1. CSE 2813 Discrete Structures Solving Recurrence Relations Section 6.2

2. CSE 2813 Discrete Structures Degree of a Recurrence Relation The degree of a recurrence relation is k if the sequence { a n } is expressed in terms of the previous k terms: a n  c 1 a n -1 + c 2 a n -2 + … + c k a n - k where c 1, c 2, …, c k are real numbers and c k  0 What is the degree of a n  2 a n -1 + a n -2 ? What is the degree of a n  a n -2 + 3 a n -3 ? What is the degree of a n  3 a n -4 ?

3. CSE 2813 Discrete Structures Linear Recurrence Relations A recurrence relation is linear when a n is a sum of multiples of the previous terms in the sequence Is a n  a n -1 + a n -2 linear ? Is a n  a n -1 + a 2 n -2 linear ?

4. CSE 2813 Discrete Structures Homogeneous Recurrence Relations A recurrence relation is homogeneous when a n depends only on multiples of previous terms. Is a n  a n -1 + a n -2 homogeneous ? Is P n  (1.11) P n -1 homogeneous ? Is H n  2 H n -1 + 1 homogeneous ?

5. CSE 2813 Discrete Structures Solving Recurrence Relations Solving 1 st Order Linear Homogeneous Recurrence Relations with Constant Coefficients (LHRRCC) –Derive the first few terms of the sequence using iteration –Notice the general pattern involved in the iteration step –Derive the general formula –Now test the general formula on some previously calculated (by iteration) terms

6. CSE 2813 Discrete Structures Solving 2 nd Order LHRRCC Form: a n  c 1 a n -1 + c 2 a n -2 with some constant values for a 0 and a 1 Assume that the solution is a n  r n, where r is a constant and r  0

7. CSE 2813 Discrete Structures Step 1 Solve the characteristic quadratic equation r 2 – c 1 r – c 2 = 0 to find the characteristic roots r 1 and r 2

8. CSE 2813 Discrete Structures Step 2 Case I: The roots are not equal a n =  1 r 1 n +  2 r 2 n Case II: The roots are equal ( r 1 = r 2 = r 0 ) a n =  1 r 0 n +  2 nr 0 n

9. CSE 2813 Discrete Structures Step 3 Apply the initial conditions to the equations derived in the previous step. –Case I: The roots are not equal a 0 =  1 r 1 0 +  2 r 2 0 =  1 +  2 a 1 =  1 r 1 1 +  2 r 2 1 =  1 r 1 +  2 r 2 –Case II: The roots are equal a 0 =  1 r 0 0 +  2  0  r 0 0 =  1 a 1 =  1 r 0 1 +  2  1  r 0 1 = (  1 +  2 ) r 0

10. CSE 2813 Discrete Structures Step 4 Solve the appropriate pair of equations for  1 and  2.

11. CSE 2813 Discrete Structures Step 5 Substitute the values of  1,  2, and the root(s) into the appropriate equation in step 2 to find the explicit formula for a n.

12. CSE 2813 Discrete Structures Example Solve the recurrence relation: a n  4 a n -1  4 a n -2 where a 0  a 1  1 Solve the recurrence relation: a n  a n -1 + 2 a n -2 where a 0  2 and a 1  7

13. CSE 2813 Discrete Structures Exercises 1, 3