Concrete mathematics : a foundation for computer science

(Divisibility ofpolynomials
is analogous
to divisibility of
integers. For example,
(k + a)\q(kl
means that the quotient
q(k)/(k+ a)
is a polynomial.
It’s well known that
(k + a)\q(k)
if and only if
q(-or) = 0.)
5.7 PARTIAL HYPERGEOMETRIC SUMS 225
where p, q, and r are polynomials subject to the following condition:
(k + a)\q(k) and (k + B)\r(k)
==+ a - /3 is not a positive integer. (5.118)
This condition is easy to achieve: We start by provisionally setting p(k) = 1,
q(k)=(k+a,)...(k+a,)z,andr(k)=(k+bl-l)...(k+b,-l)k;then
we check if (5.118) is violated. If q and r have factors (k + a) and (k + (3)
where a - (3 = N > 0, we divide them out of q and r and replace p(k) by
p(k)(k+oL-l)N-‘= p(k)(k+a-l)(k+a-2)...(k+fi+l).
The new p, q, and r still satisfy (5.117), and we can repeat this process until
(5.118) holds.
Our goal is to find a hypergeometric term T(k) such that
t(k) = cT(k+ 1) -CT(k) (5.119)
for some constant c. Let’s write
CT(k) =
r(k) s(k) t(k)
p(k) ’
(5.120)
(Exercise 55 ex- where s(k) is a secret function that must be discovered somehow. Plugging
plains why we might
want to make this
( 5.120) into (5.117) and (5.119) gives us the equation that s(k) must satisfy:
magic substitution.)
p(k) = q(k)s(k+ 1) -r(k)s(k) (5.121)
If we can find s(k) satisfying this recurrence, we’ve found t t(k) 6k.
We’re assuming that T(k+ 1 )/T(k) is a rational function of k. Therefore,
by (5.120) and (5.11g), r(k)s(k)/p(k) = T(k)/(T(k + 1) -T(k)) is a rational
function of k, and s(k) itself must be a quotient of polynomials:
s(k) = f(k)/g(kl. (5.122)
But in fact we can prove that s(k) is itself a polynomial. For if g(k) # 1,
and if f(k) and g(k) have no common factors, let N be the largest integer
such that (k + 6) and (k + l3 + N - 1) both occur as factors of g(k) for some
complex number @. The value of N is positive, since N = 1 always satisfies
this condition. Equation (5.121) can be rewritten
p(k)g(k+l)g(k) = q(k)f(k+l)g(k) -r(k)g(k+l)f(k),
and if we set k = - fi and k = -6 - N we get
r(-B)g(l-B)f(-6) = 0 = q(-B-N)f(l-B-N)g(-B-N)