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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
 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
 one year ago

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agent0smithBest 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"
 one year ago

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

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

KImustaBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

surjithayerBest ResponseYou've already chosen the best response.0
\[=1x+\left( \frac{ 1 }{2 } \right)\left( x ^{2} \right)\]
 11 months ago
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