to solve the equation f(x)=0 by Newton's Method we start with a good initial guess x0 and then run the iteration xn+1=xn- [f(xn)/f'(xn)], n = 0, 1, 2,... until we get an approximation xn+1. Suppose that you want to compute the cube root of 4 by solving the equation x^3-4=0.

Since 1^3=1 and 2^3=8, let's start with x0=1.5
Then x1=?, x2=?, x3=? (with at least 6 correct digits beyond the decimal point.)

Question

to solve the equation f(x)=0 by Newton's Method we start with a good initial guess x0 and then run the iteration xn+1=xn- [f(xn)/f'(xn)], n = 0, 1, 2,... until we get an approximation xn+1. Suppose that you want to compute the cube root of 4 by solving the equation x^3-4=0.

Since 1^3=1 and 2^3=8, let's start with x0=1.5
Then x1=?, x2=?, x3=? (with at least 6 correct digits beyond the decimal point.)

anonymous · Answer

$$f(x)=x^3-4$$
$$f'(x)=3x^2$$
$$x_{n+1}=x_{n}-\frac{f(x_n)}{f'(x_n)}$$
$$x_{n+1}=x_n-\frac{x_n^3-4}{3x_n^2}$$

anonymous · Answer

we can either do the algebra first and then plug in the numbers or just plug them in, it doesn't matter.

anonymous · Answer

by "algebra" i mean write 
$$x_{n+x}=\frac{3x_n^3-x_n^3+4}{3x_n^3}=\frac{2x_n^3+4}{3x_n^2}$$

anonymous · Answer

that should be 
$$x_{n+1}$$ not 
$$x_{n+x}$$

anonymous · Answer

now plug in 1.5 and see what you get.  then do it again with the result.  a spread sheet would be  nice, but a calculator will do.

anonymous · Answer

Got it. Thanks, sat :]