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Expansion of [1+exp(x)]^-2

  1. Mar 8, 2014 #1
    In this section, they derive the Sommerfeld formula.

    In the first step it seems like they have expanded ##\frac{1}{(1+e^x))^2}##. I'm not sure why does the series taylor expand as ##e^{-nx}##?

    Also how did they get from the 2nd to the 3rd step?

    Simply by comparing terms we see they are different:

    For 2nd step we get terms of ##x^se^x(-1 + 2e^{-x} - 3e^{-2x} + ...)##.
    For 3rd step we get terms of ##e^{-x} - 2e^{-2x} + 3e^{-3x} - ....##

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  2. jcsd
  3. Mar 9, 2014 #2
    Yeah in the second step it needs to be multiplied by -1 and the e^x should be e^(-x) instead.

    To get the expansion, you could write [itex]\frac{1}{(1+e^x)^2}[/itex] as [itex]\frac{e^{-2x}}{(1+e^{-x})^2}[/itex]. Then you could do the binomial series expansion on it, or you could write it out using the geometric series formula, which gives [itex]\frac{e^{-2x}}{(1+e^{-x})^2}=e^{-2x}(1-e^{-x}+e^{-2x}-e^{-3x}+...)^2[/itex]. Collecting the terms that come from squaring it, you get the appropriate formula.
     
  4. Mar 9, 2014 #3
    Collecting the terms give the right terms, but how did they obtain a general expression ##\sum_{n=0}^{\infty} e^{-x} [(n+1)(-1)^{n+1} e^{-nx}]##?
     
  5. Mar 9, 2014 #4

    Bill_K

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    To expand ##\frac{1}{(1 + e^{-x})^2}##, do it in terms of ##y = e^{-x}##.
    [tex]\frac{1}{1 + y} = \sum{(-1)^n y^n}[/tex]
    Differentiate both sides wrt ##y##:
    [tex]- \frac{1}{(1 + y)^2} = \sum{(-1)^n n y^{n-1}} = \sum{(-1)^{n+1} (n+1) y^n}[/tex]
     
    Last edited by a moderator: Mar 9, 2014
  6. Mar 11, 2014 #5
    Actually I think it's ##e^x## instead:

    Starting, LHS is simply geometric series with factor ##e^{-x}##:

    [tex]\frac{1}{1+e^x} = \sum (-1)^n e^{-nx} [/tex]

    Differentiating both sides with respect to x,

    [tex]-\frac{e^x}{(1+e^x)^2} = -\sum (-1)^{n+1} n e^{-nx}[/tex]

    [tex] \frac{e^x}{(1+e^x)^2} = \sum (-1)^{n+1} n e^{-nx}[/tex]

    This leads to the answer immediately.
     
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