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6.3 The Dale Equation  247

          Note that the theorem cannot be applied to A directly since φ m does not
          satisfy the Appell equation for any F(x).
            Using the identity
                              |t i+j−2 a ij | n = t n(n−1) |a ij | n ,
          it is found that
                                              2
                                 P =(1 + x) −n  A,
                                              2
                                           −n +1
                                P 11 =(1 + x)   A 11 .               (6.3.4)
          Hence,

                                         2
                             (1 + x)A = n A − (c − 1)A 11 .          (6.3.5)

          Let
                                       r−1
                               α i =  x   A ,                        (6.3.6)
                                            ri
                                    r

                               β i =  (−1) A ,                       (6.3.7)
                                          r
                                            ri
                                    r

                               λ =    (−1) α r
                                          r
                                    r
                                             r s−1
                                 =       (−1) x   A rs
                                    r  s
                                       s−1
                                 =    x   β s ,                      (6.3.8)
                                    s
          where r and s =1, 2, 3,...,n in all sums.
            Applying double-sum identity (D) in Section 3.4 with f r = r and g s =
          s − 1, then (B),
                                          r+s−1
                    (i + j − 1)A =       [x     +(−1) r+s c]A A sj
                                                           ri
                              ij
                                    r  s
                                                                     (6.3.9)
                                = xα i α j + cβ i β j

                             ij             i+j−2  is  rj
                           (A ) = −        x     A A
                                     r   s
                                = −α i α j .                        (6.3.10)
          Hence,
                          ij
                      x(A ) +(i + j − 1)A = cβ i β j ,
                                         ij
                                (x i+j−1 A ) = c(x i−1 β i )(x j−1 β j ).
                                        ij
          In particular,
                                               2
                                      11
                                    (A ) = −α ,
                                               1
                                              2
                                   (xA ) = cβ .                     (6.3.11)
                                      11
                                              1
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