Concrete mathematics : a foundation for computer science

542 ANSWERS TO EXERCISES
Hence gzn+l = 0 and gzn = (-1)“(2n)!C,.1, for all n>O.
7.15 There are (L)b+k partitions with k other objects in the subset containing
n + 1. Hence B’ (z) = eZB (z). The solution to this differential equation
is e(z) = eez+c, and c = -1 since B(0) = 1. (We can also get this result by
summing (7.49) on m, since b, = 1, {t}.)
7.16 One way is to take the logarithm of
B(z) = l/((l -z)"'(l -z2)"'(1 -~~)“~(l -z4)aa . ..).
then use the formula for In & and interchange the order of summation.
7.17 This follows since s,” tnePt dt = n!. There’s also a formula that goes
in the other direction:
+X
G(z) = & G ( zem~ie ) ee” d6 .
s-x
7.18 (a) 0, the coefficient [zn] exp(xln F(z)) is a polynomial of degree n
in x that’s a multiple of x. The first convolution formula comes from equating
coefficients of Z” in F(z)"F(z)Y = F(z)~+Y. The second comes from equating
coefficients of znP’ in F'(z)F(z)"~'F(z)Y = F'(z)F(z)~+Y~', because we have
F'(z)F(z)'-' = xm 'i(F(z)") = x-' z nf,(x)znP’ .
ll>C
(Further convolutions follow by taking a/ax, as in (7.43).)
7.20 Let G(z) = Ena gnzn. Then
.zlGik)(z) = t nkg,,znPk+’ = x(n + k - l)%Jn+kPlzn
lL)O lL>O
for all k, 1 3 0, if we regard g,, = 0 for n < 0. Hence if PO(Z), . . , P,(z) are
polynomials, not all zero, having maximum degree d, then there are polynomials
PO(n), . . . , p,,,+d(n) such that
mfd
Po(Z)G(Z) +-+ P,(z)Giml(z) = 7 F Pibhn+i-dZn.
n>o j:=o
Therefore a differentiably finite G(z) implies that
m+d
t
j=O
pj(n+d)gn+j = 0, for all n > 0.