# Binomal Expansion

• Feb 1st 2011, 11:20 AM
BabyMilo
Binomal Expansion
(1+ax)/(1+bx)

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

is this correct upto x^2 terms?

thanks
• Feb 1st 2011, 11:36 AM
Ithaka
Quote:

Originally Posted by BabyMilo
(1+ax)/(1+bx)

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

is this correct upto x^2 terms?

thanks

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

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

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

$\displaystyle =1-bx + b^2 * x^2$

Hence, your initial expression will be:

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

$\displaystyle =1-bx+b^2x^2 +ax-abx^2$

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

So yes, your expansion is correct.
• Feb 1st 2011, 12:02 PM
TheCoffeeMachine
It's correct! I happen to find the following method to be better than the binomial theorem:

\displaystyle \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}