Mark's question at Yahoo Answers (Linear recurrence relation)

[Fernando Revilla]

Here is the question

Question - Find the general solution to the 2nd order homogeneous linear recurrence below, and give a necessary and sufficient condition on u0 and u1 such that the sequence defined by the recurrence is bounded.

2*x subscript(n + 1) + 3*x subscript (n) -2*x subscript (n-1) = 0

I've found the general solution using the auxiliary equation, but I'm not sure how to prove it's bounded. I know that if a sequence converges, it means that it is bounded, but I have no clue how to show whether a recurrent sequence converges. Any help will be greatly appreciated!

Here is a link to the question:

2nd order homogeneous linear recurrence? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello Mark,

From $2\lambda^2+3\lambda-2=0$ we get $\lambda=\dfrac{1}{3}$ and $\lambda=-\dfrac{4}{3}$ so, the general solution is $$x_n=C_1\left( \dfrac{1}{3}\right)^n+C_2\left(\dfrac{-4}{3}\right)^n$$ As $|1/3|<1$ $C_1(1/3)^n\to 0$ that is, $C_1(1/3)^n$ is bounded. Taking into account that $|-4/3|>1$, the sequence $C_2(-4/3)^n$ is bounded if and only if $C_2=0$, as a consequence $x_n$ is bounded if and only if $C_2=0$. Now, for $n=0$ and for $n=1$ $$\left \{ \begin{matrix} x_0=C_1+C_2\\x_1=\dfrac{C_1}{3}-\dfrac{4C_2}{3}\end{matrix}\right.$$

But $C_2=0$ if and only if $x_0=3x_1$ (necessary and sufficient condition on $x_0$ and $x_1$ such that the sequence defined by the recurrence relation is bounded).

Edit: See the following posts.

 

[TheAvenger1]

Thank you So much for your reply, I've perfectly understood how to solve it now! Just one quick question though, shouldn't the roots of the auxiliary equation be -2 and 1/2? It's not a big deal as the concept remains the same, but I just wanted to confirm whether it was a calculation mistake on your part or have I solved the auxiliary equation wrong.

Thank you very much once again!

 

[MarkFL]

You are correct, the characteristic roots are indeed:

$\displaystyle \lambda=-2,\,\frac{1}{2}$

 

[Fernando Revilla]

Thank you So much for your reply, I've perfectly understood how to solve it now! Just one quick question though, shouldn't the roots of the auxiliary equation be -2 and 1/2? It's not a big deal as the concept remains the same, but I just wanted to confirm whether it was a calculation mistake on your part or have I solved the auxiliary equation wrong.

Thank you very much once again!

All right, the concept remains the same. Out of curiosity, my mistake: $$\lambda=\dfrac{-3\pm \sqrt{9+16}}{2\cdot \color{red}3}$$ I looked at $3$ instead of $2$!

P.S. Does anyone know about a quadratic equation's tutorial? :)