Concrete mathematics : a foundation for computer science

212 BINOMIAL COEFFICIENTS
into to but disappear from t,he term ratio. Using such tricks we can predict
without further calculation t;hat the term ratio of (5.27) is
tk+l -=-
k - r k - n
k+l k+s-n+l
fk
times (--1 )’ = 1, and Vandermonde’s convolution becomes
(5.91)
We can use this equation to determine F( a, b; c; z) in general, when z = 1 and
when b is a negative integer.
Let’s rewrite (5.91) in a form so that table lookup is easy when a new
sum needs to be evaluated. The result turns out to be
F
a,b , _ T(c-a--b)T(c)
( C 1) r(c - a) T(c - b) ’
integer b 6 0
or %c >Ra+!Xb.
(5.92)
Vandermonde’s convolution (5.27) covers only the case that one of the upper
parameters, say b, is a nonpositive integer; but Gauss proved that (5.92) is A few weeks ago, we
valid also when a, b, c are complex numbers whose real parts satisfy !Xc >
were studying what
%a + %b. In other cases, the infinite series F( “;” j 1) doesn’t converge. When ~~~r~~r~e~e jn
b = -n, the identity can be written more conveniently with factorial powers Now we’re studying
instead of Gamma functions:
stuff beyond his
Ph.D. thesis.
F(a’;ni,) = k&z = (;-;s, integer n > 0. (5.93)
Is this intimidating
or what?
It turns out that all five of the identities in Table 169 are special cases of
Vandermonde’s convolution; formula (5.93) covers them all, when proper attention
is paid to degenerate situations.
Notice that (5.82) is just the special case a = 1 of (5.93). Therefore we
don’t really need to remember (5.82); and we don’t really need the identity
(5.9) that led us to (5.82), even though Table 174 said that it was memorable.
A computer program for formula manipulation, faced with the problem
of evaluating xkGn (‘+kk), could convert the sum to a hypergeometric and
plug into the general identity for Vandermonde’s convolution.
Problem 1 in Section 5.2 asked for the value of
This problem is a natural for hypergeometrics, and after a bit of practice any
hypergeometer can read off the parameters immediately as F( 1, -m; -n; 1).
Hmmm; that problem was yet another special takeoff on Vandermonde!