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4.9 Hankelians 2  119

          4.9.2  The Derivatives of Turanians with Appell and Other
                 Elements

          Let


                       T = T  (n,r)  = C r C r+1 C r+2 ··· C r+n−1 ,  (4.9.5)


                                                           n
          where

                                                       T
                           C j = φ j φ j+1 φ j+2 ··· φ j+n−1  ,
                           φ    = mFφ m−1 .
                            m
          Theorem 4.34.


                         T = rF C r−1 C r+1 C r+2 ··· C r+n−1 .


          Proof.


                                C = F jC j−1 + C ,
                                                 ∗
                                  j              j
          where

                        ∗                                   T
                      C = 0 φ j 2φ j+1 3φ j+2 ··· (n − 1)φ j+n−2  .
                        j
          Hence,
                      r+n−1


                 T =         C r C r+1 ··· C j−1 C ··· C r+n−1


                                             j
                       j=r
                       r+n−1


                   = F         C r C r+1 ··· C j−1 (jC j−1 + C ) ··· C r+n−1
                                                       ∗
                                                       j
                         j=r

                   = rF C r−1 C r+1 C r+2 ··· C r+n−1

                         r+n−1


                      +F        C r C r+1 ··· C ··· C r+n−1
                                           ∗
                                           j
                          j=r
          after discarding determinants with two identical columns. The sum is zero
          by Theorem 3.1 in Section 3.1 on cyclic dislocations and generalizations.
          The theorem follows.
            The column parameters in the above definition of T are consecutive. If
          they are not consecutive, the notation

                                                                     (4.9.6)


                           T j 1 j 2 ...j n  = C j 1  C j 2  ··· C j r  ··· C j n
          is convenient.
                                  n


                     T       = F    j r C j 1  C j 2  ··· C (j r −1) ··· C j n    .  (4.9.7)

                      j 1 j 2 ...j n
                                 r=1
          Higher derivatives may be found by repeated application of this formula,
                                    k        ) has been found. However, the
          but no simple formula for D (T j 1 j 2 ...j n
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