Recurrence relations.

Presentasi berjudul: "Recurrence relations."— Transcript presentasi:

1 Recurrence relations

2 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

3 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

4 Linear Homogeneous Recurrence Relation
Linear Homogeneous Recurrence Relation dengan derajat k dan dengan koefisien konstan adalah Recurrence Relation yang berbentuk : Dimana c1,c2,…,ck adalah bilangan real, ck ≠ 0

5 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

6 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 : Untuk n=0,1,2,… dimana 1 dan 2 adalah konstanta.

7 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 : Untuk n=0,1,2,… dimana 1 dan 2 adalah konstanta.

8 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 {an} merupakan solusi recurrence relation : Untuk n=0,1,2,… dimana 1, 2,… k adalah konstanta.

9 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.

10 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 two equations gives us 1= 3 and 2 = -1. Therefore, the solution to the recurrence relation and initial conditions is the sequence {an} with an = 32n – (-1)n.

11 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

12 Solving Recurrence Relations
Therefore, the Fibonacci numbers are given by 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:

13 Solving Recurrence Relations
The unique solution to this system of two equations and two variables is So finally we obtained an explicit formula for the Fibonacci numbers:

14 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.

15 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.

16 Selamat Belajar