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4.6 Hessenbergians  95

                                   b r = a r ,  r > 1.

                                     B n (0) = A n ,
                                     (n)      (n)
                                   B   (0) = A  .                   (4.6.11)
                                     ij       ij
          Theorem 4.23.
          a. B = −nB n−1 .

              n
             n
                 (n)
          b.    A   = nA n−1 .
                 rr
             r=1
                   n
                       n
          c. B n =        A r (−x) n−r .
                       r
                  r=0
          Proof.
                            B 1 = −x + A 1 ,
                                  2
                            B 2 = x − 2A 1 x + A 2 ,
                                   3
                                           2
                            B 3 = −x +3A 1 x − 3A 2 x + A 3 ,       (4.6.12)
          etc., which are Appell polynomials (Appendix A.4) so that (a) is valid for
          small values of n. Assume that
                            B = −rB r−1 ,  2 ≤ r ≤ n − 1,

                              r
          and apply the method of induction.
            From (4.6.10),
                           n−2

               B n =(n − 1)!   a n−r B r  +(a 1 − x)B n−1 ,
                                 r!
                           r=0
                            n−2
                                a n−r rB r−1

               B = −(n − 1)!              − (n − 1)(a 1 − x)B n−2 − B n−1

                                    r!
                n
                             r=1
                            n−2
                                a n−r B r−1

                  = −(n − 1)!            − (n − 1)(a 1 − x)B n−2 − B n−1
                                 (r − 1)!
                             r=1
                            n−3

                  = −(n − 1)!   a n−1−r B r  − (n − 1)(a 1 − x)B n−2 − B n−1
                                    r!
                             r=0
                            n−2

                  = −(n − 1)!   b n−1−r B r  − B n−1
                                    r!
                             r=0
                  = −(n − 1)B n−1 − B n−1
                  = −nB n−1 ,
          which proves (a).
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