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# hw5-7 - HW SOLUTIONS WEEKS 5-6 1 Disclaimer Even though...

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HW SOLUTIONS WEEKS 5-6 1. Disclaimer Even though I’ve made every effort to make these solutions clear and readable, I am only human and can make mistakes. If you find a mistake, please let me know ASAP so I can correct it. Also, I have taken liberty of omitting some details that aren’t necessary for me to repeat. However, if you don’t understand a step I have omitted, ask for clarification. 2. Week 5 1. Suppose f n f uniformly on S . Let > 0. Then N N such that n > N = ⇒ | f n ( x ) - f ( x ) | < / 2 x S Then for m, n > N , we get | f n ( x ) - f m ( x ) | ≤ | f n ( x ) - f ( x ) | + | f ( x ) - f m ( x ) | < / 2 + / 2 = which implies { f n } is uniformly Cauchy. 2. Part (a). Let x [0 , 1). Then x n 1+ x n x n . The comparison test implies then that x n 1 + x n x n = 1 1 - x Therefore the series converges. Part (b). Let a [0 , 1). For any x [0 , a ], observe that x n 1 + x n a n 1 + x n a n = 1 1 - a Therefore, by the Weierstrass M-test (with M n = a n ), the series converges uni- formly on [0 , a ]. Part (c). The series does NOT converge uniformly on [0 , 1)! Remember the definition of uniform convergence of a series: n =1 g n g uniformly if and only if the sequence of partial sums s k (= k n =1 g n ) g uniformly. Then using Remark 24.4, we can say that g n g uniformly if and only if lim n →∞ sup x S {| g ( x ) - s n ( x ) |} = 0 In our case where g n ( x ) = x n 1+ x n and g ( x ) = n =0 x n 1+ x n , observe that sup x [0 , 1) {| g ( x ) - s n ( x ) |} = sup x [0 , 1) k = n +1 x k 1 + x k = 1 / 2 = Therefore the supremum is infinite for all n , so the limit is note 0, hence does not converge uniformly. 3. Part (a): Simply differentiate term by term. Be careful to make sure you index your summation correctly.

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hw5-7 - HW SOLUTIONS WEEKS 5-6 1 Disclaimer Even though...

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