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Math Help - Binomial Series Question

  1. #1
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    Binomial Series Question

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

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

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

    hence, or otherwise, show that

     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

    \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.

    \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 ....)

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

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

    giving c^2_m = \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m +2}}{(a-b)^2}<br />
 .

    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

    \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.

    \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 ....)

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

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

    giving c^2_m = \frac{a^{2m+2} - 2(ab)^{m+1} + b^{2m +2}}{(a-b)^2}<br />
 .

    I am just struggling on the next part.

    Bobak
    You could try exanding:

    <br />
\frac{1 + abx}{(1-abx)(1-a^2x)(1-b^2x)}<br />

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

    RonL
    Last edited by CaptainBlack; March 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:

    <br />
\frac{1 + abx}{(1-abx)(1-a^2x)(1-b^2x)}<br />

    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 x^m in the series is equal to 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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