1873_solutions

0 e x dx 1 2 from the fact that 374 n0 1n 2n 1

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Unformatted text preview: ions guarantees that f is analytic on 1, 1 . 1, 1 the composition theorem for analytic 6. Suppose that a n x n and b n x n are two power series that converge in an interval suppose that the set of numbers x r, r for which the equation Ý r, r where r  0, and Ý anxn  n0 bnxn n0 has a limit point in r, r . Prove that a n  b n for every n. The desired result follows at once from the fact that the sum of a power series is analytic and that two analytic functions on an open interval must be identical if they agree on a set that has a limit point in that interval. 7. Given that f is analytic and nonconstant on an open interval U and that c positive integer n such that f n c 0. Since the equation Ý fx  n0 f n c n! x c U, prove that there exists a n must hold for all x sufficiently close to c the desired result follows at once. 8. Given that f is analytic and nonconstant on an open set U and that K is a closed bounded subset of U, prove that the set x K fx 0 is finite. If the set 382 x K fx 0 were infinite then it would have to have a limit point in K and so f would have to be the constant function 0. 9. Given that f is analytic and nonconstant on an open interval U prove that the set x U fx 0 is countable. This solution makes use of the distance function of a set. For each positive integer n we define 1. K n  n, n x U R U x n Each set K n is closed and bounded and U Ý  Kn. n1 Since x fx 0  U Ý  x Kn fx 0 n1 we deduce that the set x U fx 0 is countable. 16 Integration of Functions of Two Variables Some Exercises on Iterated Riemann Integrals 1. Given that fx  Þ 0 exp x gx  t 2 dt Þ0 1 exp x 2 t 2  1 t2  1 dt and given h x  f x 2g x for every real number x, prove that the function h must be constant. What is the value of this constant? Deduce that Ý  Þ 0 exp x 2 dx  2 . For every number x we see from the theorem on differentiating a partial integral that h x  2f x f x  g x x 1 2x t 2  1 exp x 2 t 2  1  2 exp x 2 Þ exp t 2 dt  Þ dt 0 0 t2  1  2 exp x 2 Þ 0 exp x t 2 dt 2 exp x 2 Þ 0 x exp 1 x 2 t 2 dt In the latter integral we make the substitution u  tx and we deduce that h x  2 exp x 2 Þ 0 exp x t 2 dt 2 exp x 2 Þ 0 exp x for every number x. Therefore the function h is constant. Now since 383 u 2 du  0 h0  Þ 0 exp 0 2 t 2 dt Þ 1 0 exp 0 2 t 2  1 t2  1 dt   4 We deduce that h x   for every number x. 4 Now we use the fact that h n   for every positive integer n. From the 4 bounded convergence theorem we see that 1 exp n 2 t 2  1 1 lim Þ dt  Þ 2 0 dt  0 21 nÝ 0 0 t 1 t and so Ý Þ 0 exp t 2 dt t 2 dt    lim h n  nÝ 4 2  . 2 0 from which we deduce that Ý Þ 0 exp 2. Given that gy  Þ0 Ý exp x 2 cos 2xydx f y  exp y 2 g y and for every real number y, prove that the function f must be constant. What is the value of this constant? Since |exp x 2 cos 2xy | exp x 2 and |D 2 exp x 2 cos 2xy |  | 2x exp x 2 sin 2xy | 2x exp x 2 for all x and y we see that the hypotheses of the theorem on differentiation...
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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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