1873_solutions

See in a moment e x sin e x the integral dx is

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Unformatted text preview: n Ý. Note that the sequence sup f n is decreasing. To obtain a contradiction, suppose that the sequence sup f n fails to converge to 0 and, using this assumption, choose a number  0 such that the inequality sup f n  holds for every n. For each n, choose a number x n S such that f n x n  and, using the fact that x n is a sequence in the closed bounded set S, choose a partial limit x of the sequence x n . Using the fact that f n x 0 as n Ý, choose an integer N such that f N x  . Now, using the fact that the function f N is continuous at the number x, choose a number   0 such that the inequality f N t  holds for every number t S x , x   . Using the fact that x is a partial limit of the sequence x n we now choose an integer n  N such that xn x , x   . Then we have fn xn fN xn  contradicting the way in which the number x n was chosen. Some Exercises on the Tests for Uniform Convergence 364 1. Prove that the series 1 converges uniformly in x on the interval n1 sin 1  n x n 1, 1 and converges pointwise on the entire line R. If x 0 then, starting at a sufficiently large value of n, the sequence of numbers sin 1  decreasing (with limit sin 1) and so the sequence x sin 1  n n x n is is a decreasing sequence of positive numbers with limit 0 and it follows from Dirichlet’s test that the series x 1 n 1 sin 1  n n converges. If p is any given positive number and if we define x sin 1  n fn x  n then starting at a sufficiently large integer N the sequence of functions decreases uniformly to 0 on the interval 0, p as n Ý. By Dirichlet’s test for uniform convergence the series x 1 n 1 sin 1  n n converges uniformly in x on the interval 0, p . Now suppose that x  0. It is clear that sin 1  lim nÝ n x n 0 but, before we can use Dirichlet’s test, we need to show that this sequence is decreasing. We write sin 1  x t ft  t for t  0 and we observe that for each t we have t x cos 1  x sin 1  x t t 2 ft t 5/2 and it is clear that f t  0 if t is sufficiently large. A similar argument shows that if p  0 then it is possible to find an integer N such that starting at n  N, the sequence x sin 1  n n decreases to zero for every number x p, 0 and it is clear that this convergence is uniform. It therefore follows from Dirichlet’s test for uniform convergence that x 1 n 1 sin 1  n n converges uniformly on the interval p, 0 . We conclude that the series converges uniformly on every bounded set of real numbers, which is more than the exercise requested. 2. Prove that if we define Ý fx  n1 1 n1 sin 1  n x n for every real number x then the function f is differentiable at every real number and for every number x we have 365 Ý 1 fx n1 n1 cos 1  n 3/2 x n The solution of this exercise follows at once from the theorem on term by term differentiation of series, from Exercise 1 and from the uniform convergence in x of the series x 1 n 1 cos 1  n . 3/2 n Exercises on Continuity of a Limit Function 1. Suppose that a and b are real numb...
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