1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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 ?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Derivative problem
  1. Derivative problem (Replies: 4)

  2. Derivative Problem (Replies: 5)

  3. Derivative problem (Replies: 3)

  4. Derivative problem (Replies: 10)

  5. Derivative problem (Replies: 9)