Expand Denominator of $\frac{x^3}{e^x-1}$ as Power Series

[pivoxa15]

I have a function, \frac{x^3}{e^x-1}

The question than says expand the denominator as a power series in e^{-x}.I don't understand this question. How do I start doing that? It is not suggesting to approximate \frac{1}{e^x-1} as e^{-x} is it?

 

[jpr0]

If x is large then you can do this:

\frac{x^3}{e^x-1}=\frac{e^{-x} x^3}{1-e^{-x}}

and then expand the denominator as a series in e^{-x}.

 

[pivoxa15]

I still don't understand. I don't think we can assume x to be anything. Ultimately we like to integration the function.

 

[Benny]

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.

<br /> \frac{{x^3 }}{{e^x - 1}} = x^3 \left( {\frac{{e^{ - x} }}{{1 - e^{ - x} }}} \right)<br />

<br /> = x^3 e^{ - x} \left( {\frac{1}{{1 - e^{ - x} }}} \right)<br />

<br /> = x^3 e^{ - x} \sum\limits_{k = 0}^\infty {\left( {e^{ - x} } \right)^k } <br /> 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.

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

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.

 

[pivoxa15]

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.

 

[Benny]

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.

 

[pivoxa15]

True. My memory has faded somewhat about this series but I think you are right.

 

[matt grime]

nice one. e^x is a polynomial although an infinite one

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.

 

[pivoxa15]

What really got me about this question was the wording.

The phrase "...expand the denominator as a power series in e^{-x}." From the answers it seems it wants me to find a suitable power series for a function and than substitute e^{-x} into the variable of the function. I'll keep it in mind for next time.