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4.9 Hankelians 2  117

          2. The Yamazaki–Hori determinant A n is defined as follows:
                              A n = |φ m | n ,  0 ≤ m ≤ 2n − 2,

             where
                        1      2  2   m+1   2  2    m+1      2   2
                 φ m =       p (x − 1)    + q (y − 1)    ,  p + q =1.
                      m +1
             Let
                              B n = |ψ m | n ,  0 ≤ m ≤ 2n − 2,
             where

                                  ψ m =     φ m     .
                                          2
                                              2 m+1
                                        (x − y )
             Prove that
                                        = mFψ m−1 ,
                                    ∂ψ m
                                     ∂x
             where
                                             2
                                         2x(y − 1)
                                   F = −   2    2 2  .
                                         (x − y )
             Hence, prove that

                                           (n)
                                      = FB 11  ,
                                 ∂B n
                                  ∂x
                               2
                                            2
                                                    2
                           2

                         (x − y )  ∂A n  =2x n A n − (y − 1)A (n)    .
                                                          11
                                  ∂x
             Deduce the corresponding formulas for ∂B n /∂y and ∂A n /∂y and hence
             prove that A n satisfies the equation
                             x − 1         y − 1         2
                           	  2    
     	  2
                                     z x +        z y =2n z.
                               x             y
          3. If A n = |φ m | n ,0 ≤ m ≤ 2n−2, where φ m satisfies the Appell equation,
             prove that
                            i1
             a. (A ) = −φ A A  j1  − (iA i+1,j  + jA i,j+1 ),  (i, j)  =(n, n),
                  ij

                          0
                  n         n  n      n        n
             b. (A nn        0  1n 2
                    ) = −φ (A ) .
                  n           n
          4. Apply Theorem 4.33 and the Jacobi identity to prove that
                                                     2

                                	     
        A (n)
                                         = φ   0  1n   .
                                  A n
                                 A n−1         A n−1
             Hence, prove (3b).
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