A laboratory studies the spread of a disease in a population. A healthy individual is an individual who has never been affected by the disease. A sick individual is an individual who has been affected by the disease and is not cured. A recovered individual is an individual who has been affected by the disease and has recovered. Once recovered, an individual is immunized and cannot become sick again.

The first observations show that, from one day to the next:
\begin{itemize}
  \item $5\%$ of individuals become sick;
  \item $20\%$ of individuals recover.
\end{itemize}

For every natural number $n$, we denote by $a_n$ the proportion of healthy individuals $n$ days after the start of the experiment, $b_n$ the proportion of sick individuals $n$ days after the start of the experiment, and $c_n$ that of recovered individuals $n$ days after the start of the experiment.\\
We assume that at the start of the experiment, all individuals are healthy, that is $a_0 = 1$, $b_0 = 0$ and $c_0 = 0$.

\begin{enumerate}
  \item Calculate $a_1$, $b_1$ and $c_1$.
  \item a) What is the proportion of healthy individuals who remain healthy from one day to the next? Deduce $a_{n+1}$ as a function of $a_n$.\\
b) Express $b_{n+1}$ as a function of $a_n$ and $b_n$.
\end{enumerate}

We admit that $c_{n+1} = 0.2b_n + c_n$.\\
For every natural number $n$, we define $U_n = \left(\begin{array}{c} a_n \\ b_n \\ c_n \end{array}\right)$\\
We define the matrices $A = \left(\begin{array}{ccc} 0.95 & 0 & 0 \\ 0.05 & 0.8 & 0 \\ 0 & 0.2 & 1 \end{array}\right)$ and $D = \left(\begin{array}{ccc} 0.95 & 0 & 0 \\ 0 & 0.8 & 0 \\ 0 & 0 & 1 \end{array}\right)$\\
We admit that there exists an invertible matrix $P$ such that $D = P^{-1} \times A \times P$ and that, for every natural number $n$ greater than or equal to 1, $A^n = P \times D^n \times P^{-1}$.

\begin{enumerate}
  \setcounter{enumi}{2}
  \item a) Verify that, for every natural number $n$, $U_{n+1} = A \times U_n$.\\
We admit that, for every natural number $n$, $U_n = A^n \times U_0$.\\
b) Prove by induction that, for every non-zero natural number $n$,
$$D^n = \left(\begin{array}{ccc} 
0.95^n & 0 & 0 \\
0 & 0.8^n & 0 \\
0 & 0 & 1
\end{array}\right)$$
\end{enumerate}

We admit that $A^n = \left(\begin{array}{ccc} 0.95^n & 0 & 0 \\ \frac{1}{3}(0.95^n - 0.8^n) & 0.8^n & 0 \\ \frac{1}{3}(3 - 4 \times 0.95^n + 0.8^n) & 1 - 0.8^n & 1 \end{array}\right)$

\begin{enumerate}
  \setcounter{enumi}{3}
  \item a) Verify that for every natural number $n$, $b_n = \frac{1}{3}(0.95^n - 0.8^n)$\\
b) Determine the limit of the sequence $(b_n)$.\\
c) We admit that the proportion of sick individuals increases for several days, then decreases. We wish to determine the epidemic peak, that is, the moment when the proportion of sick individuals is at its maximum. To this end, we use the algorithm given in appendix 2 (to be returned with the answer sheet), in which we compare successive terms of the sequence $(b_n)$. Complete the algorithm so that it displays the rank of the day when the epidemic peak is reached and complete the table provided in appendix 2. Conclude.
\end{enumerate}