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Homogeneous Linear Recurrences

Homogeneous Linear Recurrences. To solve such recurrences we must first know how to solve an easier type of recurrence relation:

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Homogeneous Linear Recurrences

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  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. • an = 2an-1 is homogeneous • an = 2an-1 + 2n-3 - an-3 is not. • bk= 2bk-1 - bk-2 + 4bk-3 + 2is 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. • an = 2an-1 is homogeneous • an = 2an-1 + 2n-3 - an-3 is not. • bk= 2bk-1 - bk-2 + 4bk-3 + 2is not.

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

  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: Let be the solution. 1 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

  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 r1 = (1+5)/2 r2 = (1-5)/2 General solution: an = A [(1+5)/2]n +B [(1-5)/2]n Characteristic equation

  9. Fibonacci Recurrence Use initial conditions a0= 0, a1= 1 to find A,B andobtain specific solution. 0=a0= A [(1+5)/2]0+B [(1-5)/2]0 = A +B That is A +B = 0 --- (1) 1=a1= 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 2nd: 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. The given recurrence is satisfied by the following sequences: b0, b1, b2, …. 1, (2)1, (2)2, (2)3 , … 1, (5)1, (5)2, (5)3 , … Characteristic equation Or 1, 2, 4, 8, … 1, 5, 25, 125, …

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

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

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

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