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Math Help - partial fractions cubic base

  1. #1
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    partial fractions cubic base

    i still cant get the correct answer for this,i simply equated coefficients etc anad got as far as the first two terms in the answer. The question is basically
    Q express \frac{x^5-1}{x^2(x^3+1)} in partial fractions ?[/quote] \frac{x^5-1}{x^5+x^2}
    supposed working,,
    \frac{x^5-1+x^2-x^2}{x^5+x^2}

    \frac{x^5+x^2}{x^5+x^2}-\frac{1+x^2}{x^5+x^2}

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

    I'd start it that way. You can try and finish it.

    In this case:

    1+x^2\equiv(Ax+B)(x^3+1)+(Cx^2+Dx+E)(x^2)

    Since this is an identity, it is true for all values of x (this is a great help in determing the values of A,B,C,D and E!).[/quote] i miust be doing something awfully wrong
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  2. #2
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    You can make the last part a bit simpler by factorising 1+x^3 as (1+x)(1-x+x^2). Then you'll want \frac{1+x^2}{x^2(1+x)(1-x+x^2)} = \frac{Ax+B}{x^2} + \frac C{1+x} + \frac{Dx+E}{1-x+x^2}.
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  3. #3
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    Hi, thanks for your reply and your method works but i can't help to not understand why it doesnt work when i use the partial fractions, \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{Dx+E  }{x^2-x+1}
    Using this u can quickly deduce B=1,,,and C=2/3 by elimiting their factors but then when i put in x=1,,i get -2/3=2(A+D+E)
    (x^4),,,E=-1/3
    (x^5),,,A+D=-2/3 which then gives me(putting these together):2A+2D=0

    I know A must be zero and D cant be zero as the answer to this question is
    1-\frac{1}{x^2}-\frac{2}{3(x+1)}-\frac{1-2x}{3(x^2-x+1)}
    NB previous post was 1 - p(x)/q(x) IF SOMEONE COULD HELP THEYD BE REALLY COOL!!!!XD
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  4. #4
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    You seem to be doing it mostly correctly. There must be a simple arithmetic error somewhere. Starting from

    1+x^2\equiv (Ax+B)(1+x^3)+ Cx^2(1-x+x^2) +(Dx+E)x^2(1+x),

    you get B=1 and C=2/3 (by putting x=0 and x=-1). So the identity becomes

    1+x^2\equiv (Ax+1)(1+x^3)+ \tfrac23x^2(1-x+x^2) +(Dx+E)x^2(1+x).

    Now compare the coefficients of x on both sides, to get A=0, so the identity becomes

    1+x^2\equiv 1+x^3 + \tfrac23x^2(1-x+x^2) +(Dx+E)x^2(1+x).

    Next, compare the coefficients of x^4 on both sides, to get 0=\tfrac23+D, so the identity becomes

    1+x^2\equiv 1+x^3 + \tfrac23x^2(1-x+x^2) +(\tfrac23x+E)x^2(1+x).

    Now you can wrap it up by putting x=1 (say) to get an equation for E.
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  5. #5
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    my bad i got it thanks
    Last edited by oxrigby; October 11th 2008 at 02:16 PM.
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