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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 : u2022 Fourier series (5) and their coefficients (6) u2022 Calculation of Fourier coefficients by integration (Example 1) u2022 Reason why (6) gives the Fourier coefficients (Theorem 1) u2022 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 ue0a2 x u002b p ue0a3 u003d f ue0a2 x ue0a3 , g ue0a2 x u002b p ue0a3 u003d g ue0a2 x ue0a3 , (A) form a linear combination of them, say, h u003d af u002b bg ue0a2 a , b constant ue0a3 and show that h is periodic with period p ; that is, you must show that h ue0a2 x u002b p ue0a3 u003d h ue0a2 x ue0a3 . This follows by calculating and using the definition of h and then (A): h ue0a2 x u002b p ue0a3 u003d af ue0a2 x u002b p ue0a3 u002b bg ue0a2 x u002b p ue0a3 u003d af ue0a2 x ue0a3 u002b bg ue0a2 x ue0a3 u003d h ue0a2 x ue0a3 . 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 u2212 u03c0 /2 to u03c0 /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 u2212 u03c0 to u03c0 ue0a3 . Indeed, you obtain
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110 Fourier Analysis. Partial Differential Equations (PDEs) Part C a 0 u003d 1 2 u03c0 u222b u2212 u03c0 /2 u03c0 /2 1 u2219 dx u003d 1 2 . Next you use (6b) to get a n u003d 1 u03c0 u222b u2212 u03c0 /2 u03c0 /2 1 u2219 cos nx dx u003d 1 n u03c0 sin nx u2212 u03c0 /2 u03c0 /2 u003d 1 n u03c0 sin n u03c0 2 u2212 sin u2212 n u03c0 2 u003d 2 n u03c0 sin n u03c0 2 u003d 2/ n n u003d 1,5,9, u22ef 0 n u003d 2,4,6, u22ef u2212 2/ n n u003d 3,7,11, u22ef . Hence the sequence of the cosine coefficients is 2 u03c0 , 0, u2212 2 3 u03c0 , 0, 2 5 u03c0 , 0, u22ef . Finally, show that the sine coefficients are zero, b n u003d 1 u03c0 u222b u2212 u03c0 /2 u03c0 /2 1 u2219 sin nx dx u003d u2212 1 n u03c0 cos nx u2212 u03c0 /2 u03c0 /2 u003d 1 n u03c0 u2212 cos n u03c0 2 u002b cos u2212 n u03c0 2 u003d 0. Conclude that the Fourier series is a “Fourier cosine series” f ue0a2 x ue0a3 u003d 1 2 u002b 2 u03c0 cos x u2212 1 3 cos 3 x u002b 1 5 cos 5 x u2212 u002b u22ef .
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