SSM-Ch11 - Chap 11 Sec 11.1 Fourier Series Integrals and...

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Chap . 11 Fourier Series , Integrals , and Transforms Sec . 11 . 1 Fourier Series Content : u Fourier series (5) and their coefficients (6) u Calculation of Fourier coefficients by integration (Example 1) u Reason why (6) gives the Fourier coefficients (Theorem 1) u Great generality of Fourier series (Theorem 2) Problem Set 11 . 1 . Page 485 5 . Linear combinations of periodic functions . Addition of periodic functions with the same period p gives a periodic function with that period p , and so does the multiplication of such a function by a constant. Thus all functions of period p form an important example of a vector space. This is what you are supposed to prove. Now, by assumption, given any periodic functions f and g that have period p , that is, f u x U p U ± f u x U , g u x U p U ± g u x U , (A) form a linear combination of them, say, h ± af U bg u a , b constant U and show that h is periodic with period p ; that is, you must show that h u x U p U ± h u x U . This follows by calculating and using the definition of h and then (A): h u x U p U ± af u x U p U U bg u x U p U ± af u x U U bg u x U ± h u x U . 13 . Fourier series . In Probs. 13-20 you first have to find a formula for the given function. Then you obtain the Fourier coefficients from (6) by integration. In Probs. 13 and 14 the integrands of (6) are constants times cosines or sines. In Probs. 15-20 they are t times cosine or sine. Such integrals are evaluated by integration by parts, just as in calculus. Of course you have noticed that they depend on n , which gives factors 1/ n , 1/ n 2 etc. in the coefficients; these are essential with respect to convergence. In Prob. 13 you have to integrate from u u /2 to u /2 only. (Why?) From (6a) you obtain the constant term of the series (which is the mean value of the given function over the interval from u u to u U . Indeed, you obtain
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110 Fourier Analysis. Partial Differential Equations (PDEs) Part C a 0 u 1 2 u u U u /2 u /2 1 ± dx u 1 2 . Next you use (6b) to get a n u 1 u u U u /2 u /2 1 ± cos nx dx u 1 n u sin nx U u /2 u /2 u 1 n u sin n u 2 U sin U n u 2 u 2 n u sin n u 2 u 2/ n n u 1,5,9, u 0 n u 2,4,6, u U 2/ n n u 3,7,11, u . Hence the sequence of the cosine coefficients is 2 u , 0, U 2 3 u , 0, 2 5 u , 0, u . Finally, show that the sine coefficients are zero, b n u 1 u u U u /2 u /2 1 ± sin nx dx u U 1 n u cos nx U u /2 u /2 u 1 n u U cos n u 2 U cos U n u 2 u 0. Conclude that the Fourier series is a “Fourier cosine series” f U x ± u 1 2 U 2 u cos x U 1 3 cos 3 x U 1 5 cos 5 x U U u . Sec . 11 . 2 Functions of Any Period p u 2 L No new ideas in the transition to an arbitrary period, just more complicated formulas. The notation p u 2 L is practical for later applications, as is mentioned in the text. Note that Example 2 is closely related to Example 1 in the last section.
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This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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SSM-Ch11 - Chap 11 Sec 11.1 Fourier Series Integrals and...

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