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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    Can someone help me with this question?

    1. Using the result x^3 + 1 = (x+1)(x^2 - x + 1), express \frac{x^5 - 1}{x^2(x^3 + 1)} in partial fractions.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kleyzam View Post
    Can someone help me with this question?

    1. Using the result x^3 + 1 = (x+1)(x^2 - x + 1), express \frac{x^5 - 1}{x^2(x^3 + 1)} in partial fractions.
    You would want to polynomial long divide here, not PFD.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kleyzam View Post
    Can someone help me with this question?

    1. Using the result x^3 + 1 = (x+1)(x^2 - x + 1), express \frac{x^5 - 1}{x^2(x^3 + 1)} in partial fractions.
    \frac{x^5 - 1}{x^2(x^3 + 1)}=\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{Dx + E}{x^2 - x + 1}
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    You would want to polynomial long divide here, not PFD.
    i think, after that long division, s/he would still have a remainder in 2nd degree
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  5. #5
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    I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
     -x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1) + Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    i think, after that long division, s/he would still have a remainder in 2nd degree
    Yes, they will, but wow will it be much easier.
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kleyzam View Post
    I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
     -x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1) + Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)
    completely solve your RHS.. then factor every powers of x.
    x^5:
    x^4:
    x^3:
    x^2:
    x:
    constant:
    Last edited by kalagota; July 14th 2008 at 04:16 AM.
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  8. #8
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    Quote Originally Posted by kleyzam View Post
    I do arrive to this stage of the problem yet i'm finding it very hard to eliminate the coefficients:
     -x^2 - 1 = Ax(x+1)(x^2-x+1) + B(x+1)(x^2-x+1) + Cx^2(x^2 - x + 1) + (Dx + E)x^2(x^2 - x + 1)
    In your LHS, the (-x^2 -1),
    should that not be (x^5 -1)?

    Then in your last term in the the RHS, the (Dx +E)(x^2)(x^2 -x +1),
    should that not be (Dx +E)(x^2)(x +1)?
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  9. #9
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ticbol View Post
    In your LHS, the (-x^2 -1),
    should that not be (x^5 -1)?
    he already divided it..

    and yes, the last term should be the one what ticbol noted.
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  10. #10
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    Quote Originally Posted by kalagota View Post
    he already divided it..
    He divided, or simplified, the original LHS, but he did not do that in the original RHS?

    Then his final "equation or identity", even if the last term in the new RHS is corrected, is wrong.
    Because if he multiplied the original RHS by (x^2)(x^3 +1), then he should have multiplied the original LHS by the same amount.

    Clearly, (-x^2 -1) is not [(x^5 -1) / (x^2)(x^3 +1)] * [(x^2)(x^3 +1)].
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ticbol View Post
    He divided, or simplified, the original LHS, but he did not do that in the original RHS?

    Then his final "equation or identity", even if the last term in the new RHS is corrected, is wrong.
    Because if he multiplied the original RHS by (x^2)(x^3 +1), then he should have multiplied the original LHS by the same amount.

    Clearly, (-x^2 -1) is not [(x^5 -1) / (x^2)(x^3 +1)] * [(x^2)(x^3 +1)].
    of course it is not equal to that..

    what he meant was..

    \frac{x^5 -1}{(x^2)(x^3 +1)} = 1 + \frac{-x^2-1}{(x^2)(x^3 +1)}

    and \frac{-x^2-1}{(x^2)(x^3 +1)} is the one he is solving now..
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  12. #12
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    Quote Originally Posted by kalagota View Post
    of course it is not equal to that..

    what he meant was..

    \frac{x^5 -1}{(x^2)(x^3 +1)} = 1 + \frac{-x^2-1}{(x^2)(x^3 +1)}

    and \frac{-x^2-1}{(x^2)(x^3 +1)} is the one he is solving now..
    I see.
    Then what about the "1" in the new LHS? How would he incorporate that back into the final decomposition?
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  13. #13
    MHF Contributor kalagota's Avatar
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    well, given the situation he can always replace it by

    \frac{x}{x}, \frac{x^2}{x^2}, etc and add it to where it should be added..
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  14. #14
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    Quote Originally Posted by kalagota View Post
    well, given the situation he can always replace it by

    \frac{x}{x}, \frac{x^2}{x^2}, etc and add it to where it should be added..
    You lost me there.

    So the final decomposition might be:
    x/x + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
    or, (x^2 / x^2) + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
    Etc....

    And so the the final integral would be
    = x +{INT[the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]dx} +C?

    Umm...
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  15. #15
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by ticbol View Post
    You lost me there.

    So the final decomposition might be:
    x/x + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
    or, (x^2 / x^2) + [the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]?
    Etc....
    yes..

    Quote Originally Posted by ticbol View Post
    And so the the final integral would be
    = x +{INT[the decomposition of (-x^2 -1) / (x^2)(x^3 +1)]dx} +C?

    Umm...
    i don't think he wants to integrate the whole thing.. but if so, yes, it would be like that..
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