# When does Newton-Raphson Converge/Diverge?

Is there an analytical way to know an interval where all points when used in Newton-Raphson will converge/diverge?

I am aware that Newton-Raphson is a special case of fixed point iteration, where:

$$g(x) = x - \frac{f(x)}{f'(x)}$$

Also I've read that if $|f(x)\cdot f''(x)|/|f'(x)^2| \lt 1$, then convergence is assured. I am just not sure on how to use this fact? Could anyone give me some examples? Thanks.

• In general, no. – copper.hat Feb 13 '16 at 8:07
• The "fact" can be used to show that the underlying fixed point iteration is a contraction mapping. Have you looked at it in this perspective? Is this of interest? – hardmath Feb 13 '16 at 8:15
• Explain Newton’s method for f(x) = x 3+x−2 = 0. Show that Newton’s method converges if x0 ∈ [1− 1/30 , 1+1/30 ] to a limit L. Find an error estimate for the error en = |xn−L|. (Hint. x 3 −3x 2 +2 = (x−1)(x 2 −2x−2) and |x 2 − 2x − 2| ≤ 10 if 0 ≤ x ≤ 2.) – Anonymous Gal Feb 13 '16 at 8:21
• What do you mean? I really didn't understand what the book said about that part it just said that an interval exist such that which ever x i get from [r-e,r+e] will give me convergence...but what the hell is e??? – Anonymous Gal Feb 13 '16 at 8:24
• from the above question. e=1/30. how was that obtained? – Anonymous Gal Feb 13 '16 at 8:26

Consider the solution of $$f(x) = 0,$$ where $f : \mathbb{R} \rightarrow \mathbb{R}$ is at least two times differentiable with continuous derivatives and has a single root $x=r$ of multiplicity $1$. This last assumption ensures $$f'(r) \not = 0$$ which will be needed later. Let $x_n$ denote an approximation of $r$ obtained by any means necessary. Then by doing a Taylor expansion at $x=x_n$ we obtain $$0 = f(r) = f(x_n) + f'(x_n)(r-x_n) + \frac{f''(\xi)}{2}(r-x_n)^2$$ or equivalently $$- f(x_n) = f'(x_n)(r-x_n) + \frac{f''(\xi)}{2}(r-x_n)^2$$ for at least one $\xi_n$ between $r$ and $x_n$. This allows us to express Newton's iteration as $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x_n + \frac{f'(x_n)(r-x_n) + \frac{f''(\xi_n)}{2}(r-x_n)^2}{f'(x_n)}$$ While obscure, this representation allows us to immediately conclude that $$r- x_{n+1} = -\frac{f''(\xi_n)}{2 f'(x_n)}{(r-x_n)^2}$$ This is the equation which you can use to show convergence of Newton's method. Let us define the error at the $n$th step as $$e_n = r - x_n$$ then we can write $$e_{n+1} = - \frac{f''(\xi_n)}{2 f'(x_n)} e_n^2$$ Now since $f'(r) \not = 0$ we can find an interval $I = [r-\delta,r+\delta]$ surrounding the root and determine a constant $M > 0$ such that $$\forall : x, y \in I \: : \: \left| \frac{f''(x)}{2 f'(y)} \right| \leq M.$$ Here the continuity of $f'$ and $f''$ is critical. Then we can write $$|e_{n+1}| \leq M |e_n|^2 = (M|e_n|) |e_n|.$$ It follows that if $x_0 \in I$ is picked such that $$M|e_n| \leq \rho < 1$$ then not only will the error decrease, but (and this is critical) $x_1$ will belong to $I$, allowing the argument to be repeated, leading to the (pessimistic) estimate $$e_n \leq \rho^n |e_0|$$ which nevertheless establishes (local) convergence of Newton's method.
As the iteration converges, it will sooner rather than later do so quadratically, as $$\frac{e_{n+1}}{e_n^2} = - \frac{f''(\xi_n)}{2 f'(x_n)} \rightarrow -\frac{f''(r)}{2 f'(r)}, \quad n\rightarrow \infty, \quad n \in \mathbb{N}.$$ Here it is critical that Taylor's theorem ensure that $\xi_n$ is betweeen $x_n$ and $r$. Since $x_n \rightarrow r$ the squeeze lemma will ensure that $\xi_n \rightarrow r$ as $n \rightarrow \infty$.
A theoretically nice but practically nearly useless answer is provided by the Newton-Kantorovich theorem: If $L=M_2$ is an upper bound for the magnitude of the second derivative over some interval $I$, and with $x_0\in I$ and the first step $s_0=-\frac{f(x_0)}{f'(x_0)}$ the "ball" $B(x_0+s_0,|s_0|)=(x_0+s_0-|s_0|,x_0+s_0+|s_0|)$ is contained in $I$ and $$L·|f'(x_0)^{-1}|^2·|f(x_0)|\le\frac12$$ then there is a unique root inside that ball and Newton's method converges towards it.