104_Fa08_hw_2_sol

104_Fa08_hw_2_sol - Homework 2 for MATH 104 Brief Solutions...

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Unformatted text preview: Homework 2 for MATH 104 Brief Solutions Problem 1 [5P] (a) Prove the inequality: For every x R , x >- 1, and n N , (1 + x ) n 1 + nx . Hint: Induction. Solution. n = 1 : Obviously, (1 + x ) 1 + x holds. n y n + 1 : We have (1 + x ) n + 1 = (1 + x ) n (1 + x ), which, by inductive hypothesis, is (1 + nx )(1 + x ). (Note that here it is important that 1 + x 0, i.e. x - 1.) The last expression evaluates to 1 + ( n + 1) x + nx 2 . Since nx 2 0 for all n , it follows that (1 + x ) n + 1 1 + ( n + 1) x , as desired. (b) Use (a) to give an - N-argument that shows that x n = 1- 1 n 2 ! n converges to 1. (Use the Archimedean property of R .) Solution. Let > 0. By the Archimedean property, there exists an N N such that N > 1 / , hence 1 / N < . We claim that for all n N , | 1- x n | < . Since x n < 1 for all n , we have, using part (a), that for each n N | 1- x n | = 1- 1- 1 n 2 !...
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104_Fa08_hw_2_sol - Homework 2 for MATH 104 Brief Solutions...

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