Binomal Expansion

**BabyMilo** · February 1st 2011, 11:20 AM

(1+ax)/(1+bx)

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

is this correct upto x^2 terms?

thanks

**Ithaka** · February 1st 2011, 11:36 AM

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

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

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

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

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

Hence, your initial expression will be:

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

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

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

So yes, your expansion is correct.

**TheCoffeeMachine** · February 1st 2011, 12:02 PM

It's correct! I happen to find the following method to be better than the binomial theorem:

$\begin{aligned}\displaystyle \frac{1+ax}{1+bx} & = (1+ax)\sum_{k\ge 0}(-1)^kb^kx^k = \sum_{k\ge 0}(-1)^k\left(ab^kx^{k+1}+b^kx^k\right) \\& = (ax+1)-(abx^2+bx)+(ab^2x^3+b^2x^2)- \cdots \\& = 1+(a-b)x+(b^2-ab)x^2+\cdots\end{aligned}$

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