Download PDF by DAVID ALEXANDER BRANNAN: A First Course in Mathematical Analysis


ISBN-10: 0511348576

ISBN-13: 9780511348570

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2n ! n; for n ¼ 1; 2; . n ; for n ! 4: Problem 1 For each of the following statements, find a number X such that the statement is true: 1 3 (a) 1n < 100 ; for all n > X; (b) 1n < 1000 ; for all n > X: Problem 2 For each of the following statements, find a number X such that the statement is true:  n  n  Þ  Þ 1 3 (a) ðÀ1 (b) ðÀ1 n2  < 100 ; for all n > X; n2  < 1000 ; for all n > X: The solutions of Problems 1 and 2 both suggest that the larger and larger n we o ÈÉ Þn choose n, the closer and closer to 0 the terms of the sequences 1n and ðÀ1 n2 become.

We can think of the least upper bound of a set, when it exists, as a kind of ‘generalised maximum element’. If a set does not have a maximum element, but is bounded above, then we may be able to guess the value of its least upper bound. As in the case E ¼ [0, 2), there may be an obvious ‘missing point’ at the upper end of the set. However it is important to prove that your guess is correct. We now show you how to do this. Prove that the least upper bound of [0, 2) is 2. Example 3 Solution We know that M ¼ 2 is an upper bound of [0, 2), because x 2; for all x 2 ½0; 2Þ: To show that 2 is the least upper bound, we must prove that each number M0 < 2 is not an upper bound of [0, 2).

We also define We call this number b the nth root of a, and we write bp¼ffiffiffiffiffiffiffiffiffiffi p ffiffi ffi pffiffiffi n 0, since 0n ¼ 0, and if n is odd we define n ðÀaÞ ¼ À n a, since 0p¼ ffiffi ffi n ðÀ n aÞ ¼ Àa if n is odd. Let us illustrate Theorem 1 with the special case a ¼ 2 and n ¼ 2. In this case, Theorem 1 asserts the existence of a real number b such that b2 ¼ 2. In other pffiffiffi words, it asserts the existence of a decimal b which can be used to define 2 precisely. Here is a direct proof of Theorem 1 in this special case.

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A First Course in Mathematical Analysis by DAVID ALEXANDER BRANNAN

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