OK. Midway through post #4

So your matrix equation $\displaystyle

\bf{x}_{n+1} = \mathbb{A} \bf{x}_n

$ becomes

Quote:

Originally Posted by

**garymarkhov** What about the $\displaystyle y_{n}$?

The $\displaystyle \begin{color}{blue}y_n \end{color}$

is here. We define the vector $\displaystyle \begin{color}{blue}\bf{x_n} =

\left( \begin{array}{c} x_n\\ y_n \end{array} \right)\end{color}$

$\displaystyle \mathbb{T}^{-1} \bf{x}_{n+1} = \mathbb{D}\; \mathbb{T}^{-1} \bf{x}_n$.

If you let

$\displaystyle

\bf{u}_{n} = \mathbb{T}^{-1} \bf{x}_n

$

Quote:

Originally Posted by

**garymarkhov** Why are we doing this?

In order to de-couple the system. Notice that
$\displaystyle \begin{color}{blue} \underbrace{\mathbb{T}^{-1} \bf{x}_{n+1}}_{\bf{u_{n+1}}} = \mathbb{D}\; \underbrace{\mathbb{T}^{-1} \bf{x}_n}_{\bf{u_n}}\end{color}$.

then your sysem reduces to
$\displaystyle

\begin{color}{blue} \bf{u}_{n+1} = \mathbb{D} \bf{u}_n \end{color}

$

a de-coupled system. Quote:

Originally Posted by

**garymarkhov** What does that mean?

A de-coupled system is when the equations don't depend on one another. Here
$\displaystyle

\begin{color}{blue} \bf{u}_{n+1} = \mathbb{D} \bf{u}_n \end{color}

$

gives
$\displaystyle \begin{color}{blue} \left( \begin{array}{c} u_{n+1}\\v_{n+1} \end{array}\right) = \left( \begin{array}{cc} -1 & 0 \\ 0 & 2 \end{array} \right) \left( \begin{array}{c} u_n\\v_n \end{array}\right) \end{color}$

or expanding
$\displaystyle \begin{color}{blue}u_{n+1} = - u_n,\;\;\;v_{n+1} = 2 v_n\end{color}$

Each are easily solved, these give

$\displaystyle u_n = u_0 (-1)^n,\;\;\;v_n = v_0 2^n$, where $\displaystyle u_0$ and $\displaystyle v_0$ are constants.

Quote:

Originally Posted by

**garymarkhov** Why are we solving for those things?

We de-coupled the system by introucing a change of variables. We now need to get back to the original variables. This is where the
$\displaystyle

\begin{color}{blue} \bf{u}_{n} = \mathbb{T}^{-1} \bf{x}_n \end{color}

$

or $\displaystyle

\begin{color}{blue} \bf{x}_{n} = \mathbb{T} \bf{u}_n \end{color}

$

comes in - it related that two sets of variables. The $\displaystyle \begin{color}{blue} x_n\end{color}$

and $\displaystyle \begin{color}{blue} y_n\end{color} $

and the $\displaystyle \begin{color}{blue} u_n\end{color} $

and $\displaystyle \begin{color}{blue} v_n\end{color} $.