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The University of Sydney
School of Mathematics and Statistics
Solutions to MATH2069 Assignment
MATH2069: Discrete Mathematics and Graph Theory
1.
Semester 1, 2017
(a) Solve the recurrence relation
an = 12 an−2 − 16 an−3 + 9 · 2n+1 + 25 n,
n > 3,
where a0 = 20, a1 = 31 and a2 = −62.
Solution: The corresponding homogeneous relation takes the form
bn = 12 bn−2 − 16 bn−3 ,
n > 3.
The characteristic equation is x3 − 12x + 16 = 0 with the roots x = 2 of
multiplicity 2 and x = −4 of multiplicity 1. Hence the general solution of
the homogeneous relation is bn = A 2n + B n 2n + C (−4)n , where A, B and
C are arbitrary constants.
Now find a particular solution of the non-homogeneous relation
an = 12 an−2 − 16 an−3 + 18 · 2n .
Since 2 is a root of the characteristic equation of multiplicity two, we take
pn = Dn2 2n . Substituting into the relation we get
Dn2 2n = 12 D(n − 2)2 2n−2 − 16 D(n − 3)2 2n−3 + 18 · 2n .
Dividing both sides by 2n−3 we get
8 Dn2 = 24 D(n − 2)2 − 16 D(n − 3)2 + 144.
The coefficients of n2 and n on both sides agree, whereas comparing the
constant terms we obtain
0 = 96 D − 144 D + 144
so that D = 3. Thus, pn = 3 n2 2n .
Next, find a particular solution of the non-homogeneous relation
an = 12 an−2 − 16 an−3 + 25 n.
Since 1 is not a root of the characteristic equation, we can look for a particular
solution in the form qn = En + F . Substituting into the relation we get
En + F = 12(E(n − 2) + F ) − 16(E(n − 3) + F ) + 25 n.
Comparing the constant terms and the coefficients of n on both sides we
obtain the relations
F = 12(−2E + F ) − 16(−3E + F ),
c 2017 The University of Sydney
Copyright 1
E = 12E − 16E + 25.
Hence, E = 5 and F = 24. Thus, qn = 5n + 24. Therefore, by a result from
lectures, the general solution of the initial relation is found by
an = A 2n + B n 2n + C (−4)n + 3 n2 2n + 5n + 24.
To find the values of A, B and C, use the initial conditions. They give
A + C + 24 = 20,
2A + 2B − 4 C + 35 = 31,
4A + 8B + 16C + 82 = −62.
Solving these equations simultaneously, we get A = 0, B = −10 and C = −4.
The solution of the initial recurrence relation is
an = −10 n 2n + (−4)n+1 + 3 n2 2n + 5n + 24.
(b) Write down the generating function of the sequence an in part (a) in a closed
form.
Solution: By results from lectures,
∞
X
1
z =
,
1
−
z
n=0
n
∞
X
z
nz =
,
2
(1
−
z)
n=0
Since
n
∞ X
n
n=0
2
zn =
z2
.
(1 − z)3
n
,
n =n+2
2
2
we also have
∞
X
n2 z n =
n=0
z
2 z2
+
.
(1 − z)2 (1 − z)3
Hence, using the substitutions z 7→ 2z and z 7→ −4z for the generating
function of the sequence an we find
∞
X
n=0
an z n = −
4
6z
24z 2
5z
24
20z
−
+
+
+
+
2
2
3
2
(1 − 2z)
1 + 4z (1 − 2z)
(1 − 2z)
(1 − z)
1−z
=−
14z
4
24z 2
5z
24
−
+
+
+
.
2
3
2
(1 − 2z)
1 + 4z (1 − 2z)
(1 − z)
1−z
(5 marks)
2. Recall that the Fibonacci and Lucas sequences Fn and Ln are (0, 1, 1, 2, . . . ) and
(2, 1, 3, 4, . . . ), respectively.
(a) Derive closed formulas for the generating functions F (z) and L(z).
Solution:
The recurrence relation for the Fibonacci numbers is Fn =
Fn−1 + Fn−2 for n > 2 with F0 = 0 and F1 = 1. Hence,
F (z) = z +
∞
X
(Fn−1 + Fn−2 )z n = z + zF (z) + z 2 F (z),
n=2
2
which gives
z
.
1 − z − z2
Similarly, the recurrence relation for the Lucas numbers is Ln = Ln−1 + Ln−2
for n > 2 with L0 = 2 and L1 = 1. Hence,
F (z) =
L(z) = 2 + z +
∞
X
(Ln−1 + Ln−2 )z n = 2 + z + z(L(z) − 2) + z 2 L(z),
n=2
which gives
L(z) =
2−z
.
1 − z − z2
(b) Use your answers in part (a) to prove the relation
Fn =
L0
Ln−1 Ln−2
+ 2 +···+ n
2
2
2
for all n > 1.
Solution: We have
∞
X zn
z
z/2
F (z)
=
=
=
.
L(z)
2−z
1 − z/2 n=1 2n
Hence,
F (z) = L(z)
∞
X
zn
n=1
2n
.
Comparing the coefficients of z n on both sides we get
Fn =
n
X
Ln−m
,
m
2
m=1
as required.
(c) Find and prove a formula expressing Ln as a linear combination of the Fibonacci numbers.
Solution: We have
F (z)
− F (z).
z
Comparing the coefficients of z n on both sides we get
L(z) = 2
Ln = 2 Fn+1 − Fn
for n > 0.
An alternative solution is to derive this from part (b).
(5 marks)
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