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:

(a) The derivative f^{′} is either positive throughout I or negative throughout I.
(b) The derivative f^{′} is either steadily increasing in I or steadily decreasing in I.
(c) There are points a, b in I for which f(a) and f(b) are both non-zero with opposite sign.

Then:

  1. There is one and only one point z in I for which f(z) = 0.

  2. If x is on the convex side of the graph of f in I, then so is x^{′} = x - f(x)/f^{′}(x), 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 f(x)/f^{′}(c).

Proof: If f^{′} is positive in I one has f(x_{1}) < f(x_{2}) whenever x_{1} < x_{2} in I. If instead f^{′} is negative in I, then one has f(x_{1}) > f(x_{2}) for x_{1} < x_{2}. For this reason there is at most one root z in I with f(z) = 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 = f(z) < f(x). 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

 f(x) - f(z)  =  f^{′}(u)(x - z)    . 

Since f(z) = 0 and f^{′}(u) > 0, one obtains

 x - z  =  {f(x)}/{f^{′}(u)}    . 

Since f^{′} is increasing, we find f^{′}(u) < f^{′}(x), and, therefore, f(x)/f^{′}(x) < f(x)/f^{′}(u). Consequently, z < x^{′} < x.

In view of (2) one has

 z < … < x_{n} < … < x_{2} < x_{1}    . 

Letting

 x_{*}  =  inf_{(n  >=  1)}  
{x_{n}}
  , 

one has

 x_{*}  =  lim_{n  -->  INFTY}  x_{n}  , 

and, therefore, taking the limit as n --> INFTY on both sides of the relation

 x_{n+1}  =  x_{n} - {f(x_{n})}/{f^{′}(x_{n})}  , 

one finds that f(x_{*}) = 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 f(x)/f^{′}(u), 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, f^{′}(c) < f^{′}(u). Consequently,

 the error  =  x - z  =  {f(x)}/{f^{′}(u)}  <=   {f(x)}/{f^{′}(c)}    .