Defn: A linear recurrence relation has constant coefficients if the a i s are constant. Ch 14: 1, 4, 5, 10, 13, 18, 22, 24, 25 and

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1 7.4: linear homogeneous recurrence relation: Defn: A recurrence relation is linear if h n = a 1 (n)h n 1 + a (n)h n a k (n)h n k + b(n) A recurrence relation has order k if a k 0 Ex: Derangement D n = (n 1)D n 1 + (n 1)D n, D 1 = 0, D = 1 D n = nd n 1 + ( 1) n, D 1 = 0 Fibonacci: f n = f n 1 + f n, f(0) = 0, f(1) = 1 Defn: A linear recurrence relation is homogeneous if b = 0 Defn: A linear recurrence relation has constant coefficients if the a i s are constant. Tentative HW 1: Ch 14: 1, 4, 5, 10, 13, 18,, 4, 5 and A.) Suppose the sequences r n, s n, and t n satisfy the homogeneous linear recurrence relation, h n = a 1 (n)h n 1 + a (n)h n + a 3 (n)h n 3 (**). Show that the sequence, c 1 r n +c s n +c 3 t n also satisfies this homogeneous linear recurrence relation (**). B.) Suppose the sequence ψ n satisfies the linear recurrence reln, h n = a 1 (n)h n 1 + a (n)h n + a 3 (n)h n 3 + b(n) (*). Show that the sequence, c 1 r n + c s n + c 3 t n + ψ n also satisfies this linear recurrence relation. C.) How many terms of the sequence are needed to find a unique sequence with these terms satisfying (*). What linear system of equations can be used to determine c 1, c, c 3.

2 7.4: linear homogeneous recurrence relation w/constant coefficients: Ex: Solve the recurrence relation: h n + h n = 0, h 0 = 3, h 1 = 5 Guess q n is a solution. q n + q n = q n (q + 1) = 0 q + 1 = 0 implies q = ±i Thus the general solution is h n = c 1 i n + c ( i) n i.e., this function satisfies the recurrence relation. Now need to find c i s resulting in initial conditions: h 0 = 3: c 1 + c = 3 implies c = 3 c 1 h 1 = 5: c 1 i c i = 5 implies c 1 + c = 5i c c 1 = 5i. Thus c = 5i Hence c 1 = 3 5i and c = 3 ( 3 5i ) = 3+5i h n = ( 3 5i )in + ( 3+5i )( i)n satisfies the recurrence relation and the initial conditions. h n = i n [( 3 5i ) + ( 3+5i )( 1)n ] = i n [( 3 )(1 + ( 1)n ) + ( 5i )( 1 + ( 1)n )] h j = ( 3 5i )ij + ( 3+5i )( i)j = 3( 1) j h j+1 = ( 3 5i )ij+1 + ( 3+5i )( i)j+1 = 5(i) j+ = 5( 1) j Thus starting with h 0, we have the sequence: 3, 5, -3, -5, 3, 5, -3, -5, 3, 5, -3, -5, 3, 5,...

3 Ex: Solve the recurrence relation, h n h n 1 + h n 3 h n 4 = 0, h 0 = 3, h 1 = 3, h = 7, h 3 = 15. Guess q n is a solution. q n q n 1 + q n 3 q n 4 = q n 4 (q 4 q 3 + q 1) = 0, q n 4 (q 3 3q + 3q 1)(q + 1) = q n 4 (q 1) 3 (q + 1) = 0 q = 1, 1, 1, -1 Note: 1 is a repeated root Note n j (1) n, j = 0, 1,, are solutions to the recurrence relation. Check: If h n = (1) n = 1: = 0. Check: If h n = n(1) n = n: n (n 1) + (n 3) (n 4) = n n + n n = 0 Check: If h n = n (1) n = n : n (n 1) + (n 3) (n 4) = n (n n + 1) + (n 6n + 9) (n 8n + 16) = 0 General solution h n = c 1 (1) n + c n(1) n + c 3 n (1) n + c 4 ( 1) n = c 1 + c n + c 3 n + c 4 ( 1) n Now need to find c i s resulting in initial conditions: h 0 = 3 = c 1 + c 4 h 1 = 3 = c 1 + c + c 3 c 4 h = 7 = c 1 + c + 4c 3 + c 4 h 3 = 15 = c 1 + 3c + 9c 3 c 4

4 h 0 = 3 = c 1 + c 4 h 1 = 3 = c 1 + c + c 3 c 4 h = 7 = c 1 + c + 4c 3 + c 4 h 3 = 15 = c 1 + 3c + 9c 3 c Thus c 1 = 3, c =, c 3 =, c 4 = h n = c 1 + c n + c 3 n + c 4 ( 1) n = 3 n + n Hence h n = 3 n + n Check Initial Conditions: h 0 = 3, h 1 = 3, h = 7, h 3 = 15 h 0 = = 3 h 1 = 3 + = 3, h = = 7 h 3 = = 15.

