The 5F4 Summation. (Part I).

In the first paper in this series [1], we used Pfaff's method to prove that

For example [1, eq.(5.12),]

To summarize, the Pfaff method is cumbersome for this problem. No single step is substantially more difficult than the steps in Pfaff's own work (i.e. Section 2). However, the proof of twenty theorems simultaneously requires a good deal more than twenty times the effort required in Section 2. First the twenty results like (5.1) and (5.3)--(5.5) must be found empirically. Then twenty recurrences must be selected from a multitude in order to prove all twenty results.

In contrast, the application of the WZ method to prove (5.1) and (5.5) is ongoing. The question of whether WZ could prove (5.1) was raised at the Ann Arbor Conference on Combinatorics in June, 1994. Theoretically there is no doubt that the WZ method can produce a proof of both (5.1) and (5.5) eventually. After the conference, Zeilberger developed a modification of the WZ method designed to exploit the shifts of parameters that are so effective in the Pfaff approach.

The following is a summary of accomplishment of the WZ method in treating (5.1) and (5.5).

First we let

Zeilberger discovered that the shifts , greatly improve the effectiveness of the WZ-method. He was then able to obtain the following results for

Using his shift modification of the WZ method, Zeilberger has found a recurrence for the full thus giving a new proof of (5.1), and recently Peter Paule has obtained the relevant certificate without the shift modification. Both the recurrence and the certificate function are too long to be included here. Suffice it to say that the large polynomial (analogous to (5.11)) in n,k,x and z arising in Zeilberger's certificate function has 944 terms by my count and occupies twelve pages of printout. At this moment, (5.5) has still not yielded to the WZ method or its modifications. However it is surely only a matter of time before improvements in MAPLE combined with larger machines will produce a recurrence satisfied by (5.5).

Suppose that we actually had in hand the mega-massive general certificate function. We would then have a proof of (5.5) with little insight gained and by means of a recurrence so huge that no hand calculation could check it.

In contrast, Pfaff's method actually succeeds in proving (5.1) and (5.5) plus eighteen other identities. The proof is checkable by hand at each step. Also we learn that there is a cluster of identities (the computer suggests more than fifty, see [1, Section 6,]) intimately tied up with (5.1); each of these other identities conveys more information about its structure than (5.1) because (5.1) is the only one for which the right hand side is identically zero.