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book   Mobk070    March 22, 2007  11:7








                     36  ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
                       3.2.4  Examples of Fourier Series
                       Example 3.1. Determine a Fourier sine series representation of f (x) = x on the interval
                       (0, 1). The series will take the form


                                                            ∞

                                                       x =     c j sin( jπx)                        (3.28)
                                                            j=0

                       since the sin( jπx) forms an orthogonal set on (0, 1), multiply both sides by sin(kπx)dx and
                       integrate over the interval on which the function is orthogonal.

                                           1                    1
                                                            ∞

                                             x sin(kπx)dx =       c j sin( jπx)sin(kπx)dx           (3.29)
                                                            k=0
                                         x=0                  x=0

                       Noting that all of the terms on the right-hand side of (2.20) are zero except the one for which
                       k = j,


                                                1                    1

                                                                         2
                                                  x sin( jπx)dx = c j  sin ( jπx)dx                 (3.30)
                                              x=0                  x=0

                       After integrating we find


                                                          (−1) j+1  c j
                                                                  =                                 (3.31)
                                                            jπ       2

                       Thus,


                                                         ∞      j+1
                                                            (−1)

                                                    x =            2sin( jπx)                       (3.32)
                                                              jπ
                                                        j=0
                       This is an alternating sign series in which the coefficients always decrease as j increases, and
                       it therefore converges. The sine function is periodic and so the series must also be a periodic
                       function beyond the interval (0, 1). The series outside this interval forms the periodiccontinuation
                       of the series. Note that the sine is an odd function so that sin( jπx) =− sin(− jπx). Thus the
                       periodic continuation looks like Fig. 3.2. The series converges everywhere, but at x = 1itis
                       identically zero instead of one. It converges to 1 − ε arbitrarily close to x = 1.
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