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