Concrete mathematics : a foundation for computer science

514 ANSWERS TO EXERCISES
because al - a2 = (a1 + k) -~ (al + k). If al - bl is a nonnegative integer d,
this second identity allows us to express F(al , , . , a,,,; bl , . . . , b,; z) as a linear
combination of F( a2 + j, a3, . . . , a,,,; b2, . , b,; z) for 0 6 j 6 d, thereby
eliminating an upper parameter and a lower parameter. Thus, for example,
we get closed forms for F( a, b; a - 1; z), F( a, b; a - 2; z), etc.
Gauss [116, $71 derived analogous relations between F(a, b; c;z) and
any two “contiguous” hypergeometrics in which a parameter has been changed
by fl . Rainville [242’] gene:ralized this to cases with more parameters.
5.26 If the term ratio in the original hypergeometric series is tkfl /tk = r(k),
the term ratio in the new one is tk+I/tk+l = r(k + 1). Hence
al, . . . , a, al+l,...,a,+l,l
F
( bl, . . . . b, 1) Z
= 1 + "-'amzF b +,
bl . ..b. (
1)
1 ,...! b,+l,2 ’ ’
5.27 This is the sum of the even terms of F(2a1,. . . ,2a,; 2bl,. . . ,2b,; z).
We have (2a)=/(2a)X = 4(k+ a)(k+ a + i), etc.
5.28 WehaveF(“;b]z)= (I-z)~"F(~~~~"~~)= (l-zzpuF(' ,"sals)=
(, mz)c a bF(C-yb 1~). (Euler proved the identity by showing that both
sides satisfy the same differential equation. The reflection law is often attributed
to Euler, but it does not seem to appear in his published papers.)
5.29 The coefficients of 2” are equal, by Vandermonde’s convolution. (Kummer’s
original proof was different: He considered lim,,, F(m, b - a; b; z/m)
in the reflection law (5.101).)
5.30 Differentiate again to get z(1 - z)F"(z) + (2 - 3z)F'(z) - F(z) = 0.
Therefore F(z) = F(l,l;2;z) 'by (5.108).
5.31 The condition f(k) = cT(k+ 1) - CT(k) implies that f(k+ 1)/f(k) =
(T(k+2)/T(k+ 1) - l)/(l --T(k)/T(k+ 1)) is a rational function of k.
5.32 When summing a polynomial in k, Gosper’s method reduces to the
“method of undetermined coefficients!’ We have q(k) = r(k) = 1, and we
try to solve p(k) = s(k+ 1) - s(k). The method suggests letting s(k) be a
polynomial whose degree is cl = deg(p) + 1.
5.33 The solution to k = (k- l)s(k+ 1) - (k+ l)s(k) is s(k) = -k+ 5;
hence the answer is (1 - 2k)/‘2k(k - 1) + C.
5.34 The limiting relation holds because all terms for k > c vanish, and
E - c cancels with -c in the limit of the other terms. Therefore the second
partial sum is lim,,o F(-m,--n; e-mm;l) = lim,,~(e+n-m)m/(e-m)m =
(-l)yy).
5.35 (a) 2m"3n[n>0]. (b) 11 - i)PkP’[k>,O] =2k+‘[k>0].