Derivative problem

  • Thread starter Moneer81
  • Start date
  • #1
159
2
Hello,
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:

Answers and Replies

  • #2
0rthodontist
Science Advisor
1,230
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Well, when n is 0 the term is 0 so that explains the limit change.
 
  • #3
159
2
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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