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# Class Note for MATH 425A with Professor Laetsch at UA

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This 1 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 29 views.

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Date Created: 02/06/15
Math 4251 Intermediate Value Theorem 2008 Suppose f is a realvalued continuous function de ned on a closed interval 61 b g R Suppose fa i fb and let y be a number between fa and fb Then there exists a number 0 in a b such that f c y Proof using the bisection method and the Nested Intervals Theorem For de niteness we assume that f a lt y lt f b The proof in the other case is similar To use the Nested Intervals Theorem we need a sequence of closed nested intervals and as a start in this direction we de ne a1 a and b1 b So f01 lty ltfb1 Let cl be the midpoint of the interval 611 bl There are three possibilities If f c1 y we are done Otherwise we continue If fcl gty choose b2 cl so fb2 gty and choose a2 a1 DRAWA PICTURE If fcl lty choose a2 cl so fa2 lty and choose b2 b1 DRAWA PICTURE In either case f02 lty ltfb2 Now we repeat the procedure for the interval 612 b2 Let 02 be the midpoint ofthe interval 612 b2 There are three possibilities If f c2 y we are done Otherwise we continue If f02 gty choose b3 02 so fb3 gty and choose a3 a2 If f02 lty choose a3 02 so fa3 lty and choose b3 b2 In either case f03 lty ltfb3 Continue in this way if necessary If we get to the nth stage we have an interval an bn such that fan ltyltfbn Let on be the midpoint of the interval an b n There are three possibilities If f cn y we are done Otherwise we continue If fcngty choose bn1 0 so fbn1gty and choose an an If fcnlty choose an 0 so fan1lty and choose bn1 bn In either case fanl lty ltfbnl Thus either the process terminates when f has the desired value y at some midpoint of one of our intervals or we get a nested sequence of closed intervals an b n with the property that for every n f an lt y lt f b n Furthermore since each term in the sequence is an interval of half the length of the preceding term in the sequence b n an converges to 0 Thus by the Nested Interval Theorem these sequences have an intersection which consists of one point say c with the property that an converges to c and bn converges to c By the continuity of f an converges to fc and f b n converges to f c From the inequalities above f C 3y 370 and so fc y as desired

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