1873_solutions

# Function f g and so sup f g sup f sup g 5 give an

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Unformatted text preview: n, the inequality Ý  Aj jn 86 Ý  Aj jn1 n  1 for follows at once from the observation that if x  jÝ n A j then x belongs to A j for every integer  j n and so x certainly belongs to A j for every integer j n  1. c. Prove that if we define n  Aj Bn  j1 for each n then the sequence B n is an contracting sequence of sets. Given any positive integer n, the inequality n 1 n  Aj  Aj j1 j1 follows by the same sort of argument that was given for parts a and b of this exercise. d. Prove that if we define n  Aj Bn  j1 for each n then the sequence B n is an expanding sequence of sets. Given any positive integer n, the inequality n 1 n  Aj  Aj j1 j1 follows by the same sort of argument that was given for parts a and b of this exercise. 6. Prove that if A n is a sequence of subsets of R then Ý R Ý  An  R n1 An . n1 Hint: Given any number x that belongs to the set Ý  An R n1 we know from the fact that x does not belong to the union Ý  An n1 that, for each n, x must fail to belong to the set A n . Thus, whenever Ý x R  An n1 we know that x R An for every n and so Ý x R An . n1 You can use an almost identical argument to show that whenever a number x belongs to the set Ý R n1 we must have 87 An Ý x  An R n1 and we have therefore shown that Ý Ý n1 R n1  An   R An . Exercises on Mathematical Induction 1. Given that f x  log 1  x for all x  1, prove that the identity 1  x f n1 x  nf n x  0 holds whenever n is a positive integer and x  1. For each positive integer n we define P n to be the assertion that the equation 1  x f n1 x  nf n x  0 holds. The assertion P 1 says that 1 1 1 0 1x 1x 2 and this assertion is obviously true. Now suppose that n is any positive integer for which the statement P n is true. From the equation 1  x f n1 x  nf n x  0 we have 1f n1 x  1  x f n2 x  nf n1 x  0 which we can write as 1  x f n2 x  n  1 f n1 x  0. Since P 1 is true and since the condition P n  P n1 holds for every positive integer n it follows from the principle of mathematical induction that P n is true for every positive integer n. 1x 2. Given that f x  arctan 1  x for every number x, prove that the identity x 2  2x  2 f n1 x  2n x  1 f n x  n n 1 f n1 x 0 holds whenever n is a positive integer and x is a real number. For each positive integer n we define P n to be the assertion that the equation x 2  2x  2 f n1 x  2n x  1 f n x  n n 1 f n 1 x  0 holds. The assertion P 1 says that x 2  2x  2 2x 2 x 2  2x  2 2 x1 2 1 2x  x 2  2 0 and this assertion is obviously true. Now suppose that n is any positive integer for which the statement P n is true. From the equation x 2  2x  2 f n1 x  2n x  1 f n x  n n 1 f n 1 x  0 we have 2x  2 f n 1 x  x 2  2x  2 f which we can write as x 2  2x  2 f n 3 n 3 x  2nf n x  2nf n x  2n x  1 f x 2 n1 x1 f n 1 n 1 x n n x  n  1 nf n 1f n x 0 x 0 Since P 1 is true and since the condition P n  P n1 hold...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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