This is qEKHAD, one of the Maple packages

                    accompanying the forthcoming book "A=B"

                   (soon to be published by A.K.Peters, Ltd.)

             by Marko Petkovsek, Herb Wilf, and Doron Zeilberger.

             Also try out the other packages: EKHAD and multiEKHAD



            The latest versions are always available by anon. ftp to

                              ftp.math.temple.edu

                    in the directory: /pub/zeilberg/programs

            or, on the Web, at http://www.math.temple.edu/~zeilberg

            For general help, and a list of the available functions,

         type "ezra();". For specific help type "ezra(procedure_name);"


                        Warning: q is a global variable

                 qfac(k) means (1-q)*(1-q^2)*...*(1-q^k), and

                       NOT (1)(1+q)*...*(1+q+...+q^(k-1))



                             Version of  July 1995

 

                            The form in 1-1 is WZ

A PROOF OF THE   NAME   IDENTITY

by Shalosh B. Ekhad, Temple University, ekhad@math.temple.edu

I will give a short proof of the following result(   REF   ).

Theorem:Let   F(n,k)   be

                                     qfac(k)
                               - ---------------
                                 qfac(n + k + 1)


and let   SUM(n)   be the sum of   F(n,k)    with
respect to   k   .

SUM(n)    satisfies the following linear recurrence equation

                       n                        n
               (- 1 + q  q) a0 SUM(n)   (- 1 + q  q) a0 SUM(n + 1)
             - ---------------------- + --------------------------
                         n                          n
                        q  q                       q  q

                                      =0.


PROOF: We cleverly construct   G(n,k)   :=

                                      k
                              (- 1 + q  q) a0 qfac(k)
                       - --------------------------------
                                 k  n  2
                         (- 1 + q  q  q ) qfac(n + k + 1)

with the motive that

                      n                         n
              (- 1 + q  q) a0 F(n, k)   (- 1 + q  q) a0 F(n + 1, k)
            - ----------------------- + ---------------------------
                         n                           n
                        q  q                        q  q

=   G(n,k)-G(n,k-1)   (check!)

and the theorem follows upon summing with respect to   k   .
                                   This took

                                      .583

                              Seconds of CPU time


To get the form given to the companion of F in 1-1 replace k by k-1 in G(n,k).