Orthogonal Polynomials and Special Functions

October 1996 Orthogonal Polynomials and Special Functions Newsletter 18
> sumrecursion(
> (hyperterm([-n,-n+1/4],[2*n+5/4],1/9,k))/
> (pochhammer(5/4,2*n)/pochhammer(2/3,n)/
> pochhammer(13/12,n)*(2^6/3^5)^n),k,s(n));
s( n ) − s( n + 1 ) = 0
which means that s(n) is a constant, namely 1, which is found
by substituting n = 0.
The other functions loaded with sumtools are shorthands
for the functions already described above. For example
hyperrecursion has the same effect as sumrecursion(
hyperterm(...)) and hypersum is a shorthand for sum with
hypergeometric-term summands, but works more powerful:
> hypersum([-n,b],[c],1,n);
pochhammer( −b + c, n )
pochhammer( c, n )
Without any doubt the package sumtools can be of great value
for anyone working in the field of special functions. Hopefully in
the near future this package will be extended with routines that
calculate differential/difference equations and that can handle
the q-case.
References
[1] Gosper, R.W.: Decision procedure for indefinite hypergeometric
summation. Proc. Nat. Acad. Sci. USA 75 (1978),
40–42.
[2] Gessel, I. and Stanton, D.: Strange evaluations of hypergeometric
series. SIAM J. Math. Anal. 13 (1982), 295–308.
[3] Koepf, W.: Algorithms for m-fold hypergeometric summation.
J. Symbolic Computation 20 (1995), 399–417.
[4] Zeilberger, D.: A fast algorithm for proving terminating hypergeometric
identities. J. Comput. Appl. Math. 32 (1990),
207–211.
Note On a Problem of Koornwinder
by Wolfram Koepf
(koepf@zib.de)
Zeilberger’s algorithm ([5]–[6], see also [2], [1]) determines recurrence
equations for hypergeometric functions
α1 α2 · · · αp
S(n) :=
pFq
x
=
k=0
β1 β2 · · · βq
∞
Ak x k ∞ (α1)k · (α2)k · · · (αp)k
=
(β1)k · (β2)k · · · (βq)k
k=0
whose upper parameters αk and lower parameters βk are
rational-linear in a variable n, whenever the term ratio
Ak+1
Ak
=
x k
k!
(k + α1) · (k + α2) · · · (k + αp)
∈ Q(k, n)
(k + β1) · (k + β2) · · · (k + βq) · (k + 1)
is a rational function in both n and k. As usual (a)k =
a(a + 1) · · · (a + k − 1) denotes the Pochhammer symbol. We
call the summand Akx k a hypergeometric term. The resulting
recurrence equation has polynomial coefficients with respect to
n. If it is of first order, the sum has a rational term ratio with
respect to n, and hence itself is a hypergeometric term.
In [3] Koornwider asked the question whether an application
of Zeilberger’s algorithm might generate a hypergeometric term
(1)
whose upper an lower parameters are not rational assuming the
parameters of the input summand are rational:
Problem 6.1. If Zeilberger’s algorithm succeeds,
can S(n)/S(n − 1) then always be factorized as a
quotient of products of linear forms over Z in n and
the parameters?
In this note, we will answer Koornwinder’s question in the negative,
by providing a counterexample.
Note that Koornwinder’s question in principle is independent
of Zeilberger’s algorithm, and asks whether there are hypergeometric
sums that can be represented by hypergeometric terms
with nonrational parameters. None example of this type can
be found in the literature, see in particular the rather extensive
mathematical dictionary on hypergeometric function identities
[4]. Nevertheless we have used Zeilberger’s algorithm to find
our counterexamples.
The identity
=
3F2
k
j=0
3/4, 5/4 − k, −k
1/4 − k, −1/4 −1
(3/4)j (5/4 − k)j (−k)j
(1/4 − k)j (−1/4)j
(−1) j
(
=
√ 3/2)k (− √ 3/2)k
( √ 3/2 − 1)k (− √ 3/2 − 1)k (1 − 4k) 2k
constitutes such an example. With the sumtools package of
Maple V.4 (see René Swarttouw’s article on p. 17) this result is
deduced by an application of Zeilberger’s algorithm using the
commands
> with(sumtools):
> readlib(‘sum/simpcomb‘):
> sumrecursion(
hyperterm([3/4,5/4-k,-k],[1/4-k,-1/4],-1,j),j,S(k));
with the result
−2 (−5 + 4 k) 4 k 2 − 8 k + 1 S(k − 1)+
S(k) (−1 + 4 k) 4 k 2 − 16 k + 13 = 0 .
We finish this note with a rather simple family of examples. If
j ∈ N then for the polynomials
a + j, −k
2F1
a x
one gets a recurrence equation of first order with respect to
k with polynomial coefficients in k of degree j that have no
factorization over Q. As a particular case one has for j = 2
> sumrecursion(hyperterm([a+2, -k],[a],x,j),j,S(k));
(x 2 k + x 2 a 2 + x 2 k 2 − 2 x k a + a − 2 x k + a 2 + 2 x 2 k a
− 2 x a + x 2 a − 2 x a 2 )S( k + 1 ) + S( k ) ( x − 1 )
(x 2 k 2 + x 2 a 2 + 3 x 2 k + 3 x 2 a + 2 x 2 k a + 2 x 2
− 4 x a − 2 x − 2 x k − 2 x k a − 2 x a 2 + a 2 + a) = 0 .
A more detailed discussion will be given in a forthcoming paper.
j!