Concrete mathematics : a foundation for computer science

226 BINOMIAL COEFFICIENTS
Now f(-b) # 0 and f(l - 6 -N) # 0, because f and g have no common
roots. Also g(1 - l3) # 0 and g(-(3 - N) # 0, because g(k) would otherwise
contain the factor (k+ fi - 1) or (k+ (3 +N), contrary to the maximality of N.
Therefore
T--f') = q(-8-N) = 0.
But this contradicts condition (5.118). Hence s(k) must be a polynomial.
The remaining task is to decide whether there exists a polynomial s(k)
satisfying (5.121), when p(k), q(k), and r(k) are given polynomials. It’s easy
to decide this for polynomials of any particular degree d, since we can write
s(k) = cXdkd + (xdp, kdm~’ -1- *. . + olo , Kd # 0
for unknown coefficients (&d, . . . , o(o) and plug this expression into the defining
equation. The polynomial s(k) will satisfy the recurrence if and only if
the a’s satisfy certain linear equations, because each power of k must have
the same coefficient on both sides of (5.121).
But how can we determine the degree of s? It turns out that there
actually are at most two possibilities. We can rewrite (5.121) in the form
&(k) = Q(k)(s(k+ 1) +s(k)) + R(k)(s(k+ 1) -s(k)),
where Q(k) = q(k) -r(k) and R(k) = q(k) +r(k).
(5.123)
If s(k) has degree d, then the sum s(k + 1) + s(k) = 2adkd + . . . also has
degree d, while the difference s(k + 1) - s(k) = As(k) = dadkd-’ + . . . has
degree d - 1. (The zero polynomial can be assumed to have degree -1.) Let’s
write deg(p) for the degree of a polynomial p. If deg(Q) 3 deg(R), then
the degree of the right-hand side of (5.128) is deg(Q) + d, so we must have
d = deg(p) - deg(Q). On the other hand if deg(Q) e: deg(R) = d’, we can
write Q(k) = @kd’-’ f. . . and R(k) = ykd’ +. . . where y # 0; the right-hand
side of (5.123) has the form
(2,-?% + yd ,d)kd+d’-’ + . . . .
Ergo, two possibilities: Either 28 + yd # 0, and d = deg(p) - deg(R) + 1;
or 28 + yd = 0, and d > deg(p) - deg(R) + 1. The second case needs to be
examined only if -2B/y is an integer d greater than deg(p) - deg(R) + 1.
Thus we have enough facts to decide if a suitable polynomial s(k) exists.
If so, we can plug it into (5.120) and we have our T. If not, we’ve proved that
t t(k) 6k is not a hypergeometric term.