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# cse41 - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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CHAPTER 4 FOURIER SERIES AND INTEGRALS 4.1 FOURIER SERIES FOR PERIODIC FUNCTIONS This section explains three Fourier series: sines, cosines, and exponentials e ikx . Square waves (1 or 0 or 1) are great examples, with delta functions in the derivative. We look at a spike, a step function, and a ramp—and smoother functions too. Start with sin x .I t ha s pe r iod 2 π since sin( x +2 π )= s in x . It is an odd function since sin( x )= sin x , and it vanishes at x =0 and x = π . Every function sin nx has those three properties, and Fourier looked at inﬁnite combinations of the sines : Fourier sine series S ( x )= b 1 sin x + b 2 sin 2 x + b 3 sin 3 x + ··· = ± n =1 b n sin nx (1) If the numbers b 1 ,b 2 ,... drop oﬀ quickly enough (we are foreshadowing the im- portance of the decay rate) then the sum S ( x ) will inherit all three properties: Periodic S ( x +2 π )= S ( x ) Odd S ( x )= S ( x ) S (0) = S ( π )=0 200 years ago, Fourier startled the mathematicians in France by suggesting that any function S ( x ) with those properties could be expressed as an inﬁnite series of sines. This idea started an enormous development of Fourier series. Our ﬁrst step is to compute from S ( x )the number b k that multiplies sin kx . Suppose S ( x )= b n sin nx . Multiply both sides by sin kx . Integrate from 0 to π : ² π ² π ² π S ( x )sin kx dx = b 1 sin x sin kx dx + ··· + b k sin kx sin kx dx + ··· (2) 0 0 0 On the right side, all integrals are zero except the highlighted one with n = k . This property of orthogonality will dominate the whole chapter. The sines make 90 angles in function space, when their inner products are integrals from 0 to π : Orthogonality ² π 0 sin nx sin kx dx =0 if n
± ² 318 Chapter 4 Fourier Series and Integrals ³ ´ µ π Zero comes quickly if we integrate cos mx dx = sin m mx 0 =0 0. So we use this: 1 1 Product of sines sin nx sin kx = cos( n k ) x cos( n + k ) x. (4) 2 2 Integrating cos mx with m = n k and m = n + k proves orthogonality of the sines. The exception is when n = k . Then we are integrating (sin kx ) 2 = 1 2 1 2 cos 2 kx : π π π 1 1 π sin kx sin kx dx = dx cos 2 kx dx = . (5) 2 2 2 The highlighted term in equation (2) is b k π/ 2 . Multiply both sides of (2) by 2 : 0 0 0 Sine coeﬃcients S ( x )= S ( x ) b k = 2 π π 0 S ( x )sin kx dx = 1 π π π S ( x )sin kx dx. (6) Notice that S ( x )sin kx is even (equal integrals from π to 0 and from 0 to π ).

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cse41 - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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