m425b-ex1soln-s08

m425b-ex1soln-s08 - MATH 425b MIDTERM 1 SOLUTIONS SPRING...

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MATH 425b MIDTERM 1 SOLUTIONS SPRING 2008 Prof. Alexander (1)(a) Given ± > 0 let δ = ± 1 . Then f E, | y - x | < δ = ⇒ | f ( y ) - f ( x ) | ≤ | y - x | α < δ α = ±, i.e. this δ “works” uniformly over E . This shows E is equicontinuous. (b) Suppose f n E for all n and f n f uniformly. Then f (0) = lim n f n (0) = 0 and for all x,y [0 , 1], | f ( x ) - f ( y ) | = lim n | f n ( x ) - f n ( y ) | ≤ | y - x | α , so f E . (2)(a) a n 0 means | a n | 1 /n < 1 for large n , so lim sup n | a n | 1 /n 1. Therefore the radius of convergence R = (lim sup n | a n | 1 /n ) - 1 1. (b) If n a n < then by Theorem 8.2, lim x 1 f ( x ) = n a n < . But lim x 1 f ( x ) = lim x 1 (1 - x ) - 1 / 2 = , a contradiction. Therefore n a n = . (3)(a) For - N k N , the k th Fourier coefficient of f is c k = 1 2 π Z π - π f ( x ) e - ikx dx = 1 2 π Z π - π N X n = - N a n e inx ! e
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This note was uploaded on 06/23/2008 for the course MATH 245 taught by Professor Alexander during the Spring '07 term at USC.

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m425b-ex1soln-s08 - MATH 425b MIDTERM 1 SOLUTIONS SPRING...

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