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5.3 The Matsuno Identities  189

          Hence,
                                    A n = xA n−1 .
                    2
          But A 2 = x . The theorem follows.

          5.3.2  Particular Identities

                                                   n
          It is shown in the previous section that A n = x provided only that the x i
          are distinct. It will now be shown that the diagonal elements of A n can be
          modified in such a way that A n = x as before, but only if the x i are the
                                         n
          zeros of certain orthogonal polynomials. These identities supplement those
          given by Matsuno.
            It is well known that the zeros of the Laguerre polynomial L n (x), the
          Hermite polynomial H n (x), and the Legendre polynomial P n (x) are dis-
          tinct. Let p n (x) represent any one of these polynomials and let its zeros be
          denoted by x i ,1 ≤ i ≤ n. Then,
                                          n

                                p n (x)= k  (x − x r ),              (5.3.3)
                                         r=1
          where k is a constant. Hence,

                                             n

                           log p n (x) = log k +  log(x − x r ),
                                            r=1
                             p (x)         1

                                      n
                                   =           .                     (5.3.4)
                              n
                             p n (x)
                                     r=1  x − x r
          It follows that
                                1     (x − x i )p (x) − p n (x)
                           n

                                    =         n           .          (5.3.5)
                                          (x − x i )p n (x)
                          r=1  x − x r
                          r =i
          Hence, applying the l’Hopital limit theorem twice,

                            1            (x − x i )p (x) − p n (x)
                       n

                                 = lim           n
                                   x→x i    (x − x i )p n (x)
                      r=1  x i − x r
                      r =i
                                         (x − x i )p (x)+ p (x)

                                 = lim           n      n
                                   x→x i (x − x i )p (x)+2p (x)


                                                 n       n
                                    p (x i )

                                 =   n    .                          (5.3.6)
                                   2p (x i )

                                     n
          The sum on the left appears in the diagonal elements of A n . Now redefine
          A n as follows:
                                     A n = |a ij | n ,
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