Concrete Mathematics, Lesson 13: Exercises - Cs.ioc.ee

Concrete Mathematics, Lesson 13: Exercises
Silvio Capobianco
Exercise on the Rational Expansion Theorem
Solve the recurrence:
gn = 6gn−1 − 9gn−2 , n ≥ 2 (1)
with the initial conditions g0 = 1, g1 = 9.
Solution. We recall the Rational Expansion Theorem: if G(z) is the generating
function for the sequence 〈g0, g1, . . .〉 and P (z), Q(z) are two polynomials
such that G(z) = P (z)/Q(z), deg P < deg Q, and Q(z) = (1 − ρ1z) d1 · · · (1 −
ρℓz) dℓ with the ρk’s pairwise distinct, then there exist polynomials f1, . . . , fℓ
such that
gn = f1(n)ρ n 1 + . . . + fℓ(n)ρ n ℓ for n ≥ 0 ;
moreover, for every 1 ≤ k ≤ ℓ, deg fk = dk − 1, and the leading coefficient of
fk is
ak = (−ρk) dkP (1/ρk)dk
Q (dk) (1/ρk)
=
P (1/ρk)
(dk − 1)!
. dj
j=k (1 − ρj/ρk)
From Equation (1) follows that gnzn = 6gn−1zn − 9gn−2zn for every n ≥ 2:
thus,
gnz n = 6
gn−1z n − 9
gn−2z n ,
that is,
n≥2
n≥2
n≥2
G(z) − g0 − g1z = 6z(G(z) − g0) − 9z 2 G(z) :
as g0 = 1 and g1 = 9, this yields
G(z) − 1 − 9z = 6z(G(z) − 1) − 9z 2 G(z) ,
1

that is,
G(z) =
1 + 3z 1 + 3z
= .
1 − 6z + 9z2 (1 − 3z) 2
In the notation of the Rational Expansion Theorem, we have P (z) = 1 + 3z,
Q(z) = (1 − 3z) 2 , ρ1 = 3, d1 = 2. Therefore, gn = (a1n + c1) · 3 n where
a1 =
(1 + 3/3)
(2 − 1)! · 1
= 2 .
For n = 0 we find 1 = g0 = (0 + c1) · 1, yielding c1 = 1. Therefore,
gn = (2n + 1) · 3 n .
If we prefer, we can use expansion into partial fractions. As (1−3z) 2 appears
in the denominator, we must find A and B such that
1 + 3z A
=
1 − 6z + z2 1 − 3z +
B
:
(1 − 3z) 2
this means A · (1 − 3z) + B = 1 + 3z, which translates into the two linear
equations in two unknowns A, B,
A + B = 1 ; −3A = 3 :
then clearly A = −1, B = 2. Moreover, 1/(1 − 3z) =
n≥0 3nzn , while
1
(1 − 3z) 2 =
1 d 1
3 dz 1 − 3z
= 1 d
3
3 dz
n≥0
n z n
= 1
n · 3
3
n≥1
n · z n−1
= 1
(n + 1) · 3
3
n≥0
n+1 · z n
=
(n + 1) · 3 n · z n .
n≥0
Then gn = (−1) · 3 n + 2 · (n + 1) · 3 n = (2n + 1) · 3 n , as we had found before.
2

Exercise 7.11, point 1
Let an = bn = cn = 0 for n < 0, and
A(z) =
anz n ; B(z) =
bnz n ; C(z) =
cnz n
n
Express C in terms of A and B when cn =
j+2k≤n ajbk.
Solution. We know that, if an = [zn ]G(z), then
k≤n ak = [zn ]G(z)/(1 − z).
Then we can solve the exercise as soon as we find G(z) such that [zn
]G(z) =
j+2k=n ajbk. But the latter is the coefficient of index n of the convolution
of A with a power series whose odd-indexed coefficients are 0, and whose
coefficient of index 2k is bk: such function is precisely B(z2 ). Therefore,
n
C(z) = A(z)B(z2 )
1 − z
3
n