Newton's method converges linearly from sufficiently close to a root of finite multiplicity greater than one

From Calculus
Jump to: navigation, search

Statement

Suppose f is a function of one variable that is at least one time continuously differentiable at a root \alpha. Further, suppose f'(\alpha) = 0, so that \alpha is root of multiplicity greater than 1. Then, there exists \varepsilon > 0 such that for any x_0 \in (\alpha - \varepsilon,\alpha + \varepsilon), the sequence obtained by applying Newton's method either reaches the root in finitely many steps or has linear convergence to the root \alpha. The convergence rate is 1/r where r is the multiplicity of the root \alpha (i.e., the order of \alpha as a zero).

Related facts

Proof

Based on the information about the order of the root \alpha being r, there exists \varepsilon > 0 and constants A such that:

A(x - \alpha)^r \le f(x) \le B(x - \alpha)^r