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5.3 The Matsuno Identities  191

          Hence, if
                                      u ij ,    j  = i

                               a ij =      x i
                                      x −   2 ,j = i,
                                            i
                                          1−x
          then
                                  A n = |a ij | n = x .             (5.3.13)
                                                n

          Exercises

          1. Let A n denote the determinant defined in (5.3.9) and let
                                      B n = |b ij | n ,

             where

                                          2  ,  j  = i
                                 b ij =  x i −x j
                                        x +  1  ,j = i,
                                            x i
             where, as for A n (x), the x i denote the zeros of the Laguerre polynomial.
             Prove that
                                                   x
                                                  1 2
                                              n
                                                   2
                                 B n (x − 1)=2 A n
             and, hence, prove that
                                    B n (x)=(x +1) .
                                                  n
          2. Let

                                     A (p)  = |a (p) | n ,
                                       n     ij
             where
                                     u ,          j  = i
                                       p
                                        ij
                                    
                                (p)        n !
                               a  =    x −    u ,j = i,
                                               p
                                ij
                                               ir
                                          r=1
                                    
                                    
                                          r =i
                                       1
                               u ij =  x i − x j  = −u ji
             and the x i are the zeros of the Hermite polynomial H n (x). Prove that
                                      n

                               A (2)  =  [x − (r − 1)],
                                 n
                                      r=1
                                      n
                                              1  2
                                 (4)
                               A   =      x − (r − 1) .
                                              6
                                 n
                                      r=1
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