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Unformatted text preview: MATH 425b ASSIGNMENT 4 SOLUTIONS SPRING 2008 Prof. Alexander Chapter 8: (9)(a) Define f on [0 , ∞ ) by f ( x ) = 1 /n for x ∈ [ n 1 ,n ), for all n ≥ 1. Then f ( x ) ≤ 1 x ≤ f ( x 1) for all x ≥ 1 . (1) Also s N = Z N +1 1 f ( x 1) dx = Z N f ( x ) dx = 1 + Z N 1 f ( x ) dx. (2) Hence log N s N = Z N 1 1 x dx 1 Z N 1 f ( x ) dx = Z N 1 1 x f ( x ) dx 1 . Since the integrand on the right side is positive, this shows log N s N is increasing in N . But (6) and the first equality in (2) show that s N ≥ log( N + 1) so log N s N ≤ log N log( N + 1) = log( N/ ( N + 1)) → 0, so log N s N is bounded. Since the sequence is bounded and increasing, it has a finite limit. (11) Let > 0. By assumption there exists M such that x ≥ M implies  f ( x ) 1  < . This lets us deal with the contribution to the integral from [ M, ∞ ), showing it is near 1, since for suffiently small t we have t Z ∞ M e tx f ( x ) dx 1 = t Z ∞ M e tx ( f ( x ) 1) dx + t Z ∞ M e tx dx 1 (3) ≤ t Z ∞ M e tx  f ( x ) 1  dx +  e tM 1  ≤ t Z ∞ M e tx dx +  e tM 1  < +  e tM 1  < 2 . Now we need to show the contribution to the integral from [0 ,M ] is small. Since f ∈ R on [0 ,M ], f must be bounded on [0 ,M ], say  f ( x )  ≤ B for all x ∈ [0 ,M ]. Therefore for sufficiently small t , t Z M e tx f ( x ) dx ≤ Bt Z M e tx dx = B (1 e tM ) < . 1 Using the triangle inequality, this and (3) show that t Z ∞ e tx f ( x ) dx 1 ≤ t Z M e tx f ( x ) dx + t Z ∞ M e tx f ( x ) dx 1 < + 2 = 3 . Since is arbitrary, this shows that lim t → t Z M e tx f ( x ) dx = 1 . (12)(a) For n 6 = 0, c n = 1 2 π Z π π f ( x ) e inx dx = 1 2 π Z δ δ (cos nx + i sin nx ) dx = 1 2 π Z δ δ cos nx dx (since sin nx is odd) = 1 2 πn sin nx δ δ = sin nδ nπ , and c = 1 2 π Z δ...
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This note was uploaded on 06/23/2008 for the course MATH 425B taught by Professor Alexander during the Fall '07 term at USC.
 Fall '07
 Alexander
 Math

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