Math 774 Homework 4
Austin Mohr
November 10, 2010
Problem 1
Proposition 1. The recurrence ar − 7ar−1 + 10ar−2 = 3r with initial conditions a0 = 0 and a1 = 1 admits the closed formula
1
ar = (2r+4 + 11 · 5r − 3r+3 ).
6
Proof. We first solve the related homogenous recurrence
ar − 7ar−1 + 10ar−2 = 0.
It has characteristic equation
x2 − 7x + 10 = 0,
(h)
which has as its roots x = 2 and x = 5. Letting ar
the homogenous recurrence, we have
denote the solution of
r
r
a(h)
r = A2 + B5
for some coefficients A and B to be determined.
(p)
Next, we guess a particular solution, denoted ar . Suppose the particular solution has the form
r
a(p)
r = C3
for some coefficient C to be determined. It follows that
C3r − 7(C3r−1 ) + 10(C3r−2 ) = 3r ,
from which we determine that C = − 29 .
Now, the solution ar to the original recurrence is given by
(p)
ar = a(h)
r + ar
1
9
= A2r + B5r − 3r .
2
From the initial conditions, we know that
0=A+B−
and
1 = 2A + 5B −
Solving this system gives A =
ar =
10
3
9
2
27
.
2
and B = 76 . Thus, we have
10 r 7 r 9 r
2 + 5 − 3 ,
3
6
2
or, after some algebra,
1
ar = (2r+4 + 11 · 5r − 3r+3 ).
6
Proposition 2. The recurrence ar +6ar−1 +9ar−2 = 3 with initial conditions
a0 = 0 and a1 = 1 admits the closed formula
3
r
3
ar = − −
(−3)r + .
16 12
16
Proof. We first solve the related homogenous recurrence
ar + 6ar−1 + 9ar−2 = 0.
It has characteristic equation
x2 + 6x + 9 = 0,
(h)
which has as the root x = 2 of multiplicity 2. Letting ar denote the solution
of the homogenous recurrence, we have
r
a(h)
r = (A + Br)(−3)
for some coefficients A and B to be determined.
(p)
Next, we guess a particular solution, denoted ar . Suppose the particular solution has the form
a(p)
r =C
2
for some coefficient C to be determined. It follows that
C + 6C + 9C = 3,
3
from which we determine that C = 16
.
Now, the solution ar to the original recurrence is given by
(p)
ar = a(h)
r + ar
= (A + Br)(−3)r +
3
.
16
From the initial conditions, we know that
0=A+
3
16
and
3
.
16
3
1
Solving this system gives A = − 16
and B = − 12
. Thus, we have
3
r
3
ar = − −
(−3)r + .
16 12
16
1 = −3A − 3B +
Proposition 3. The recurrence ar + ar−2 = 0 with initial conditions a0 =
0, a1 = 2 admits the closed formula
3πr
ar = −2 sin
.
2
Proof. The homogenous recurrence above has characteristic equation
x2 + 1 = 0,
π
which has as its roots x = i and x = −i, which we write instead as ei 2 and
3π
ei 2 , respectively. Thus, we have
πr 3πr
ar = A cos
+ B sin
2
2
for some coefficients A and B to be determined.
From the initial conditions, we know that
0=A
3
and
2 = −B.
Thus, we have
ar = −2 sin
3πr
2
.
Problem 2
Proposition 4. The recurrence ar = ar−1 − ar−2 with conditions a1 = 1
and a2 = 0 admits the closed formula
πr √3
5πr
ar = cos
−
sin
.
3
3
3
Proof. Observe first that a0 must satisfy
0 = 1 − a0 ,
and so a0 = 1.
The homogenous recurrence above has characteristic equation
x2 − x + 1 = 0,
√
√
which has as its roots x = 1+i2 3 and x = 1−i2 3 , which we write instead as
5π
π
ei 3 and ei 3 , respectively. Thus, we have
πr 5πr
ar = A cos
+ B sin
3
3
for some coefficients A and B to be determined.
