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5.2 The Generalized Cusick Identities  185

            From (5.2.4),

                                           = ψ i .
                                     ∂a i,2n
                                    ∂φ 2n−1
          Also,
                                          = H n−1 .
                                    ∂H n
                                   ∂φ 2n−1
          Hence,
                            2n−1
                                    i+1  (n)
                                (−1)   Pf  ψ i = H n−1 K n .        (5.2.24)
                                         i
                             i=1
          Similarly,
                            2n−1
                                    i+1  (n)
                                (−1)   Pf  φ i = H n K n−1 .        (5.2.25)
                                         i
                             i=1
          The following three theorems express modified forms of |a ij | n in terms of
          the Hankelians.
            Let B n (φ) denote the determinant which is obtained from |a ij | n by
          replacing the last row by the row

                                   φ 1 φ 2 φ 3 ...φ n .
          Theorem 5.6.
                                 2
          a. B 2n−1 (φ)= H n−1 H n K n−1 ,
                         2
          b. B 2n−1 (ψ)= H n−1 K n−1 K n ,
                         2
          c. B 2n (φ)= −H K n−1 K n ,
                         n      2
          d. B 2n (ψ)= −H n−1 H n K .
                                n
          Proof. Expanding B 2n−1 (φ) by elements from the last row and their
          cofactors and referring to (5.2.13), (5.2.14), and (5.2.25),
                                    2n−1
                                           (2n−1)
                         B 2n−1 (φ)=    φ j A
                                           2n−1,j
                                    j=1
                                          2n−1
                                      (n)         i+1  (n)
                                  =Pf         (−1)   Pf
                                      2n−1             i  φ i
                                           i=1
                                  =Pf n−1 H n K n−1 .               (5.2.26)
          Part (a) now follows from Theorem 5.1 and (b) is proved in a similar
          manner.
            Expanding B 2n (φ) with the aid of Theorem 3.9 on bordered determinants
          (Section 3.7) and referring to (5.2.11) and (5.2.25),
                          2n−1 2n−1
                                          (2n−1)
                B 2n (φ)= −        a i,2n φ j A
                                          ij
                           i=1  j=1
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