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Thread: Binomial Series Question

  1. #1
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    Binomial Series Question

    Given that $\displaystyle b > a > 0$, find by using the binomial theorem, coefficients $\displaystyle c_m$ such that.

    $\displaystyle \frac{1}{(1-ax)(1-bx)} = c_0 + c_1 x + c_2 x^2 + .... c_m x^m + .....$, for $\displaystyle b|x| < 1$

    Show that $\displaystyle c^2_m = \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m +2}}{(a-b)^2}$

    hence, or otherwise, show that

    $\displaystyle c^2_0 + c^2_1 x + c^2_2 x^2 + .... c^2_m x^m + .... = \frac{1 + abx}{(1-abx)(1-a^2x)(1-b^2x)}$

    for x in suitable interval which you should determine



    For the first part I first split into partial fraction using "the cover up rule" to get

    $\displaystyle \frac{1}{(1-ax)(1-bx)} = \frac{a}{a-b} \times \frac{1}{1-ax} + \frac{b}{b-a} \times \frac{1}{1-bx} $

    then using the binomial expansion.

    $\displaystyle \Rightarrow \frac{a}{a-b} \times (1 +ax + a^2x^2 + a^3x^3 ....) + \frac{b}{b-a} \times (1 +bx + b^2x^2 + b^3x^3 ....) $

    $\displaystyle \Rightarrow \frac{a}{a-b} + \frac{b}{b-a} \ \ + \ \ \left( \frac{a}{a-b} \times a + \frac{b}{b-a} \times b \right ) x \ \ +\ \ \left( \frac{a}{a-b} \times a^2 + \frac{b}{b-a} \times b^2 \right) x^2 ....$

    giving $\displaystyle c_m = \frac{a^{m+1}}{a-b} + \frac{b^{m+1}}{b-a} $

    $\displaystyle \Rightarrow \frac{a^{m+1} - b^{m+1}}{a-b}$

    giving $\displaystyle c^2_m = \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m +2}}{(a-b)^2}
    $.

    I am just struggling on the next part.

    Bobak
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by bobak View Post
    For the first part I first split into partial fraction using "the cover up rule" to get

    $\displaystyle \frac{1}{(1-ax)(1-bx)} = \frac{a}{a-b} \times \frac{1}{1-ax} + \frac{b}{b-a} \times \frac{1}{1-bx} $

    then using the binomial expansion.

    $\displaystyle \Rightarrow \frac{a}{a-b} \times (1 +ax + a^2x^2 + a^3x^3 ....) + \frac{b}{b-a} \times (1 +bx + b^2x^2 + b^3x^3 ....) $

    $\displaystyle \Rightarrow \frac{a}{a-b} + \frac{b}{b-a} \ \ + \ \ \left( \frac{a}{a-b} \times a + \frac{b}{b-a} \times b \right ) x \ \ +\ \ \left( \frac{a}{a-b} \times a^2 + \frac{b}{b-a} \times b^2 \right) x^2 ....$

    giving $\displaystyle c_m = \frac{a^{m+1}}{a-b} + \frac{b^{m+1}}{b-a} $

    $\displaystyle \Rightarrow \frac{a^{m+1} - b^{m+1}}{a-b}$

    giving $\displaystyle c^2_m = \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m +2}}{(a-b)^2}
    $.

    I am just struggling on the next part.

    Bobak
    You could try exanding:

    $\displaystyle
    \frac{1 + abx}{(1-abx)(1-a^2x)(1-b^2x)}
    $

    and equating coefficients (rather show that the coefficient of $\displaystyle x^m$ in the series you find is in fact $\displaystyle c_m^2$).

    RonL
    Last edited by CaptainBlack; Mar 6th 2008 at 07:58 AM.
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    You could try exanding:

    $\displaystyle
    \frac{1 + abx}{(1-abx)(1-a^2x)(1-b^2x)}
    $

    and equating coefficients.

    RonL
    Isn't that the "otherwise" method? any other methods we can apply here ?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by bobak View Post
    Isn't that the "otherwise" method? any other methods we can apply here ?
    Well you will be showing that the coefficient of $\displaystyle x^m$ in the series is equal to $\displaystyle c_m^2$, so you will be using the result.

    (and the interval will be an interval of convergence of the series)

    RonL
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