In this chapter we consider the problem of finding zeros of a continuous

In this chapter we consider the problem of finding

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In this chapter we consider the problem of finding zeros of a continuous function f , i.e. solving f ( x ) = 0 for example e x - x = 0 or a system of nonlinear equations: f 1 ( x 1 , x 2 , · · · , x n ) = 0 , f 2 ( x 1 , x 2 , · · · , x n ) = 0 , . . . f n ( x 1 , x 2 , · · · , x n ) = 0 . (12.1) We are going to write this generic system in vector form as f ( x ) = 0 , (12.2) where f : U R n R n . Unless otherwise noted the function f is assumed to be smooth in its domain U . We are going to start with the scalar case, n = 1 and look a very simple but robust method that relies only on the continuity of the function and the existence of a zero. 12.2 Bisection Suppose we are interested in solving a nonlinear equation in one unknown f ( x ) = 0 , (12.3) 193
194 CHAPTER 12. NON-LINEAR EQUATIONS where f is a continuous function on an interval [ a, b ] and has at least one zero there. Suppose that f has values of different sign at the end points of the interval, i.e. f ( a ) f ( b ) < 0 . (12.4) By the Intermediate Value Theorem, f has at least one zero in ( a, b ). To locate a zero we bisect the interval and check on which subinterval f changes sign. We repeat the process until we bracket a zero within a desired accuracy. The Bisection algorithm to find a zero x * is shown below. Algorithm 7 The Bisection Method 1: Given f , a and b ( a < b ), TOL , and N max , set k = 1 and do: 2: while ( b - a ) > TOL and k N max do 3: c = ( a + b ) / 2 4: if f ( c ) == 0 then 5: x * = c . This is the solution 6: stop 7: end if 8: if sign( f ( c )) == sign( f ( a )) then 9: a c 10: else 11: b c 12: end if 13: k k + 1 14: end while 15: x * ( a + b ) / 2 12.2.1 Convergence of the Bisection Method With the bisection method we generate a sequence c k = a k + b k 2 , k = 1 , 2 , . . . (12.5) where each a k and b k are the endpoints of the subinterval we select at each bisection step (because f changes sign there). Since b k - a k = b - a 2 k - 1 , k = 1 , 2 , . . . (12.6)
12.3. RATE OF CONVERGENCE 195 and c k = a k + b k 2 is the midpoint of the interval then | c k - x * | ≤ 1 2 ( b k - a k ) = b - a 2 k (12.7) and consequently c k x * , a zero of f in [ a, b ]. 12.3 Rate of Convergence We now define in precise terms the rate of convergence of a sequence of approximations to a value x * . Definition 21. Suppose a sequence { x n } n =1 converges to x * as n → ∞ . We say that x n x * of order p ( p 1 ) if there is a positive integer N and a constant C such that | x n +1 - x * | ≤ C | x n - x * | p , for all n N. (12.8) or equivalently lim n →∞ | x n +1 - x * | | x n - x * | p = C. (12.9) Example 32. The sequence generated by the bisection method converges lin- early to x * because | c n +1 - x * | | c n - x * | b - a 2 n +1 b - a 2 n = 1 2 . Let’s examine the significance of the rate of convergence. Consider first, p = 1, linear convergence. Suppose | x n +1 - x * | ≈ C | x n - x * | , n N. (12.10) Then | x N +1 - x * | ≈ C | x N - x * | , | x N +2 - x * | ≈ C | x N +1 - x * | ≈ C ( C | x N - x * | ) = C 2 | x N - x * | . Continuing this way we get | x N + k - x * | ≈ C k | x N - x * | , k = 0 , 1 , . . . (12.11)
196 CHAPTER 12. NON-LINEAR EQUATIONS and this is the reason of the requirement C < 1 for p = 1. If the error at the N step, | x N - x * | , is small enough it will be reduced by a factor of C k after k more steps. Setting C k = 10 - d k , then the error | x N - x * | will be reduced approximately d k = b log 10 1 C c k (12.12) digits.around Let us now do a similar analysis for p = 2, quadratic convergence. We

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