5 7.4: linear homogeneous recurrence relation: h n a 1 h n 1 a h n... a k h n k = 0 Suppose ϕ(n) and ψ(n) are solns to the above recurrence relation, then Claim 1: cϕ(n) is a solution for any constant c Claim : ϕ(n) + ψ(n) is also a solution. Hence if ϕ i (n) are solns, then Σc i ϕ i (n) is a soln for any constants c i.

6 Thm 7.4.1: Suppose a i are constants and q 0. Then q n is a solution to h n a 1 h n 1 a h n... a k h n k = 0 iff q is a root of the polynomial equation x k a 1 x k 1 a x k... a k = 0 If this characteristic equation has k distinct roots, q 1, q,..., q k, then h n = c 1 q n 1 + c q n c k q n k is the general solution. I.e, given any initial values for h 0, h 1,..., h k 1, there exists c 1, c,..., c k such that h n = c 1 q1 n + c q n c k qk n satisfies the recurrence relation and the initial conditions. Thm 7.4.: Suppose q i is an s i -fold root of the characteristic equation. Then H i (n) = c 1 q n i + c nq n i c s i n s i 1 q n i is a solution to the recurrence relation. If the characteristic equation has t distinct roots q 1,..., q t with multiplicity s 1,..., s t, respectively, then h n = H 1 (n) +...H t (n) is a general solution.

7 7.5: Non-homogeneous Recurrence Relations. h n a 1 h n 1 a h n... a k h n k = b Let k(h) = h n a 1 h n 1 a h n... a k h n k Suppose ϕ is a solution to the recurrence relation k(h) = 0 and β is a solution to the recurrence relation k(h) = b. Claim: ϕ + β is a solution to To solve a non-homogeneous recurrence relation. Step 1: Solve homogeneous equation. Recall if constant coeffficents, guess h n = q n for homogeneous eq n. Step : Guess a solution to non-homogeneous equation, by guessing a solution β n similar to b(n). Step 3a: Note general solution is c i ϕ i (n) + β(n). Step 3b: Find c i using initial conditions. Ex: Solve the recurrence relation: h n + h n = 14n, h 0 = 3, h 1 = 5 Step 1: Guess q n is a solution to homogeneous equation: h n + h n = 0. q n + q n = q n (q + 1) = 0 q + 1 = 0 implies q = ±i Thus the general solution to homogeneous equation is h n = c 1 i n + c ( i) n

8 Step : Guess a solution to non-homogeneous equation: Guess β n = xn + y. h n + h n = 14n Plug β n into non-homogeneous equation: [xn + y] + [x(n ) + y] = 14n Solve for x and y: xn + y x = 14n implies x = 7 and y = 7. Thus a solution to non-homogeneous equation is β(n) = 7n + 7. Step 3a: Note general soln to non-homogeneous equation is h n = c 1 i n + c ( i) n + 7n + 7 Step 3b: Find c i using initial conditions. h n + h n = 14n, h 0 = 3, h 1 = 5 h 0 = 3: c 1 i 0 + c ( i) 0 + 7(0) + 7 = 3 implies c 1 + c = 4 h 1 = 5: c 1 i 1 + c ( i) 1 + 7(1) + 7 = 5 implies ic 1 ic = 9 c 1 + c = 4 c 1 + c = 9i implies c 1 = 4+9i = + 9i and c = 4 9i = 9i h n = ( + 9i )in + ( 9i )( i)n + 7n + 7 = (i n )[( )(1 + ( 1) n ) + ( 9i )(1 ( 1)n )] + 7n + 7 h j = ( 1) j ( 4) + 7(j) + 7 = 4( 1) j j h j+1 = (i j+1 )9i+7(j+1)+7 = (i j+ )9+14j+14 = 9( 1) j+1 +14j+14 Thus the sequence is 3, 5, 5, 37, 31, 33, 53, 65, 59, 61, 81, 93,...