MoreSol12and13

# MoreSol12and13 - Mathematics 421 More Solutions Chapters 12...

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Unformatted text preview: Mathematics 421 More Solutions Chapters 12 and 13 Spring 2008 Introduction To aid preparation for the second exam, some complete solutions of problems have been prepared. The solutions of 12.2#9 and 12.3#22, as well as the solutions of the problems in the second supplement, have been posted previously. The solutions included here are: 12.1#14; 12.4#10; 12.6#16; and 13.1#14. Problem 14 of 12.1 To show that L .x/ D 1; L 1 D x C 1; L 2 .x/ D 1 = 2 x 2 2x C 1 are orthogonal on .0; 1 / with the weight function e x , one must show that the following definite integrals are zero: Z 1 e x L .x/L 1 .x/dx D Z 1 e x .1 x/dx Z 1 e x L .x/L 2 .x/dx D Z 1 e x 1 = 2 x 2 2x 1 dx Z 1 e x L 1 .x/L 2 .x/dx D Z 1 e x 1 3x C 5 = 2 x 2 1 = 2 x 3 dx Note that each pair of different functions must be considered; if one more function were added to the list, three more integrals (for a total of six) would need to be computed, and if two more functions were added, we would need to check a total of ten integrals. In fact, there is an infinite sequence of orthogonal polynomials for this weight, one of each degree. They are known as the Laguerre polynomials . As an aid to computing the integrals, It is useful to recall that Z 1 e x x n dx is the value of the Laplace transform of t n at s D 1 , which is nŠ . Alternatively, the calculation of that transform may be mimicked using the integration by parts formula Z 1 e x x n dx D n Z 1 e x x n 1 dx OEx n e x Łt 1 D n R 1 e x x n 1 dx for n > 0 1 for n D The explicit integrals can be evaluated term by term using these values. Further calculations may be done to show that these functions also have h L i ;L i i D 1 for i D 0;1;2 . Problem 10 of 12.4 We want the complex Fourier series of f.x/ D cos x if 0 < x < =2 if =2 < x < with f.x C / D f.x/ . A graph of two periods of the function (on the interval OE ; Ł ) is shown below Mathematics 421 More Solutions Chapters 12 and 13, p. 2 The period is , so if we wanted to give the function on our usual symmetric interval, that interval would be OE =2; =2Ł , so the p of the usual formulas is =2 . Thus, we seek the c n giving f.x/ D 1 X 1 c n e 2inx : The coefficients c n are found by averaging f.x/e 2inx over a full period P which may be taken to be OE =2; =2Ł or OE0; Ł , or any convenient interval of length . However, the function is defined by cases , so the integral will be written as a sum of the two “case integrals” with domain of integration equal to an interval on which f.x/ is given by a particular formula, with f.x/ replaced by that formula. Since f.x/ D on part of the interval, that part of the integral will be zero and need not be written. However, the average must still be taken over a period of...
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MoreSol12and13 - Mathematics 421 More Solutions Chapters 12...

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