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Math Help - Finding the Constants in partial fractions

  1. #1
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    Finding the Constants in partial fractions

    Find constants A, B, and C such that:


    1/(x^3 - x) = A/(x-1) + B/(x) + C/(x+1)




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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    Find constants A, B, and C such that:


    1/(x^3 - x) = A/(x-1) + B/(x) + C/(x+1)




    Thanks

    Note: 1/(x^3 -x)=1/x(x^2 - 1) = 1/x(x+1)(x-1) = A/(x-1) + B/x + C/(x+1)

    The Lcm of A/(x-1) + B/x + C/(x+1) is x(x+1)(x-1), so we can rewrite A/(x-1) + B/x + C/(x+1) as [A(x^2 + x) + B(x^2 -1) + C(x^2 -x)]/x(x+1)(x-1)

    Expanding the numerator we get:
    (Ax^2 + Ax + Bx^2 - B + Cx^2 - Cx)/x(x+1)(x-1)
    Grouping like powers of x we get:
    [(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)

    So now we have 1/x(x+1)(x-1) = [(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)
    Since the denominators are equal, the numerators must be equal to maintain the equation.

    => 1 = (A+B+C)x^2 + (A-C)x - B
    Now we equate like powers of x (Note we can think of 1 as 0x^2 + 0x +1)

    Then A+B+C=0-------------(1)
    A-C=0----------------(2)
    B = -1 ----------------(3)

    => A+C=1 -------------------Plugged in the value of B
    A-C=0 --------------------rewrite equation 2

    Adding these equations eliminates C, so we end up with:

    2A=1 => A=1/2.
    But A+C = 1, => C=1/2.

    Thus the constants are A=1/2, B=-1, C=1/2

    So 1/(x^3 - x) = (1/2)/(x-1) - 1/(x) + (1/2)/(x+1), and you can check this.


    There is another major method used to come up with the 3 equations, but this is the mainstream one I think.
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  3. #3
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    Hello, Mr_Green!

    This is a problem in Partial Fractions.


    Find constants A, B, and C such that:
    . . 1/(x - x) = A/(x - 1) + B/(x) + C/(x + 1)

    . . . . . . . . . . . . . .1 . . . . . . . .A . . . .B . . . . C
    We have: . ----------------- . = . --- + ------ + -------
    . . . . . . . . x(x - 1)(x + 1) . . . . x . . .x - 1 . . x + 1


    Multiply through by x(x - 1)(x + 1):

    . . 1 .= .A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)


    Let x = 0: . 1 .= .A(-1)(1) + B(0)(1) + C(0)(-1) . . A = -1

    Let x = 1: . 1 .= .A(0)(2) + B(1)(2) + C(1)(0) . . B =

    Let x = -1: .1 .= .A(-2)(0) + B(-1)(0) + C(-1)(-2) . . C =

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Yup, that's a much nicer way to do it, gets to the point quickly, good for exams. (I think you switched the values for A and B though).
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