Newton exact for linear functions

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Explain why the Newton iteration is exact in a single step for linear functions $f(x)=ax+b$ (with $a\ne 0$).

Solution

For $f(x)=ax+b$ we have $f'(x)=a$. Starting from any $x_0$, one Newton step gives $x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = x_0 - \frac{ax_0+b}{a} = x_0 - x_0 - \frac{b}{a} = \boxed{-\frac{b}{a}}.$ But $-\dfrac{b}{a}$ is exactly the zero of $f$ (solving $ax+b=0$). So Newton finds the root in a single step, whatever the initial estimate.

The geometric reason is immediate: Newton’s method replaces the function with its tangent line and looks for its zero. If the function is already a line, the tangent coincides with the function itself, and its zero is the exact solution.

Links

Topics: Continuity
Concepts: Newton-Raphson method · Line · Tangent line
Methods: Newton-Raphson
Skills: Proving
Exercise types: Proof