Discrete Math -- thanks in advance!! We will find the solution to the following

Question

Discrete Math --

thanks in advance!!

We will find the solution to the following the recurrence: a_n = -1a_n - 1 + 12a_n - 2 for n reaterthanorequalto 2 with initial conditions a_0 = 1, a_1 = 8. The first step in any problem like this is to find the characteristic equation by trying a solution of the "geometric" format a_n -= r^n. r^n = -1r^n - 1 + 12_r^n - 2 Since we are assuming r notequalto 0 we can divide by the smallest power of r, i.e., r^n - 2 to get the characteristic equation: r^2 = -1r + 12. Find the two roots of the characteristic equation r_1 and r_2. When entering your answers use r_1 and r_2: Since the roots are distinct, the general theory (Theorem 1 in Section 8.2 of Rosen) tells us that the general solution to our recurrence looks like: a_n = a_1 (r_1)^n + a_2(r_2)^n for suitable constants a_1, a_2. To find the values of these constants we have to use the initial conditions a_ 0 1, a_1 = 8. These yield by using n = 0 and n = 1 in the formula above: 1 = a_1 (r_1)^0 + a_2(r_2)^0 8 = a_1(r_1)^1 + a_2(r_2)^1 By plugging in your previously found numerical values for r_1 and r_2 and doing some algebra, find a_1, a_2. (Be careful to note that (-x)^n notequalto -(x^n) when n is even, for example (-3)^2 notequalto -(3^2). where the numbers r_i, a_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 = -1*r + 12

r^2 + r - 12 = 0

r^2 + 4*r - 3*r - 12 = 0

(r + 4)*(r - 3) = 0

So,

Since given that r1 < r2

r1 = -4 & r2 = 3

Now given that:

1 = a1*r1^0 + a2*r2^0

a1 + a2 = 1

a1 = 1 - a2

&

8 = a1*r1^1 + a2*r2^1

8 = -4*a1 + 3*a2

8 = -4*(1 - a2) + 3*a2

8 = -4 + 4*a2 + 3*a2

7*a2 = 12

a2 = 12/7

So a1 will be:

a1 = 1 - a2

a1 = 1 - 12/7

a1 = -5/7

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