Univ at Albany: W. F. Hammond: Honors Calculus Notes on Newton<apos/>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: () The derivative f is either positive throughout I or negative throughout I () The derivative f is either steadily increasing in I or steadily decreasing in I () There are points a, b in I for which f(a) and f(b) are both nonzero with opposite sign Then: There is one and only one point z in I for which f(z) 0 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 Successive iterations of Newtons method beginning with a point x on the convex side of the graph of f in I will converge to z 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(x1) f(x2) whenever x1 x2 in I If instead f is negative in I, then one has f(x1) f(x2) for x1 x2 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 xz or both Under this assumption the convex side of the graph of f is the right side Suppose that z x: then 0f(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 xn x2 x1 Letting x inf(n 1) xn one has x limn xn and, therefore, taking the limit as n on both sides of the relation xn1 xn f(xn)f(xn) 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 zux 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)