Page 461 - A First Course In Stochastic Models
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456                           APPENDICES

                is sought that agrees with f at n equally spaced points x l = 2πℓ/n between 0 and
                2π. More specifically, we look for complex numbers c 0 , . . . , c n−1 such that

                                n−1
                                      ik(2πℓ/n)
                                   c k e     = f ℓ ,  ℓ = 0, . . . , n − 1.   (D.1)
                                k=0
                It is convenient to write these linear equations in matrix notation as
                                              Fc = f.

                Here F is a complex-valued matrix whose (ℓ, k)th element (F) ℓk is given by
                                           kℓ
                                  (F) ℓk = w ,  ℓ, k = 0, . . . , n − 1,
                where the complex number w is defined by
                                            w = e 2πi/n .

                Let F be the matrix whose elements are the complex conjugates of the elements
                of the matrix F. The matrix F has the nice property that

                                          FF = FF = nI,                       (D.2)
                where I is the identity matrix (the column vectors of the symmetric matrix F
                form an orthogonal system). To verify this, let w = e −2πi/n  denote the complex
                conjugate of w. The inproduct of the rth row of F and the sth column of F is
                given by
                                   0
                                 0
                                           s
                                               2r
                                        r
                          γ rs = w w + w w + w w  2s  + · · · + w (n−1)r w (n−1)s .
                                       0
                For r = s each term equals e = 1 and so the sum γ rs is n. For r 	= s the sum γ rs
                                                                             s
                                                       n
                                                                          r
                can be written as 1 + α + · · · + α n−1  = (1 − α )/(1 − α) with α = w w (	= 1).
                                                              n
                                       n
                      n
                Since w = e 2πi  = 1 and w = e −2πi  = 1, we have α = 1 and so γ rs = 0 for
                r 	= s. This gives (D.2). By (D.2), we have F −1  = (1/n)F. It now follows that
                the vector c of Fourier coefficients is given by c = (1/n)Ff . Componentwise, we
                have
                                     n−1
                                   1       −2πiℓk/n
                               c k =    f ℓ e     ,  k = 0, . . . , n − 1.    (D.3)
                                   n
                                     ℓ=0
                This inversion formula parallels the formula c k = (2π) −1     π  f (x) e −ikx  dx in
                                                                  −π
                continuous Fourier analysis. Notice that (D.3) inherits the structure of (D.1).
                  In many applications, however, we proceed in reverse order: we know the
                Fourier coefficients c k and wish to calculate the original coefficients f j . By the
                formula (D.1) we can transform c back into f . The matrix multiplications in (D.1)
                                    2
                would normally require n multiplications. However, the discrete FFT method per-
                forms the multiplications in an extremely fast and ingenious way that requires
                                                   2
                only n log (n) multiplications instead of n . The key to the method is the simple
                        2
                observation that the discrete Fourier transform of length n (n even) can be written
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