Final Practice Solutions

# Final Practice Solutions - SOLUTIONS FOR PRACTICE PROBLEMS...

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Unformatted text preview: SOLUTIONS FOR PRACTICE PROBLEMS FOR FINAL The comprehensive final will be given on Wednesday, March 17, 8-10 am , in our usual classroom (MST 124). It will be a closed-book test . No textbooks, notes, or electronic devices will be allowed. Scratch paper will be provided to those needing it. You only need to bring your pens, papers, and erasers. In preparing for the test, you can practice solving the problems from the list below. In addition, take a look at the homework problems (at least one problems on the midterm will come more or less directly from the homework), and at the examples given in the textbook. The list below contains problems of different levels of difficulty. Extra hard (in my opinion) problems are marked by (!). 1. (!) Prove that, for any positive integer n , 2 n summationdisplay k =1 ( − 1) k +1 k = 2 n summationdisplay j = n +1 1 j . We prove this statement by induction on n . For n ∈ N , let P n be the following statement: 2 n summationdisplay k =1 ( − 1) k +1 k = 2 n summationdisplay j = n +1 1 j . The basic step consists of verifying P 1 . This is easy: 1 − 1 / 2 = 1 / 2. For the inductive step, we have to show that, if 2 n summationdisplay k =1 ( − 1) k +1 k = 2 n summationdisplay j = n +1 1 j , then 2 n +2 summationdisplay k =1 ( − 1) k +1 k = 2 n +2 summationdisplay j = n +2 1 j . By the induction hypothesis, 2 n summationdisplay k =1 ( − 1) k +1 k = 1 n + 1 + 2 n summationdisplay j = n +2 1 j , hence 2 n +2 summationdisplay k =1 ( − 1) k +1 k = 2 n summationdisplay k =1 ( − 1) k +1 k + 1 2 n + 1 − 1 2 n + 2 = 2 n summationdisplay j = n +2 1 j + 1 2 n + 1 + 1 n + 1 − 1 2 n + 2 = 2 n summationdisplay j = n +2 1 j + 1 2 n + 1 + 1 2 n + 2 , as desired. 1 2 SOLUTIONS FOR PRACTICE PROBLEMS FOR FINAL 2. (!) Suppose p is a polynomial of degree n , and x ∈ R . Prove that there exist a ,a 1 ,... ,a n ∈ R s.t. p ( x ) = ∑ n k =0 a k ( x − x ) k for any x ∈ R . Hint . Use induction on n . The statement is obvious for n = 0. This is the base of induction. To handle the step, suppose the statement is true for n , and p is a polynomial of degree n + 1. We have to find a ,a 1 ,... ,a n ,a n +1 ∈ R s.t. p ( x ) = ∑ n +1 k =0 a k ( x − x ) k for any x ∈ R . Write p ( x ) = ∑ n +1 k =0 b k x k . Let a n +1 = b n +1 , and q ( x ) = p ( x ) − a n +1 ( x − x ) n +1 = n +1 summationdisplay k =0 b k x k − a n +1 n +1 summationdisplay k =0 ( − 1) n +1- k...
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Final Practice Solutions - SOLUTIONS FOR PRACTICE PROBLEMS...

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