104hw9sum06

# 104hw9sum06 - MATH 104 SUMMER 2006 HOMEWORK 9 SOLUTION...

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Unformatted text preview: MATH 104, SUMMER 2006, HOMEWORK 9 SOLUTION BENJAMIN JOHNSON Due August 7 Assignment: Section 25: 25.3, 25.15 Section 26: 26.2, 26.5 Section 28: 28.2(a), 28.6(b), 28.8 Section 25 25.3 Let f n ( x ) = n + cos x 2 n + sin 2 x for all real numbers x . (a) Show that h f n i converges uniformly on R . Proof. Let ± > 0. Choose N = 1 ± . Let n ∈ N * . Assume n > N . Let x ∈ R . Then | f n ( x )-1 2 | = | n + cos x 2 n + sin 2 x-1 2 | = | n + cos x-( n + 1 2 sin 2 x ) 2 n + sin 2 x | = | cosx-1 2 sin 2 x 2 n + sin 2 x | < | 2 2 n | = 1 n < 1 N = ± So f n → 1 2 uniformly on R . I.e., f n → f uniformly on R , where f is the constant function f ( x ) = 1 2 . ± (b) Calculate lim n →∞ R 7 2 f n ( x ) dx . Since f n → 1 2 uniformly on R , lim n →∞ R 7 2 f n ( x ) dx = R 7 2 1 2 dx = [ 1 2 x ] 7 2 = 7 2-2 2 = 5 2 . 25.15 Let h f n i be a sequence of continuous functions on [ a , b ]. Suppose that, for each x ∈ [ a , b ], h f n ( x ) i is a non-increasing sequence of real numbers. (a) Prove that if f n → 0 pointwise on [ a , b ], then f n → 0 uniformly on [ a , b ]. Proof. Assume that f n → 0 pointwise on [ a , b ]. Together with the hypothesis about non-increasing, this implies that for each x and n , we must have f n ( x ) ≥ 0. We’ll proceed by contradiction. (A direct proof using ideas from section 13 is sketched later). Suppose that f n does not converge to 0 uniformly on [ a , b ]. Then there is a ﬁxed ± > 0 such that for each integer k , there exist n k > k and x k ∈ [ a , b ] with | f n k ( x k ) | ≥ ± . Let z = lim inf x k . Then z ∈ [ a , b ]. Date : August 7, 2006. 1 2 BENJAMIN JOHNSON Since h f n ( z ) i → 0, there is an N ∈ R so that ( ∀ n > N )( | f n ( z ) | < ± 2 )....
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104hw9sum06 - MATH 104 SUMMER 2006 HOMEWORK 9 SOLUTION...

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