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

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The discussion centers on solving a second-order homogeneous linear recurrence relation and determining the conditions for boundedness of the sequence. The general solution is derived using the auxiliary equation, yielding roots that indicate boundedness depends on the coefficient C2 being zero. It is established that the sequence is bounded if and only if the initial conditions satisfy x0 = 3x1. A participant clarifies a calculation error regarding the roots of the auxiliary equation, confirming that the correct roots are -2 and 1/2. The conversation concludes with a request for resources on quadratic equations.
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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.
 
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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.
 
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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!
 
You are correct, the characteristic roots are indeed:

$\displaystyle \lambda=-2,\,\frac{1}{2}$
 
TheAvenger said:
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? :)
 
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