By DAVID ALEXANDER BRANNAN

**Read Online or Download A First Course in Mathematical Analysis PDF**

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**Extra resources for A First Course in Mathematical Analysis**

**Example text**

1: Numbers 28 The strategies for proving that M is the least upper bound or m the greatest lower bound of ƒ on I are similar to the corresponding strategies for the least upper bound or the greatest lower bound of a set E. Strategies Let ƒ be a function defined on an interval I R. Then: To show that m is the greatest lower bound, or infimum, of f on I, check that: 1. f(x) ! m, for all x 2 I; 2. if m0 > m, then there is some x 2 I such that f(x) < m0 . To show that M is the least upper bound, or supremum, of f on I, check that: 1.

N n 2 n2 1 2 3 4 5 2 1 4 4 8 9 16 16 32 25 This assumption is just P(k). Since P(k þ 1) is: 2kþ1 ! (k þ 1)2. 1: Numbers 20 Three important inequalities in Analysis Our first inequality, called Bernoulli’s Inequality, will be of regular use in later chapters. Theorem 1 Bernoulli’s Inequality For any real number x ! À1 and any natural number n, (1 þ x)n ! 1 þ nx. Remark The value of this result will come from making suitable choices of x and n for particular purposes. In part (a) of Example 6, you saw that (1 þ x)n !

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, . , an and b1, b2, . , bn, we have À 2 ÁÀ Á ða1 b1 þ a2 b2 þ Á Á Á þan bn Þ2 a1 þ a22 þ Á Á Á þ a2n b21 þ b22 þ Á Á Á þ b2n : Proof If all the as are zero, the result is obvious; so we need only examine the case when not all the as are zero. It follows that, if we denote the Pn 2 2 2 2 sum k¼1 ak ¼ a1 þ a2 þ Á Á Á þ an by A, then A > 0. Also, denote Pn 2 2 2 2 k¼1 bk ¼ b1 þb2 þÁÁÁþbn Pn by B and k¼1 ak bk ¼ a1 b1 þ a2 b2 þÁÁÁþan bn by C.