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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 
Last edited:

TheAvenger

New member
Feb 24, 2013
4
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

Administrator
Staff member
Feb 24, 2012
13,775
You are correct, the characteristic roots are indeed:

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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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? :)