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Unformatted text preview: Introductory Course: Fourier Analysis and its many uses Solutions  Exercises 16.116.5, 16.9, 16.12, 16.13, 16.15, 16.16, 16.17, 16.20, 16.21, 16.23 from “A First Look at Fourier Analysis” by T.W. K¨ orner (prepared by Mihai Stoiciu) Exercise 16.1. i) Take P ( t ) = f ( n ) (0) n ! t n + f ( n 1) (0) ( n 1) ! t n 1 + ··· + f (0) . ii) Note first that g (0) = g (0) = ··· = g ( n ) (0) = 0. Since g (0) = g ( t ) = 0 it follows from Rolle’s theorem that there exists a c 1 ∈ (0 , t ) such that g ( c 1 ) = 0. Observe now that g (0) = g ( c 1 ) = 0 so Rolle’s theorem implies that there exists a c 2 ∈ (0 , c 1 ) such that g 00 ( c 2 ) = 0. Repeating this procedure ( n + 1) times we get that there exists c = c n +1 ∈ (0 , t ) such that g ( n +1) ( c ) = 0. This immediately implies f ( t ) = P ( t ) + f ( n +1) ( c ) ( n +1)! t n +1 . Exercise 16.2. A polynomial P of degree at most (2 n + 1) with prescribed val ues for P (0) , P (0) , ... P ( n ) (0) , P (1) , P (1) , ... P ( n ) (1) can be obtained by taking a linear combination of the polynomials 1 k ! x k ( x 1) n +1 , 0 ≤ k ≤ n and 1 k ! x n +1 ( x 1) k , 0 ≤ k ≤ n . We prove now that P is unique. Let Q be a polynomial of degree at most (2 n + 1) such that P (0) = Q (0) , P (0) = Q (0) , ...P ( n ) (0) = Q ( n ) (0) , P (1) = Q (1) , P (1) = Q (1) , ... P ( n ) (1) = Q ( n ) (1). R = P Q is a polynomial of degree at most 2 n + 1 and R (0) = R (0) = ··· = R ( n ) (0) = R (1) = R (1) = ··· = R ( n ) (1) = 0. Therefore the polynomials X n +1 and ( X 1) n +1 divide R and since deg R ≤ 2 n + 1 we obtain R = 0. Let P be the unique polynomial of degree at most (2 n + 1) such that P (0) = f (0) , P (0) = f (0) , ... P ( n ) (0) = f ( n ) (0) , P (1) = f (1) , P (1) = f (1) , ... P ( n ) (1) = f ( n ) (1). For a fixed y ∈ (0 , 1) let g ( x ) = f ( x ) P ( x ) f ( y ) P ( y ) y n +1 ( y 1) n +1 x n +1 ( x 1) n +1 Let’s observe that g (0) = g (0) = ··· = g ( n ) (0) = 0 and g (1) = g (1) = ··· = g ( n ) (1) = 0. Furthermore g (0) = g ( y ) = g (1) = 0. Therefore, from Rolle’s theorem it follows that there exists two points c 1 1 ∈ (0 , y ) and c 1 2 ∈ ( y, 1) such that g ( c 1 1 ) = g ( c 1 2 ) = 0. Since g (0) = g (1) = 0 we can apply Rolle’s theorem again and we get three points c 2 1 ∈ (0 , c 1 1 ) , c 2 2 ∈ ( c 1 1 , c 1 2 ) and c 2 3 ∈ ( c 1 2 , 1) such that g 00 ( c 2 1 ) = g 00 ( c 2 2 ) = g 00 ( c 2 3 ) = 0. Repeating this procedure ( n + 1) times we get ( n + 2) points c n +1 1 , c n +1 2 , ... c n +1 n +2 ∈ (0 , 1) such that g ( n +1) ( c n +1 1 ) = g ( n +1) ( c n +1 2 ) = ··· = g ( n +1) ( c n +1 n +2 ) = 0. Now we can apply Rolle’s theorem again and we get ( n + 1) points c n +2 1 ∈ ( c n +1 1 , c n +1 2 ) , c n +2 2 ∈ ( c n +1 2 , c n +1 3 ) , ... c n +2 n +1 ∈ ( c n +1 n +1 , c n +1 n +2 ) 1 such that g ( n +2) ( c n +2 1 ) = g ( n +2) ( c n +2 2 ) = ··· = g ( n +2) ( c n +2 n +1 ) = 0. Repeating this procedure) = 0....
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This note was uploaded on 01/31/2011 for the course MATH 201b taught by Professor Groah during the Fall '08 term at UC Davis.
 Fall '08
 Groah
 Math

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