Calculate the value of the square root of 3 to four decimal places using the Newton-Raphson process

Newton-Raphson is used for finding a numerical solution to equations of the form f(x)=0. The question asks for the square root of 3, which algebraically looks like this: x = sqrt(3) This can be rewritten in the form f(x)=0, giving: x^2 - 3 = 0 (Note that in doing this you also create the negative solution by squaring it but this can be ignored in this question) You can now calculate the Newton function for f(x), which is: xn+1 = xn - (xn^2 - 3)/(2xn) Now choose a sensible x0, 1.5 is a good place to start as you know the answer will be in (1,2) as 1^2 = 1 and 2^2 = 4 So now you just apply the iteration until the first four decimal places do not change: x0 = 1.5 x1 = 1.75 x2 = 1.73214... x3 = 1.73205... x4 = 1.73205... So after 4 iterations the first 4 decimal places do not change. So the answer is 1.7321