Concrete mathematics : a foundation for computer science

512 ANSWERS TO EXERCISES
5.7 Yes, because rs = (--1 ) "/( -r - 1)". We also have
rqr + i)" = (2r)9;!%
5.8 f(k) = (k/n - 1)” is a polynomial of degree n whose leading coefficient
is nn. By (5.40)~ the sum is n!/nn. When n is large, Stirling’s approximation
says that this is approximately &/en. (This is quite different from
(1 - l/e), which is what we get if we use the approximation (1 -k/n)” N eek,
valid for fixed k as n + oo.)
5 . 9 E,(z)t = tksO t(tk + t)k-‘zk/k! = tk.Jk + l)k '(tz)k/k! = 1, (tz),
by (5.60).
5’1o tk>O 2zk/(k + 2) = F(2,l; 3; z), since tk+l/tk = (k + 2)z/(k + 3).
5.11 The first is Besselian and the second is Gaussian: But not Imbesselian.
z-~‘sinz = tka,(-l)kz2k/(2k+1)! = F(l;l,i;-z2/4);
z- ’ arcsin 2 = tkZo z2k(;)k/(2k+ l)k! = F(;, ;; 5;~~).
5.12 (a) Yes, the term ratio is n. (b) No, the value should be 1 when
k = 0; but (k + 1)" works, if n is an integer. (c) Yes, the term ratio is
(k+l)(k+3)/(k+2). (d) No, the term ratio is 1 +l/(k+l)Hk; and Hk N Ink
isn’t a rational function. (e) Yes, the term ratio is
t(k+ 1) T(n - k)
t(k) I T(n - k - .I) ’
(f) Not always; e.g., not when t(k) = 2k and T(k) = 1. (g) Yes, the term ratio
can be written
at(k+l)/t(k) + bt(k-t2)/t(k) + ct(k+3)/t(k)
a+bt(k+l)/t(k) +ct(k+2)/t(k) ’
and t(k+m)/t(k) = (t(k+m)/t(k+m-1)) . . . (t(k+ 1)/t(k)) is arational
function of k.
5.13 R, = n!n+‘/Pi = Qn/P,, = Qi/n!“+‘.
5.14 The first factor in (5.25) is (,‘i k,) when k < 1, and this is (-1 )Lpkpm x
(r-r::). The sum for k 6 1 is the sum over all k, since m 3 0. (The condition
n 3 0 isn’t really needed, although k must assume negative values if n < 0.)
To go from (5.25) to (5.26), first replace s by -1 -n - q.
5.15 If n is odd, the sum is zero, since we can replace k by n-k. If n = 2m,
the sum is (-1)“(3m)!/m!3, by (5.29) with a = b = c = m.