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Unformatted text preview: n, the inequality
Ý Aj
jn 86 Ý Aj jn1 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 j1 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 j1 j1 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 j1 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 j1 j1 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 n1 An . n1 Hint: Given any number x that belongs to the set
Ý An R n1 we know from the fact that x does not belong to the union
Ý An
n1 that, for each n, x must fail to belong to the set A n . Thus, whenever
Ý x R An
n1 we know that
x R An for every n and so
Ý x R An . n1 You can use an almost identical argument to show that whenever a number x belongs to the set
Ý R
n1 we must have 87 An Ý x An R n1 and we have therefore shown that
Ý Ý n1 R n1 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 n1 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 n1 x nf n x 0
holds. The assertion P 1 says that
1
1
1
0
1x
1x 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 n1 x nf n x 0
we have
1f n1 x 1 x f n2 x nf n1 x 0
which we can write as
1 x f n2 x n 1 f n1 x 0.
Since P 1 is true and since the condition P n P n1 holds for every positive integer n it follows from
the principle of mathematical induction that P n is true for every positive integer n.
1x 2. Given that f x arctan 1 x for every number x, prove that the identity
x 2 2x 2 f n1 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 n1 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 x1 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 n1 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 n1 x1 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 n1 hold...
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 Fall '08
 STAFF
 Math, Calculus

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