Generating functions solve recurrence

Solving
Recurrences
using
Generating
Functions

To solve the recurrence relation by
generating functions, we have numeric
function for the closed form expression of
generating function
A(z) = a0 + a1z + a2z2 + a3z3 + …

Generating function Numeric function
A(z) =
𝟏
𝟏−𝒂𝒛
A(z) =
1
1−𝑧 2
A(z) =
𝒛
𝟏−𝒛 𝟐
A(z) =
𝒂𝒛
𝟏−𝒂𝒛 𝟐
A(z) = 𝒆 𝒛
A(z) = 𝟏 + 𝒛 𝒏
𝒂 𝒓
(r + 1)
r
rar
𝟏
𝒏!
𝒏 < 𝒓, 𝟎 ≤ 𝒓 ≤ 𝒏
𝟎, 𝒓 > 𝒏

Example 1:
Solve recurrence relation by generating function
1. Multiply both sides by zr
Steps to follow
ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
a r - 2 1
r
- 3
2
r
= 0r
z ar-
z ar-
z

Solving Recurrence relation using Generating Functions
ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
2. Since r  2, summing for all r, we get
arzr – 2 ar-1zr – 3 ar-2zr = 0
r = 2

r = 2

r = 2

Solving individually
For the 1st term:
 arzr = a2z2 + a3z3 + a4z4 + r = 2

A(z) = a0 + a1z + a2z2 + a3z3 + …
 arzr = A(z) - a0 - a1zr = 2

  

ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
For the 2nd term:
 ar-1zr = a1z2 + a2z3 + a3z4 + 
r = 2

Factoring out z:
 ar-1zr = z[A(z) – a0 ]r = 2

A(z) = a0 + a1z + a2z2 + a3z3 + …
Z(a1z + a2z2 + a3z3 + )

ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
For the 3rd term:
 ar-2zr = a0z2 + a1z3 + a2z4 + 
r = 2

Factoring out z2:
 ar-2zr = z2[A(z)]r = 2

A(z) = a0 + a1z + a2z2 + a3z3 + …
z2( a0 + a1z + a2z2 + )

ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
Substitute
 ar-2zr = z2[A(z)]r = 2
 arzr – 2 ar-1zr – 3  ar-2zr = 0
r = 2

r = 2

r = 2

by
 arzr = A(z) - a0 - a1zr = 2


 ar-1zr = z[A(z) – a0 ]r = 2


ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
We get
z2[A(z)]A(z) - a0 - a1z z[A(z) – a0 ]– 2 – 3
Substitute the value of a0 = 3; a1 = 1
A(z) – 3 - (1)z z[A(z) – 3 ]– 2 – 3z2[A(z)]
simplify
A(z) =
𝟑 −𝟓𝒛
𝟏 −𝟐𝒛 −𝟑𝒛 𝟐

ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
By partial fraction
A(z) =
𝟑 −𝟓𝒛
𝟏 −𝟐𝒛 −𝟑𝒛 𝟐
A(z) =
𝟑 −𝟓𝒛
(𝟏 −𝟑𝒛)(𝟏+𝒛)
𝟑 −𝟓𝒛
(𝟏 −𝟑𝒛)(𝟏+𝒛)
=
𝟏
(𝟏 −𝟑𝒛)
𝟐
(𝟏 + 𝒛)
+

ar - 2ar-1 - 3ar-2 = 0, for r  2, a0 = 3; a1 = 1
𝟑 −𝟓𝒛
(𝟏 −𝟑𝒛)(𝟏+𝒛)
=
𝟏
(𝟏 −𝟑𝒛)
𝟏
(𝟏 + 𝒛)
+
From the table previously presented:
A(z) =
𝟏
𝟏−𝒂𝒛
𝒂 𝒓
ar = 1  3r + 2  (-1)r

Example 2
G(x) = a0 + a1x + a2x2 +    =  anxn:
Let a0 = 1; a1 = 5, and an = an-1 - 6an-2 for n  2.
Let G(x) be the generating function for the sequence
a0; a1; a2; : : :. That is,
We will use generating functions to obtain a formula
for an.

n=0

Solving Recurrences using Generating Functions
The first step in the process is to use the
recurrence relation to replace
This can only be done when n  2,
an by an-1 - 6an-2.
so the first two terms (arising form the initial
conditions) need to be separated from the
sigma notation to obtain:

