Recurrence relations

Recurrence relations
Pengertian
 Suatu Recurrence Relation untuk deret {an} adalah
formula yang menyatakan an dalam bentuk satu
atau lebih deret sebelumnya.
 Misalkan a0, a1, a2,…, an-1 untuk setiap bilangan
bulat n dimana n≥ n0,dimana n0 adalah bilangan
bulat positif.
 Suatu deret disebut solusi dari Recurrence Relation
jika deret tersebut memenuhi Recurrence Relation
Contoh :
 {an}
adalah deret yang memenuhi
Recurrence Relation an = an-1 – an-2 untuk
n=2,3,4,… dan diketahui a0 = 3 dan a1 = 5.
Berapa a2 dan a3?
 a2 = a1 – a0 = 5 – 3 = 2
 a3 = a2 – a1 = 2 – 5 = -3
Linear Homogeneous Recurrence Relation
 Linear Homogeneous Recurrence Relation dengan derajat k
dan dengan koefisien konstan adalah Recurrence Relation
yang berbentuk :
an  c1.an1  c2 .an2  ...  ck .ank
Dimana c1,c2,…,ck adalah bilangan real, ck ≠ 0
Linear Homogeneous Recurrence Relation
 Ciri-ciri :
 Linear : suku-suku pada ruas kanannya merupakan
penjumlahan dari deret sebelumnya
 Homogeneous : tidak ada elemen yang bukan
kelipatan dari aj
 Constant : Koefisien c tidak tergantung dari n
Teorema 1
 Jika c1 dan c2 adalah bilangan real dan r2-c1r-c2=0 memiliki
dua akar berbeda yaitu r1 dan r2, maka deret {an}
merupakan solusi recurrence relation :
an  c1.an1  c2 .an2  an   .r   2 .r2
n
1 1
 Untuk n=0,1,2,… dimana 1 dan 2 adalah konstanta.
n
Teorema 2
 Jika c1 dan c2 adalah bilangan real dengan c2≠0 dan r2-c1r-c2=0
memiliki 1 akar r0, maka deret {an} merupakan solusi
recurrence relation :
an  c1.an1  c2 .an2  an  1.r0   2 .r0 .n
n
 Untuk n=0,1,2,… dimana 1 dan 2 adalah konstanta.
n
Teorema 3
 Jika
c1,c2,…,ck adalah bilangan bulat dan persamaan
karakteristik rk-c1rk-1-…-ck=0 memiliki k akar yang berbeda
yaitu r1, r2,… rk maka deret
recurrence relation :
{an}
merupakan solusi
an  c1.an1  c2 .an2  ...  ck .ank
 an  1.r1   2 .r2  ...   k .rk
n
n
n
 Untuk n=0,1,2,… dimana 1, 2,… k adalah konstanta.
Solving Recurrence Relations
Example: What is the solution of the recurrence
relation an = an-1 + 2an-2 with a0 = 2 and a1 = 7 ?
Solution: The characteristic equation of the
recurrence relation is r2 – r – 2 = 0.
Its roots are r = 2 and r = -1.
Hence, the sequence {an} is a solution to the
recurrence relation if and only if:
an = 12n + 2(-1)n for some constants 1 and 2.
Solving Recurrence Relations
Given the equation an = 12n + 2(-1)n and the initial
conditions a0 = 2 and a1 = 7, it follows that
a0 = 2 = 1 + 2
a1 = 7 = 12 + 2 (-1)
Solving
these
1= 3 and 2 = -1.
two
equations
gives
us
Therefore, the solution to the recurrence relation and initial
conditions is the sequence {an} with
an = 32n – (-1)n.
Solving Recurrence Relations
Example: Give an explicit formula for the
Fibonacci numbers.
Solution: The Fibonacci numbers satisfy the
recurrence relation fn = fn-1 + fn-2 with initial
conditions f0 = 0 and f1 = 1.
The characteristic equation is r2 – r – 1 = 0.
Its roots are
1 5
1 5
r1 
, r2 
2
2
Solving Recurrence Relations
Therefore, the Fibonacci numbers are given by
n
1 5 
1 5 




f n  1 


2

 2 
 2 


n
for some constants α1 and α 2.
We can determine values for these constants so that the
sequence meets the conditions f0 = 0 and f1 = 1:
f 0  1   2  0
1 5 
1 5 
  2
 1
f1  1 

 2 
2




Solving Recurrence Relations
The unique solution to this system of two
equations and two variables is
1
1
1 
, 2  
5
5
So finally we obtained an explicit formula
for the Fibonacci numbers:
n
1 1 5 
1 1 5 

 


fn 
5  2 
5  2 
n
Solving Recurrence Relations
Example:
What is the solution of the recurrence relation an
= 6an-1 – 9an-2 with a0 = 1 and a1 = 6?
Solution:
The only root of r2 – 6r + 9 = 0 is r0 = 3.
Hence, the solution to the recurrence relation is
an = 13n + 2n3n for some constants 1 and 2.
Solving Recurrence Relations
To match the initial condition, we need
a0 = 1 = 1
a1 = 6 = 13 + 23
Solving these equations yields
1 = 1 and 2 = 1.
Consequently, the overall solution is given by
an = 3n + n3n.
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