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Asymptotic Expansion

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate an asymptotic expansion to all orders of I(x) as x-->infinity.

    I(x)=[itex]\int^{\infty}_{0}[/itex]e^{-xt} ln(1+[itex]\sqrt{t})[/itex]dt


    2. Relevant equations



    3. The attempt at a solution
    I'm not exactly sure how to begin with this. I've read through the section on my book on asymptotics but am still confused. I think I need to expand the integrand as a Taylor Series.
     
  2. jcsd
  3. Oct 25, 2012 #2

    Zondrina

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    Do you remember the Taylor series for ex and ln(1+x)?
     
  4. Oct 25, 2012 #3
    Yes.

    e[itex]^{x}[/itex]=1+x+[itex]\frac{x^{2}}{2!}[/itex]+[itex]\frac{x^{3}}{3!}[/itex]+[itex]\cdots[/itex]

    ln(1+x)=x-[itex]\frac{x^{2}}{2}[/itex]+[itex]\frac{x^{3}}{3}[/itex]-[itex]\cdots[/itex]
     
  5. Oct 25, 2012 #4

    Zondrina

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    Write them as sums for ease, remember though that the series for ln(1+x) does not exist at n=0 so start your series at n=1.

    Now how can you re-write your integral so you can integrate term by term?
     
  6. Oct 25, 2012 #5
    How do I know where to truncate the Taylor series so that I can multiply them together?
     
  7. Oct 25, 2012 #6

    Zondrina

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    So :

    [itex]e^{-xt} = - \sum_{n=0}^{∞} \frac{xt}{n!}[/itex]

    What happens when you evaluate the first sum? Can you re-write your sum then :)?

    And :

    [itex]ln(1 + \sqrt{t}) = \sum_{n=1}^{∞} (-1)^{n+1} \frac{t^{n/2}}{n}[/itex]
     
  8. Oct 26, 2012 #7
    Hmm, I'm not exactly sure what you mean. Could you clarify?
     
  9. Oct 26, 2012 #8

    Zondrina

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    What is the first term of the series of e-xt
     
  10. Oct 26, 2012 #9
    I believe it is 1...
     
  11. Oct 26, 2012 #10

    Zondrina

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    Oh a bit of a fumble earlier. Sorry it's 2am.

    [itex]e^{-xt} = \sum_{n=0}^{∞} \frac{(-xt)^n}{n!}[/itex]

    If the first term is 1, then re-write :

    [itex]e^{-xt} = 1 + \sum_{n=1}^{∞} \frac{(-xt)^n}{n!}[/itex]
     
  12. Oct 26, 2012 #11
    No worries!

    OK, I've got that. I'm just not sure where to go from there.
     
  13. Oct 26, 2012 #12

    Zondrina

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    Multiply the two sums of the integral together and then integrate term by term.
     
  14. Oct 26, 2012 #13
    OK, simple enough. It looks like this series diverges when x tends to infinity, though? Is that expected?
     
  15. Oct 26, 2012 #14
    Thinking about this, it seems to make sense. When x goes to infinity, the e^(-xt) becomes small quickly, and so the integral is dominated by the log term--but integrating that term from 0 to infinity diverges, so it makes sense that the expansion would diverge as x tends to infinity. Right?
     
  16. Oct 26, 2012 #15

    Zondrina

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    I'm not sure, I didn't actually compute it, but by the looks of it I believe the integral will diverge.
     
  17. Oct 27, 2012 #16
    How exactly do I multiply these two together if they have infinite terms?
     
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