1. ## Recursive Sequence Convergence

$\displaystyle x_0 > x_1$ and r and s are positive numbers with p + q = 1. $\displaystyle x_n = px_{n-1} + qx_{n-2}$ for all $\displaystyle n \geq 2$. Prove that $\displaystyle (x_n)$ is convergent and find limits in terms of $\displaystyle x_0, x_1$, p and q.

Okay I began by letting p = 1 - q so I could eliminate that variable. Then I began working it out up to $\displaystyle x_7$ but I couldn't find any pattern in terms of coefficients of the polynomial. It's a polynomial with respect to p. I am trying to get this to work out to something that I can put into summation notation, prove it is monotone, and then find an upper and/or lower bound but I am unsure how to go about putting it in summation notation - it's entirely possible this is not the way to do it but it seems most logical to me. The first term (in order of descending exponent on p) of the expansion is always $\displaystyle p^{n-1}*a_1$. The last term is $\displaystyle a_0$ when n is even and $\displaystyle a_1$ when n is odd. These are the only real patterns I've been able to discern though. Any insight is much appreciated. Thanks.

2. Here is an idea for you The sequence can be written as follows

$\displaystyle \begin{bmatrix}0 && 1 \\ (1-p) && p \end{bmatrix}\begin{bmatrix} x_{n-2} \\x_{n-1}\end{bmatrix}=\begin{bmatrix} x_{n-1} \\x_{n}\end{bmatrix}$

This matrix can be diagonalized and solved explicitly

For the two eigenvalues I got

$\displaystyle \lambda = p-1 \text{ or } \lambda =1$

I hope this helps.

3. Originally Posted by valerian
$\displaystyle x_0 > x_1$ and r and s are positive numbers with p + q = 1. $\displaystyle x_n = px_{n-1} + qx_{n-2}$ for all $\displaystyle n \geq 2$. Prove that $\displaystyle (x_n)$ is convergent and find limits in terms of $\displaystyle x_0, x_1$, p and q.

Okay I began by letting p = 1 - q so I could eliminate that variable. Then I began working it out up to $\displaystyle x_7$ but I couldn't find any pattern in terms of coefficients of the polynomial. It's a polynomial with respect to p. I am trying to get this to work out to something that I can put into summation notation, prove it is monotone, and then find an upper and/or lower bound but I am unsure how to go about putting it in summation notation - it's entirely possible this is not the way to do it but it seems most logical to me. The first term (in order of descending exponent on p) of the expansion is always $\displaystyle p^{n-1}*a_1$. The last term is $\displaystyle a_0$ when n is even and $\displaystyle a_1$ when n is odd. These are the only real patterns I've been able to discern though. Any insight is much appreciated. Thanks.
The sequence is defined by the 'recursive relation'...

$\displaystyle \displaystyle x_{n} - p\ x_{n-1} - q\ x_{n-2}=0$ (1)

... that is 'linear and homogeneous' and the solution of which is...

$\displaystyle \displaystyle x_{n} = c_{1}\ \alpha_{1}^{n} + c_{2}\ \alpha_{2}^{n}$ (2)

... where$\displaystyle c_{1}$ and $\displaystyle c_{2}$ are 'arbitrary constants' that depend from $\displaystyle x_{0}$ and $\displaystyle x_{1}$ , $\displaystyle \alpha_{1}$ and $\displaystyle \alpha_{2}$ the solution of the 'characteristic equation'...

$\displaystyle \displaystyle \alpha^{2}- p\ \alpha - q= 0 \implies \alpha= \frac{p \pm \sqrt{p^{2}+4\ p -4}}{2}$ (3)

Now for $\displaystyle 0< p= 1-q <1$ is $\displaystyle |\alpha_{1}|<1$ and $\displaystyle |\alpha_{2}|<1$ , so that in any case is...

$\displaystyle \displaystyle \lim_{n \rightarrow \infty} x_{n} = 0$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Put x_n=m^n... and try to solve the quadratic equation... (hmmm...)

[EDIT: 4 MINUTES LATE]