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178   5. Further Determinant Theory

          6. Prove that the determinant A n in (5.1.10) satisfies the relation


                               A n+1 = vA +(u − nv )A n .
                                         n
             Put v = 1 to get

                                   A n+1 = A + A 1 A n
                                            n
             where
                                   u   u    u     u      ···

                                   −1   u  2u    3u     ···

                                       −1   u   3u   ···   .

                            A n =
                                            −1   u

                                                     ···

                                                 ···  ···
                                                         n
             These functions appear in a paper by Yebbou on the calculation of
             determining factors in the theory of differential equations. Yebbou uses
             the notation α [n]  in place of A n .
          5.2 The Generalized Cusick Identities

          The principal Cusick identity in its generalized form relates a particular
          skew-symmetric determinant (Section 4.3) to two Hankelians (Section 4.8).


          5.2.1  Three Determinants
          Let φ r and ψ r , r ≥ 1, be two sets of arbitrary functions and define three
          power series as follows:
                                     ∞

                               Φ i =   φ r t r−i ,  i ≥ 1;
                                    r=i
                                     ∞

                               Ψ i =   ψ r t r−i ,  i ≥ 1;
                                    r=i
                               G i =Φ i Ψ i .                        (5.2.1)
          Let
                                    ∞
                                          j−i−1
                             G i =     a ij t  ,  i ≥ 1.             (5.2.2)
                                  j=i+1
          Then, equating coefficients of t j−i−1 ,

                                  j−i

                             a ij =  φ s+i−1 ψ j−s ,  i < j.         (5.2.3)
                                  s=1
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