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4.6 Hessenbergians  93

            In the next theorem, φ m and ψ m are functions of x.
          Theorem 4.22. If

                           φ    =(m + a)Fφ m−1 ,  F = F(x),
                            m
          then
                               ψ     =(a +2 − m)Fψ m−1 .
                                m
          Proof. It follows from (4.6.4) that

                                     n
                                           i+1
                               ψ n =   (−1)   φ i ψ n−i .            (4.6.6)
                                    i=1
          It may be verified by elementary methods that


                                  ψ =(a +1)Fψ 0 ,
                                   1

                                  ψ = aFψ 1 ,
                                   2
                                  ψ =(a − 1)Fψ 2 ,

                                   3
          etc., so that the theorem is known to be true for small values of m. Assume
          it to be true for 1 ≤ m ≤ n − 1 and apply the method of induction.
          Differentiating (4.6.6),
                     n
                           i+1
               ψ =    (−1)   (φ ψ n−i + φ i ψ    )


                n              i          n−i
                    i=1
                      n
                            i+1
                 = F    (−1)   [(i + a)φ i−1 ψ n−i +(a +2 − n + i)φ i ψ n−1−i ]
                      i=1
                 = F(S 1 + S 2 + S 3 ),
          where
                               n
                                     i+1
                         S 1 =   (−1)   (i + a)φ i−1 ψ n−i ,
                              i=1
                                         n
                                               i+1
                         S 2 =(a +2 − n)   (−1)   φ i ψ n−1−i ,
                                        i=1
                               n
                                     i+1
                         S 3 =   (−1)   iφ i ψ n−1−i .
                              i=1
          Since the i = n terms in S 2 and S 3 are zero, the upper limits in these sums
          can be reduced to (n − 1). It follows that
                                S 2 =(a +2 − n)ψ n−1 .
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