1、 We give some important basic facts from calculus as follows. Theorem 1.1.1(Intermediate Value Theorem) Let f be a continuous function on the interval a, b(that is f Ca. b). Then f realizes every value between f(a) and f(b). More precisely, if y is a number between f(a) and f(b), then there exists a
2、 number c with a c b such that f(c) = y.Theorem 1.1.2(Continuous Limits) Let f be a continuous function in a neighborhood of x0, and assume = x0. Then In other words, limits may be brought inside continuous functions.Theorem 1.1.3 (Mean Value Theorem) Let f be a continuously differentiable function
3、on the interval a, b (that is f C1a, b). Then there exists a number between a and b such that f() = (f(b) f(a) / (b a).Theorem 1.1.4 (Rolles Theorem) Let f C1a. b) and assume that f(a) = f(b). Then there exists a number between a and b such that f() = 0.Theorem 1.1.5 (Taylors Theorem with Remainder)
4、 Let x and x0 be real numbers, and let f be k+1 times continuously differentiable on the interval between x and x0 (that is f Ck+1a, b). Then there exists a number between x and x0 such that1.2 The Bisection MethodThis process involves finding a root, or solution, of an equation of the form (1.1.1),
5、 for a given function f. A root of this equation is also called a zero of the function f. The first technique, based on the Intermediate Value Theorem, is called the Bisection, or Binary-search, method. Theorem 1.2.1 Let f Ca. b, satisfying f(a)f(b) 0. Then f has a root between a and b, that is, the
6、re exists a number c satisfying a c b and f(c) = 0.Proof.Suppose f Ca, b, with f(a) and f(b) of opposite sign. By the Intermediate Value Theorem (Theorem 1.1.1), there exists a number c in (a, b) with f(c) = 0. Although the procedure will work when there is more than one root in the interval (a, b),
7、 we assume for simplicity that the root in this interval is unique. The method calls for a repeated halving of subintervals of a, b and, at each step, locating the half containing c.To begin, set a1=a and b1=b, and let c1 be the midpoint of a, b; that is,If f(c1) = 0, then c=c1, and we are done. If
8、f(c1)0, then f(c1) has the same sign as either f(a1) or f(b1). When f(c1) and f(a1) have the same sign, c (c1,b1), and we set a2=c1 and b2 = b1. When f(c1) and f(a1) have opposite signs, c (a1,c1) and we set a2=a1 and b2=c1. We then reapply the process to the interval a2, b2. This produces the metho
9、d described in Algorithm 1.2.1Algorithm 1.2.1 : Bisection To find a solution to f(x) = 0 given the continuous function f on the interval a, b, where f(a) and f(b) have opposite signs:INPUT endpoints a, b; tolerance TOL; maximum number of iterations N0,OUTPUT approximate solution p or message of fail
10、ure.Step 1 set i = 1; FA = f(a);Step 2 While i N0 do Steps 3-6 Step 3 Set c = (a+b) / 2 FC = f(c); Step 4 If FC = 0 or (b-a) / 2 0 then set a = c; FA = FC; else set b = c;Step 7 OUTPUT(Method failed after N0 iterations, N0 = ,N0); STOP;Other stopping procedure can be applied in Step 4 of Algorithm 1
11、.2.1 or in any of the iterative techniques in this chapter. For example, we can select a tolerance 0 and generate c1, c2, , cN until one of the following conditions is met: |cN - cN-1| , (1.2.1) |cN - cN-1| / | cN | , cN 0, or (1.2.2) |f(cN)| N0. Note that to start the Bisection Algorithm, an interval a, b must be found with f(a)f(b) 0. At each step
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