We find the solution to the following lhcc recurrence: a_n = -8_a_n-1 - 16 a_n_2

Question

We find the solution to the following lhcc recurrence: a_n = -8_a_n-1 - 16 a_n_2 for n greaterthanorequalto 2 with initial conditions a_0 = 4, a_1 = 6. The first step as usual is to find the characteristic equation by trying a solution of the:geometric" format a_n = r^n. (We assume also r notequalto 0). In this case we get: r^n = -8r^n-1 - 16 5^n-2. Since we assuming r notequalto 0 we can divide by the smallest power of r, i.r., r^n-2 to get the characteristic equation: r^2 = -8r - 16. (Notice since our lhcc recurrence was degree 2, the characteristic equation is degree 2.) This characteristic equation has a single root r. (We say the root has multiplicity 2). Find r. r = Since the root is repeated, the general theory (Theorem 2 in Section 8.2 of Rosen) tells us that the general solution to our lhcc recurrence looks like: a_n = alpha_1 (r)^n + alpha_2 n(r)^n for suitable constants alpha_1, alpha_2. To find the values of these constants we have to use the initial conditions a_0 = 4, a_1 = 6. These yield by using n = 0 and n=1 in the formula above: 4 = alpha_1 (r)^0 + alpha_2 0(r)^0 and 6 = alpha_1 (r)^1 + alpha_2(r)^1 By plugging in your previously found numerical value for r and doing some algebra, find alpha_1, alpha_2: alpha_1 = alpha_2 = Note the final solution of the recurrence is: a_n = alpha_1(r)^n + alpha_2 n(r)^n where the numbers r, alpha_i have been found by your work. This gives an explicit numerical formula in terms of n for the a_n.

Explanation / Answer

r^2 = -8r - 16

r^2 + 8r + 16 = 0

(r + 4)^2 = 0

r = -4

Characteristic equation's single root will be

r = -4

Now using given constants:

4 = a1*r^0 + a2*0*r^0

6 = a1*r^1 + a2*1*r^1

r^0 = 1

using eq. 1

4 = a1*1 + a2*0*1

4 = a1

Now using this value of a1 and r in eq. 2

6 = 4*r^1 + a2*1*r^1

r = -4

6 = 4*(-4)^1 + a2*1*(-4)^1

6 = -16 - 4*a2

4*a2 = -22

a2 = -22/4

a2 = -11/2 = -5.5

So,

an = 4*(-4)^n - 5.5*n*(-4)^n

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