OFFS_12-19

# OFFS_12-19 - Tutorial OFFS Orthogonal Functions and Fourier...

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Tutorial OFFS: Orthogonal Functions and Fourier Series (the proof of the pudding is in the eating), we require ¡ Á 2 ; x 0 ¢ = a 0 ¡ x 0 ; x 0 ¢ + a 1 ¡ x 1 ; x 0 ¢ + a 2 ¡ x 2 ; x 0 ¢ = a 0 ¢ 2+ a 1 ¢ 0 + a 2 ¢ 2 3 = 0 ; ¡ Á 2 ; x 1 ¢ = a 0 ¡ x 0 ; x 1 ¢ + a 1 ¡ x 1 ; x 1 ¢ + a 2 ¡ x 2 ; x 1 ¢ = a 0 ¢ 0+ a 1 ¢ 2 3 + a 2 ¢ 0 = 0 : The latter yields a 1 = 0 : The former gives the relationship a 2 = ¡ 3 a 0 : Thus, Á 2 = a 0 ¡ 1 ¡ 3 x 2 ¢ : The constant a 0 can be determined by imposing a normalization condition on Á 2 . Exercise 12. Three references are: (1) M. L. Boas, Mathematical Methods in the Physical Sciences (Second Edition), Wiley, 1983, p. 313; (2) C.R. Wylie and L.C.Barrett, Advanced En- gineering Mathematics (Sixth Edition), McGraw-Hill, 1995, p. 504; G. Arfken, Mathematical Methods for Physicists (Third Edition), Academic Press, 1985, p. 760. Exercise 13. We multiply f ( x )= a 0 + P 1 n =1 a n cos k n x + b n sin k n x by cos k m x and integrate from x 0 to x 0 + ¸ : Z x 0 + ¸ x 0 f ( x )cos k m xdx = Z x 0 + ¸ x 0 a 0 cos k m xdx + 1 X n =1 a n Z x 0 + ¸ x 0 cos k m x cos k n xdx + 1 X n =1 b n Z x 0 + ¸ x 0 cos k m x sin k n xdx: For m = 0 , cos k m x = 1 , and the second and third terms on the right vanish = ) a o ¸ = R x 0 + ¸ x 0 f ( x ) dx: For m 6 =0 ; the first and third terms vanish and the second becomes Z x 0 + ¸ x 0 f ( x )cos k m xdx = 1 X n =1 a n ¢ ¸ 2 ± mn = ¸ 2 a m : Multiplying f ( x ) by sin k n x and integrating yields the respective result for b m : Exercise 14. Take f ( x ) to be,say, symmetric about the point x = a:

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## This note was uploaded on 07/13/2008 for the course PHY 201 taught by Professor Covatto during the Spring '08 term at ASU.

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OFFS_12-19 - Tutorial OFFS Orthogonal Functions and Fourier...

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