Download A first course of mathematical analysis by David Alexander Brannan PDF

By David Alexander Brannan

Mathematical research (often known as complex Calculus) is mostly came across through scholars to be one among their toughest classes in arithmetic. this article makes use of the so-called sequential method of continuity, differentiability and integration to help you comprehend the subject.Topics which are quite often glossed over within the commonplace Calculus classes are given cautious research right here. for instance, what precisely is a 'continuous' functionality? and the way precisely can one provide a cautious definition of 'integral'? The latter query is frequently one of many mysterious issues in a Calculus path - and it's relatively tricky to provide a rigorous therapy of integration! The textual content has lots of diagrams and valuable margin notes; and makes use of many graded examples and routines, usually with entire recommendations, to lead scholars throughout the tough issues. it's compatible for self-study or use in parallel with a customary college path at the topic.

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An, we have 1 a1 þ a2 þ Á Á Á þ a n : ð a1 a2 . . an Þ n n We give the proof of Theorem 3 at the end of the sub-section. Problem 11 Use Theorem 3 with the n þ 1 positive numbers 1, 1 þ 1n, 1 þ 1n, . , 1 þ 1n to prove that, for any positive integer n    nþ1 1 n 1 1þ 1þ : n nþ1 For example, with n ¼ 3   1 3 64 ¼ 2:37 . . 1þ ¼ 3 27  5 1þ 1 4 4 ¼ 625 ¼ 2:44 . . 256 Proofs of Theorems 2 and 3 You may omit these proofs at a first reading. Theorem 2 Cauchy–Schwarz Inequality For any real numbers a1, a2, .

J3j À jbj   ¼ 3 À jbj ! 3 À jbj: Now jbj < 1, so that À jbj > À1. Thus 3 À jbj > 3 À 1 ¼ 2; Again, we use the Transitive Rule. and we can then deduce from the previous chain of inequalities that & j3 À bj > 2, as desired. Remarks 1. The results of Example 1 can also be stated in the form: (a) j3 þ a3j 4, for jaj 1; (b) j3 À bj > 2, for jbj < 1. 2. The reverse implications   3 þ a3 4 ) jaj 1 and j 3 À bj > 2 ) j bj < 1 are FALSE. For example, try putting a ¼ À32 and b ¼ À2! Problem 1 (a) jaj 1 2 Use the Triangle Inequality to prove that:   3 ; (b) jbj < 1 ) b3 À 1 > 7.

Solution First, ƒ is increasing on the interval 12 , 3Þ, so that since 12 x < 3 it follows that 14 Àf ÁðxÞ < 9. Hence ƒ is bounded above and bounded below. Â Next, since f 12 ¼ 14 and 14 is a lower bound for ƒ on the interval 12 , 3Þ, it follows that ƒ has a minimum value of 14 on this interval. Â Finally, 9 is an upper bound for ƒ on the interval 12 , 3Þ but there is no point Â1 x in 2 , 3Þ for which ƒ(x) ¼ 9. So 9 cannot be a maximum of ƒ on the interval. pffiffiffi Àpffiffiffi Á However, if y is any number in (8, 9) there is a number x > y in y, 3 & À pffiffiffi Á Â1 Á 2 2 2, 3 & 2 , 3 such that ƒ(x) ¼ x > y, so that no number in (8, 9) will serve Strictly speaking, M and m are the upper bound and lower bound of the image set {ƒ(x): x 2 I}.

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