Page 175 - Determinants and Their Applications in Mathematical Physics
P. 175

160   4. Particular Determinants

          Reverting to x and referring to (4.12.17),

                                    G n−1    F n G n−2
                                           =    2   ,              (4.12.28)
                               xD x
                                    F n−1     F
                                                n−1
          where the elements in the determinants are now ψ m (x), m =0, 1, 2,....
            The difference formula
                            ∆ ψ 0 = xψ m ,  m =1, 2, 3,...,        (4.12.29)
                              m
          is proved in Appendix A.8. Hence, applying the theorem in Section 4.8.2
          on Hankelians whose elements are differences,

                            E n = |ψ m | n ,  0 ≤ m ≤ 2n − 2
                                    m
                               = |∆ ψ 0 | n

                                   ψ 0  xψ 1  xψ 2  ···

                                   xψ 1  xψ 2  xψ 3  ···
                                                       .           (4.12.30)
                               =
                                   xψ 2  xψ 3  xψ 4  ···
                                   ....................

                                                      n
          Every element except the one in position (1, 1) contains the factor x. Hence,
          removing these factors and applying the relation
                          ψ 0 /x = ψ 0 +1,

                                      ψ 0 +1 ψ 1  ψ 2  ···

                                       ψ 1  ψ 2  ψ 3
                                  n                  ···
                            E n = x

                                       ψ 2  ψ 3  ψ 4
                                                     ···
                                     ....................

                                                        n
                                           (n)
                               = x n  E n + E 11  .                (4.12.31)
          Hence

                              (n)           1 − x n
                             E   = G n−1 =          E n .          (4.12.32)
                              11
                                              x n
          Put
                                    u n =  G n  ,
                                          F n
                                          E n−1
                                     v n =  E n  .                 (4.12.33)
          The theorem is proved by deducing and solving a differential–difference
          equation satisfied by u n :
                                        E n−1 E n+1
                                      =      2    .
                                  v n
                                 v n+1      E
                                             n
          From (4.12.32),
                               G n−1   x(1 − x )v n+1
                                              n
                                     =              .              (4.12.34)
                                          1 − x n+1
                                G n
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