. I know that if it had been...$$\frac{\partial}{\partial p_{j}}$$ 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

 

[uart]

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.

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

In particular, think about what happens when $j = i$ and what happens when $j \neq i$.

 

[Radiohannah]

Hey, thanks, butin 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 my indices, but would leave me with the second term being $$\frac{p_{j}}{p_{i}}$$ which doesn't correlate with my notes, it has

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

?
Thanks

Hannah

 

[uart]

Hannah. $p_i$ is one variable in your system, $p_j$ is another different variable (for $i \neq j$). What do you get when you take the partial derivative of a variable with respect to a different variable?

 

[Radiohannah]

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 :-)

 

[uart]

^^^ 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.

 

[Stephen Tashi]

but would leave me with the second term being $$\frac{p_{j}}{p_{i}}$$

I don't think so. For example, what is $$x \frac{\partial}{\partial y} ln(x)$$ ?

 

[Mute]

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

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

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

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

is that although you recognize that


$$\frac{\partial p_{j}}{ \partial p_{i} } = \delta _{ij}$$

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

$$\frac{\partial f(p_j)}{\partial p_i}$$

you must use the chain rule:

$$\frac{\partial f(p_j)}{\partial p_i} = \frac{\partial f(p_j)}{\partial p_j}\frac{\partial p_{j}}{ \partial p_{i} } = \frac{\partial f(p_j)}{\partial p_j} \delta_{ij}$$

In particular,

$$\frac{\partial \ln(p_j)}{\partial p_i} = \frac{1}{p_j} \delta_{ij}$$

 

[Radiohannah]

*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

 

[Radiohannah]

Mute! I see! Woops! Thank you that makes sense :-)

 

[uart]

*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

Yes you've got it. :) And that's all the discrete delta function $\delta_{i,j}$ 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 $i \neq j$ separately.