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Suppose we are employing Newton's method: $$ x_{k+1}=x_k - \frac{f(x_k)}{f'(x_k)}. $$

Suppose $f$ is twice differentiable, $f(c)=0$, $f'(x) \neq 0$ on $(c-h, c+h)$, and $x_1 \in (c-h, c+h)$. Let $\epsilon_k : = x_k - c$. We know that $$ |\epsilon_{k+1}| = \left| \frac{f''(\xi_k)}{2f'(x_k)} \right| {\epsilon_k} ^2, $$ for some $\xi_k$ between $c$ and $x_k$.

Therefore if $\displaystyle |\epsilon_1| \leq \frac{\sup\limits_{x\in (c-h,c+h)}|f''(x)|}{\inf\limits_{x\in (c-h, c+h)}|f'(x)|}$, then we will have convergence (in fact, quadratic convergence).

Another sufficient condition (I believe) is that $f$ is twice differentiable, $f'(x) \neq 0$ on $[c,c+h)$, $x_1 \in [c, c+h)$, and $ff'' > 0$ on $(c,c+h)$ (or similarly in a left-neighborhood), since in this case, we can show that $x_k$ is monotone increasing (decreasing) and bounded above (below) by $c$.

What are some other sufficient conditions for convergence of Newton's Method on $\mathbb{R}$?

Eric Auld
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1 Answers1

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If $x_{n+1}=g(x_n)$ is the iteration, it is sufficient that $x$ belong to a closed interval $I\subset\mathbb{R}$ such that $$ g(I)\subset I, \qquad\text{and}\qquad |g'(x)|<1 \text{ for all }x\in I. $$ This then allows you to use a fixed-point theorem to show that $g$ has a unique fixed point in $I$ and $x_n$ must converge to it.

One difference between this condition and the one you wrote down in the question is that it says nothing about $\epsilon_1$. And indeed you won't ever know $\epsilon_1$ because that amount to knowing the exact root.

Kirill
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