# Thread: Rational functions and approximate expansions

1. ## Rational functions and approximate expansions

Hi,
I know how to do normal binomial expansions but I can't remember questions quite like this which I've just found in my textbook, by exam being in just over a day.

The first bit is to show that (1+x)/(2+x) - (1-x)/(2-x) = 2x/(4-x^2). I did that fine. The second bit says 'Hence or otherwise show that for small x f(x) = 1/2x + 1/8x^3 + 1/32x^5 and I'm stuck. Any help would be much appreciated.

2. Originally Posted by JeWiSh
Hi,
I know how to do normal binomial expansions but I can't remember questions quite like this which I've just found in my textbook, by exam being in just over a day.

The first bit is to show that (1+x)/(2+x) - (1-x)/(2-x) = 2x/(4-x^2). I did that fine. The second bit says 'Hence or otherwise show that for small x f(x) = 1/2x + 1/8x^3 + 1/32x^5 and I'm stuck. Any help would be much appreciated.
Does it really say "for small x f(x)= that"? Or does it say it is approximately that?
I would be inclined to note that $\frac{2x}{4- x^2}= \frac{2x}{4(1- \frac{x^2}{4})}$ $= \left(\frac{x}{2}\right)\frac{1}{1- \frac{x^2}{4}}$
and recognize that final fraction as the sum of a geometric series:
$\sum_{n=0}^\infty r^n= \frac{1}{1- r}$
(as long as |r|< 1)
with $r= \frac{x^2}{4}$
That is, $\frac{2x}{4- x^2}= \frac{x}{2}\sum_{n=0}^\infty\left(\frac{x^2}{4}\ri ght)^n$ as long as $x^2/4< 1$ or x< 2.

If x is sufficiently small, higher powers of x can be neglected and
$\frac{2x}{4- x^2}= {x/2}(1+ (x^2/4)+ (x^2/4)^2)= x/2+ x^3/8+ x^5/32$ approximately.

3. Yeah it did literally say what I wrote..
Thanks for the help

4. Then it's not true. Even for small values of x those are not equal. For sufficiently small x, they are approximately equal.

(Unless, of course, by "small x" they mean "x= 0"!)