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Is this true?

Newton's method converges quadratically to any root

No — it is false.

The counterexample

$f (x) = x^{2}$ at the root 0

The root is double: $x_{n}$ ₊ $_{1} = x_{n} /2$ , only linear convergence.

The kind of mistake this is

Numerical convergence

A scheme assumed to converge, or to converge at its advertised rate.

Drill statements like this

Others that fail the same way

From Numerical Analysis › Root finding: bisection, Newton–Raphson, fixed point, order of convergence

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