# Tricky differentiation

1. May 26, 2008

### raintrek

1. The problem statement, all variables and given/known data

Find $$\frac{dU_{ave}}{d\beta}$$

where

$$U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right)$$

2. Relevant equations

3. The attempt at a solution

My answer is supposed to be $$-(U_{ave}^{2})+(U_{ave})^{2}$$

However I keep getting zero. I can only assume that the summation sign has something to do with the final form of the answer, but I have no idea how to deal with it. Can anyone offer a suggestion? Thanks

2. May 26, 2008

### Mute

I suspect you've written the problem incorrectly here. As you've written it,

$$U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right) = \sum_k U_k$$

What it should be is

$$U_{ave} = \frac{\sum_k U_k e^{-U_k \beta}}{\sum_k e^{-U_k \beta}}$$

This is a standard integral in statistical mechanics, and there's a neat trick that always goes along with solving it. Supposing U_k is indepent of beta, write the denominator as

$$-\sum_k \frac{\partial}{\partial \beta} e^{-U_k \beta}$$

Pull the derivative outside the sum, and you note that what you have looks like (1/y) dy/dx = d(ln y)/dx. Then, take the second derivative of that result:

$$-\frac{\partial}{\partial \beta} \frac{\partial \ln Z}{\partial \beta} = \mbox{answer}$$

where $Z = \sum_k \exp(-U_k \beta)$.

Last edited: May 26, 2008
3. May 26, 2008

### raintrek

Hmm, I've transcribed the problem straight from the question sheet. I too had that thought.

I'm not sure I follow you with your methodology though. If I try to solve the equation using the standard quotient rule, it gives me zero - is there a reason for that?