Concrete mathematics : a foundation for computer science

y Vandermonde’s convolution (5.92).
A ANSWERS TO EXERCISES 517
5.51 (a) Reflection gives F(a, -n; 2a; 2) = (-1 )“F( a, -n; 2a; 2). (Incidentally,
this formula implies the remarkable identity A2”‘+’ f(0) = 0, when
f(n) = 2nxc/(2x)“.>~
(b) The term-by-term limit is &kSm (r) m(-2)k plus an additional
term for k = 2m - 1: the additional term is
(-m)... (-1) (1)...(m) (-2m+ 1) . . . (-1)22m+’
I:-2m). (-1) (2m - l)!
,I ,I pm+1
= (-ltm+'* =-
-2
(CL') '
hence, by (5.104), this limit is -l/( y2), the negative of what we had.
5.52 The terms of both series are zero for k > N. This identity corresponds
to replacing k by N - k. Notice that
5.53 When b = -i, the left side of (5.110) is 1 - 22 and the right side is
(1 -42+422)"2, independent of a. The right side is the formal power series
l/2
l+ 1
( )
42(2-l)+ l/2
2 16z2(z-1)2+~~~,
( 1
which can be expanded and rearranged to give 1 - 22+ Oz2 + Oz3 f. ; but the
rearrangement involves divergent series in its intermediate steps when z = 1,
so it is not legitimate.
5.54 If m + n is odd, say 2N - 1, we want to show that
N-m-;, -N+c
lim F 1 =o.
E'O ( -m+e 1)
Equation (5.92) applies, since -m + c > -m - i + E, and the denominator
factor T(c-b) = T(N-m) is infinite since N < m; the other factors are finite.
Otherwise m + n is even; setting n = m ~ 2N we have
-N, N-m-i+e
1 = (N-1/21N
fi,mo F
( -m+c 1) rnN
by (5.93). The remaining job is to show that
(N - l/2)! (m-N)!
-(-l/2)! m! =