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

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

• Newton's method converges quadratically from sufficiently close to a root of multiplicity one

Proof

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

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