Page 272 - Determinants and Their Applications in Mathematical Physics
P. 272

6.5 The Toda Equations  257

          where A n and B n are Hankelians defined as
                           A n = |φ m | n ,  0 ≤ m ≤ 2n − 2,
                           B n = |φ m | n ,  1 ≤ m ≤ 2n − 1,
                           φ    =(m +1)φ m+1 .
                            m
          Proof.
                                   (n)      n+1  (n)
                                  B   =(−1)    A 11  ,
                                   1n
                                 (n+1)
                                A 1,n+1  =(−1) B n .                (6.5.10)
                                            n
          It follows from Theorems 4.35 and 4.36 in Section 4.9.2 on derivatives of
          Turanians that
                                                (n+1)
                              D(A n )= −(2n − 1)A    ,
                                                n+1,n
                                           (n+1)
                             D(B n )= −2nB      ,
                                           n,n+1
                                (n)             (n+1)
                            D(A    )= −(2n − 1)A       ,
                                11              1,n+1;1n
                                (n)        (n+1)
                            D(B 11  )= −2nB 1,n+1;1,n .             (6.5.11)
          The algebraic identity in Theorem 4.29 in Section 4.8.5 on Turanians is
          satisfied by both A n and B n .
                                        (n)    (n)
                       B y     = B n D(B   ) − B  D(B n )
                         2
                         n 2n−1         11     11
                                       (n)  (n+1)    (n+1)
                               =2n B    B      − B n B
                                      11  n,n+1      1,n+1;1n
                                     (n)  (n+1)
                               =2nB     B
                                     1n  1,n+1
                                       (n)  (n+1)
                               = −2nA    A     .
                                       11  11
          Applying the Jacobi identity,
                (n)  (n+1)                  (n)      (n+1)
              A   A     (y 2n − y 2n−2 )= A n+1 A  − A n A
                11  11                      11       11
                                            (n+1)        n+1     (n+1)
                                    = A n+1 A         − A       A
                                            1,n+1;1,n+1  n+1,n+1  11

                                           (n+1) 2
                                    = − A
                                          1,n+1
                                         2
                                    = −B .
                                         n
          Hence,
                               y     (y 2n − y 2n−2 )=2n,
                                2n−1
          which proves the theorem when n is odd.
                  (n+1) 2     (n+1)                  (n+1)

                A      y   = A     D(A n+1 ) − A n+1 D(A  )
                 1,n+1  2n    11                     11
                                           (n+2)        (n+1)  (n+2)
                           =(2n +1) A n+1 A 1,n+2;1,n+1  − A 11  A n+2,n+1
                                       (n+1)  (n+2)
                           = −(2n +1)A     A     .
                                       1,n+1  1,n+2
   267   268   269   270   271   272   273   274   275   276   277