Semester 2 Tutorial Solutions Week 10 2007 1. Consider the gen

THE UNIVERSITY OF SYDNEY
Math1904 Discrete Mathematics (Advanced)
Semester 2 Tutorial Solutions Week 10 2007
1. Consider the generating function
G(z) = a0 + a1z + a2z 2 + · · · .
Verify the following table of closed forms for G corresponding to the given choices for an.
Solution
1
an
β n
nβ n
n 2 β n
G(z)
1
1 − z
1
1 − βz
βz
(1 − βz) 2
β2z2 + βz
(1 − βz) 3
We know from Example 14.5 of the textbook that
1 + z + z 2 + z 3 + · · · = 1
1 − z
and on substituting βz for z in this equation we have
1 + βz + β 2 z 2 + β 3 z 3 1
+ · · · =
1 − βz .
Now differentiate to obtain
and then multiply by z to get
Differentiate again to get
β + 2β 2 z + 3β 3 z 2 + 4β 4 z 3 + · · · =
βz + 2β 2 z 2 + 3β 3 z 3 + 4β 4 z 4 + · · · =
β
(1 − βz) 2
βz
(1 − βz) 2.
β + 2 2 β 2 z + 3 2 β 3 z 2 + 4 2 β 4 z 3 β
+ · · · =
(1 − βz) 2 + 2β2z (1 − βz) 3 = β2z + β
(1 − βz) 3.
On multiplying by z we get the last line of the table.

2. Show that the coefficient of z n in
Solution
We have
1
1 − 7z + 12z 2 is 4n+1 − 3 n+1 .
1
1
=
1 − 7z + 12z2 (1 − 3z)(1 − 4z)
= 1
� �
1 1
−
z 1 − 4z 1 − 3z
= 1
z
∞�
n=0
[(4z) n − (3z) n ] .
and from this we see that the coefficient of z n is 4 n+1 − 3 n+1 , as claimed.
3. Determine the closed form of the generating function corresponding to the given sequence
(i) an = n(n + 1)(n + 2).
� �
n + 5
(ii) an = .
5
Solution
Both parts can be done by using the formula from lectures:
∞�
� �
n + p − 1
z
n
n=0
n 1
=
(1 − z) p.
� �
n + 3
(i) Here an+1 = 6 and so
n
∞�
an+1z n 6
=
(1 − z) 4.
(ii)
n=0
Since a0 = 0, multiplying by z we get the answer
1
(1 − z) 6.
6z
(1 − z) 4.
4. Solve the recurrence relations
(i) an = 4an−1 − 4an−2 for n ≥ 2, a0 = 6, a1 = 8.
(ii) xn+2 + 2xn+1 − 3xn = 0 for n ≥ 0, x0 = 2, x1 = −2.
Solution
(i) The characteristic equation is λ 2 − 4λ + 4 = 0; that is, (λ − 2) 2 = 0 and therefore
the general solution is an = A 2 n + Bn 2 n for some constants A and B. Using the
given initial conditions we have A = 6 and 2A + 2B = 8, hence B = −2. Thus the
solution is an = 6 · 2 n − 2n · 2 n .
(ii) The characteristic equation is λ 2 +2λ−3 = 0. The roots are 1 and −3, and therefore
the general solution is xn = A + B (−3) n for some constants A and B. Using the
given initial conditions we have A + B = 2 and A − 3B = −2, hence A = B = 1.
Thus the solution is xn = 1 + (−3) n .

5. Find the general solution to the recurrence relations
(i) an+3 = 3an+2 − 4an.
(ii) xn+3 + 6xn+2 + 12xn+1 + 8xn = 0.
Solution
(i) The characteristic equation is λ 3 − 3λ 2 + 4 = 0 and this factorizes as
Thus the general solution is
(λ − 2) 2 (λ + 1) = 0.
an = A 2 n + Bn 2 n + C (−1) n .
(ii) The characteristic equation is λ 3 + 6λ 2 + 12λ + 8 = 0 which factorizes as
Thus the general solution is
(λ + 2) 3 = 0.
xn = (A + Bn + Cn 2 )(−2) n .
6. For each natural number n ≥ 1, let an be the number of n digit numbers which contain no
0’s and an even number of 9’s.
(i) Show that a1 = 8 and that for n ≥ 2, an = 7an−1 + 9 n−1 .
(ii) Solve the recurrence relation of (i) using the method of generating functions.
Solution
(i) There are 8 digits other than 0 and 9, hence a1 = 8. For n ≥ 2, a number of n digits
of the required form can be obtained by adjoining a single digit (other than 0 or 9)
to one of length n − 1 (this can be done in 8 ways) or by adding a 9 to the end of a
number of length n −1 which has no 0’s and an odd number of 9’s (this can be done
in 9 n−1 − an−1 ways. Thus an = 8an−1 + (9 n−1 − an−1) = 7an−1 + 9 n−1 .
(ii) Suppose that
G(z) = a0 + a1z + a2z 2 + · · ·
where a1 = 8 and an = 7an−1 + 9 n−1 for n > 0. Then a0 = 1 and we have
G(z) = a0 + a1z + a2z 2 + · · · + anz n + · · ·
zG(z) = a0z + a1z 2 + · · · + an−1z n + · · ·
G(z) − 7zG(z) = 1 + 9 0 z + 9 1 z 2 + · · · + 9 n−1 z n + · · ·
hence G(z) − 7zG(z) = 1 + z/(1 − 9z) and it follows that
1 − 8z
G(z) =
(1 − 7z)(1 − 9z)
= 1
� �
1 1
+
2 1 − 7z 1 − 9z
= 1
∞�
(7
2
n + 9 n )z n
and therefore an = 1
2 (7n + 9 n ).
n=0

THE UNIVERSITY OF SYDNEY
Math1904 Discrete Mathematics (Advanced)
Semester 2 Further Exercises Solutions Week 10 2007
1. (i) Write down a recurrence relation for the number an of sequences of length n that
can be formed from the strings 00 and 1. For example, the strings of length 4 are
1111, 1100, 1001, 0011 and 0000.
(ii) Solve the recurrence you found in (i).
Solution
(i) A string of length n + 1 either ends with 1 or 0. If it ends with 1, then it can be
obtained in just one way by adding a 1 to the end of a string of length n. If it
ends with 0, then it actually ends with 00 and it can be obtained by adding 00 to
a uniquely determined string of length n − 1. Thus an+1 = an + an−1, provided we
have n ≥ 2. If n = 1, then the recurrence holds provided we define a0 to be 1.
(ii) The recurrence found in (i) defines the Fibonacci numbers 1, 1, 2, 3, 5, ... and the
solution is given on page 105 of the text book.
2. (i) Show that for any numbers a and b, the sequence defined by xn = a3 n + bn3 n is a
solution to the recurrence equation
xn+2 = 6xn+1 − 9xn.
(ii) Find the partial solution of this equation with the initial conditions x0 = 15 and
x1 = 30.
Solution
(i) With the given value of xn we have
6xn+1 − 9xn = 6(a3 n+1 + b(n + 1)3 n+1 ) − 9(a3 n + bn3 n )
= 2a3 n+2 + 2b(n + 1)3 n+2 − a3 n+2 − bn3 n+2
= a3 n+2 + b(n + 2)3 n+2
= xn+2.
(ii) We have a = 15 and 3(a + b) = 30. This gives b = −5. The partial solution is
xn = (15 − 5n) 3 n .