Solving Recurrence Relations with Generating Functions - faculty

Solving Recurrence Relations with Generating Functions The sequence of numbers 1,1,2,3,5,8,13, .... is known as the Fibonacci sequence. We can describe the sequence in terms of a recurrence relation. Let an denote the n’th term in this sequence (where n = 0, 1, 2, 3, . . . ). Then we can define the sequence as: a0 = 1, a1 = 1 and an = an−1 + an−2 for n ≥ 2. How do we find a formula for an ? Let G(x) = a0 + a1 x + a2 x2 + · · · + an xn + · · · be the generating function for the sequence. Then

G(x) = 1 + x + = 1+x+ = 1+x+

∞ X

an xn

n=2 ∞ X

(an−1 + an−2 )xn

n=2 ∞ X

n

an−1 x +

n=2

∞ X

an−2 xn

n=2 2

3

= 1 + x + (a1 x + a2 x + · · · ) + (a0 x2 + a1 x3 + a2 x4 + · · · ) = 1 + x + x(a1 x + a2 x2 + · · · ) + x2 (a0 + a1 x + a2 x2 · · · ) = 1 + x + x(G(x) − 1) + x2 G(x)

Therefore, G(x) = 1 + x + x(G(x) − 1) + x2 G(x). From this, we obtain 1 −1 that G(x) = 1−x−x 2 = x2 +x−1 . −1 What is the expansion for x2 +x−1 ? The roots of −x2 + x + 1 = 0 are x=

√ −1± 5 . 2

Let α1 =

√ −1+ 5 2

and α2 =

√ −1− 5 . 2

Then

−1 −1 1 1 1 = = ( − ). +x−1 (x − α1 )(x − α2 ) α1 − α2 x − α2 x − α1 √ 1 1 Given that α1 − α2 = 5 it follows that G(x) = √15 ( x−α − x−α ). We also 2 1 have G(x) =

x2

1 x − α1

= −

1 1 1 x x2 ( ) = − (1 + + + ···) α1 1 − αx1 α1 α1 α12

1 x − α2

= −

1 1 1 x x2 ( ) = − (1 + + + · · · ). α2 1 − αx2 α2 α2 α22 1

Since G(x) = 0, 1, 2, 3, . . . .

−1 (x−α1 )(x−α2 ) ,

it follows that an =

√1 5



1 αn+1 1

−

1



αn+1 2

, n =

Example Solve for an given that a0 = 0, a1 = 6 and an = −3an−1 + 10an−2 + 3 · 2n , for n ≥ 2. Solution Let G(x) = Then G(x) = 6x + = 6x +

∞ X n=2 ∞ X

P∞

n=0 an x

n

be the generating function for a0 , a1 , a2 , . . . .

an xn −3an−1 + 10an−2 + 3 · 2n )xn

n=2 ∞ X

= 6x − 3

an−1 xn + 10

n=2

∞ X

an−2 xn − 3

n=1

∞ X

2n xn

n=2

= 6x − 3x(a1 x + a2 x2 + · · · ) + 10x2 (a0 + a1 x + a2 x2 + · · · ) + 3((2x)2 + (2x)3 + (2x)4 + · · · ) = 6x − 3xG(x) + 10x2 G(x) + 3 · (2x)2 (1 + (2x) + (2x)2 + · · · ). From the above, we see that G(x) = 6x + G(x)(−3x + 10x2 ) + 12x2 · Therefore,

12x2 1 − 2x 12x2 6x + . 1 + 3x − 10x2 (1 − 2x)(1 + 3x − 10x2 )

G(x)(1 + 3x − 10x2 ) = 6x + G(x) =

Note that 1 + 3x − 10x2 = (1 − 2x)(1 + 5x). Therefore G(x) =

1 1−2x .

6x 12x2 + . (1 − 2x)(1 + 5x) (1 − 2x)2 (1 + 5x)

Using partial fractions, we obtain that 1 2 1 5 1 = · + · . (1 − 2x)(1 + 5x) 7 (1 − 2x) 7 1 + 5x From this, we also obtain that 1 2 1 10 1 25 1 = · + · + . (1 − 2x)2 (1 + 5x) 7 (1 − 2x)2 49 1 − 2x 49 1 + 5x 2

Therefore 

   2 1 1 10 1 5 1 25 1 2 2 G(x) = 6x +12x + . + + 7 1 − 2x 7 1 + 5x 7 (1 − 2x)2 49 1 − 2x 49 1 + 5x   1 1 The coeff. of xn in 6x 72 1−2x + 57 1+5x equals 6× coeff. of xn−1 in
5 1 2 1 6 27 · 2n−1 + 57 · (−5)n−1 . 7 1−2x + 7 1+5x which equals   1 10 1 25 1 The coeff. of xn in 12x2 72 (1−2x) equals 12× coeff. 2 + 49 1−2x + 49 1+5x
1 10 1 25 1 2 n−2 + 10 2n−2 + 25 (−5)n−2 . of xn−2 in 27 (1−2x) 2 + 49 1−2x + 49 1+5x which equals 12 7 (n − 1) · 2 49 49 Putting the above together, we obtain that     2 10 n−2 25 2 n−1 5 n−1 n−2 n−2 ·2 + · (−5) +12 (n − 1) · 2 + 2 + (−5) , n ≥ 2. an = 6 7 7 7 49 49

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