Solving RecurrencesAndreas Klappenecker
Generating Functions
Given a sequence (a0, a1, a2, a3,...) of real numbers, one can form its generating function, an infinite series given by
The generating functions is a formal power series, meaning that we treat it as an algebraic object, and we are not concerned with convergence questions of the power series.
∞�
k=0
akxk
Example 1
Suppose that the sequence is given by
(k+1)k>=0
Then its generating function is given by ∞�
k=0
(k + 1)xk = 1 + 2x+ 3x2 + · · ·
Example 2
The generating function of the sequence (1, 1, 1, 1, ...)is given by
1 + x+ x2 + x3 + · · · = 1
1− x.
Example 3
The generating function of the sequence (1, a, a2, a3, . . .)is given by
1 + ax+ a2x2 + a3x3 + · · · = 1
1− ax.
Example 4
The generating function of (1, 1, 1, 1, 1, 1, 0, 0, 0, . . .)is given by
1 + x+ x2 + x3 + x4 + x5.
x6 − 1
x− 1= 1 + x+ x2 + x3 + x4 + x5.
Sum of Sequences
Let (ak)k≥0 and (bk)k≥0 be sequences with generat-ing functions a(x) and b(x), respectively.
Then (ak + bk)k≥0 has the generating function
a(x) + b(x).
Products
Let a(x) =∞�
k=0
akxk and b(x) =∞�
k=0
bkxk.
Then
a(x)b(x) =∞�
k=0
k�
j=0
ajbk−j
xk
Example
Recall that the constant sequence (1, 1, 1, 1, . . .) hasthe generating function 1/(1− x).
What is the sequence corresponding to the gen-erating function
1
(1− x)2=
1
1− x
1
1− x?
1
1− x
1
1− x=
∞�
k=0
k�
j=0
1
xk =∞�
k=0
(k + 1)xk
Table of Generating Functions
Study the table on page 542!
Solving Recurrences
Shifting Sequences
Let G(x) be the generating function of the sequence(ak)k. Then
xG(x) =∞�
k=0
akxk+1 =
∞�
k=1
ak−1xk.
Solving a RecurrenceSuppose we have the recurrence system with
initial condition a0 = 2 and
recurrence ak = 3ak−1 for k ≥ 1.
G(x)− 3xG(x) =
∞�
k=0
akxk − 3
∞�
k=1
ak−1xk
= a0 +∞�
k=1
(ak − 3ak−1)xk
= 2.
Thus, G(x)− 3xG(x) = (1− 3x)G(x) = 2. Hence,
G(x) =2
1− 3x.
Solving a Recurrence
Since we know that 1/(1-ax)=1+ax+a2x2+...,
we have
G(x) = 2(1+3x+32x2+...).
Therefore, a sequence solving the recurrence is given by
(2,2x3,2x32,...)=(2x3k)
Fibonacci Numbers
The Fibonacci numbers satisfy the recurrence:
f0 = 0f1 = 1fn = fn−1 + fn−2 for n ≥ 2
Fibonacci Numbers
The Fibonacci numbers satisfy the recurrence:
f0 = 0f1 = 1f2 = f1 + f0f3 = f2 + f1f4 = f3 + f2
...
Fibonacci Numbers
Let F (x) =�∞
k=0 fkxk be the generating function of
the Fibonacci sequence. The recurrence
fn = fn−1 + fn−2
orfn − fn−1 − fn−2 = 0
suggests that we should consider
F (x)− xF (x)− x2F (x)
If it were not for the initial conditions, the latter sumwould be identically 0.
Fibonacci Numbers
The value of F (x)− xF (x)− x2F (x) is
f0 + f1x+ f2x2+ f3x3
+ · · ·−(f0x+ f1x2
+ f2x3+ · · · )
−(f0x2+ f1x3
+ · · · )
f0 + (f1 − f0)x
Since f0 + (f1 − f0)x = x, we get
F (x)− xF (x)− x2F (x) = F (x)(1− x− x2) = x.
Hence,
F (x) =x
1− x− x2
Fibonacci Numbers
We found the generating function
F (x) =x
1− x− x2
This function is of a simple form.If we can find the coefficient of xn in the power
series of F (x), we have found a closed form solutionto the Fibonacci numbers.
Fibonacci Numbers
Since x/(1−x−x2) is a rational function, we canuse the method of partial fractions, which you mightknow from calculus.
Factor the denominator
1− x− x2 = (1− α1x)(1− α2x),
where
α1 =1 +
√5
2, α2 =
1−√5
2.
Then
x
1− x− x2=
A1
1− α1x+
A2
1− α2x.
Fibonacci Numbers
We can plug in various values for x to obtainlinear equations that the real numbers A1 and A2
must satisfy. Solving these equations, we get
x
1− x− x2=
A1
1− α1x+
A2
1− α2x
with
A1 =1√5, A2 = − 1√
5.
so
F (x) =x
1− x− x2=
1√5
�1
1− α1x− 1
1− α2x
�
Fibonacci Numbers
We can get the closed form by observing that
F (x) = 1√5
�1
1−α1x− 1
1−α2x
�
= 1√5
�(1 + α1x+ α2
1x2 + · · · )− (1 + α2x+ α2
2x2 + · · · )
�
Therefore,
fn =1√5(αn
1−αn2 ) =
1√5
��1 +
√5
2
�n
−�1−
√5
2
�n