Newton-Raphson method

The reference iterative method: linearize f at the current iterate and jump to the tangent's root. Full derivation along three routes and proof of quadratic order with its error equation.

The formula

xk+1=xkf(xk)f(xk),k=0,1,2,x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)},\qquad k=0,1,2,\dots
Newton-Raphson method.

Geometrically, each iteration replaces the curve by its tangent line at (xk,f(xk))(x_k,f(x_k)) and takes as next approximation the point where that tangent crosses the axis. It is a fixed-point method with ϕ(x)=xf(x)f(x)\phi(x)=x-\frac{f(x)}{f'(x)}: one-point, memoryless and using derivatives.

Derivation and order

DerivationDerivation: Newton-Raphson and its quadratic orderView as its own page →

Route 1: the tangent line

  1. At the current iterate xkx_k, the tangent line to the curve y=f(x)y=f(x) is

    y=f(xk)+f(xk)(xxk)y=f(x_k)+f'(x_k)(x-x_k)
  2. The tangent is the best linear approximation of ff near xkx_k, so we take as next iterate the point where the tangent vanishes (y=0y=0):

    0=f(xk)+f(xk)(xk+1xk)    xk+1=xkf(xk)f(xk)0=f(x_k)+f'(x_k)(x_{k+1}-x_k)\;\Rightarrow\; x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}

Route 2: Fundamental Theorem of Calculus

  1. Write ff via the Fundamental Theorem of Calculus from xkx_k and approximate the integral with a rectangle (constant integrand f(xk)f'(x_k), the same idea as in the Euler derivation):

    f(x)=f(xk)+xkxf(t)dt    f(xk)+f(xk)(xxk)f(x)=f(x_k)+\int_{x_k}^{x}f'(t)\,dt\;\approx\; f(x_k)+f'(x_k)(x-x_k)
  2. Evaluate at x=αx=\alpha, where f(α)=0f(\alpha)=0, and solve for α\alpha:

    0f(xk)+f(xk)(αxk)    αxkf(xk)f(xk)0\approx f(x_k)+f'(x_k)(\alpha-x_k)\;\Rightarrow\;\alpha\approx x_k-\frac{f(x_k)}{f'(x_k)}
  3. That approximation of α\alpha is the next iterate. Approximating the integral with richer quadratures (trapezoid, midpoint, Simpson) produces higher-order methods along this same route.

Proof of order 2 (error equation)

  1. Let α\alpha be a simple root (f(α)=0f(\alpha)=0, f(α)0f'(\alpha)\ne 0), ek=xkαe_k=x_k-\alpha the error and c2=12f(α)f(α)c_2=\frac{1}{2}\frac{f''(\alpha)}{f'(\alpha)}. Expand f(xk)f(x_k) by Taylor around α\alpha; the constant term f(α)f(\alpha) vanishes:

    f(xk)=f(α)ek+12f(α)ek2+O(ek3)=f(α)[ek+c2ek2]+O(ek3)f(x_k)=f'(\alpha)\,e_k+\frac{1}{2}f''(\alpha)\,e_k^2+\mathcal{O}(e_k^3)=f'(\alpha)\bigl[e_k+c_2e_k^2\bigr]+\mathcal{O}(e_k^3)
  2. Expand the derivative as well:

    f(xk)=f(α)+f(α)ek+O(ek2)=f(α)[1+2c2ek]+O(ek2)f'(x_k)=f'(\alpha)+f''(\alpha)\,e_k+\mathcal{O}(e_k^2)=f'(\alpha)\bigl[1+2c_2e_k\bigr]+\mathcal{O}(e_k^2)
  3. Divide both expansions. The factor f(α)f'(\alpha) cancels and, using 11+2c2ek=12c2ek+O(ek2)\frac{1}{1+2c_2e_k}=1-2c_2e_k+\mathcal{O}(e_k^2):

    f(xk)f(xk)=(ek+c2ek2)(12c2ek)+O(ek3)=ekc2ek2+O(ek3)\frac{f(x_k)}{f'(x_k)}=\bigl(e_k+c_2e_k^2\bigr)\bigl(1-2c_2e_k\bigr)+\mathcal{O}(e_k^3)=e_k-c_2e_k^2+\mathcal{O}(e_k^3)
  4. Subtract α\alpha in Newton's formula and substitute: the linear terms in eke_k cancel and the quadratic error equation remains. Newton's method has order p=2p=2.

    ek+1=xk+1α=ekf(xk)f(xk)=c2ek2+O(ek3)e_{k+1}=x_{k+1}-\alpha=e_k-\frac{f(x_k)}{f'(x_k)}=c_2e_k^2+\mathcal{O}(e_k^3)

Beyond Newton

Newton is the starting point of almost everything else: replacing the derivative by differences yields the secant and Steffensen methods; composing and adding weight functions builds the high-order methods (Traub, Ostrowski, Jarratt); and its vector version solves systems of nonlinear equations. It is also the standard tool for the implicit equations arising in implicit ODE methods.