1. Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type of recurrence relation: DEF: A linear recurrence relation is said to be homogeneous if it is a linear combination of the previous terms of the recurrence without an additional function of n or a constant. a n = 2a n-1 is homogeneous a n = 2a n-1 + 2 n-3 - a n-3 is not. b k = 2b k-1 - b k-2 + 4b k-3 + 2 is not.

2. Homogeneous Linear Recurrences To solve such recurrences we must first know how to solve an easier type of recurrence relation: DEF: A linear recurrence relation is said to be homogeneous if it is a linear combination of the previous terms of the recurrence without an additional function of n or a constant. a n = 2a n-1 is homogeneous a n = 2a n-1 + 2 n-3 - a n-3 is not. b k = 2b k-1 - b k-2 + 4b k-3 + 2 is not.

3. First Order Homogenous Recurrences Characteristic equation Using initial conditions: General Solution: Closed Form Solution: Quiz tomorrow from Solving Recurrence Relations

4. Solving Homogenous Recurrences Finding a closed form solution to a recurrence relation. Multiply both sides by Characteristic equation roots

5. Example continued… The closed form solution is given by: Where constants A and B are to determined by employing initial conditions. Plug in n = 0 in equation 1 to obtain: Plug in n = 1 in equation 1 to obtain: Adding the two equation, we get 1 Let be the solution.

6. Example continued… By substituting A = 3 and B = -2 in equation 1, we get the final solution in the closed form: Verification Test Data

7. Example 2 The closed form solution is given by: Using initial conditions:

8. Solution of Fibonacci Recurrence Solve the quadratic equation: r 2 = r + 1 r 2 - r - 1 = 0 to obtain r 1 = (1+  5)/2 r 2 = (1-  5)/2 General solution: a n = A [(1+  5)/2] n +B [(1-  5)/2] n Characteristic equation

9. Fibonacci Recurrence Use initial conditions a 0 = 0, a 1 = 1 to find A,B and obtain specific solution. 0=a 0 = A [(1+  5)/2] 0 +B [(1-  5)/2] 0 = A +B That is A +B = 0 --- (1) 1=a 1 = A [(1+  5)/2] 1 +B [(1-  5)/2] 1 = A(1+  5)/2 +B (1-  5)/2 = (A+B )/2 + (A-B )  5/2 That is (A+B )/2 + (A-B )  5/2 = 1 or (A+B ) + (A-B )  5 = 2 -- (2) First equation give B = -A. Plug into 2 nd : A = 1/  5, B = -1/  5 Final answer:

10. Case where the roots are identical Finding a closed form solution to a recurrence relation when the roots are identical. Test Data The closed form solution is given by:

11. Where constants A and B are to determined by employing initial conditions. I Plug in k = 0 in equation 1 to obtain: Plug in k= 1 in equation 1 to obtain:

12. A variation of the previous Example Finding a closed form solution to a recurrence relation when the roots are identical. Test Data The closed form solution is given by:

13. Where constants A and B are to determined by employing initial conditions. I Plug in k = 0 in equation 1 to obtain: Plug in k= 1 in equation 1 to obtain:

14. Relationship between Recurrence Relation and Sequences Finding sequences that satisfy a recurrence relation. Characteristic equation The given recurrence is satisfied by the following sequences: b 0, b 1, b 2, …. 1, (2) 1, (2) 2, (2) 3, … 1, (5) 1, (5) 2, (5) 3, … 1, 2, 4, 8, … 1, 5, 25, 125, … Or

15. Verification 1, 5, 25, 125, … 1, 2, 4, 8, …

16. Practice Problem List first 5 terms of sequence generated by the recurrence. Solve the recurrence in the close form. Step 1: Write characteristic equation. Step 2: Find roots of the characteristic equation. Step 3: Write the general solution. Step 4: Find constants A and B using boundary conditions. Step 5: Substitute constants in the general solution.

17. Solving Recurrences using Spreadsheet

18. 3 rd order Homogenous Recurrences Characteristic equation Roots D = 1, D = 2, and D = 3 General Solution: Where A, B, and C are constants to be determined by using boundary conditions. Let r = 0 Let r = 1 (I) (II)

19. 3 rd order Homogenous Recurrences Let r = 2 We have to solve (I), (II), and (III) simultaneously for A, B, and C. (III) The solution is not complete.