G(x) = a0 + a1x +  anxn
= a0 + a1x + (an-1 - 6an-1)xn
= a0 + a1x +  an-1xn- 6an-1xn

n=2

n=2

n=2

n=2
= a0 + a1x + x  an-1xn-1- 6x2an-2xn-2

n=2

n=2
In the last line above, constants and powers of x have
been factored out

so that the power to which x is raised “inside"
each sum exactly matches the subscript on the
coefficient ai.
which will make it easy to recognize each
summation as being G(x), with a few terms
possibly missing.
The reason for doing this is that the next step is
to make a change of index (on the subscripts),

In the first sum
When n = 2, k = 1

n=2
Put k = n - 1
 an-1xn-1
As n goes to infinity, so does k.


k=2
 akxk = G(x) - ao
the sum becomes
With this change of index,

In the next sum

n=2
 an-2xn-2
When n = 2, k = 0
Put k = n - 2
As n goes to infinity, so does k.


k=0
 akxk = G(x)
the sum becomes
With this change of index,

Combining these steps,
we arrive at
G(x) = a0 + a1x + x(G(x) - a0) - 6x2G(x).
After plugging in the known values for
a0 and a1,
then rearranging

G(x) =
𝟏 + 𝟒𝒙
𝟏 − 𝒙 + 𝟔𝒙 𝟐
this becomes
(1 - x + 6x2)G(x) = 1 + 4x.
Hence, we obtain the closed form

Notice the similarity of the coefficients in
1 - x + 6x2 and an – an-1 + 6an-2.
The next step is to use partial fractions to
determine the power series representation
of
1
1 − 𝑥 + 6𝑥2

We will eventually want the
sum of coefficient of xn and
four times the coefficient of xn-1
in this series.

Next, we want to write
1- x + 6x2
If  and  are integers,
they are positive among the
divisors of 6 and their negatives.
(1- x)(1- x).=

Multiplying
(1- x)(1- x).
we have,
1 - x - x + x2
 = 6
 = 1- 
lead to
-  - = -1 or
1- x + 6x2

 = 6
Therefore,
(1- ) = 6
- 2 = 6
and the quadratic formula gives
∝ =
1 ± 𝑖 23
2

Let,
𝛼 =
1 − 𝑖 23
2
𝛽 =
1 + 𝑖 23
2
So that
If, 1
1 − 𝑥 + 6𝑥2
=
𝐴
1 − 𝛼𝑥
+
𝐵
1 − 𝛽𝑥
=
𝐴 1 − 𝛽𝑥 + 𝐵(1 − 𝛼𝑥)
(1 − 𝛼𝑥)(1 − 𝛽𝑥)

-A - B = 0
then, equating coefficients of like powers of x
from the leftmost and rightmost expressions
gives the two equations in the two
unknowns A and B:
A + B = 1

so that ( -  )A =  .
Since  -  = -i 23,
the quantity
𝐴 =
𝛼
−i 23
=
1 − 𝑖 23
−2i 23
𝑖 23
i 23
 =
23 + 𝑖 23
46
B = 1 - A =
and, therefore,
23 − 𝑖 23
46

Hence,
1
1 − 𝑥 + 6𝑥2
=
𝐴
1 − 𝛼𝑥
+
𝐵
1 − 𝛽𝑥

n=0
A 𝛼nxn= +

n=0
B 𝛽nxn

The coefficient of xn is
A𝛼 + 𝐵𝛽 = +
23 + 𝑖 23
46
1 − 𝑖 23
2
𝑛
23 − 𝑖 23
46
1 + 𝑖 23
2
𝑛

The coefficient of xn in
G(x) =(1+𝟒𝐱)
and 4 times the coefficient of xn-1 in,
is the sum of the coefficient of xn in
1
1 − 𝑥 + 6𝑥2
𝟏
𝟏 − 𝒙 + 𝟔𝒙 𝟐
1
1 − 𝑥 + 6𝑥2

which equals
+
23 + 𝑖 23
46
1 − 𝑖 23
2
𝑛
23 − 𝑖 23
46
1 + 𝑖 23
2
𝑛
+ +
23 + 𝑖 23
46
1 − 𝑖 23
2
𝑛−1
23 − 𝑖 23
46
1 + 𝑖 23
2
𝑛−1
4  4 

Generating functions solve recurrence

Generating functions solve recurrence

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