
Today we will go through some examples. The first type of examples involves solving a linear recurrence equation to derive a general formula for a sequence. The second type of examples does the reverse process, in which, we are given a formula of a sequence and asked to determine its recurrence equation.
Suppose we have a sequence \{f_n\} that satisfies the recurrence equation a_k f_{n} + a_{k-1} f_{n-1} + a_{k-2} f_{n-2} + \dots + a_0 f_{n-k}=0 with some initial conditions on the values of f_0, f_1, \dots, f_{k-1}, we follow the following steps to determine a general formula for the sequence.
Step 1. Solve the recurrence equation.
Form the characteristic equation a_k x^k + a_{k-1} x^{k-1} + \dots + a_1 x + a_0=0 and determine its roots
a_k x^k + a_{k-1} x^{k-1} + \dots + a_1 x + a_0=a_k(x - x_1)^{j_1} (x - x_2)^{j_2} \dots (x - x_t)^{j_t}.
Suppose the characteristic equation has t roots x_1, \dots, x_t, where x_1 is a root of multiplicity j_1, x_2 is a root of multiplicity j_2, etc... Then for any \alpha_{ij}, the sequence f_n = (\alpha_{11} + \alpha_{12} n + \dots + \alpha_{1 j_1} n^{j_1 - 1}) ~ x_1^n + (\alpha_{21} + \alpha_{22} n + \dots + \alpha_{2 j_2} n^{j_2 - 1}) ~ x_2^n + \dots + (\alpha_{t1} + \alpha_{t2} n + \dots + \alpha_{t j_t} n^{j_t - 1}) ~ x_t^n satisfies the recurrence equation a_k f_{n} + a_{k-1} f_{n-1} + a_{k-2} f_{n-2} + \dots + a_0 f_{n-k}=0.
Step 2. Solve the initial condition.
Substituting the values of f_0, f_1, \dots, f_{k-1} into the formula of f_n to form a system of equations for the unknown \alpha_{ij} and solve this system of equations.
Let us do some exercises.
Problem 1: Determine a general formula for the sequence f_0 = 5, ~f_1 = 9, ~f_n = 6 f_{n-1} - 9 f_{n-2}.
Solution: From the recurrence equation f_n - 6 f_{n-1} + 9 f_{n-2}=0 we have the following characteristic equation x^2 - 6 x + 9 = 0. Solving this equation, we have x^2 - 6 x + 9 = (x-3)^2= 0, thus, it has one root x = 3 of multiplicity 2. So we determine the sequence of the following form f_n = (\alpha_{11} + \alpha_{12} ~n )~ 3^n. With n=0,1, we have f_0 = \alpha_{11} = 5, f_1 = (\alpha_{11} + \alpha_{12}) ~3 = 9. Solving these equations we obtain \alpha_{11} = 5 and \alpha_{12} = -2. Therefore, f_n = (5 - 2n) ~3^n.
Problem 2: Determine a general formula for the sequence f_0 = 1, ~f_1 = 5, ~f_2 = 22, ~f_n = 8 f_{n-1} - 21 f_{n-2} + 18 f_{n-3}.
Solution: From the recurrence equation f_n - 8 f_{n-1} + 21 f_{n-2}- 18 f_{n-3}=0 we have the following characteristic equation x^3 - 8 x^2 + 21 x - 18 = 0. Solving this equation, we have x^3 - 8 x^2 + 21 x - 18 = (x-2)(x-3)^2= 0. So we determine the sequence of the following form f_n = \alpha_{11} ~ 2^n + (\alpha_{21} + \alpha_{22} ~n )~ 3^n. With n=0,1,2, we have f_0 = \alpha_{11} + \alpha_{21} = 1, f_1 = 2 \alpha_{11} + 3 \alpha_{21} + 3 \alpha_{22} = 5, f_2 = 4 \alpha_{11} + 9 \alpha_{21} + 18 \alpha_{22} = 22. Solving these equations we obtain \alpha_{11} = 1, \alpha_{21}=0 and \alpha_{22} = 1. Therefore, f_n = 2^n + n ~3^n.
Problem 3: Determine a general formula for the sequence f_0 = 5, ~f_1 = 1, ~f_2 = 11, ~f_3 = 11, ~f_n = 2 f_{n-1} - 2 f_{n-3} + f_{n-4}.
