soln2 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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Unformatted text preview: MATH 300 Fall 2007 Advanced Boundary Value Problems I Solutions to Problem Set 2 To Be Completed by: Friday October 12, 2007 Department of Mathematical and Statistical Sciences University of Alberta Question 1. [p 95, #3.2.2 (e), (f)] For the following functions, sketch the Fourier series of f ( x ) on the interval- L x L, and determine the Fourier coefficients: (a) f ( x ) = ( 1 for | x | < L/ 2 for | x | > L/ 2 (b) f ( x ) = ( 1 if 0 < x < L if- L < x < Solution: (a) From Dirichlets theorem the Fourier series of f ( x ) converges to 1 2 f ( x + ) + f ( x- ) if- L x L, and the graph of the Fourier series of f ( x ) on the interval- L x L is shown below. 1 y x L/2-L L-L/2 Since f ( x ) is an even piecewise smooth function on the interval [- L, L ] , it has a Fourier series repre- sentation of the form f ( x ) a + X n =1 a n cos nx L where a = 1 L Z L f ( x ) dx = 1 L Z L 2 1 dx = 1 2 , and a n = 2 L Z L f ( x ) cos nx L dx = 2 L Z L 2 cos nx L dx = 2 n sin nx L L 2 = 2 n sin n 2 for n 1 . (b) Again, from Dirichlets theorem the Fourier series of f ( x ) converges to 1 2 f ( x + ) + f ( x- ) if- L x L, and the graph of the Fourier series of f ( x ) on the interval- L x L is shown below. y x-L L 1 Since f ( x )- 1 2 is an odd piecewise smooth function on the interval [- L, L ] , it has a Fourier series representation of the form f ( x )- 1 2 X n =1 b n sin nx L where b n = 2 L Z L f ( x )- 1 2 sin nx L dx = 1 L Z L sin nx L dx =- 1 n cos nx L L =- 1 n (cos n- 1) = 1 n [1- (- 1) n ] for n 1 . Question 2. [p 96, #3.2.3] Show that the Fourier series operation is linear: that is, show that the Fourier series of c 1 f ( x ) + c 2 g ( x ) is the sum of c 1 times the Fourier series of f ( x ) and c 2 times the Fourier series of g ( x ) . Solution: Suppose that the Fourier series of f and g are given by f ( x ) A + X n =1 A n cos nx L + B n sin nx L and g ( x ) C + X n =1 C n cos nx L + D n sin nx L where A = 1 2 L Z L- L f ( x ) dx, A n = 1 L Z L- L f ( x ) cos nx L dx, B n = 1 L Z L- L f ( x ) sin nx L dx for n 1 , and C = 1 2 L Z L- L g ( x ) dx, C n = 1 L Z L- L g ( x ) cos nx L dx, D n = 1 L Z L- L g ( x ) sin nx L dx for n 1 . If c 1 and c 2 are scalars, and the Fourier series of c 1 f + c 2 g is c 1 f ( x ) + c 2 g ( x ) E + X n =1 E n cos nx L + F n sin nx L , then E = 1 2 L Z L- L ( c 1 f ( x ) + c 2 g ( x )) dx = c 1 2 L Z L- L f ( x ) dx + c 2 2 L Z L- L g ( x ) dx = c 1 A + c 2 C . Also, E n = 1 L Z L- L ( c 1 f ( x ) + c 2 g ( x )) cos nx L dx = c 1 L Z L- L f ( x ) cos nx L dx + c 2 L Z L- L g ( x ) cos nx L dx = c 1 A n + c 2 C n for n 1 . Similarly, F n = 1 L Z L- L ( c 1 f ( x ) + c 2 g ( x )) sin nx L dx = c 1 L Z L- L f ( x ) sin nx L dx + c 2 L Z L- L g ( x ) sin nx L dx = c 1 B n + c 2 D n for n 1 . Therefore the Fourier series for c 1 f + c 2 g is c 1 f ( x ) + c 2 g...
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This homework help was uploaded on 04/15/2008 for the course APMA 3102 taught by Professor Courdurier during the Spring '00 term at Columbia.

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soln2 - MATH 300 Fall 2007 Advanced Boundary Value Problems...

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