Linear Algebra - Linear Constant Coefficient Difference Equations

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  • #1
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Homework Statement



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?
 

Answers and Replies

  • #2
So I assume you used the ansatz
[tex]
x[n]=\lambda^n
[/tex]
?

well, just build up a linear combination from your three lambdas
[tex]
x[n]=A+B(-1)^n+C n (-1)^n
[/tex]
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
HallsofIvy
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Homework Statement



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?

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

well, just build up a linear combination from your three lambdas
[tex]
x[n]=A+B(-1)^n+C n (-1)^n
[/tex]
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.
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 [itex]x_n= A+ Bn+ C (-1)^n[/itex].
 
  • #4
I think 0 is not even a single root. Plugging in x[n]=lambdan gives
[tex]
\lambda^{n}=-\lambda^{n-1}+\lambda^{n-2}+\lambda^{n-3}
[/tex]
or
[tex]
\lambda^{3}=-\lambda^{2}+\lambda^{1}+1
[/tex]
I think the OP's calculation of the roots was correct.
 

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