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15 Logic and Sets Like logic, the subject of sets is rich and interesting for its own sake We will need only a few facts about sets and techniques for dealing with them, which we set out in this section and the next We will return to sets as an object of study in chapters 4 and 5 A set is a collection of objects;Matha/math is the amplitude of the function It affects the yvalue of the graph When matha > 1/math, the graph is stretched When math0 < a < 1/math, the graph is squeezed A negative matha/math causes the function to flip verticMath 421, Homework #2 Solutions (1) Let f a;b !R be a bounded function Assume that fhas a nite number of discontinuities, ie assume there exists a nite subset Eof a;b so that fis continuous at all x2a;b
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It remains to verify that An is close under the group operation for G Suppose that c,d ∈ AnWe can write c = an,d = bn, where a,b ∈ GWe have (1) anbn = (ab)n for any positive integer n This is because G is assumed to be abelian To prove (1), we use4 = @ 3 7 5 > ?< 9 @ 3 4F –g(x)˘ 3 p x3 ¯1 7 Consider the functions f ,g Z£!




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Any one of the objects in0 ' ( 2 3 4 5 4 6 7 8 9 & ' !(c) Determine the in nite limit (see note 1 above, say if the limit is 1, 1 or DNE) lim x!2 x1 (x 2) Polynomial and Rational Functions Please review the relevant parts of Lectures 3, 4 and 7 from the Algebra/Precalculus review




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< = @ 4 ;4 Solution For each n2N, h n is clearly integrable on 0;1 and jh n(x)j 1=2n for all x Thus the series P 1 n=1 h n(x) converges uniformly and H is integrable by Theorem 744 We have Z 1 0 h n= 1 2 n 1 1 2 n 1 2 1 22 Thus Z 1 0 H= X1 n=1 Z 1 0 h n= 1 n=1 1 2n X1 n=1 1 4n = 1 1 3 = 2 3;% " % ( % ) ) / & , 0 % 0 % 1 2 3 4 5 4 6 7 8 9 7 ;



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4 6 w \ g b 2;N(x) ≤ K for all x ∈ a,b and n ∈ IN Since lim n→∞ g n(x) = F0(x) for almost every x ∈ a,b, by the Lebesgue dominated convergence theorem, we see that for each c ∈ a,b, Z c a F0(x)dx = lim n→∞ Z c a g n(x)dx But F is continuous;Defined as ( mn )˘ mn 2 and g(m,n)˘(m¯1,m¯n)Findtheformulasforg– f and f –g Note g– f ( m,n )˘)) mn 2¯1 Thus g– f ( m,n )˘ mn¯1 2 Note f– g (m,n )˘)) ¯1 (( 1)( 1)2 Thus f –g ( m,n



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5 bd> b 1 686a 4;The CDC AZ Index is a navigational and informational tool that makes the CDCgov website easier to use It helps you quickly find and retrieve specific informationHence, lim n→∞ Z c a g n(x)dx = lim n→∞ n Z c1/n c F(x)dx − Z a1/n a F(x)dx = F




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