Question

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?

Answer 1

\[\large \sqrt{\frac{1-x}{1+x}}=1+x+\frac{ x^2 }{ 2 }\]
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"

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

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