Solving a Recurrence Relation with Multiple Roots

[silvermane]

Homework Statement

Solve the recurrence relation
a_n = 5a_n−1 − 3a_n−2 − 9a_n−3 for n ≥ 3
with initial values a₀ = 0, a₁ = 11, and a₂ = 34.

Homework Equations

its given lol

The Attempt at a Solution

I found that the characteristic equation for this rr is x³ - 5x² + 3x + 9 and found that the characteristic roots are 3, 3, -1...because we have 2 indistinct roots, I multiplied one of the 3 terms by n to get

a_n = r3ⁿ + sn3ⁿ - t
and so plugging back into the give rr we have

r3ⁿ + sn3ⁿ - t = 5(r3^n-1 + s(n-1)3^n-1 - t) - 3(r3^n-2 + s(n-2)3^n-2 - t) - 9(r3^n-3 + s(n-3)3^n-3 - t)

I'm thinking that in order to solve this, we're going to have to set this up as a system of equations, but I'm not sure how to do that with what I have. Any hints/tips/ suggestions on where to go next would be very helpful.

 

[JSuarez]

a_n = r3ⁿ + sn3ⁿ - t

Given this, then r,s and t must be equal to what for you to have a₀ = 0, a₁ = 11, and a₂ = 34? This above expression must be valid for all n, after all, not only for $n\geq 3$.

By the way, you claim that one of your roots is -1; are you sure that the above is entirely correct?

 

[silvermane]

By the way, you claim that one of your roots is -1; are you sure that the above is entirely correct?

To see if a root exists, we would plug it into the characteristic equation. When -1 is plugged into the equation, we obtain 0, therefore it is a root of the equation. Corollary, I saw that 3 was a root in same fashion, and found that it was a root of multiplicity 2 when I plugged it into the derivative. This was how we were showed to find the roots.

 

[JSuarez]

I don't have any doubt that -1 is a root: it is. But, if 3 is a root (forget the multiplicity for a moment) and it gives rise to a term 3ⁿ in the solution, then what would be the term corresponding to -1?