Method of False Position

6.2.1 Method of False Position

Given two initial approximations a 0 = a and b 0 = b such that f (a)f (b) < 0, a nested sequence of intervals (a 0 ,b 0 ) ⊃ (a 1 ,b 1 ) ⊃ (a 2 ,b 2 ) ⊃ · · · such that f (a n )f (b n )< 0, n = 0, 1, 2, . . . , can be generated as follows. Given (a n ,b n ) , we take x n +1 to be the intersection of the x-axis and the secant through the point (a n , f (a n )) and (b n , f (b n )) . Then by Newton’s interpolation formula x n +1 satisfies

0, set a n +1 =x n +1 and b n +1 =b n ; otherwise set b n +1 =x n +1 and

a n +1 =a n . As for bisection, convergence to a root is guaranteed (in exact arithmetic) for a continuous function f (x). This is the false-position method or, in Latin, regula falsi. 179 Note that if f (x) is linear we obtain the root in just one step, but sometimes the rate of convergence can be much slower than for bisection. Suppose now that f (x) is convex on [a, b], f (a) < 0, and f (b) > 0, as in Fig- ure 6.2.1. Then the secant through x = a and x = b will lie above the curve and hence intersect the x-axis to the left of α. The same is true for all subsequent secants and therefore

the right endpoint b will be kept. The approximations x 1 ,x 2 ,x 3 ,... will all lie on the con- vex side of the curve and cannot go beyond the root α. A similar behavior, with monotone convergence and one of the points a or b fixed, will occur whenever f ′′ (x) exists and has constant sign on [a, b].

179 Regula falsi is a very old method that originated in fifth century Indian texts and was used in medieval Arabic mathematics. It got its current name from the Italian mathematician Leonardo Pisano, also known as Leonardo

Fibonacci (ca 1170–1250). He is considered to be one of the most talented mathematicians of the Middle Ages.

6.2. Methods Based on Interpolation 627

Figure 6.2.1. The false-position method.

Example 6.2.1.

We apply the method of false position to the f (x) = (x/2) 2 − sin x = 0 from Exam- ple 6.1.2 with initial approximations a 0 = 1.5, b 1 = 2. We have f (1.5) = −0.434995 < 0

and f (2.0) = +0.090703 > 0, and successive iterates are as follows. n

5 1.933753 734053 Note that f (x n )< 0 for all n ≥ 0 and consequently b n = 2 is fixed. In the limit convergence

is linear with rate approximately equal to C ≈ 0.034. If f is twice continuously differentiable and f ′′ (α)

will be reached on which f ′′ (x) does not change sign. Then, as in the example above, one of the endpoints (say b) will be retained and a n =x n in all future steps. By (6.2.1) the successive iterations are

To determine the speed of convergence subtract α and divide by ǫ n =x n − α to get

ǫ n +1

f (x n )

x n −b

x n −α f (x n ) − f (b)

628 Chapter 6. Solving Scalar Nonlinear Equations Since lim n →∞ x n = α and f (α) = 0, it follows that

which shows that convergence is linear. Convergence will be very slow if f (x) is very flat near the root α, f (b) is large, and α near b, since then (b − α)f ′ (α) ≪ f (b) and C ≈ 1.