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Why is my partial fraction decomp. wrong?

  1. Jan 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Decompose [itex]\frac{2(1-2x^2)}{x(1-x^2)}[/itex]

    I get A = 2, B =-1, C = 1, but this doesn't recompose into the correct equation, and the calculators for partial fraction decomposition online all agree that it should be A = 2, B = 1, C = 1.

    Here is one of the online calculator results with steps shown: http://www.emathhelp.net/calculators/algebra-2/partial-fraction-decomposition-calculator/?numer=2(1-2x^2)&denom=x(1-x)(1+x)&steps=on

    2. Relevant equations


    3. The attempt at a solution

    [tex]\frac{2(1-2x^2)}{x(1-x^2)} = \frac{2(1-2x^2)}{x(1-x)(1+x)}[/tex]

    [tex]\frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x} = \frac{2(1-2x^2)}{x(1-x)(1+x)}[/tex]

    [tex]A(1-x)(1+x) + Bx(1+x) + Cx(1-x) = 2(1-2x^2)[/tex]

    Let (i) x = 0, (ii) x = +1, (iii) x = -1

    [tex]x = 0, A = 2(1-0), A = 2[/tex]
    [tex]x = 1, B(1)(1+(1)) = 2(1-2) = -2, 2B = -2, B = -1[/tex]
    [tex]x = -1, C(-1)(1-(-1)) = 2(1-2) = -2, -2C = -2, C = 1 [/tex]

    This of course recomposes to

    [tex]\frac{2(1-x^2)-2x^2}{x(1-x)(1+x)}[/tex]

    not the original equation.

    The only thing the online calculator really did differently from me is change a term in the denominator from [itex]1-x^2[/itex] to [itex]x^2-1[/itex]. I used a shortcut of subbing in the roots of the denominator in order to quickly find the constant numerator values, but even if I use the long method of matching the term coefficients as the online calculator used, I still get my wrong values for A,B and C...why does changing [itex]1-x^2[/itex] to [itex]x^2-1[/itex] matter?
     
    Last edited: Jan 8, 2016
  2. jcsd
  3. Jan 8, 2016 #2

    Samy_A

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    Is there a reason you use ##x-1## instead of ##1-x## under the "B"?:
    [tex]\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1} = \frac{2(1-2x^2)}{x(1-x)(1+x)}[/tex]
     
  4. Jan 8, 2016 #3
    typo, will fix, thanks.
     
  5. Jan 8, 2016 #4

    Samy_A

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    As ##\frac{2(1-x^2)-2x^2}{x(1-x)(1+x)}=\frac{2-2x^2-2x^2}{x(1-x)(1+x)}=\frac{2-4x^2}{x(1-x)(1+x)}=\frac{2(1-2x^2)}{x(1-x²)}##, I don't see why you say that it recomposes to something different from the original expression.

    Your result looks correct to me.
     
  6. Jan 8, 2016 #5
    Selective blindness strikes again...that's worse news actually, because that means it wasn't the part frac decomp where I've messed up the solution to this larger diff eq problem.
     
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