Concrete mathematics : a foundation for computer science

A ANSWERS TO EXERCISES 529
For example, we obtain the somewhat surprising noncommutative factorizations
(abc+a+c)(l +ba) = (ab+l)(cba+a+c)
from the case k = 2, m = 0, n = 3.)
6.30 The derivative of K(xl , . . . ,x,) with respect to xm is
K(x,,... ,xm-l)K(xm+l,...,xn),
and the second derivative is zero; hence the answer is
K(x,, . . . ,x,1 +K(xl,...,xm~1)K(x,+l,...,x,)~
6.31 Since xK = (x + n - 1)s = tk (L)x”(n - I)*, we have ]z] =
(L)(n- l)“k. These coefficients, incidentally, satisfy the recurrence
= (n-l+k)/nkl~+~~~:/, integersn,k>O.
6.32 Ek,,k{nlk} = {“+,“+‘} and &k$,, {i}(m+l)” k = {G’,:}, both
of which appear in Table 251.
6.33 If n > 0, we have [‘;I = i(n- l)! (Ht~ , - Hf-I,), by (6.71); {;} =
i(3” - 3.2” + 3), by (6.19).
6.34 We have (i’) = l/(k+ l), (-,‘) = Hr!,, and in general (z) is given
by (6.38) for all integers n.
6.35 Let n be the least integer > l/e such that [HnJ > [H, -,J.
6.36 Now dk+, = (lOOf(l +dl)+...+(l+dk))/(lOO+k), and the solution
is dk+l = Hk+100 - Hlol + 1 for k 3 1. This exceeds 2 when k 3 176.
6.37 The sum (by parts) is H,, - (z + 2 + . . . + $-) = H,, - H,. The
infinite sum is therefore lnm. (It follows that
x ym(k) ~ :=
k>, k(k+ 1)
mlnm,
m-l
because v,(k) = (m- 1) xi,, (k mod mj)/mj.)
6.38 (-l)‘((“,‘)r-’ - (1: ])Hk) + C. (By parts, using (5.16).)
6.39 Write it as x,sjsn jj’ xjbksn Hk and sum first on k via (6.67), to get
(n+l)HE-(2n+l)H,+2n.