From the initial conditions for a0 and a1 , we know that
1=A
and
√
3
1
1= A−
B.
2
2
√
Solving this system gives A = 1 and B = − 33 . Thus, we have
πr √3
5πr
ar = cos
−
sin
.
3
3
3
4
Proposition 5. The recurrence ar = ar−1 − ar−2 with conditions a0 = 0
and a3 = 0 admits the closed formula
√
2c 3
5πr
ar = −
sin
3
3
for any constant c.
Proof. Let a1 = c. The recurrence forces
a2 = c − 0.
Now, a3 = 0 is satisfied, as
0 = c − c.
As before, we have
ar = A cos
πr 3
+ B sin
5πr
3
for some coefficients A and B to be determined.
From the initial conditions for a0 and a1 , we know that
0=A
and
√
1
3
c= A−
B.
2
2
√
Solving this system gives A = 0 and B = − 2c3 3 . Thus, we have
√
2c 3
5πr
ar = −
sin
.
3
3
Proposition 6. The recurrence ar = ar−1 − ar−2 is inconsistent with the
conditions a0 = 1 and a3 = 2.
Proof. Let a1 = c. From the recurrence, we see that
a2 = c − 1,
and so
a3 = (c − 1) − c
= −1,
which is inconsistent with the given constraints.
5
Problem 3
Consider a 1 × n board. Suppose we color each square of the board so that
no two adjacent squares are colored red.
Proposition 7. The number g(n) of colorings if only the colors blue and
red are available is given by
√
√
5+3 5 n 5−3 5 n
ar =
φ ,
φ +
10
10
where φ denotes the golden ratio
√
1+ 5
2
and φ denotes its conjugate
√
1− 5
2 .
Proof. Consider a properly-colored 1×n board. The first square must either
be blue or red.
If the first square is blue, then there are no restrictions on subsequent
squares other than the requirement that no two red squares are adjacent.
Thus, there are g(n − 1) ways to color the 1 × n board such that the first
square is blue.
If the first square is red, then the second must be blue (else we have
two adjacent red squares). The remaining n − 2 squares may be colored
however we please, so long as no two red squares are adjacent. Thus, there
are g(n − 2) ways to color the 1 × n board such that the first square is red.
Taken together, we have that
g(n) = g(n − 1) + g(n − 2).
By inspection, we see that g(1) = 2 (we may color a single square either blue or red) and g(2) = 3 (our options are blue/blue, blue/red, and
red/blue).
The homogenous recurrence above has characteristic equation
x2 − x − 1 = 0,
which has as its roots x = φ and x = φ. Thus, we have
ar = Aφn + Bφ
n
for some coefficients A and B to be determined.
From the initial conditions, we know that
2 = Aφ + Bφ
6
and
2
3 = Aφ2 + Bφ .
Solving this system gives A =
√
5+3 5
10
and B =
√
5−3 5
10 .
Thus, we have
√
√
5+3 5 n 5−3 5 n
g(n) =
φ +
φ .
10
10
The closed formula above agrees with the number of colorings for n = 1
and n = 2. For n = 3, we have g(3) = 5 colorings, which we list explicitly.
blue/blue/blue
blue/blue/red
blue/red/blue
red/blue/blue
red/blue/red
Proposition 8. The number h(n) of colorings if only the colors blue, white,
and red are available is given by
√
√
√ n 3−2 3
√
3+2 3
h(n) =
(1 + 3) +
(1 − 3)n .
6
6
Proof. Consider a properly-colored 1×n board. The first square must either
be blue, white, or red.
If the first square is blue or white, then there are no restrictions on
subsequent squares other than the requirement that no two red squares are
adjacent. Thus, there are 2h(n − 1) ways to color the 1 × n board such that
the first square is blue or white.
If the first square is red, then the second must be blue or white (else we
have two adjacent red squares). The remaining n−2 squares may be colored
however we please, so long as no two red squares are adjacent. Thus, there
are 2h(n − 2) ways to color the 1 × n board such that the first square is red.
