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 Group Title

$\large \sqrt{\frac{1-x}{1+x}}=1+x+\frac{ x^2 }{ 2 }$ 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 Group Title

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

3. phi Group Title

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

4. KImusta Group Title

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 Group Title

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

6. surjithayer Group Title

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