Download 78 Topics in Discrete Mathematics Example 3.5. According to these

78
Topics in Discrete Mathematics
Example 3.5. According to these rules, the sum of
F (z) = z + z 2 + 2z 3 + 3z 4 + · · · and C(z) = 1 + z + 2z 2 + 5z 3 + 14z 4 + · · ·
is just obtained by adding the coefficients term-by-term:
F (z) + C(z) = 1 + 2z + 3z 2 + 7z 3 + 17z 4 + · · · .
We obtain each coefficient of the product by adding appropriate products of
one coefficient from F (z) and one coefficient from C(z):
coefficient of z 0 in F (z)C(z) =
0×1=0
coefficient of z 1 in F (z)C(z) =
0×1+1×1=1
coefficient of z 2 in F (z)C(z) =
0×2+1×1+1×1=2
coefficient of z 3 in F (z)C(z) =
0×5+1×2+1×1+2×1=5
coefficient of z 4 in F (z)C(z) =
1 × 5 + 1 × 2 + 2 × 1 + 3 × 1 = 12
so F (z)C(z) =
z + 2z 2 + 5z 3 + 12z 4 + · · ·
It is often useful to write formal power series in sigma notation as follows.
Definition 3.6. The symbol
∞
X
an z n means a0 + a1 z + a2 z 2 + a3 z 3 + · · · .
n=0
The
P∞ lowern limit of the3 sum 4can be5 varied in the obvious way: for instance,
n=3 bn z means b3 z + b4 z + b5 z + · · · . The summation variable n can of
course be replaced with any other letter.
In this notation the rules for addition and multiplication become:
!
!
∞
∞
∞
X
X
X
n
n
an z
+
bn z
=
(an + bn ) z n ,
n=0
∞
X
n=0
!
an z
n
n=0
∞
X
n=0
!
bn z
n
=
n=0
∞
X
n
X
n=0
m=0
(3.1)
!
am bn−m
n
z .
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CHAPTER 3. GENERATING FUNCTIONS
Remark 3.18*. Theorem 3.17 is equivalent to saying that for all s ≥ 1,
∞ X
n+s−1 n
2
3
s
(1 + z + z + z + · · · ) =
z .
(3.8)
n
n=0
To get a z n term on the left-hand side of (3.8), one must select from the
various factors the z n1 , z n2 , · · · , z ns terms where n1 , n2 , · · · , ns are nonnegative integers such that n1 + n2 + · · · + ns = n. So (3.8) is equivalent to
the statement that the number
of s-tuples (n1 , n2 , · · · , ns ) ∈ Ns such that
n1 + n2 + · · · + ns = n is n+s−1
. Thus Theorem 3.17 is just a disguised form
n
of Theorem 1.60.
Multiplying Theorem 3.17 on both sides by z k gives a generalization of (3.6):
∞ X
n n
zk
z =
.
(3.9)
k
(1 − z)k+1
n=0
(You would normally change the summation sothat it started from n = k,
but in this case it makes no difference because nk = 0 for n = 0, 1, · · · , k−1.)
Since we can express nm as a linear combination of binomial coefficients nk
for various k by Theorem 1.86, we can use (3.9) to get a closed formula for
the generating function of the sequence 0m , 1m , 2m , · · · for any m, and hence
of any sequence where the nth term is a polynomial function of n.
Example 3.19. To find a closed formula for 02 + 12 z + 22 z 2 + 32 z 3 + · · · , we
first express n2 as an integer linear combination of binomial coefficients:
n2 =
n
1
1
n
2
.
2
+
Hence
∞
X
2
n
n z =
n=0
∞
X
n=0
=
1
n
1
n
z +
∞
X
2
n
2
zn
n=0
z
2z 2
z + z2
+
=
(1 − z)2 (1 − z)3
(1 − z)3
Another seemingly dubious operation which is often useful is to take a formal
power series A(z) and substitute for z another formal power series F (z),
expanding out each F (z)n and collecting terms; the result is denoted A(F (z)).
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Topics in Discrete Mathematics
2−9z
Example 3.25. To find the coefficient of z n in 1−7z+12z
2 , we need to write
it in partial-fractions form. The denominator factorizes as (1 − 3z)(1 − 4z),
so this amounts to finding constants C1 and C2 such that
C1
C2
2 − 9z
=
+
.
(1 − 3z)(1 − 4z)
1 − 3z 1 − 4z
Multiplying both sides by (1 − 3z)(1 − 4z), this equation becomes
2 − 9z = C1 (1 − 4z) + C2 (1 − 3z) = (C1 + C2 ) − (4C1 + 3C2 )z
Equating coefficients, we get two linear equations for the two unknowns of
3
which the unique solution is C1 =
desired partial-fractions expression is:
2 − 9z
=
1 − 7z + 12z 2
and C2 =
−1
. So the
1
3
−
1 − 3z 1 − 4z
We can use (3.10) to read off that
the coefficient of z n in
2 − 9z
is
1 − 7z + 12z 2
3n+1 − 4n
5−9z
Example 3.26. Let us find the coefficient of z n in 1−6z+9z
2 . The denomi2
nator factorizes as (1 − 3z) ; when we have a repeated factor like this, the
partial-fractions form should include a term for every power of the factor up
to the multiplicity, which in this case means
C1
C2
5 − 9z
=
+
.
