# Proof that Newton's Method gives better and better approximations with each iteration?

I've seen this question and answer: Why does Newton's method work?

It gives some geometric intuition as to what is going on when applying Newton's method, but what I really need to know is why it works that way.

How can you prove that each approximation is closer to the correct value than the previous one?

And also, if the following is indeed true, how can you prove that the "gap" between each two subsequent approximations gets smaller and smaller as you keep going up in the sequence of approximations?

• If you start close enough to a non degenerate root $r$, you can show that the iterates satisfy $|x_{n+1}-r| \le C |x_n-r|^2$. Hence if you start with $C|x_0-r| <1$, then you have the desired conclusion. This is true in higher dimensions as well. It is not too surprising in that each iteration solves the linearised problem, so what is left over is second order. – copper.hat May 27 '16 at 19:13

• @jeremyradcliff To be a bit more precise than Farnight's comment: Newton's method's theory says if you use an exact derivative (rather than a numerical approximation) and start close enough to a non-degenerate root, then you get quadratic convergence. You essentially never get better than that except for linear problems. But you often get something pretty good when you use a numerical approximation to the derivative and/or start far away from the root. For example, Newton's method for calculating $\sqrt{x}$ is the iteration $y_{k+1}=\frac{y_k+x/y_k}{2}$. This converges whenever $x_0>0$. – Ian May 27 '16 at 19:49