# This complex form of the fourier series is completely

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This complex form of the Fourier series is completely equivalent to the original series. Given the a n ’s and b n ’s we can compute the c n ’s using the formula above, and conversely, given the c n ’s we can solve for a 0 = 2 c 0 a n = c n + c - n for n > 0 b n = ic n - ic - n for n > 0 III.4.2 Inner product for a space of functions We will consider the space of complex valued functions f ( x ) on the interval [0 , 1] that obey Z 1 0 | f ( x ) | 2 dx < This space is called L 2 ([0 , 1]) and is an example of a Hilbert space. If f and g are two functions in this space, then we define the inner product to be h f, g i = Z 1 0 f ( x ) g ( x ) dx Here f ( x ) denotes the complex conjugate of f . III.4.3 An orthonormal basis Now we will show that the complex exponential functions appearing in the Fourier expansion are an orthonormal set. Let e n ( x ) = e i 2 πnx for n = 0 , ± 1 , ± 2 , . . . . We must compute h e n , e m i . Since e n ( x ) = e - i 2 πnx , we find h e n , e m i = Z 1 0 e - i 2 πnx e i 2 πmx dx = Z 1 0 e i 2 π ( m - n ) x dx 90
III.4 Fourier series If n = m then e i 2 π ( m - n ) x = 1 so the integral equals 1. On the other hand if n 6 = m then e i 2 π ( m - n ) x has an anti-derivative e i 2 π ( m - n ) x / 2 π ( m - n ) that takes on the same value (namely 1 / 2 π ( m - n )) at both endpoints x = 0 and x = 1. Hence the integral is zero in this case. Thus the functions { e n ( x ) } form an orthonormal set. In fact, they are a basis for our space of functions. The fact that they span the space, i.e., that every function can be written as an infinite linear combination of the e n ’s, is more difficult to show. (For a start, it would require a discussion of what it means for an infinite linear combination to converge!) However, if you accept the fact that the e n ’s do indeed form an infinite basis, then it is very easy to compute the coefficients. Starting with f ( x ) = X n = -∞ c n e n ( x ) we simply take the inner product of both sides with e m . The only term in the infinite sum that survives is the one with n = m . Thus h e m , f i = X n = -∞ c n h e m , e n i = c m and we obtain the formula c m = Z 1 0 e - i 2 πmx f ( x ) dx III.4.4 An example Let’s compute the Fourier coefficients for the square wave function f ( x ) = 1 if 0 x 1 / 2 - 1 if 1 / 2 < x 1 If n = 0 then e - i 2 πnx = e 0 = 1 so c 0 is simply the integral of f . c 0 = Z 1 0 f ( x ) dx = Z 1 / 2 0 1 dx - Z 1 1 / 2 1 dx = 0 91
III Orthogonality Otherwise, we have c n = Z 1 0 e - i 2 πnx f ( x ) dx = Z 1 / 2 0 e - i 2 πnx dx - Z 1 1 / 2 e - i 2 πnx dx = e - i 2 πnx - i 2 πn x =1 / 2 x =0 - e - i 2 πnx - i 2 πn x =1 x =1 / 2 = 2 - 2 e iπn 2 πin = 0 if n is even 2 /iπn if n is odd Thus we conclude that f ( x ) = X n = -∞ n odd 2 iπn e i 2 πnx To see how well this series is approximating f ( x ) we go back to the real form of the series. Using a n = c n + c - n and b n = ic n - ic - n we find that a n = 0 for all n , b n = 0 for n even and b n = 4 /πn for n odd. Thus f ( x ) = X n =1 n odd 4 πn sin(2 πnx ) = X n =0 4 π (2 n + 1) sin(2 π (2 n + 1) x ) We can use MATLAB/Octave to see how well this series is converging. The file ftdemo1.m contains a function that take an integer N as an argument and plots the sum of the first 2 N + 1 terms in the Fourier series above. Here is a listing: