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Math Help - Series representation of x^2/(a^3-x^3)

  1. #1
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    Series representation of x^2/(a^3-x^3)

    Hi all,

    I'm having trouble with the series representation of a function.

    f(x) = \frac{x^2}{a^3-x^3}

    First I tried breaking this up into partial fractions, then I realized I'm not very good at doing partial fractions... But I think I found another path that maybe I'm supposed to use:

    Consider that:
    \int \frac{x^2}{a^3-x^3} dx = -\frac{1}{3} ln(a^3 - x^3) + C

    I think this somehow relates to the integral of the alternating geometric series (switching variable for clarity):
    \int \frac{1}{1+u} du = ln(1 + u) + C

    I know I can integrate the terms of the geometric series to get the natural log, but I also know I can't just plug u = a^3 - x^3 - 1, and now I'm not sure what to do. Am I headed down the right tree or barking up the wrong path? Do I need to return to the partial fractions approach instead?

    Thanks again for any help,

    Brian
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  2. #2
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    log(1+x) can be easily expanded. it was somewht like this log(1+x)=1+x^2/2-x^3/3+.....(i am not sure please correct me if i am wrong). since u have got an expression in log u should be able to expand it......
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  3. #3
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    then differentiate it once again and u should get the series representation as my common sense says!.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by buckeye1973 View Post
    Hi all,

    I'm having trouble with the series representation of a function.

    f(x) = \frac{x^2}{a^3-x^3}

    First I tried breaking this up into partial fractions, then I realized I'm not very good at doing partial fractions... But I think I found another path that maybe I'm supposed to use:

    Consider that:
    \int \frac{x^2}{a^3-x^3} dx = -\frac{1}{3} ln(a^3 - x^3) + C

    I think this somehow relates to the integral of the alternating geometric series (switching variable for clarity):
    \int \frac{1}{1+u} du = ln(1 + u) + C

    I know I can integrate the terms of the geometric series to get the natural log, but I also know I can't just plug u = a^3 - x^3 - 1, and now I'm not sure what to do. Am I headed down the right tree or barking up the wrong path? Do I need to return to the partial fractions approach instead?

    Thanks again for any help,

    Brian
    The integration method works ok (don't forget to differentiate).

    How about this though? f(x)=\frac{x^2}{a^3-x^3}=\frac{x^2}{a^3}\frac{1}{1-\left(\frac{x}{a}\right)^3}. Now, just for the sake of clarity try calling z=\left(\frac{x}{a}\right)^3. Look familiar?
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  5. #5
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    Quote Originally Posted by Pulock2009 View Post
    log(1+x) can be easily expanded. it was somewht like this log(1+x)=1+x^2/2-x^3/3+.....(i am not sure please correct me if i am wrong). since u have got an expression in log u should be able to expand it......
    Actually, it's 1-x^2/2+x^3/3-x^4/4+... I already knew this part, you've basically restated my question.

    Quote Originally Posted by buckeye1973 View Post
    I know I can integrate the terms of the [alternating] geometric series to get the natural log
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    How about this though? f(x)=\frac{x^2}{a^3-x^3}=\frac{x^2}{a^3}\frac{1}{1-\left(\frac{x}{a}\right)^3}. Now, just for the sake of clarity try calling z=\left(\frac{x}{a}\right)^3. Look familiar?
    Ah yes, that does it. I wasn't seeing the basic algebra to get my original expression into the known geometric series form. Thank you very much!

    Brian
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  7. #7
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    ok! i get u: u were talking about expanding it as a geometric series with a common ratio and all that......
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