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Asymptotic series as x approaches infinity

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the first 3 (non-zero) terms of the asymptotic series as x→∞ for ln(ex+1)

    2. Relevant equations

    ln(1+ε) ~ ε-ε2/2+ε3/3-O(ε4) (x→0)

    3. The attempt at a solution

    I found
    x+ln(1+e-x) (x→∞) = ln(ex+1) (x→∞)​
    ε=e-x
    which allows me to use the Maclaurin series above (since now the ε→0 as x→∞) yielding
    x+[e-x-e-2x/2+e-3x/3+O(e-4x)]

    My question
    Is there some expansion I should be using for e-x (x→∞) or some rearrangement of the ex (x→0) that I can use as replacement for e-x to find the next two terms of the series

    Thanks
    Clay
     
  2. jcsd
  3. Nov 29, 2011 #2
    You could set x'=1/x to express ln(1+e^(-x)) in terms of x' for x'->0
     
  4. Nov 29, 2011 #3

    Mute

    User Avatar
    Homework Helper

    No. The asymptotic expansion of exp(-x) for large x is just exp(-x). The taylor series expansion is valid only for small x, so there's nothing else you can do. If you're looking for an expansion that looks like

    [tex]\ln(1+e^x) \sim f(x)\left( 1 + \frac{a_1}{x} + \frac{a_2}{x^2} + \mathcal O \left(\frac{1}{x^3}\right) \right),[/tex]

    then I don't think that's possible. (Mathematica, seems to agree, suggesting there is no expansion of the function around x = infinity). I'd say the best you could do is define y = e^x and then modify the asymptotic series you already derived that's expressed as ln(y) + inverse powers of e^x = y.
     
    Last edited: Nov 29, 2011
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