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# CHAPTER 6 Max, Min, Sup, Inf We would like to begin by asking for the maximum of the function f (x) = (sin x)/x. An approximate graph is indicated...

i want the solution of ma301 chapter 6 #6,7,8,9,10,13

CHAPTER 6 Max, Min, Sup, Inf We would like to begin by asking for the maximum of the function f ( x ) = (sin x ) /x . An approximate graph is indicated below. Looking at the graph, it is clear that f ( x ) 1 for all x in the domain of f . Furthermore, 1 is the smallest number which is greater than all of f ’s values. o y=(sin x)/x 1 Figure 1 Loosely speaking, one might say that 1 is the ‘maximum value’ of f ( x ). The problem is that one is not a value of f ( x ) at all. There is no x in the domain of f such that f ( x ) = 1. In this situation, we use the word ‘supremum’ instead of the word ‘maximum’. The distinction between these two concepts is described in the following de±nition. Definition 1 . Let S be a set of real numbers. An upper bound for S is a number B such that x B for all x S . The supremum, if it exists, (“sup”, “LUB,” “least upper bound”) of S is the smallest 81
82 6. MAX, MIN, SUP, INF upper bound for S . An upper bound which actually belongs to the set is called a maximum. Proving that a certain number M is the LUB of a set S is often done in two steps: (1) Prove that M is an upper bound for S –i.e. show that M s for all s S . (2) Prove that M is the least upper bound for S . Often this is done by assuming that there is an ǫ > 0 such that M ǫ is also an upper bound for S . One then exhibits an element s S with s > M ǫ , showing that M ǫ is not an upper bound. Example 1 . Find the least upper bound for the following set and prove that your answer is correct. S = { 1 2 , 2 3 , 3 4 ,... , n n + 1 ... } Solution: We note that every element of S is less than 1 since n n + 1 < 1 We claim that the least upper bound is 1. Assume that 1 is not the least upper bound,. Then there is an ǫ > 0 such that 1 ǫ is also an upper bound. However, we claim that there is a natural number n such that 1 ǫ < n n + 1 . This inequality is equivalent with the following sequence of inequali- ties 1 n n + 1 < ǫ 1 n + 1 < ǫ 1 ǫ < n + 1 1 ǫ 1 < n.
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