# Linear Algebra - Linear Constant Coefficient Difference Equations

1. Dec 6, 2008

### symsane

1. The problem statement, all variables and given/known data

x[n] = -x[n-1] + x[n-2] + x[n-3] and x[0]= 0, x[-1]= 0, x[-2]= -1

We want to find "x[n]". When I started to solve this problem I acquired that lambda1= 1 and lambda2,3= -1. I couldn't do anything after lambda's are found. So, can anyone solve the remaining part?

2. Dec 6, 2008

### Pere Callahan

So I assume you used the ansatz
$$x[n]=\lambda^n$$
?

well, just build up a linear combination from your three lambdas
$$x[n]=A+B(-1)^n+C n (-1)^n$$
Because there is this double root at -1 you need to multiply one of the terms by n. Do you know this type of thing from ODE's?

Then plug in your initial values and solve for A,B,C.

This is a linear system of equations with three unknowns and solution A=-1/4, B=1/4, C=-1/2.

3. Dec 6, 2008

### HallsofIvy

Staff Emeritus
But it is 0 that is a double root so it is not (-1)n that you need to multiply by n! You should have $x_n= A+ Bn+ C (-1)^n$.

4. Dec 6, 2008

### Pere Callahan

I think 0 is not even a single root. Plugging in x[n]=lambdan gives
$$\lambda^{n}=-\lambda^{n-1}+\lambda^{n-2}+\lambda^{n-3}$$
or
$$\lambda^{3}=-\lambda^{2}+\lambda^{1}+1$$
I think the OP's calculation of the roots was correct.