1873_solutions

Sec w which means that the given integral diverges f

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Unformatted text preview: nverges boundedly to the constant function 0. Finally we observe that fn 359 Þ 0 f n x dx  Þ 0 nxe n x dx  1 as n 1 2 1en 1 n 2 22 0 Ý. d. f n x  nx exp nx 2 for each x 0, 1 and each positive integer n. 3 2.5 2 1.5 1 0.5 0 0.2 0.4 x 0.6 0.8 1 For each n and x the equation f n x  0 says that 2n 2 x 2 exp nx 2  0 n exp nx 2 which gives us x  1 2n and we observe that 1  n exp 1 2 2 2n from which we deduce that the sequence f n fails to be boundedly convergent on the interval 0, 1 . As in the earlier examples the sequence f n converges pointwise on 0, 1 to the constant function 0. Finally we observe that fn Þ 0 f n x dx  Þ 0 nx exp 1 as n 1 nx 2 dx  1e 2 n 1 2 1 2 Ý. e. f n x  nx exp n 2 x for each x 0, 1 and each positive integer n.x exp x 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0.2 0.4 x 0.6 0.8 1 For each n and x, the equation f n x  0 says that n exp n 2 x n 3 x exp n 2 x  0 which gives us x  1 n2 and we observe that 1 1  ne . n2 Ý and therefore the sequence f n converges uniformly to the fn We deduce that sup f n 0 as n constant function 0 on 0, 1 . Finally we observe that Þ 0 f n x dx  Þ 0 nx exp 1 1  as n 1 n 3 exp n 2 Ý. 360 n 2 x dx 1 n exp n 2 1 n3 0 2. Given that f n x  x n for all x and n, prove that the series f n converges pointwise, but not uniformly, on the interval 0, 1 and that f n converges uniformly on the interval 0,  whenever 0   1. The fact that f n converges pointwise on the interval 0, 1 follows from the fact that Ý xn  n0 whenever 0 1 1 x x  1. Now given any n we have n 1 1 x xn x1 0 1 1  sup 1  sup sup x n 1 1x j0 x x n 1 1x 0 x1 0 x1 Ý and so, certainly, the sequence of numbers n 1 sup does not approach 0 as n 3. 1 xn x 0 x1 j0 Ý. Given that f n x  sin nx /n 2 for all n and x, prove that the series f n converges uniformly on R. Use Scientific Notebook to sketch some of the graphs of these functions to motivate your conclusions. 1 0.8 0.6 0.4 0.2 -4 -2 0 -0.2 2x 4 -0.4 -0.6 -0.8 -1 Since 1 n2 f n follows at once from the comparison test. |f n x | for all n and x the uniform convergence of 4. Prove that the series x n /n! converges uniformly in x on every bounded interval but does not converge uniformly in x on the entire line R. Hint: You do not need to know that the sum of this series is e x for each x in order to answer this question. Note that whenever x  0 and n is a positive integer we have Ý j0 xj j! n x n 1 n1 ! xj j! j0 and therefore if n is a positive integer we have Ý sup j0 xj j! n xj j! j0 x0 5. Prove that the series 2n ! n x n! 2 4n 361  Ý. does not converge uniformly in x on the interval ,  whenever 0   1. Hint: Whenever 0 1, 1 but that it does converge uniformly on the interval x  1 and n is a positive integer we have Ý j1 n 2j ! j x 4 j j! 2 j1 2j ! j x 4 j j! 2 Ý jn1 2j ! j x. 4 j j! 2 Now suppose that n is any given positive integer. Using the fact that the series choose an integer k  n  1 such that k 2j ! 4 j j! 2 diverges, we 2j !  1. 4 j...
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