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Math Help - Completing the Square

  1. #1
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    Completing the Square

    By factoring x^3 +1 as a sum of cubes, rewrite the integral in part a, then express
    1/ (x^3 +1) as the sum of a power series and use it to prove the formula for pi.

    n= [3(sq. rt 3)] / 4 , the sum of the series from n=0 to infinity [(-1) ^n] / 8^n [(2/3n+1) + (1/3n+2)]
    Last edited by mr fantastic; August 16th 2010 at 09:14 PM. Reason: Deleted question already asked in another thread.
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  2. #2
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    Rewrite what integral? Are you asked to find \int{\frac{1}{x^3 + 1}\,dx}?


    (x^3 + 1) = (x + 1)(x^2 - x + 1), so

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

    Now rewriting using partial fractions...

    \frac{A}{x + 1} + \frac{Bx + C}{x^2 - x + 1} = \frac{1}{(x + 1)(x^2 - x + 1)}

    \frac{A(x^2 - x + 1) + (Bx + C)(x + 1)}{(x + 1)(x^2 - x + 1)} = \frac{1}{x^2 - x + 1}

    A(x^2 - x + 1) + (Bx + C)(x + 1) = 1

    Ax^2 - Ax + A + Bx^2 + Bx + Cx + C = 1

    (A + B)x^2 + (-A + B + C)x + A + C = 0x^2 + 0x + 1

    which gives

    A + B = 0
    -A + B + C = 0
    A + C = 1

    Solving these equations simultaneously gives A = \frac{1}{3}, B = -\frac{1}{3}, C = \frac{2}{3}, so that means

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


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