Determin a formula for x_n in a discrete dynamical system.

Question

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Determine a formula for $x_n$ given the following discrete dynamical system with initial values $x_0 = 1$ and $y_0$ = 4.

$x_{n+1} = 5x_n + y_n$

$y_{n+1} = 2x_n + 4y_n$

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i think you have to calculate this with help of the formula:

my $x_k$ formula will be: $x_k = 2 *6^k [1,1] + 1 * 3^k * [-1 , 2]$

the answer of the questioin is:

can someone explain me the following:

*what did i calculate? i want to know this so when they ask for this, i know what to answer to the question.

*how do i come to the solution of $x_n$?

thanks in advance.

Answer 1

As mentioned you are solving a linear system, in this case the system: $$\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix}=A\begin{pmatrix}x_{n-1}\\ y_{n-1}\end{pmatrix}$$ where $$A=\begin{pmatrix}5 & 1\\ 2 & 4\end{pmatrix}.$$ Note that $A$ has eigenvalues $3$ and $6$ and respective corresponding eigenvectors $(-1,2)$ and $(1,1)$. The initial vector is given by $$\begin{pmatrix}x_{0}\\ y_{0}\end{pmatrix}=\begin{pmatrix}1\\ 4\end{pmatrix}=2\begin{pmatrix}1\\ 1\end{pmatrix}+\begin{pmatrix}-1\\ 2\end{pmatrix}$$ hence $$\begin{pmatrix}x_{n}\\ y_{n}\end{pmatrix}=A\begin{pmatrix}x_{n-1}\\ y_{n-1}\end{pmatrix}=2A^{n}\begin{pmatrix}1\\ 1\end{pmatrix}+A^{n}\begin{pmatrix}-1\\ 2\end{pmatrix}=2\cdot6^{n}\begin{pmatrix}1\\ 1\end{pmatrix}+3^{n}\begin{pmatrix}-1\\ 2\end{pmatrix}.$$ It follow that $x_{n}=2\cdot6^{n}-3^{n}$.

This is the underlying mathematics that gives you your formula.

Answer 2

$$x_{n+1}=5 x_n+ y_n, ~~~(1); ~~~ y_{n+1}=2 x_n+4 y_n ~~~(2)$$ Add them as $$x_{n+1}+ty_{n+1}=(5+2t) x_n + (1+4t) y_n. ~~~~(3)$$ Now let $$t=\frac{1+4t}{5+2t} \Rightarrow t=-1,1/2 \Rightarrow 5+2t=3,6.$$ From (3) this means $$x_{n+1}-y_{n+1}=3(x_n-y_n) \Rightarrow f_{n+1}=3 f_n~~~(4)$$ $$x_{n+1}+\frac{1}{2} y_{n+1} =6 (x_n+\frac{1}{2}y_n) \Rightarrow g_{n+1}=6g_n~~~(5).$$ The solution of (4) is nothing but a GP: $$f_n= A ~3^n=x_n - y_n~~~(6)$$ and that ogf (5) is $$g_n=B~ 6^n =x_n+\frac{1}{2}y_n, ~~~(7).$$ Using $x_0=1, y_0=4$, you get $A=-3, B=3$ Finally solve On (6) and (7) for $x_n$ and $y_n$ to get $$x_n=2~~ 6^n-3^n,~~~y_n=2~~ 3^n+2 ~~6^n.$$

These simple ideas lie behind this solution.