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Math Help - Binomal Expansion

  1. #1
    Senior Member
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    Binomal Expansion

    (1+ax)/(1+bx)

    = 1+(a-b)x+(b^2-ab)x^2

    is this correct upto x^2 terms?

    thanks
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  2. #2
    Junior Member
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    Quote Originally Posted by BabyMilo View Post
    (1+ax)/(1+bx)

    = 1+(a-b)x+(b^2-ab)x^2

    is this correct upto x^2 terms?

    thanks
    \frac{(1+ax)}{(1+bx)}=
    =(1+ax) * (1+bx)^{-1}

    Now we expand =(1+bx)^{-1}

    =1+\frac{-1}{1} * (bx)^1 + \frac{(-1) * (-2)}{1 * 2} * (bx)^2

    =1-bx + b^2 * x^2

    Hence, your initial expression will be:

    =(1+ax) * (1-bx + b^2 x^2)

    =1-bx+b^2x^2 +ax-abx^2

    =1+(a-b)x+(b^2-ab)x^2

    So yes, your expansion is correct.
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  3. #3
    Super Member
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    It's correct! I happen to find the following method to be better than the binomial theorem:

    \begin{aligned}\displaystyle \frac{1+ax}{1+bx} & = (1+ax)\sum_{k\ge 0}(-1)^kb^kx^k = \sum_{k\ge 0}(-1)^k\left(ab^kx^{k+1}+b^kx^k\right) \\& = (ax+1)-(abx^2+bx)+(ab^2x^3+b^2x^2)- \cdots \\& = 1+(a-b)x+(b^2-ab)x^2+\cdots\end{aligned}
    Last edited by TheCoffeeMachine; February 1st 2011 at 03:07 PM.
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