Counterexample

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

Why it fails

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

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Keep practicing

Numerical convergence

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

Drill statements like this

Go deeper· Explained in full in Root finding and orders of convergence· Asked in CSIR NET December 2025· From Numerical Analysis › Root finding: bisection, Newton–Raphson, fixed point, order of convergence

Published 23 Aug 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

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