How do you prove that the square root of a number #S# can be approximated by using the recurrence relation: #x_(n+1) = 1/2(x_n+S/x_n)# ?

Not sure how to prove this or why it works?

1 Answer
sente
Mar 8, 2016

This is an application of Newton's method.

Let #f(x) = x^2 - S => f'(x) = 2x#

Then the zeros of #f# are #+-sqrt(S)# and thus, applying Newton's method, can be approximated by selecting an #x_0# that is "close" to the desired root and applying the recurrence relation

#x_(n+1) = x_n - f(x_n)/(f'(x_n))#

#=x_n - ((x_n^2-S)/(2x_n))#

#=x_n - x_n/2 + S/(2x_n)#

#=(x_n)/2+S/(2x_n)#

#=1/2(x_n+S/x_n)#

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