1873_solutions

# F gx and f gx gx given that f and g are riemann

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Unformatted text preview: ber x in the interval a, b and choose an integer N such that Ý kJ , y n  nN 4 . Using the fact that f is continuous at each of the numbers y n for n N, choose   0 such that whenever n N and |t y n |   we have . |f t f yn |  2 var , a, b Now suppose that P  x 0 , x 1 , , x p x j 1 , x j for each j, and define g to be the is any partition of a, b for which P   and that t j step function that takes the value f x j at each number x j and that takes the constant value f t j in each interval x j 1 , x j . We see that Þa b gd Þa b Ý Ý g y n J , y n fd  n1 f y n J , y n n1 N Ý N g y n J , y n n1 f y n J , y n n1 Ý |g y n |J , y n   nN1 |f y n |J , y n nN1 Ý N |g y n f y n |J , y n  2 n1 kJ , y n nN1 N  n1 2 var , a, b J , y n  2 3. By combining the preceding two exercises, obtain an analog of Exercise 2 that does not require the assumption that  varies discretely on a, b . 318 The desired extension to the case in which  is an arbitrary increasing function follows at once when we split  into its continuous and discrete parts. 12 Infinite Series Some Elementary Exercises on Series 1. Find the nth partial sum of the series n3 . n n1 n3 Deduce that this series is convergent and find its sum. Solution: Given any positive integer j we have j3 j j1 j3 2 j1  1 j 1 j3 and therefore n j1 j3 j j1 j3 n 1 j 2 j1  j1 n 1 1 j3 n 3 n 2 j  j2 1 j j1 j4  22 2 2 3 n1 22 1 1 2 3 1 2 as n 2. 1 j 1 1 1 2 1 3 1 3 1 n1 1 n2 1 6 Ý. xn. a. Find the derivative of the nth partial sum of the series If n is any positive integer and x 1 then n xj  j1 xn . x x1 1 Differentiating we obtain n jx j 1 1 xn j1 xn nx n  nx n1  1 1 x2 nx n 1 1 x2 nx n 1 . Deduce that if |x |  1 we have b. Find the nth partial sum of the series Ý nx n 1 nx n 1  n1 1 1 x 2 . In order to deduce the identity Ý  n1 we need to take the limit as n 1 1 x 2 Ý of each side of of the identity n jx j 1 1 j1 319 xn 1 nx n  nx n1 x2 x 1 n3 and for this purpose we need to know that lim nx n  0 nÝ whenever |x |  1. Later in this chapter, we shall see some simple ways of finding the latter limit. Perhaps the simplest way to find it right now is to use L’Hôpital’s rule. Suppose that |x |  1. To show that n|x | n 0 as n Ý we use the fact that n n|x | n  1/|x | n We define c  log 1/|x | . Note that c  0 and n|x | n  n e cn n and the fact that n|x | 0 as n Ý follows from the fact that lim tct  0 tÝe which follows easily from L’Hôpital’s rule. An alternative to L’Hôpital’s rule is to give the students the assignment of proving the inequality 22 e ct 1  ct  c t 2 for all t 0. This inequality follows easily from the mean value theorem and from it we obtain the above limit easily. 3. Given that |x |  1 and that n sn  1 x 2j 3j j1 for every positive integer n, obtain the identity 3x 4 s n 1 x 2  2x 2  3x 2n4 3n 1 x 2n2 1 x2 x 2 n 4 and deduce that Ý 3x 4 1 x2 1 x 2j  2x 2  3j j1 We observe that n sn...
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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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