Page 128 - Determinants and Their Applications in Mathematical Physics
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4.8 Hankelians 1  113
                       2n−2

                                        A A jq  =(i + j − 2)A .       (D 2 )
                                                           ij
                                         ip
                           mφ m
                       m=1      p+q=m+1
            These can be proved by putting a ij = φ i+j−2 and f r = g r = r − 1 in (C)
          and (D), respectively, and rearranging the double sum, but they can also
          be proved directly by taking advantage of the second kind of homogeneity
          of Hankelians and applying the modified Euler theorem in Appendix A.9.
          Proof. A n and A ij  are homogeneous functions of degree n(n − 1) and
                           n
          (2 − i − j), respectively, in the suffixes of their elements. Hence, denoting
          the sums by S 1 and S 2 . respectively,
                                      2n−2
                                               ∂A
                               AS 1 =     mφ m
                                      m=1     ∂φ m
                                   = n(n − 1)A,
                                       2n−2
                                                ∂A ij

                                 S 2 = −    mφ m
                                        m=1     ∂φ m
                                   = −(2 − i − j)A .
                                                  ij
          The theorem follows.
          Theorem 4.32.

                            n   n

                                  (r + s − 2)φ r+s−3 A rs  =0,         (E)
                           r=1 s=1
          which can be rearranged in the form

                             2n−2

                                 mφ m−1         A pq  =0              (E 1 )
                             m=1        p+q=m+2
          and

                   n  n
                                                           i,j+1
                        (r + s − 2)φ r+s−3 A A sj  = iA i+1,j  + jA
                                         ir
                  r=1 s=1                                              (F)
                                              =0,     (i, j)=(n, n).
          which can be rearranged in the form

                   2n−2
                                                          i,j+1
                                       ip
                       mφ m−1         A A jq  = iA i+1,j  + jA
                   m=1        p+q=m+2                                 (F 1 )
                                             =0,     (i, j)=(n, n).
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