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

Question

Cameron Alexander · Accepted Answer

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

Christopher Agresta · Answer

undefined To use linear approximation to estimate the square root of 3.60, you start with a known value close to 3.60. Let's take $ x = 3.6 $. Then, we find the square root of this known value $ f(x) = \sqrt{x} $.

Next, we find the derivative of $ f(x) $ at $ x = 3.6 $, denoted as $ f'(x) $.

After finding $ f(x) $ and $ f'(x) $, we use the linear approximation formula:

$$ f(a + \Delta x) ≈ f(a) + f'(a) * \Delta x $$

where $ a $ is the known value (in this case, 3.6), $ f(a) $ is the known square root of that value, and $ \Delta x $ is the difference between the value you're estimating and the known value (in this case, $ \Delta x = 3.60 - 3.6 = 0.0 $).

Plugging in the values, you can approximate $ \sqrt{3.60} $ using linear approximation.