Question

Calculate one iterations of Newton's Method to approximate a zero for f(x) = x ^ 2 - 3 where x 1 =2 Write your answer in decimal form.

Answer 1

Recall that
x = x - f/f'
So, starting with x1=2
x2 = x1 - f(x1)/f'(x1) = 2 - f(2)/f'(2) = 2 - (2^2-3)/(2*2) = 2 - 1/4 = 7/4
f(x2) = 1/16 ... getting closer, eh?

Answer 2

f(x) = x^2 - 3
f ' (x) = 2x

x_new = x - f(x)/f ' (x)
= x - (x^2 - 3)/2x
= (2x^2 - x^2 + 3)/2x
= (x^2 + 3)/(2x)

x_new x
2 1.75 <===== one iteration
1.75 1.7321
1.7321 1.73205
1.73205 1.73205 ---- repeat to 5 decimals

(2 more iterations gave the answer correct to 9 decimals on my calculator)

so x^2 - 3 = 0 has a root at x = 1.73205

actual answer:
x^2 - 3 = 0
x^2 = 3
x = ± √3 = appr ± 1.73201..