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5.1 Determinants Which Represent Particular Polynomials  173
                      1
             φ n =
                  (1 − x) n+1
                     1 − x

                                                                  x
                       1        1 − x

                                                                  0
                      1/2!        1     1 − x

                                                                  0
                      1/3!      1/2!      1    1 − x              0   (5.1.8)



                    .....................................................
                    1/(n − 1)! 1/(n − 2)! ..............  1  1 − x  0

                     1/n!     1/(n − 1)! ..............  1/2!  1  0

                                                                    n+1
          After expanding the determinant by the single nonzero element in the last
          column, the theorem follows from (5.1.2) and (5.1.4).
          Exercises
          Prove that


                             α 0  α 1  α 2  α 3  ··· α n−1  α n
                             −y  x

             n

                                −y    x
          1.    α r x n−r r                                    .
                                     −y   x
                      y =
             r=0

                                            .................
                                                   −y    x

                                                            n+1
                        1    x   x 2  x 3  ···  x n−1  x n

                                      2
                        −1   y   xy  x y  ··· x n−2 y  x n−1

                                                          y
                            −1   y    xy  ··· x    y  x
                                                n−3    n−2
                                                          y
                   n                                           .
                                 −1   y   ··· x n−4 y  x n−3
          2. (x + y) =
                                                          y

                                   ...........................
                                                −1      y

                                                            n+1

                       x      b
                n
          3. (−b) 2 F 0  , −n; −
                       a      a
                        b
                   −c 1
                   a   −c 2  b

                        2a         b

                            −c 3
                             3a   −c 4                    ,

               =

                                  ···  ···    ···


                                            −c n−1
                                                      b
                                           (n − 1)a −c n n

             where
                         c r =(r − 1)a + b + x.         (Frost and Sackfield)
             and 2 F 0 is the generalized hypergeometric function.
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