DAVID ALEXANDER BRANNAN's A First Course in Mathematical Analysis PDF


ISBN-10: 0511348576

ISBN-13: 9780511348570

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Bn ; we have ða1 b1 þ a2 b2 þ Á Á Á þ an bn Þ2 À 2 ÁÀ Á a1 þ a22 þ Á Á Á þ a2n b21 þ b22 þ Á Á Á þ b2n : We give the proof of Theorem 2 at the end of the sub-section. 3 Proving inequalities Problem 10 21 Use Theorem 2 to prove that  for any positive real numbers a1 , a2 , . n : ð1 þ 2 þ 3Þ   1 1 1 11 þ þ ¼6Â 1 2 3 6 Our final result also has many useful applications. In Example 3 you proved Àa þ bÁ2 that ab , for a, b 2 R; it follows that, if a and b are positive, then 2 ¼ 11 ! 32 : 1 ðabÞ2 aþb 2 .

It follows that ƒ has no maximum value & on 12 , 3Þ. Problem 3 Let ƒ be the function defined by f ð xÞ ¼ x12 ; x 2 ½À3; À2Þ. Determine whether ƒ is bounded above or below, and any maximum or minimum value of ƒ. 2 Least upper bounds and greatest lower bounds We have seen that the interval [0, 2] has a maximum element 2, but [0, 2) has no maximum element. However, the number 2 is ‘rather like’ a maximum element of [0, 2), because 2 is an upper bound of [0, 2) and any number less than 2 is not an upper bound of [0, 2).

Problem 7 We decrease the sum by omitting all but the first two terms. À Án 1 Prove the inequality 1 þ 1n ! 52 À 2n ; for n ! 1. Hint: consider the first three terms in the binomial expansion. Example 7 Solution Prove that 2n ! n2, for n ! 4. Let P(n) be the statement PðnÞ : 2n ! n2 : First we show that P(4) is true: 24 ! 42. STEP 1 Since 24 ¼ 16 and 42 ¼ 16, P(4) is certainly true. STEP 2 We now assume that P(k) holds for some k ! 4, and deduce that P(k þ 1) is then true. So, we are assuming that 2k !

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A First Course in Mathematical Analysis by DAVID ALEXANDER BRANNAN

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