Page 367 - Schaum's Outline of Theory and Problems of Advanced Calculus
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358                               FOURIER SERIES                          [CHAP. 13


                                                        ! 2                     ! 2
                                                                        1
                                           ð L  r ffiffiffiffi 1  m x      ð L  r ffiffiffiffi  m x
                              Then                sin     dx ¼ 1;         cos    dx ¼ 1
                                             L   L    L              L  L    L
                                                 L                   L   1
                                                ð                   ð        2
                                                     2
                              Also,                ð1Þ dx ¼ 2L  or      p ffiffiffiffiffiffi  dx ¼ 1
                                                  L                   L  2L
                              Thus the required orthonormal set is given by
                                             1   1    x  1     x  1    2 x  1    2 x
                                             ffiffiffiffiffiffi ; p sin  ffiffiffiffi cos  ; p sin  ffiffiffiffi cos  ; ...
                                           p     ffiffiffiffi   ; p        ffiffiffiffi   ; p
                                             2L  L    L   L    L   L    L   L     L


                     MISCELLANEOUS PROBLEMS

                     13.26. Find a Fourier series for f ðxÞ¼ cos  x;    @ x @  , where   6¼ 0;  1;  2;  3; ... .
                              We shall take the period as 2  so that 2L ¼ 2 ; L ¼  .  Since the function is even, b n ¼ 0 and
                                                  2  ð L         2  ð
                                                                    cos  x cos nx dx
                                              a n ¼  f ðxÞ cos nx dx ¼
                                                  L 0              0
                                                  1  ð
                                                     fcosð    nÞx þ cosð  þ nÞxg dx
                                                ¼
                                                    0
                                                  1 sinð    nÞ   sinð  þ nÞ     2  sin    cos n

                                                          n       þ n       ð    n Þ
                                                ¼           þ          ¼     2   2
                                                  2 sin

                                                0 ¼
                          Then
                                             sin     2  sin     X  cos n
                                                          1
                                                                  cos nx
                                                                n
                                      cos  x ¼   þ           2   2
                                                          n¼1

                                             sin    1  2          2           2
                                                         1           2           3
                                           ¼           2  2  cos x þ  2  2  cos 2x    2  2  cos 3x þ
                                                 !         !        !
                                                2        2         2
                                               x        x         x
                                                                          .
                     13.27. Prove that sin x ¼ x 1    2  1    2  1    2

                                                       ð2 Þ     ð3 Þ
                              Let x ¼   in the Fourier series obtained in Problem 13.26.  Then

                                                   sin    1  2       2      2
                                                                1      2      3
                                             cos   ¼      þ  2  2  þ  2  2  þ  2  2  þ
                              or
                                                       1    2      2       2
                                                               1      2      3
                                                  cot       ¼  2  2  þ  2  2  þ  2  2  þ             ð1Þ
                          This result is of interest since it represents an expansion of the contangent into partial fractions.
                              By the Weierstrass M test, the series on the right of (1)converges uniformly for 0 @ j j @ jxj < 1 and
                          the left-hand side of (1) approaches zero as   ! 0, as is seen by using L’Hospital’s rule.  Thus, we can
                          integrate both sides of (1) from 0 to x to obtain
                                             x               ð  x       ð  x
                                            ð
                                                       1        2          2
                                                 cot       d  ¼  2  d  þ   2  2  d  þ
                                             0                0     1    0     2
                                                                  !          !
                                                          x       2         2
                                                   sin           x         x
                          or                    ln         ¼ ln 1    þ ln 1    þ
                                                          0      1 2       2 2
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