Asked by Paul
                Engineers and physicists are fond of using the approximation Square root (1+x) = 1+(x/2)?
What approximation do you suppose the engineers and physicists would use for the quantity (1+x)^(3/2)
            
        What approximation do you suppose the engineers and physicists would use for the quantity (1+x)^(3/2)
Answers
                    Answered by
            drwls
            
    The approximation is only good when x is much less than 1.
It is obtained from the Taylor series rule
f(x) = f(0) + f'(0) * x
= 1 + (3/2)*(1+x)^(1/2)*x + ...
(evaluated at x=0)
= 1 + (3/2)x
    
It is obtained from the Taylor series rule
f(x) = f(0) + f'(0) * x
= 1 + (3/2)*(1+x)^(1/2)*x + ...
(evaluated at x=0)
= 1 + (3/2)x
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