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Tricky differentiation

  • Thread starter raintrek
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  • #1
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Homework Statement



Find [tex]\frac{dU_{ave}}{d\beta}[/tex]

where

[tex]U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right)[/tex]


Homework Equations





The Attempt at a Solution



My answer is supposed to be [tex]-(U_{ave}^{2})+(U_{ave})^{2}[/tex]

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
 

Answers and Replies

  • #2
Mute
Homework Helper
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I suspect you've written the problem incorrectly here. As you've written it,

[tex]U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right) = \sum_k U_k [/tex]

What it should be is

[tex]U_{ave} = \frac{\sum_k U_k e^{-U_k \beta}}{\sum_k e^{-U_k \beta}}[/tex]

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

[tex]-\sum_k \frac{\partial}{\partial \beta} e^{-U_k \beta}[/tex]

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:

[tex]-\frac{\partial}{\partial \beta} \frac{\partial \ln Z}{\partial \beta} = \mbox{answer}[/tex]

where [itex]Z = \sum_k \exp(-U_k \beta)[/itex].
 
Last edited:
  • #3
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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?
 

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