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Teylor expantion of ln(1-e^-x)^-1

  1. Jul 17, 2009 #1
    Dears
    (ln(1-e^(-x))^(-1))^2=?
    thankyou very much
    http://www.uploadgeek.com/share-69EF_4A600E7F.html [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 17, 2009 #2

    HallsofIvy

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    Taylor's expansion about what point?
     
  4. Jul 17, 2009 #3
    and is the (-1) inside or outside the logarithm?
     
  5. Jul 17, 2009 #4
    (-1) is inside ln
    ln((1-e^-x)^-1)
    thank u
     
  6. Jul 17, 2009 #5
    Do, the taylor expansions, of ln(.), e^-x and (.)^-1separately, then compose (function composition) the results. Use the multinomial theorem to expand the terms in brackets. Then collect terms. As opposed to using the multinomial theorem you may be able to also use the fact that polynomial multiplication is the convolution of the coeficients. Have fun. It is going to be a mess. If you are skilled you may be able to write a computer algebra program which will give you each coefficient of the resulting polynomial.
     
  7. Jul 17, 2009 #6
    He still didn't answer "about what point"... x=0 is NOT a candidate.

    Simplify what Johns said by using the exponent identity for logarithms.
    log(A^(-1)) = -log(A)
     
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