Notes on Newton's Method

Supplementary Material for Honors Calculus
Originally prepared in the Fall Semester 1995

Revised for the Web: July 27, 2004

Proposition: Let f be a function that is differentiable on an interval I and assume further:


  1. There is one and only one point z in I for which fz=0.

  2. If x is on the convex side of the graph of f in I, then so is x=xfxfx, and x lies between x and z.

  3. Successive iterations of Newton's method beginning with a point x on the convex side of the graph of f in I will converge to z.

  4. Error control principle. If c is any point in I on the concave side of the graph of f and x is on the convex side, then the distance between x and z is at most the absolute value of fxfc.

Proof: If f is positive in I one has fx1<fx2 whenever x1<x2 in I. If instead f is negative in I, then one has fx1>fx2 for x1<x2. For this reason there is at most one root z in I with fz=0. The Intermediate Value Theorem for Continuous Functions guarantees that there is at least one root between a and b.

We shall assume that f is positive and increasing. One may reduce each of the other three cases to this case by reflecting either in the horizontal axis or in the vertical line x=z or both. Under this assumption the convex side of the graph of f is the right side. Suppose that z<x: then 0=fz<fx. We apply the Mean Value Theorem to f on the interval z,x to conclude that there is a number u with z<u<x for which fxfz=fuxz. Since fz=0 and fu>0, one obtains xz=fxfu. Since f is increasing, we find fu<fx, and, therefore, fxfx<fxfu. Consequently, z<x<x.

In view of (2) one has z<<xn<<x2<x1. Letting x*=infn1xn, one has x*=limnxn, and, therefore, taking the limit as n on both sides of the relation xn+1=xnfxnfxn, one finds that fx*=0. Since by (1) there is only one root of f in I, it follows that x*=z.

In the proof of (2) we saw that for x in I on the right side of z the distance from x to z is fxfu, where z<u<x. Since c is on the concave side of the graph of f, i.e., c<z, we find also c<u, hence, fc<fu. Consequently, the error=xz=fxfufxfc.