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332   Appendix

          Illustration. The function
                                        2        2   5     8
                                  Bx 0 x 2 x x 5  Cx x 1 x + Dx 2
                                                 0
                                        3
                                                     3
                 f = Ax 0 x 2 x 4 x 6 +     +     3       4         (A.9.4)
                                     x 1       Ex x 4 + Fx
                                                  0       1
          is homogeneous of degree 4 in its variables and homogeneous of degree 12
          in the suffixes of its variables. Hence,
                                    6
                                         ∂f
                                            =4f,
                                      x r
                                   r=0  ∂x r
                                   6
                                         ∂f

                                            =12f.
                                     rx r
                                  r=1   ∂x r
          A.10 Formulas Related to the Function
                         √
                                  2 2n
                  (x +     1+ x )
          Define functions λ nr and µ nr as follows. If n is a positive integer,
                                n                   n
                   +                       +              2r−1
              (x +   1+ x )  =     λ nr x 2r  +  1+ x 2  µ nr x  ,  (A.10.1)
                         2 2n
                               r=0                 r=1
          where
                                      n  	 n + r
                              λ nr =              2 ,              (A.10.2)
                                                   2r
                                    n + r    2r
                              µ nr =  rλ nr  .                     (A.10.3)
                                      n
          Define the function ν i as follows:
                                             ∞

                                      −1/2
                                (1 + z)   =     ν i z .            (A.10.4)
                                                  i
                                             i=0
          Then
                                       (−1) i  	  2i
                                  ν i =
                                        2 2i  i
                                    = P 2i (0),
                                  ν 0 =1,                          (A.10.5)
          where P n (x) is the Legendre polynomial.

          Theorem A.7.
                        n

                          λ n−1,j−1 ν i+j−2 =  δ in  ,  1 ≤ i ≤ n.
                                          2 2(n−1)
                       j=1
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