1873_solutions

# Any interval that is included in the cantor set must

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Unformatted text preview: sin 2x dx x 2x 0 w  2 w Ý Þ sin 2x dx lim 2x 0 and the substitution u  2x yields Þ0 2w lim wÝ b. Þ0 Ý sin 2 x dx x2 Þ0 w lim wÝ c. Þ0 Ý 2 sin 2 x cos 2 x dx x2 sin u du  u 2 sin x cos x dx  lim x wÝ Ý w Þ0 Ý Ý sin x dx. x sin 2 x dx  lim sin 2 w  w wÝ x2 2 sin 2 x cos 2 x dx  lim wÝ x2 Þ0 Þ0 Ý sin 2 x dx. x2 Þ0 wÝ lim d. Þ0 Þ0 Þ0 w 2 sin 2 x cos 2 x dx x2 w 2 x  w Ý 2 Þ sin 22 dx lim 0 2x 2 sin 4 x dx x2 357 2w sin 2 u du  u2 Þ0 Ý sin 2 x dx. x2 Ý 2 sin 4 x dx  x2 Þ0 Ý  Þ0 Ý  Þ0 Þ0 Ý 2 sin 2 x 1 cos 2 x dx x2 2 sin 2 x dx Þ Ý 2 sin 2 x cos 2 x dx 0 x2 x2 Ý 2 2 2 sin x dx Þ sin x dx  Þ Ý sin 2 x dx. 2 0 0 x x2 x2 14 Sequences and Series of Functions Exercises on Convergence of Sequences of Functions 1. For each of the following definitions of the function f n on the interval 0, 1 prove that the sequence f n converges pointwise to the function 0 on 0, 1 and determine whether the sequence converges boundedly and whether it converges uniformly. In each case, determine whether or not we have Þ 0 f n  0. 1 lim nÝ In each case, use Scientific Notebook to sketch some graphs of the given function and ask yourself whether your conclusion is compatible with what you see in the graphs. a. f n x  nx exp nx for each x 0, 1 and each positive integer n. Hint: To see that the sequence converges boundedly but not uniformly to the constant 0, observe that each function f n has its maximum when n exp nx n 2 x exp nx  0 which occurs when x  1/n. Note that f n 1/n  1/e for each n. 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 When x  0 the condition lim nx exp nx  0 nÝ is obvious and when x 0 the condition lim nx exp nx  0 nÝ follows from the fact that lim u  0. eu converges pointwise and boundedly but not uniformly to the uÝ Therefore the sequence f n constant function 0. Finally we observe that Þ 0 f n x dx  Þ 0 nxe nx dx  1 as n 1 Ý. 358 1 n 1 en 1 ne n 0 b. f n x  n 2 x exp nx for each x 0, 1 and each positive integer n. 5 4 3 2 1 0 0.2 0.4 x 0.6 0.8 1 Arguing as we did in Part a, we see that each function f n has its maximum value n at the e number 1 . Since sup f n Ý as n Ý we see that f n fails to converge boundedly. n When x  0 the condition lim n 2 x exp nx  0 nÝ is obvious and when x 0 the condition 2x2 lim n 2 x exp nx  1 n Ý n nx  0 x lim e nÝ follows from the fact that 2 lim uu  0. e converges pointwise to the constant function 0. Finally we observe uÝ Therefore the sequence f n that Þ 0 f n x dx  Þ 0 n 2 xe nx dx  1 1 as n 1 n 1 1 en en converges pointwise to the constant function 0 we Ý. So even though the sequence f n do not have Þ0 fn  Þ0 0 1 lim nÝ c. f n x  nx exp n 2 x 2 for each x 1 0, 1 and each positive integer n. 0.4 0.3 0.2 0.1 0 0.2 0.4 x 0.6 0.8 1 For each x and n we have f n x  n exp n 2 x 2 and so the equation f n x  0 holds when x  2n 3 x 2 exp n 2 x 2 1 2n . Since 1  1. 2e 2n From this observation we see that the sequence f n co...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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