1. ## 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}
$
.

I am just struggling on the next part.

Bobak

2. Originally Posted by bobak
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}
$
.

I am just struggling on the next part.

Bobak
You could try exanding:

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

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

RonL

3. Originally Posted by CaptainBlack
You could try exanding:

$
\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 ?

4. Originally Posted by bobak
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