We define the sequences $(u_n)$ and $(v_n)$ on the set $\mathbb{N}$ of natural numbers by:
$$u_0 = 0 ; v_0 = 1, \text{ and } \left\{\begin{array}{l} u_{n+1} = \dfrac{u_n + v_n}{2} \\ v_{n+1} = \dfrac{u_n + 2v_n}{3} \end{array}\right.$$

The purpose of this exercise is to study the convergence of sequences $(u_n)$ and $(v_n)$.

\begin{enumerate}
  \item Calculate $u_1$ and $v_1$.
  \item We consider the following algorithm:

Variables: $u$, $v$ and $w$ real numbers; $N$ and $k$ integers\\
Initialization: $u$ takes the value 0; $v$ takes the value 1\\
Start of algorithm\\
Enter the value of $N$\\
For $k$ varying from 1 to $N$\\
\quad $w$ takes the value $u$\\
\quad $u$ takes the value $\dfrac{w + v}{2}$\\
\quad $v$ takes the value $\dfrac{w + 2v}{3}$\\
End of For\\
Display $u$\\
Display $v$\\
End of algorithm

a. We execute this algorithm by entering $N = 2$. Copy and complete the table given below containing the state of variables during the execution of the algorithm.

\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
$k$ & $w$ & $u$ & $v$ \\
\hline
1 & & & \\
\hline
2 & & & \\
\hline
\end{tabular}
\end{center}

b. For a given number $N$, what do the values displayed by the algorithm correspond to with respect to the situation studied in this exercise?

  \item For all natural numbers $n$ we define the column vector $X_n$ by $X_n = \binom{u_n}{v_n}$ and the matrix $A$ by
$$A = \left(\begin{array}{ll} \dfrac{1}{2} & \dfrac{1}{2} \\ \dfrac{1}{3} & \dfrac{2}{3} \end{array}\right).$$
a. Verify that, for all natural numbers $n$, $X_{n+1} = A X_n$.\\
b. Prove by induction that $X_n = A^n X_0$ for all natural numbers $n$.

  \item We define matrices $P$, $P'$ and $B$ by
$$P = \left(\begin{array}{cc} \dfrac{4}{5} & \dfrac{6}{5} \\ -\dfrac{6}{5} & \dfrac{6}{5} \end{array}\right), \quad P' = \left(\begin{array}{cc} \dfrac{1}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{1}{3} \end{array}\right)$$
\end{enumerate}