Newton’s method

Newton’s method

Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

June 14, 2009

Suppose that f : R → R and f ∈ C²[a, b], i.e., f⁰⁰ exists and is continuous.

If f (x^∗) = 0 and x^∗= x + h where h is small, then byTaylor’stheorem 0 = f (x^∗) = f (x + h)

= f (x ) + f⁰(x )h +1

2f⁰⁰(x )h²+ 1

3!f⁰⁰⁰(x )h³+ · · ·

= f (x ) + f⁰(x )h + O(h²).

Since h is small, O(h²) is negligible. It is reasonable to drop O(h²) terms.

This implies

f (x ) + f⁰(x )h ≈ 0 and h ≈ −f (x )

f⁰(x ), if f⁰(x ) 6= 0.

Hence

x + h = x − f (x ) f⁰(x ) is a better approximation to x^∗.

This sets the stage for the Newton-Rapbson’smethod, which starts with an initial approximation x₀ and generates the sequence {x_n}^∞_n=0 defined by

xn+1 = xn− f (xn) f⁰(x_n).

Since the Taylor’s expansion of f (x ) at x_k is given by f (x ) = f (x_k) + f⁰(x_k)(x − x_k) +1

2f⁰⁰(x_k)(x − x_k)²+ · · · . At x_k, one uses thetangent line

y = `(x ) = f (xk) + f⁰(xk)(x − xk)

to approximate the curveof f (x ) and uses the zero of the tangent line to approximate the zero of f (x ).

Newton’s Method

Given x₀, tolerance TOL, maximum number of iteration M.

Set i = 1 and x = x₀− f (x₀)/f⁰(x₀).

While i ≤ M and |x − x0| ≥ TOL

Set i = i + 1, x0 = x and x = x0− f (x₀)/f⁰(x0).

End While

Problem

The equation f (x ) ≡ x²− 10 cos x = 0 has two solutions ±1.3793646. Use Newton’s method to approximate the solutions with initial values ±25.

Requirements

1 Write two MATLAB functions, said fun f and fun df, to compute the values of f and f⁰, respectively.

2 Implement the Newton’s algorithm as a MATLAB function:

I Input arguments: fun f, fun df, initial value

I Output arguments: approximated solution of the equation