ADVANCED CALCULUS I
ADVANCED CALCULUS I MATH 409
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This 4 page Class Notes was uploaded by Vivien Bradtke V on Wednesday October 21, 2015. The Class Notes belongs to MATH 409 at Texas A&M University taught by Staff in Fall. Since its upload, it has received 13 views. For similar materials see /class/226048/math-409-texas-a-m-university in Mathematics (M) at Texas A&M University.
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Date Created: 10/21/15
Math 409 502 Harold P Boas b0astamuedu Compactness De nition An interval is called compact if the interval is both closed and bounded Examples 0 The interval 71 1 is compact 0 The interval 01 is not compact It is bounded but not closed 0 The interval 0 00 is not compact It is closed but not bounded The key property is that a sequence of points in a compact interval must have a cluster point that is still in the interval Math 409502 October 22 2004 i slide 2 Continuous functions on compact intervals Theorem If f is a continuous function on a compact interval then the range of f is again a compact interval The conclusion of the theorem has three parts 1 The range is an interval 2 The range is bounded 3 The range contains the endpoints of the interval 0 The first part is the Intermediate Value Theorem 0 The second part says that the function has a finite supremum and a finite infimum 0 The third part says that the function attains a maximum value and a minimum value Math 409502 October 22 2004 i slide 3 Why did the chicken cross the road Theorem Intermediate Value Theorem If f is a continuous function on the compact interval a h then every number between fa and is in the range off Example Show that the function x5 7 4x2 ex has a zero between x 0 and x 1 Solution At 3 0 the function has the value 1 and at x 1 the function has the value 73 e lt 0 By the theorem the function takes the intermediate value 0 somewhere in the interval 0 1 Using the bisection method we could locate the zero of the function more precisely Math 409502 October 22 2004 i slide 4 Proof of the intermediate value theorem The book gives a proof page 173 using the Nested Interval Theorem Here is a different proof Without loss of generality it may be assumed that f a lt f We need to show that if k is a number such that f a lt k lt f h then the number k is in the range of f Let 8 denote the set of points c with the property that f x lt k for all x in the interval a c The set S is non empty because f a lt k and f is continuous Let d denote the supremum of 8 Claim k Because f is continuous lim fd 7 so g k HHOO by the limit location theorem It cannot be that f d lt k strict inequality because then there would be points to the right of d in the set 8 again by the continuity of f 80 the claim holds Math 409502 October 22 2004 i slide 5 Homework 0 Read sections 121 and 122 pages 172 177 0 D0 Exercises 1211 and 1215 on page 180 Math 409502 October 22 2004 i slide 6
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