Let f(x) = x^2 - 3 (so a solution to f(x) = 0 is x =sqrt3

Let f(x) = x^2 - 3 (so a solution to f(x) = 0 is x =sqrt3). Apply the Newton-
Raphson method to f(x), with a starting value x0 = 2, to obtain an approximation
top sqrt3 which is accurate to 6 decimal places. Check your answer by evaluating
sqrt3 on your calculator.

The Newton-Raphson formula is
x₁ = x₀ - (f(x₀))/f'(x₀)
f'(x) = 2x hence
f(x₀) = f(2) = 1; f'(x₀) = 4

x₁ = 2 - 1/4 = 1.75
Now replace x₀ by 1.75.
x₁ = 1.75 - [1.75² - 3] / (2*1.75)
... = 1.7321
which now becomes x₀

x₁ = 1.7321 - [1.7321² - 3] / (2*1.7321)
... = 1.732051 correct to six decimal places.
(actually it's 1.7320508 to seven decimal places, which still agrees with the calculator value)