Solution: From the recurrence equation f_n - 2 f_{n-1} + 2 f_{n-3}- f_{n-4}=0 we have the following characteristic equation x^4 - 2 x^3 + 2x - 1 = 0. Solving this equation, we have x^4 - 2 x^3 + 2x - 1 = (x+1) (x-1)^3 = 0. So we determine the sequence of the following form f_n = \alpha_{11} ~ (-1)^n + (\alpha_{21} + \alpha_{22} ~n + \alpha_{23} ~n^2) ~1^n. With n=0,1,2,3, we have f_0 = \alpha_{11} + \alpha_{21} = 5, f_1 = - \alpha_{11} + \alpha_{21} + \alpha_{22} + \alpha_{23} = 1, f_2 = \alpha_{11} + \alpha_{21} + 2 \alpha_{22} + 4 \alpha_{23} = 11, f_3 = - \alpha_{11} + \alpha_{21} + 3 \alpha_{22} + 9 \alpha_{23}= 11. Solving these equations we obtain \alpha_{11} = 3, \alpha_{21}=2, \alpha_{22} = 1 and \alpha_{23} = 1. Therefore, f_n = 3 (-1)^{n} + 2 + n + n^2 .
Problem 4: Determine a recurrence condition for the sequence f_n = n ~3^n.
Solution: The characteristic equation must have the form (x-3)^2 = x^2 - 6 x + 9 = 0. So the recurrence equation is f_n - 6 f_{n-1} + 9 f_{n-2} = 0. Combining with the initial condition, we have f_0= 0, ~f_1 = 3, ~f_n = 6 f_{n-1} - 9 f_{n-2}.
Problem 5: Determine a recurrence condition for the sequence f_n = (2n + 1) 3^n - 2^n.
Solution: The characteristic equation must have the form (x-2)(x-3)^2 = x^3 - 8 x^2 + 21 x - 18 = 0. So the recurrence equation is f_n - 8 f_{n-1} + 21 f_{n-2}- 18 f_{n-3}=0. Combining with the initial condition, we have f_0= 0, ~f_1 = 7, ~f_2 = 41, ~f_n = 8 f_{n-1} - 21 f_{n-2} + 18 f_{n-3}.
Problem 6: Determine a recurrence condition for the sequence f_n = n^2 + n + (-1)^n.
Solution: We have f_n = (n^2 + n) ~ 1^n + (-1)^n, so the characteristic equation must have the form (x+1) (x-1)^3 = x^4 - 2 x^3 + 2x - 1 = 0. So the recurrence equation is f_n - 2 f_{n-1} + 2 f_{n-3}- f_{n-4} = 0. Combining with the initial condition, we have f_0= 1, ~f_1 = 1, ~f_2 = 7, ~f_3 =11 , ~f_n = 2 f_{n-1} - 2 f_{n-3} + f_{n-4}.
Solution: We have f_n = (n^2 + n) ~ 1^n + (-1)^n, so the characteristic equation must have the form (x+1) (x-1)^3 = x^4 - 2 x^3 + 2x - 1 = 0. So the recurrence equation is f_n - 2 f_{n-1} + 2 f_{n-3}- f_{n-4} = 0. Combining with the initial condition, we have f_0= 1, ~f_1 = 1, ~f_2 = 7, ~f_3 =11 , ~f_n = 2 f_{n-1} - 2 f_{n-3} + f_{n-4}.
Problem 7: Determine a recurrence condition for the sequence f_n = (2n + 1) (-1)^n + (n-1) ~2^n.
Solution: The characteristic equation must have the form (x+1)^2 (x-2)^2 = x^4 - 2 x^3 - 3 x^2 + 4x + 4 = 0. So the recurrence equation is f_n - 2 f_{n-1} - 3 f_{n-2} + 4 f_{n-3} + 4 f_{n-4} = 0. Combining with the initial condition, we have f_0= 0, ~f_1 = -3, ~f_2 = 9, ~f_3 =9 , ~f_n = 2 f_{n-1} + 3 f_{n-2} - 4 f_{n-3} - 4 f_{n-4}.
Let us stop here for now, hope to see you again in our next post.
Homework.
1. Determine a general formula for the sequence f_0 = 3, ~f_1 = 6, ~f_n = 4 f_{n-1} - 4 f_{n-2}.
2. Determine a general formula for the sequence f_0 = 0, ~f_1 = 9, ~f_n = 6 f_{n-1} - 9 f_{n-2}.
3. Determine a general formula for the sequence f_0 = 0, ~f_1 = 1, ~f_2 = 5, ~f_n = 8 f_{n-1} - 21 f_{n-2} + 18 f_{n-3}.
4. Determine a general formula for the sequence f_0 = 1, ~f_1 = 0, ~f_2 = 3, ~f_3 = 2, ~f_n = 2 f_{n-1} - 2 f_{n-3} + f_{n-4}.
5. Determine a recurrence condition for the sequence f_n = n~ 2^n.
6. Determine a recurrence condition for the sequence f_n = 2n^3 - n^2 + 3n -1.
Solution.
1. f_n = 3 \times 2^n
2. f_n = n \times 3^{n+1}
3. f_n = 3^n - 2^n
4. f_n = (-1)^{n} + n