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November 28, 2014

November 28, 2014

Posted by **Anonymous** on Thursday, March 6, 2008 at 9:33pm.

(1+x)^k=1+kx to find an approximation for the function f(x)=1/square root of (4+x) for values of x near zero.

- calculus -
**Count Iblis**, Thursday, March 6, 2008 at 9:43pmLet's go astep further and also work out the quadratic term using the general formula:

(1+x)^k=1+kx + k(k-1)/2 x^2 +

k(k-1)(k-2)/3! x^3 + ....

f(x)= (4 + x)^(-1/2) =

4^(-1/2) (1 + x/4)^(-1/2) =

1/2 [1 - x/8 + 3/128 x^2 + ...] =

1/2 - x/16 + 3/256 x^2 + ...

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