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A.8 Differences  329

          where

                                    n             2r+1
                                             n   x
                             f(x)=    (−1) r           ,
                                              r  2r +1
                                   r=0

                                    n
                                             n

                            f (x)=    (−1) r     x 2r
                                              r
                                   r=0
                                 =(1 − x ) .
                                         2 n
                                   ,
                                     x
                             f(x)=    (1 − t ) dt,
                                           2 n
                                     0
                                   ,  1
                             f(1) =   (1 − t ) dt
                                           2 n
                                     0
                                     π/2
                                   ,
                                 =      cos 2n+1  θdθ
                                     0
                                      1

                                    Γ   Γ(n +1)
                                 =    2     3    .
                                     2Γ n +
                                            2
          The proof is completed by applying the Legendre duplication formula for
          the Gamma function (Appendix A.1). This result is applied at the end of
          Section 4.10.3 on bordered Yamazaki–Hori determinants.
          Example A.3. If
                                        x 2r+2  − c
                                   u r =        ,
                                          r +1
          then
                                   2
                                 (x − 1) n+1  − (−1) (c − 1)
                                                  n
                          ∆ u 0 =                         .
                           n
                                           n +1
          Proof.

                              n                2r+2
                                         n   x     − 1   c − 1
                     ∆ u 0 =    (−1) n−r              −
                       n
                                         r     r +1      r +1
                             r=0
                           =(−1) [S(x)+(c − 1)S(0)],
                                 n
          where
                          	 
   2r+2
                  n
                           n
                               x    − 1
           S(x)=    (−1) r
                            r    r +1
                  r=0
                    1          	 n +1  
  2r+2
                        n
               =          (−1) r        (x    − 1)
                  n +1           r +1
                       r=0
                       n+1
                    1         r+1  n +1
               =          (−1)            (x 2r  − 1), (The r = 0 term is zero)
                  n +1               r
                       r=0
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