Chapter12Solutions

Chapter12Solutions - Mathematics 421 Solutions 12.2 and...

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Mathematics 421 Solutions 12.2 and 12.3 Spring 2008 Introduction A student asked about Problem 9 in Section 12.2 in office hours. The solution has some unusual features, so it is useful to share it with the whole class. Problem 22 of Section 12.3 gave more trouble than it should have, so details will be presented here. Problem 9 of 12.2 The Fourier series of f.x/ D ± 0 if ± ± < x < 0 sin x if 0 < x < ± on the interval Œ ± ±;±Ł is to be found. That is the heavier line in the graph below (the significance of the lighter line will appear in the solution). Since the interval is Œ ± ±;±Ł , the series is a 0 =2 C 1 X n D 1 a n cos nx C b n sin nx To get the general formula for a n , integrate each case of the definition of f.x/ over the interval on which that case is used. The interval where f.x/ D 0 contributes zero, so we need only write the integral for the other case to get a n D 1 ± Z ± ± ± f.x/ cos nx dx D 1 ± Z ± 0 sin x cos nx dx D 1 Z ± 0 sin .n C 1/x ± sin .n ± 1/x dx D 1 ² 1 n ± 1 cos .n ± 1/x ± 1 n C 1 cos .n C 1/x ³ ± 0 D 1 ´ cos .n ± 1/± ± 1 n ± 1 ± cos .n C 1/± ± 1 n C 1 µ Both of the cos .n ˙ 1/± quantities in the last line may be written as ± . ± 1/ n . Thus, a n D 1 C . ± 1/ n ´ 1 n C 1 ± 1 n ± 1 µ D ± 1 C . ± 1/ n ±.n 2 ± 1/
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Mathematics 421 Solutions 12.2 and 12.3, p. 2 The presence of n 2 ± 1 in the denominator of this expression indicates that the formula cannot be valid for n D 1 . Looking more closely at the computation, we see that this denominator arose from
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This note was uploaded on 09/29/2009 for the course 650 421 taught by Professor Bumby during the Spring '08 term at Rutgers.

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Chapter12Solutions - Mathematics 421 Solutions 12.2 and...

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