Taken together, we have that
h(n) = 2h(n − 1) + 2h(n − 2).
By inspection, we see that h(1) = 3 (we may color a single square either
blue, white, or red) and h(2) = 8 (our options are blue/blue, blue/white,
white/blue, white/white, red/blue, red/white, blue/red, and white/red).
7
The homogenous recurrence above has characteristic equation
x2 − 2x − 2 = 0,
√
√
√
which has
√ as its roots x = 1 + 3 and x = 1 − 3. Let α = 1 + 3 and
α = 1 − 3. Thus, we have
ar = Aαn + Bα
n
for some coefficients A and B to be determined.
From the initial conditions, we know that
3 = Aα + Bα
and
8 = Aα2 + Bα 2 .
Solving this system gives A =
√
3+2 3
6
and B =
√
3−2 3
.
6
Thus, we have
√
√
3+2 3 n 3−2 3 n
h(n) =
α +
α .
6
6
The closed formula above agrees with the number of colorings for n = 1
and n = 2. For n = 3, we have h(3) = 22 colorings, which we list explicitly.
8
blue/blue/blue
blue/blue/white
blue/white/blue
white/blue/blue
blue/white/white
white/blue/white
white/white/blue
white/white/white
blue/blue/red
blue/red/blue
red/blue/blue
red/blue/red
white/white/red
white/red/white
red/white/white
red/white/red
blue/white/red
blue/red/white
white/blue/red
white/red/blue
red/blue/white
red/white/blue
Problem 4
Proposition 9. The number ar of partitions of a set of r elements satisfies
the recurrence
r X
r
ar+1 =
ai ,
i
i=0
where a0 = 1.
Proof. Let [r + 1] be the set to be partitioned. For a given parition, let A
denote the subset containing the element r + 1. Now, we have 0 ≤ |Ac | ≤ r.
Let us first enumerate the number of partitions of [r + 1] satisfying|Ac | = i,
for some fixed value of i. We may choose the elements of Ac in ri distinct
ways. Each such selection of i elements can be further partitioned in ai ways.
The remaining
elements are grouped into the subset A. Thus, for |Ac | = i,
there are ri ai distinct partitions of [r + 1]. Summing over 0 ≤ i ≤ r gives
the desired result.
9
Problem 5
There are two kinds of particles in a reactor. In every second, an α particle
will split into three β particles, while a β particle will split into an α particle
and two β particles.
Proposition 10. Let α(t) and β(t) denote the number of α and β particles,
respectively, in the reactor at time t. If there is a single α particle in the
reactor at t = 0, then
3
α(t) = (3t−1 − (−1)t−1 )
4
and
3
β(t) = (3t − (−1)t )
4
Proof. At time t + 1, the number of α particles is precisely the same as the
number of β particles at time t (since each β particle decays into exactly
one α particle). In other words,
α(t + 1) = β(t).
Similarly, the number of β particles at time t + 1 is equal to three times the
number of α particles at time t (since each α particle decays into exactly
three β particles) and two times the number of β particles at time t (since
each β particle decays into two β particles). In other words,
β(t + 1) = 3α(t) + 2β(t).
Equivalently,
β(t + 1) = 3β(t − 1) + 2β(t).
The homogenous recurrence above has characteristic equation
x2 − 2x − 31 = 0,
which has as its roots x = 3 and x = −1. Thus, we have
β(t) = A3t + B(−1)t
for some coefficients A and B to be determined.
Since there is one α particle in the reactor at time t = 0, we know that
β(0) = 0 and β(1) = 3. Thus,
0=A+B
10
and
3 = 3A − B.
Solving this system gives A =
3
4
and B = − 34 . Thus, we have
3
3
β(t) = 3t − (−1)t ,
4
4
or, after some algebra,
3
β(t) = (3t − (−1)t ).
4
Using the fact that α(t + 1) = β(t), we immediately deduce that
3
α(t) = (3t−1 − (−1)t−1 ).
