Question

We will find the solution to the following lhcc recurrence: an=8an−1−16an−2 for n≥2 with initial conditions a0=4,a1=7. The first step as usual is to find the characteristic equation by trying a solution of the "geometric" format an=rnan=rn. (We assume also r≠0). In this case we get: rn=8r^n−1−16r^n−2. Since we are assuming r≠0r≠0 we can divide by the smallest power of r, i.e., rn−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 rr. (We say the root has multiplicity 2). Find r.
r=?

Answer 1

The root of the characteristic equation r² = 8r - 16 is r = 4. This is a repeated root with multiplicity 2, which affects the form of the general solution to the recurrence relation.

Step-by-step explanation:

We are tasked with finding the root r of the characteristic equation r² = 8r - 16 derived from the linear homogeneous recurrence relation with constant coefficients an = 8an-1 - 16an-2. To solve for r, we rearrange the equation into standard quadratic form:

r² - 8r + 16 = 0

Then, we factor the quadratic equation:

(r - 4)(r - 4) = 0

This implies:

r = 4

Since r is a repeated root with multiplicity 2, this means the general solution of the recurrence relation will be in the form an = (A + Bn)rⁿ, where A and B are constants determined by the initial conditions.

Answer 2

idk try to figure it pouy

Step-by-step explanation: