How do you use linear approximation to the square root function to estimate square roots #sqrt 3.60#?

1 Answer
Jan 29, 2017

#1.9-0.01/(2 xx 1.9)#
#1.8973684 ...#
compared to exact value of
#1.8973665... #

Explanation:

Noting that #2.0^2=4.0# exactly and #1.9^2=3.61# exactly, do a Taylor Series or Binomial Series to get a linear approximation to #sqrt(x)#around #3.61#:

Using the Taylor series we get
#f(a+h) = f(a)+hf'(a)...#
Setting #f(x)=sqrt(x)#, #f prime(x)=(1/2) xx 1/sqrt(x)#, #a=3.60#, #h=-0.01# and truncating after the term in #h#we get.
#sqrt(3.61-0.01)approx sqrt(3.61) - 0.01 xx (1/2) xx 1/(sqrt(3.61))#
#sqrt(3.60) approx 1.9-0.01/(2 xx 1.9)#
which is about #.0001%# too high.

If you prefer the Binomial Series:
#(3.61-0.01)^(1/2)#
#=1.9 xx ( 1-0.01/3.61)^(1/2)#
#=1.9 xx (1-(1/2)xx(0.01/3.61)...)#
#approx1.9-(1/2)xx(0.01/1.9)#
as before.