1. ## Approximation of 1/(2+x)

I've just learned that 4) $\displaystyle \frac{1}{1-x}=1+x+x^2+...+x^n$ where $\displaystyle |x|<1$. I am looking at an approximation of:

$\displaystyle \frac{1}{2+x}$ which makes use of a technique called scaling and also applies 4) to reach an approximation. The whole solution looks like:

$\displaystyle \frac{1}{2+x}=\frac{1/2}{1+x/2} \approx \frac{1}{2}(1-\frac{x}{2})$

The denominator of the equation goes from $\displaystyle 2+x$ to $\displaystyle 1+x/2$?

It is said that the approximation is developed by scaling and using 4). I am trying to determine what happened to the original equation to reach the approximation.

Thanks for any responses...

2. ## Re: Approximation of 1/(2+x)

Originally Posted by sepoto
I've just learned that 4) $\displaystyle \frac{1}{1-x}=1+x+x^2+...+x^n$ where $\displaystyle |x|<1$. I am looking at an approximation of:

$\displaystyle \frac{1}{2+x}$ which makes use of a technique called scaling and also applies 4) to reach an approximation. The whole solution looks like:

$\displaystyle \frac{1}{2+x}=\frac{1/2}{1+x/2} \approx \frac{1}{2}(1-\frac{x}{2})$

The denominator of the equation goes from $\displaystyle 2+x$ to $\displaystyle 1+x/2$?

It is said that the approximation is developed by scaling and using 4). I am trying to determine what happened to the original equation to reach the approximation.

Thanks for any responses...
$\displaystyle \frac{1}{2 \left(\frac{x}{2}+1\right)}$ = $\displaystyle \frac{1}{2}\left(\frac{1}{1--\frac{x}{2}}\right)$

and now they take the first couple of terms of the series representation

= $\displaystyle \frac{1}{2} \left(1-\frac{x}{2}\right)$