# expansion of infinite series

• February 25th 2009, 11:11 PM
thereddevils
expansion of infinite series
$f(x)=\frac{x^2+5x}{(1+x)(1-x)^2}$

If the expansion of f(x) in ascending power of x is

$c_0+c_1x+c_2x+c_3x+...+c_rx^r+...$

find $c_0,c_1$ and $c_2$ and show that $c_3=11$
• February 26th 2009, 09:57 AM
Jester
Quote:

Originally Posted by thereddevils
$f(x)=\frac{x^2+5x}{(1+x)(1-x)^2}$

If the expansion of f(x) in ascending power of x is

$c_0+c_1x+c_2x+c_3x+...+c_rx^r+...$

find $c_0,c_1$ and $c_2$ and show that $c_3=11$

First notice that

$\frac{x^2+5x}{(1+x)(1-x)^2} = \frac{3}{(1-x)^2} - \frac{2}{1-x} - \frac{1}{1+x}$

If you consider the following geometric power series

$\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 \cdots$

so

$\frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - + \cdots$

$\frac{d}{dx} \left( \frac{1}{1-x}\right) = \frac{1}{(1-x)^2} = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 \cdots$

Substitute above giving

$3\left( 1 + 2 x + 3 x^2 + 4 x^3\right) - 2 \left( 1 + x + x^2 + x^3\right) - \left( 1 - x + x^2 - x^3\right) \cdots$
$=(3-2-1) + (6 - 2+1)x + (9 - 2-1)x^2 +(12 - 2 +1) x^3 \cdots$
$=5x + 6x^2 + 11 x^3 \cdots$