Page 368 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 368

CHAP. 13]                         FOURIER SERIES                                359

                                                               !         !             !
                                                              2         2             2
                                             sin  x          x         x             x
                           i.e.,          ln      ¼ lim ln 1    þ ln 1    þ     þ ln 1
                                               x    n!1      1 2       2 2           n 2
                                                         (     2 !    2  !     2 !)
                                                              x      x        x
                                                  ¼ lim ln  1     1         1
                                                              1 2    2 2      n 2
                                                    n!1
                                                     (         2  !   2 !      2  !)
                                                              x      x        x
                                                  ¼ ln lim  1     1         1
                                                              1 2    2 2      n 2
                                                       n!1
                           so that
                                                       !      !        !        !      !
                                        sin  x        x 2   x 2       x 2      x 2   x 2
                                             ¼ lim
                                          x   n!1  1    1 2  1    2 2      1    n 2  ¼ 1    1 2  1    2 2       ð2Þ
                           Replacing x by x= ,we obtain
                                                                 !        !
                                                               x 2     x 2
                                                                      ð2 Þ
                                                     sin x ¼ x 1    2  1    2                         ð3Þ
                           called the infinite product for sin x, which can be shown valid for all x.  The result is of interest since it
                           corresponds to a factorization of sin x in a manner analogous to factorization of a polynomial.
                                       2   2   4   4   6   6   8   8 ...
                     13.28. Prove that  ¼                 .
                                    2  1   3   3   5   5   7   7   9 ...
                              Let x ¼ 1=2in equation (2)ofProblem 13.27.  Then,

                                           2      1      1     1       1 3  3 5  5 7
                                              ¼ 1    2 2  1    4 2  1    6 2      ¼  2 2  4 4  6 6



                           Taking reciprocals of both sides, we obtain the required result, which is often called Wallis’ product.



                                                 Supplementary Problems

                     FOURIER SERIES
                     13.29. Graph each of the following functions and find their corresponding Fourier series using properties of even
                           and odd functions wherever applicable.
                                     80 < x < 2                          x   4 @ x @ 0

                                                  Period 4                               Period 8
                                     82 < x < 4                           x   0 @ x @ 4
                           ðaÞ f ðxÞ¼                          ðbÞ f ðxÞ¼
                                                                         2x  0 @ x < 3

                           ðcÞ f ðxÞ¼ 4x; 0 < x < 10;  Period 10  ðdÞ f ðxÞ¼            Period 6
                                                                         0   3 < x < 0
                                   16  X           n x             8  X           n x
                                                                     1
                                     1
                           Ans:  ðaÞ   ð1   cos n Þ  sin    ðbÞ 2      ð1   cos n Þ  cos
                                           n        2               2     n 2      4
                                     n¼1                            n¼1

                                      40  X 1  n x              3  X  6ðcos n    1Þ  n x  6cos n   n x
                                        1
                                                                   1
                                          n    5                2        n         3     n       3
                               ðcÞ 20       sin             ðdÞ  þ        2 2   cos          sin
                                        n¼1                        n¼1
                     13.30. In each part of Problem 13.29, tell where the discontinuities of f ðxÞ are located and to what value the series
                           converges at the discontunities.
                           Ans.  (a) x ¼ 0;  2;  4; ...; 0 ðbÞ no discontinuities  (c) x ¼ 0;  10;  20; .. .; 20
                               (d) x ¼ 3;  9;  15; .. .; 3
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