## KImusta Group Title How do we show that if x is small enough for its cube to be assumed and higher powers to be neglected, √((1-x)/(1+x))=1-x+x^2⁄2? one year ago one year ago

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1. agent0smith

$\large \sqrt{\frac{1-x}{1+x}}=1+x+\frac{ x^2 }{ 2 }$ right? Not 100% sure what the question is, you may want to rewrite this "if x is small enough for itscube to be assumed and higher powers to be neglected"

2. KImusta

You're correct, but please note the small changes to the question. I've edited

3. phi

You use a taylor series expansion around x=0 http://en.wikipedia.org/wiki/Taylor_series

4. KImusta

Thanks phi. You gave me a nice idea. I applied the Binomial Theorem to expand the expression as far as x^2 and I got the solution.

5. surjithayer

$\sqrt{\frac{ 1-X }{ 1+X }}\times \sqrt{\frac{ 1+X }{ 1+X }}=\frac{ \sqrt{1-X ^{2}} }{ 1+X }$ $\sqrt{\frac{ 1-X }{1+X }}\times \sqrt{\frac{ 1-X }{ 1-X }} =\frac{ 1-X }{\sqrt{1-X ^{2}} }$ $=\left( 1-X \right)\times \left( 1-X ^{2} \right)^{\frac{ -1 }{ 2 }}$ $=\left( 1-X \right)\left( 1+\left( \frac{ -1 }{2 } \right)\left(- X ^{2} \right) \right)$ $=1+\frac{ X ^{2} }{2 }-X+Terms containing x ^{3} & higher powers=1-x+\frac{ x ^{2} }{ 2 }$

6. surjithayer

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