4
Problem 6
Proposition 11. Let f (n) denote the number of directed paths of length n
that start from vertex a and end at vertex d, where length is the number of
steps (arcs) used. For n ≥ 3,
f (n) = χ(n) +
n
bX
3c
2n−1−3k ,
k=1
where
(
1
χ(n) =
0
if 3 | n
otherwise.
Proof. Represent a directed path as a binary string according to the figure
above. Such a string either consists entirely of 1’s or has at least one 0.
Observe, however, that the binary string consisting of all 1’s represents a
11
valid directed path precisely when 3 | n (hence the inclusion of the χ(n)
term).
It remains to enumerate the number of binary strings containing at least
one 0 that correspond to valid directed paths. (Note we are no longer imposing any constraints on n.) Consider the last 0 appearing in the binary
string. From the figure, we see that following any arc labelled 0 results in
state a. Since all the following
bits are 1, there must be 3k of them, for
some integer 1 ≤ k ≤ n3 .
Now, for some fixed value of k, we have n − 1 − 3k bits preceeding
the final 0, which can be assigned values in 2n−1−3k ways. Moreover, each
such assignment is indeed a valid path, as the final 0 will result in state a
regardless of the current state. Summing over k gives the desired result.
Problem 7
Gossip is spread among r people via telephone. Specifically, in a telephone
conversation between A and B, A tells B all of the gossip he has heard,
and B reciprocates. Let ar denote the minimum number of telephone calls
among r people so that all gossip will be known to everyone.
Lemma 1. For all r ≥ 3, ar ≥ r.
Proof. Consider the multigraph G where V (G) is the collection of gossips
and we introduce an edge between gossips if they have had a conversation.
Now, if G is disconnected, then certainly not all gossip is known to everyone.
Let us thus assume that G is connected but has exactly r − 1 edges. Under
these assumptions, G is a tree, and so possesses at least two leaves A and B.
Being leaves, we see that A and B have only one call in which to exchange
gossip. If we assume, without loss of generality, that A’s conversation takes
place first, then it follows that A will not know B’s gossip (as B has not
yet told anyone his gossip). Since A learns no new information after this
moment, we conclude that A never learns B’s gossip, and so r − 1 calls is
insufficient.
Proposition 12. a2 = 1
Proof. Certainly, at least one call must be made in general. We demonstrate
below a strategy using one call.
12
Proposition 13. a3 = 3
Proof. By the lemma, a3 ≥ 3. We demonstrate below a strategy using three
calls.
Proposition 14. a4 = 4
Proof. By the lemma, a4 ≥ 4. We demonstrate below a strategy using four
calls.
Proposition 15. For all r ≥ 1, ar ≤ ar−1 + 2.
13
Proof. Denote the gossips by Ai for 1 ≤ i ≤ r. We know there exists some
strategy for gossips A1 , . . . , Ar−1 , say, so that they can share all gossip using
ar−1 calls.
We now present a strategy for all r gossips using ar−1 + 2 calls. First, Ar
will call A1 . Next, follow a strategy for A1 , . . . , Ar−1 using ar−1 calls. Now,
A1 knows Ar ’s gossip, so Ar ’s gossip will be shared with all of A1 , . . . , Ar−1
(as A1 ’s gossip is shared with all of A1 , . . . , Ar−1 ). Finally, A1 will call Ar
again. Since A1 has since learned all the gossip, Ar now knows all the gossip.
Thus, every A1 , . . . , Ar knows all the gossip in ar−1 + 2 calls.
Proposition 16. For all r ≥ 4, ar ≤ 2r − 4.
Proof. For r = 4, we have shown that a4 = 4 = 2(4) − 4. Suppose now that
ak ≤ 2k − 4 for any k ≥ 5. It follows that
ak+1 ≤ ak + 2
(by previous proposition)
≤ 2k − 4 + 2
(by inductive hypothesis)
= 2(k + 1) − 4.
14