Numerical methodsLesson 2 of 2 in the course Root finding

Newton's Method: Convergence, Safeguards, and Extensions

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

This lesson introduces Newton's method as a way to approximate a root of a function by repeatedly improving an initial guess.

A root (or zero) of a real-valued function ff is a value α\alpha for which f(α)=0f(\alpha)=0. Newton's method starts with an estimate x0x_0 of a root and generates a sequence of improved estimates, called iterates.

The method is most effective when the initial guess is close to the root and the derivative does not vanish at the iterates. It is an iterative numerical algorithm: each new estimate is calculated from the previous one.

All flashcards

What is a root of a function?
A value α for which f(α)=0f(\alpha)=0.
What is the standard Newton iteration?
xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, provided f′(xn)≠0f'(x_n)\ne 0.
What does quadratic convergence mean?
Near the root, the next error is approximately bounded by a constant times the square of the current error.
How can Newton's method be modified for a root of known multiplicity m?
Use xn+1=xn−mf(xn)f′(xn)x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)}.
What replaces the derivative in Newton's method for a system of equations?
The Jacobian matrix, used to compute a correction by solving a linear system.

Quiz questions

  1. For a simple root α, which condition supports local convergence of Newton's method?Answer: ff is continuously differentiable near α and f′(α)≠0f'(\alpha)\ne 0.
  2. Near a suitable simple root, what error behavior characterizes quadratic convergence?Answer: ∣en+1∣|e_{n+1}| is bounded approximately by C∣en∣2C|e_n|^2.
  3. For a square system F(x)=0F(\mathbf{x})=\mathbf{0}, how is a Newton correction usually computed?Answer: Solve JF(xn) Δx=−F(xn)J_F(\mathbf{x}_n)\,\Delta\mathbf{x}=-F(\mathbf{x}_n).
  4. To use Newton's method to locate a stationary point of a differentiable function gg, which equation is solved?Answer: g′(x)=0g'(x)=0, using xn+1=xn−g′(xn)g′′(xn)x_{n+1}=x_n-\frac{g'(x_n)}{g''(x_n)}.
  5. When the known root multiplicity is mm, which update modifies ordinary Newton iteration?Answer: xn+1=xn−mf(xn)f′(xn)x_{n+1}=x_n-m\frac{f(x_n)}{f'(x_n)}.

Based on the Wikipedia article Newton's method, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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