2
(1 − 3z)
1 − 3z (1 − 3z)2
Clearing the denominators, this becomes
5 − 9z = C1 (1 − 3z) + C2 = (C1 + C2 ) − 3C1 z,
which has the unique solution C1 = 3, C2 = 2. So
5 − 9z
3
2
=
+
.
2
1 − 6z + 9z
1 − 3z (1 − 3z)2
By (3.10) and (3.11), the coefficient of z n is 3 × 3n + 2(n + 1)3n = (2n + 5)3n .
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CHAPTER 3. GENERATING FUNCTIONS
Example 3.40. Recall the tower of Hanoi sequence satisfying h0 = 0 and
the recurrence relation hn = 2hn−1 + 1 for all n ≥ 1. If H(z) denotes the
generating function, we have the following equation for H(z) in terms of itself
and the geometric series:
H(z) = h0 + h1 z + h2 z 2 + h3 z 3 + · · ·
= 0 + (2h0 + 1)z + (2h1 + 1)z 2 + (2h2 + 1)z 3 + · · ·
=
2zH(z) +
z
1−z
which shows that
H(z) =
1
1
z
=
−
(1 − z)(1 − 2z)
1 − 2z 1 − z
From this we deduce the formula hn = 2n − 1 immediately.
Example 3.41*. Recall from Example 2.47 the sequence defined by
a0 = 3, a1 = 5, an = 2an−1 + 3an−2 + 8n − 4 for n ≥ 2.
P
n
Let A(z) = ∞
n=0 an z be the generating function. Then
A(z) = a0 + a1 z +
∞
X
an z n
n=2
= 3 + 5z +
∞
X
(2an−1 + 3an−2 + 8n − 4) z n
n=2
∞
X
= 3 + 5z + 2
∞
X
n
an−1 z + 3
n=2
∞
X
= 3 + 5z + 2z
an z n + 3z
an−2 z + 8
n=2
∞
X
2
n=1
n
n=0
2
∞
X
n
nz − 4
n=2
an z n + 8
∞
X
n zn − 4
n=2
= 3 + 5z + 2z(A(z) − 3) + 3z A(z)
z
1
+8
−z −4
−1−z ,
(1 − z)2
1−z
where the last line uses (3.3) and (3.6). Tidying this up gives
(1 − 2z − 3z 2 )A(z) =
3 − 7z + 17z 2 − 5z 3
,
(1 − z)2
∞
X
zn
n=2
∞
X
n=2
zn
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Topics in Discrete Mathematics
which implies that
C(z) =
∞
X
2n (2n − 1)!!
n=0
(n + 1)!
zn.
(3.23)
From this we can read off our desired formula for the Catalan numbers:
2n (2n − 1)!!
(2n)!
cn =
=
,
(3.24)
(n + 1)!
(n + 1)!n!
where the second equality uses (2n − 1)!! = (2n)!
(see Example
1.40). Note
2n n!
2n
2n+1
1
1
that cn is not quite a binomial coefficient, but cn = n+1 n = 2n+1
.
n
Generating functions can be very helpful in finding closed formulas for sums
f (0) + f (1) + · · · + f (n). The reason is that if A(z) is the generating function
of a0 , a1 , a2 , · · · , then by (3.3) and the multiplication rule,
∞
X
A(z)
(a0 + a1 + · · · + an ) z n ,
=
1−z
n=0
so
A(z)
1−z
(3.25)
is the generating function of the sequence of partial sums of a0 , a1 , · · · .
Example 3.46. To find a closed formula for 0×20 +1×21 +· · ·+n×2n , we first
need a formula for the generating function of the sequence 0 × 20 , 1 × 21 , · · · :
by (3.11), this is
∞
X
n2n z n =
n=0
2z
(1 − 2z)2
Now divide both sides by 1 − z, and use (3.25):
0×20 +1×21 +· · ·+n×2n = coefficient of z n in
2z
(1 − z)(1 − 2z)2
We already found this coefficient in Example 3.27: it is (n − 1)2n+1 + 2.
Example 3.47. In Example 2.22 we were unable to guess a closed formula
for the sum of squares 12 +22 +· · ·+n2 . To solve this problem using generating
functions, we first need a formula for the generating function of the sequence
of squares itself, i.e. 02 + 12 z + 22 z 2 + · · · . We found this in Example 3.19:
∞
X
z
2z 2
nz =
+
.
(1 − z)2 (1 − z)3
n=0
2 n