Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Power Series?

  1. Sep 27, 2006 #1
    I have a function, [tex]\frac{x^3}{e^x-1}[/tex]

    The question than says expand the denominator as a power series in [tex]e^{-x}[/tex].

    I don't understand this question. How do I start doing that? It is not suggesting to approximate [tex]\frac{1}{e^x-1}[/tex] as [tex]e^{-x}[/tex] is it?
    Last edited: Sep 27, 2006
  2. jcsd
  3. Sep 27, 2006 #2
    If [itex]x[/itex] is large then you can do this:

    [tex]\frac{x^3}{e^x-1}=\frac{e^{-x} x^3}{1-e^{-x}}[/tex]

    and then expand the denominator as a series in [itex]e^{-x}[/itex].
    Last edited: Sep 27, 2006
  4. Sep 27, 2006 #3
    I still don't understand. I don't think we can assume x to be anything. Ultimately we like to integration the function.
  5. Sep 28, 2006 #4
    Hi there, I think this is a situation where it is best to post the question in full. You mention integration so I assume that you're working on the question where you integrate (x^3)/(exp(x)-1) from zero to infinity.

    \frac{{x^3 }}{{e^x - 1}} = x^3 \left( {\frac{{e^{ - x} }}{{1 - e^{ - x} }}} \right)

    = x^3 e^{ - x} \left( {\frac{1}{{1 - e^{ - x} }}} \right)

    = x^3 e^{ - x} \sum\limits_{k = 0}^\infty {\left( {e^{ - x} } \right)^k }
    [/tex] this equality follows since for x > 0 (you can ignore the end-point for the integration) we have exp(-x) < 1 so that a simple form of a the geometric series can be used.

    = x^3 \sum\limits_{k = 0}^\infty {\left( {e^{ - x} } \right)^{k + 1} }

    Now just interchange order of summation and integration, use parts once or twice and then use the sum that you will have evaluated in the previous part of the question.

    You should check the starting index though, when I did this question I kept on getting an extra "+1" factor. I haven't gone back to finish this question so I'm not sure if it's completely correct.
  6. Sep 28, 2006 #5
    nice one. e^x is a polynomial although an infinite one but should still be allowed to substitute the x in 1/(1-x) hence act like a geometric series. I'm not sure if you can stick functions that cannot form polynomials into the x and get a geometric series.
  7. Sep 28, 2006 #6
    I don't think that there is any restriction on what you can put into the geometric series as long as the "common ratio" between consecutive terms in the sum has an absolute value which is less than one.
  8. Sep 29, 2006 #7
    True. My memory has faded somewhat about this series but I think you are right.
  9. Sep 29, 2006 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    There is no such thing as a polynomial with infinitely many terms. They are called power series.

    Anyway, you were asked to expand as a power series in something. It doesn't matter a damn what the something is. (1-y)^-1 as a power series in y is whateever it is, and clearly it matters not one iota if y is exp(x), cos(x), ut+rs+ztverawe.
    Last edited: Sep 29, 2006
  10. Sep 29, 2006 #9
    What really got me about this question was the wording.

    The phrase "...expand the denominator as a power series in [tex]e^{-x}[/tex]." From the answers it seems it wants me to find a suitable power series for a function and than substitute [tex]e^{-x}[/tex] into the variable of the function. I'll keep it in mind for next time.
    Last edited: Sep 29, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook