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256   6. Applications of Determinants in Mathematical Physics
                              2
                             ρ B n+1 B n−1 e −2x
                           =
                                    B 2
                                     n
                             B n+1 B n−1
                           =           .
                                 B 2
                                  n
          This equation is identical in form to the equation in the corollary to
          Theorem 6.3. Hence,
                                 i+j−2
                         B n = D      g(x) ,  g(x) arbitrary,


                                 x
                                          n
          which is equivalent to the stated result.
          Theorem 6.5. The equation
                               2
                                    2
                             (D + D ) log u n =  u n+1 u n−1
                                                 u
                               x    y              2
                                                   n
          is satisfied by the function
                                           i−1  j−1
                              u n = A n = D  D   (f) ,


                                          z   ¯ z
                                                     n
          where z =  1  (x + iy), ¯ is the complex conjugate of z and the function
                              z
                    2
                 z
          f = f(z, ¯) is arbitrary.
          Proof.
                        2
                                      2
                      D (log A n )=  1    D +2D z D ¯z + D 2    log A n ,
                                   4
                        x             z             ¯ z
                        2           1     2           2
                      D (log A n )= −  D − 2D z D ¯z + D  log A n .
                                    4
                        y               z             ¯ z
          Hence, the equation is transformed into
                                            A n+1 A n−1
                             D z D ¯z (log A n )=     ,
                                                A 2
                                                 n
          which is identical in form to the equation in Theorem 6.3. The present
          theorem follows.
          6.5.3  The Milne-Thomson Equation
          Theorem 6.6. The equation

                               y (y n+1 − y n−1 )= n +1
                                n
          is satisfied by the function defined separately for odd and even values of n
          as follows:
                                        (n)
                                       B 11    11
                               y 2n−1 =    = B ,
                                               n
                                       B n
                                       A n+1     1
                                 y 2n =  (n+1)  =  11  ,
                                       A 11     A n+1
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