Download Solve recursions by the generating function method

UDE, Fakultät für Mathematik: Discrete Mathematics (C1) [Recursions]
1
Solve recursions by the generating function method
Define generating function for a sequence, defined by a recursion
Determine the sequence (an ) defined by
an+2 = 8 · 3n+1 + 9an − 8an+1 ,
Define the generating function:
A(x) :=
n ∈ N0
∞
X
a0 := 2, a1 := −6
ak x k
k=0
Separate the terms involving initial conditions
A(x) = 2 − 6x +
∞
X
ak+2 xk+2
k=0
Substitute the recursion equation
A(x) = 2 − 6x +
∞
X
8 · 3k+1 + 9ak − 8ak+1 xk+2
k=0
Rewrite using the generating function
A(x) = 2 − 6x +
∞
X
8 · 3k+1 xk+2 +
k=0
= 2 − 6x + 8x
∞
X
k=0
∞
X
9ak xk+2 −
3
x
+ 9x
2
8ak+1 xk+2
k=0
k=0
k+1 k+1
∞
X
∞
X
k
ak x − 8x
k=0
∞
X
ak+1 xk+1
k=0
1
= 2 − 6x + 8x(
− 1) + 9x2 A(x) − 8x(A(x) − 2)
1 − 3x
Solve for generating function
1
2 − 6x + 8x( 1−3x
− 1) + 16x
1 + 8x − 9x2
(2 + 10x)(1 − 3x) + 24x2
=
(1 + 8x − 9x2 )(1 − 3x)
−6x2 + 4x + 2
=
(1 + 8x − 9x2 )(1 − 3x)
A(x) =
UDE, Fakultät für Mathematik: Discrete Mathematics (C1) [Recursions]
2
(1 − x)(6x + 2)
(1 − x)(9x + 1)(1 − 3x)
6x + 2
=
(9x + 1)(1 − 3x)
=
Note: To obtain the factorization above it may be helpful to rewrite the poylnomial, e.g.:
8
1
1 + 8x − 9x2 = (−9)(x2 − x − )
9
9
Then determine the zeros:
4
y= ±
9
r
16 1
4 5
+ = ±
81 9
9 9
Hence:
1
1 + 8x − 9x2 = (−9)(x − 1)(x + ) = (1 − x)(1 + 9x)
9
Expand into a power series
For this step one may use partial fraction decomposition, Taylor-series-expansion or operations
with known power series. Here we use partial fraction decomposition:
A(x) =
a
b
+
⇔ 6x + 2 = a(1 + 9x) + b(1 − 3x) = (9a − 3b)x + (a + b)
1 − 3x 1 + 9x
The coefficients may be determined by solving the system
9a − 3b = 6,
a+b=2
⇐⇒
a = 1, b = 1
Alternatively one may insert suitable values for x, e.g. x = −1/9 and x = 1/3. Now A(x) can be
expanded with the help of the geometric series:
∞
X
1
1
3k + (−9)k xk
A(x) =
+
=
1 − 3x 1 + 9x k=0
Read off the sequence (and check if desired)
So the solution is:
an = 3n + (−9)n
The initial conditions a0 = 2 and a1 = −6 are satisfied. So is the recursion equation:
an+2 − 9an + 8an+1 = 3n+2 + (−9)n+2 − 9 (3n + (−9)n ) + 8 3n+1 + (−9)n+1
= 3n (9 − 9 + 24) + (−9)n (81 − 9 − 72) = 8 · 3n+1
D01