TITLE: Proof that Newton Raphson method has quadratic convergence
QUESTION [11 upvotes]: I've googled this and I've seen different types of proofs but they all use notations that I don't understand.
But first of all, I need to understand what quadratic convergence means, I read that it has to   do with the speed of an algorithm. Is this correct?
Ok, so I know that this is the Newton-Raphson method:
$$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}$$
How do I prove that it converges?
Thanks.

REPLY [3 votes]: Let $f$ be twice continuous differentiable on some interval $(a, b)$. Assume that $f(c) = 0$ where $a < c < b$ and that $f'(c) \neq 0$. Then there exists $0 < \epsilon < \min(c - a, b-c)$ with the following property: pick any $x_0 \in (c- \epsilon, c + \epsilon)$ and define iteratively$$x_{n+1} = x_n - {{f(x_n)}\over{f'(x_n)}},\text{ }n \ge 0.\tag*{$(1)$}$$We have that $\{x_n\}_{n=0}^\infty$ converges to $c$ in the following fashion:$$\left|x_{n+1} - c\right| \le M\left|x_n - c\right|^2 \text{ for all }n \ge 0,\tag*{$(2)$}$$where $M$ is some constant.

We may assume $c = 0$, and $\epsilon$ small enough that$$\left|f'(x)\right| > {{\left|f'(0)\right|}\over2}$$ when $\left|x\right| < \epsilon$ for some $B \in \mathbb{R}^+$. Then by Taylor's Theorem,$$f(x_n) - xf'(x_n) = {{x_n^2}\over2} f''(y_n)$$for some $y_n$ between (fix this->)$)$ and $x_n$. Thus, for $\left|x_n\right| < \epsilon$, we have$$\left|x_{n+1}\right| = \left|-x_{n+1}\right|$$$$=\left| {{f(x_n)}\over{f'(x_n)}} - x_n\right|$$$$= {1\over{\left|f'(x_n)\right|}} \cdot \left| f(x_n) - x_n f'(x_n)\right|$$$$= {1\over{\left|f'(x_n)\right|}} \cdot \left| {{x_n^2}\over2} f''(y_n)\right|$$$$\le {2\over{f'(0)}} \cdot {B\over2} x_n^2.$$What we are doing is taking $x_{n+1}$ to be the point where the tangent line to the graph of $f$ at $x_n$ hits the $x$-axis. The inequality we proved shows that for $\left|x_n\right| < \epsilon$, we have$$\left| {{x_{n+1}}\over{x_n}}\right| < Mx_n,$$so if$$\left| x_n\right| < \min\left( \epsilon, {1\over{2M}}\right),$$we have$$\left|x_{n+1}\right| < {{\left|x_n\right|}\over2}.$$Thus,$$\left|x_n\right| \to 0\text{ as } n \to \infty.$$

The rate of convergence in $(2)$ is quadratic and thus faster than in the contraction principle. There the convergence is exponential, here it is super-exponential. This plays an important role in applications, also to problems in pure mathematics (Nash embedding). If you have experience in programming, you could write a brief code which computes this Newton sequence for simple functions of your choice. You will see how rapidly the sequence stabilizes behind the comma.