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

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agent0smith
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \sqrt{\frac{1x}{1+x}}=1+x+\frac{ x^2 }{ 2 }\] 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"

KImusta
 one year ago
Best ResponseYou've already chosen the best response.0You're correct, but please note the small changes to the question. I've edited

phi
 one year ago
Best ResponseYou've already chosen the best response.0You use a taylor series expansion around x=0 http://en.wikipedia.org/wiki/Taylor_series

KImusta
 one year ago
Best ResponseYou've already chosen the best response.0Thanks 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.

surjithayer
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{\frac{ 1X }{ 1+X }}\times \sqrt{\frac{ 1+X }{ 1+X }}=\frac{ \sqrt{1X ^{2}} }{ 1+X }\] \[\sqrt{\frac{ 1X }{1+X }}\times \sqrt{\frac{ 1X }{ 1X }} =\frac{ 1X }{\sqrt{1X ^{2}} }\] \[=\left( 1X \right)\times \left( 1X ^{2} \right)^{\frac{ 1 }{ 2 }}\] \[=\left( 1X \right)\left( 1+\left( \frac{ 1 }{2 } \right)\left( X ^{2} \right) \right)\] \[=1+\frac{ X ^{2} }{2 }X+Terms containing x ^{3} & higher powers=1x+\frac{ x ^{2} }{ 2 }\]

surjithayer
 one year ago
Best ResponseYou've already chosen the best response.0\[=1x+\left( \frac{ 1 }{2 } \right)\left( x ^{2} \right)\]
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