Question

Solve the recurrence relation: \(a_n = 6a_{n-1} - 8a_{n-2}; a_0 = 1, a_1 = 0 \)

Answer 1

\(a_2 = -8, a_3 = -48, a_4 = -224\\
x^n = 6x^{n-1} - 8x^{n-2} \equiv x^n - 6x^{n-1} + 8x^{n-2} = 0\\
x^2 - 6x + 8 = 0; x = 6,4\)

So now I should just be able to find constants to fulfill \(a_n = A6^n + B4^n\) ?

Answer 2

Not sure how you got that second equaation

Answer 3

is this linear algebra?

Answer 4

Me neither!

Answer 5

Discrete math. For linear homogeneous recurrence relations with constant coefficients you're supposed to be able to find a solution of the form \(S_n = t^n\) which comes out to that.

Answer 6

Yeah, just solve for \(A\) and \(B\) and see if it works

Answer 7

ok yeah it works