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Derivative problem

  1. Jan 20, 2006 #1
    I was trying to take the derivative of the following summation function:
    [tex] f(\varepsilon)=\sum_{n=0}^\infty e^ \frac {-n\varepsilon} {kT} [/tex]

    so since the derivative of an exponential function that is raised to a power and that power is function will just be the function itself times the derivative of the power, I figured that the answer would be"

    [tex] \frac {df}{d\varepsilon} = \sum_{n=0}^\infty (\frac{-n}{kT}) . e^ \frac {-n\varepsilon}{kT} [/tex]

    and then of course we can take the constants outside the summation

    but to my surprise, the book did it this way:

    [tex] \frac {df}{d\varepsilon} = \sum_{n=1}^\infty (\frac{n}{kT}) . e^ \frac {-n\varepsilon}{kT} [/tex]

    so my question is how did the lower limit change from n=0 to n=1 and where did the minus sign (-n/kT) go? do these two have something to do with each other?

    thanks a lot
    Last edited: Jan 20, 2006
  2. jcsd
  3. Jan 20, 2006 #2


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    Science Advisor

    Well, when n is 0 the term is 0 so that explains the limit change.
  4. Jan 20, 2006 #3
    actually I knew about the lower limit of the summation I don't know why I still asked the question :)

    do you know where the minus sign went though ?
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