s3 - Math 5615H: Introduction to Analysis I. Homework #3....

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Math 5615H: Introduction to Analysis I. Fall 2011 Homework #3. Problems and Solutions. #1. Let A := { a 1 ,a 2 ,... } be a set of real numbers defined as follows: a 1 = 1 , and a k +1 = 1 + a k for k = 1 , 2 ,.... Find sup A . Solution. By induction, 1 a k < 4 for all k 1, therefore, L := sup A [1 , 4]. Since a k +1 = 1 + a k 1 + L for all k , we have L = sup { a k +1 } ≤ 1 + L . On the other hand, from 1 + a k = a k +1 L it follows 1 + L L , because in an equivalent form, a k = ( a k +1 - 1) 2 ( L - 1) 2 implies L := sup { a k } ≤ ( L - 1) 2 . Therefore, 1 + L = L [1 , 4], and L = (3 + 5) / 2. #2. Let z 1 and z 2 be complex imaginary (not real) numbers such that both z 1 + z 2 and z 1 z 2 are real . Show that z 2 = z 1 , i.e. z 2 is the conjugate of z 1 . Proof. Denote z 1 := a 1 + b 1 i, z 2 = a 2 + b 2 i . Since z 1 + z 2 = ( a 1 + a 2 ) + ( b 1 + b 2 ) i is real, we have b 2 = - b 1 6 = 0. Since z 1 z 2 is real, we also have Im(
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This document was uploaded on 02/15/2012.

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