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feat(tstmg): affine cours sur les moyennes
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This commit is contained in:
Bertrand Benjamin
2025-11-04 09:43:17 +01:00
parent c114312941
commit 670a8a1b43
5 changed files with 77 additions and 39 deletions
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Tstmg/05_Suites_et_moyennes/1B_moyennes.pdf
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Tstmg/05_Suites_et_moyennes/1B_moyennes.tex
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@@ -22,21 +22,21 @@
m = \frac{a + b}{2} = (a + b) \times \frac{1}{2} m = \frac{a + b}{2} = (a + b) \times \frac{1}{2}
\] \]
\vspace{1em} % \vspace{1em}
%
\begin{center} % \begin{center}
\begin{tikzpicture}[scale=1.5] % \begin{tikzpicture}[scale=1.5]
\draw[thick,->] (0,0) -- (4,0); % \draw[thick,->] (0,0) -- (4,0);
\draw[thick] (0.5,-0.1) -- (0.5,0.1) node[below=3pt] {$a$}; % \draw[thick] (0.5,-0.1) -- (0.5,0.1) node[below=3pt] {$a$};
\draw[thick] (3.5,-0.1) -- (3.5,0.1) node[below=3pt] {$b$}; % \draw[thick] (3.5,-0.1) -- (3.5,0.1) node[below=3pt] {$b$};
\draw[thick,red] (2,-0.15) -- (2,0.15) node[below=5pt] {$m$}; % \draw[thick,red] (2,-0.15) -- (2,0.15) node[below=5pt] {$m$};
%
\draw[<->] (0.5,0.3) -- (2,0.3) node[midway,above] {\small $\frac{b-a}{2}$}; % \draw[<->] (0.5,0.3) -- (2,0.3) node[midway,above] {\small $\frac{b-a}{2}$};
\draw[<->] (2,0.3) -- (3.5,0.3) node[midway,above] {\small $\frac{b-a}{2}$}; % \draw[<->] (2,0.3) -- (3.5,0.3) node[midway,above] {\small $\frac{b-a}{2}$};
\end{tikzpicture} % \end{tikzpicture}
\end{center} % \end{center}
%
La moyenne arithmétique est au \textbf{milieu} de $a$ et $b$. % La moyenne arithmétique est au \textbf{milieu} de $a$ et $b$.
\end{definition} \end{definition}
@@ -51,27 +51,27 @@
m = \sqrt{a \times b} = (a \times b)^{\frac{1}{2}} m = \sqrt{a \times b} = (a \times b)^{\frac{1}{2}}
\] \]
\vspace{1em} % \vspace{1em}
%
\begin{center} % \begin{center}
\begin{tikzpicture}[scale=0.8] % \begin{tikzpicture}[scale=0.8]
\draw[thick] (0,0) rectangle (4,2); % \draw[thick] (0,0) rectangle (4,2);
\draw[dashed] (0,0) -- (4,2); % \draw[dashed] (0,0) -- (4,2);
\node at (2,-0.5) {$a$}; % \node at (2,-0.5) {$a$};
\node at (-0.5,1) {$b$}; % \node at (-0.5,1) {$b$};
%
\draw[thick,red] (0,0) -- (2.828,2.828); % \draw[thick,red] (0,0) -- (2.828,2.828);
\node[red] at (1.2,1.8) {$m$}; % \node[red] at (1.2,1.8) {$m$};
%
\draw[thick,red] (2.828,0) -- (2.828,2.828); % \draw[thick,red] (2.828,0) -- (2.828,2.828);
\node[red] at (3.3,1.4) {$m$}; % \node[red] at (3.3,1.4) {$m$};
%
\draw[thick] (2.828,-0.1) -- (2.828,0.1); % \draw[thick] (2.828,-0.1) -- (2.828,0.1);
\draw[thick] (4.1,2.828) -- (3.9,2.828); % \draw[thick] (4.1,2.828) -- (3.9,2.828);
\end{tikzpicture} % \end{tikzpicture}
\end{center} % \end{center}
%
La moyenne géométrique est le côté d'un \textbf{carré} d'aire $a \times b$. % La moyenne géométrique est le côté d'un \textbf{carré} d'aire $a \times b$.
\end{definition} \end{definition}
@@ -87,6 +87,35 @@
\end{itemize} \end{itemize}
\afaire{} \afaire{}
\begin{methode}[Taux d'évolution moyen]
\begin{minipage}{0.65\textwidth}
Un quantité subit une transformation global d'un taux d'évolution $t_{global}$.
On souhaite découper cette évolution en deux évolutions identiques alors le taux d'évolution de chacune de ces évolutions se calcule avec
$$ t_{moyen} = \sqrt{(1 + t_{global})} - 1 $$
\end{minipage}
\hfill
\begin{minipage}{0.3\textwidth}
\begin{tikzpicture}[scale=0.8,
roundnode/.style={circle, draw=highlightbg, fill=green!5, very thick, minimum size=3mm},
]
\node[roundnode] (a) {};
\node[roundnode, fill=green!30] (b) [right=1.5cm of a] {};
\node[roundnode] (c) [right=1.5cm of b] {};
\path[->] (a.north) edge [bend left] node [above] {$t_{global}$} (c.north);
\path[->] (a.south) edge [bend right] node [below] {$t_{moyen}$} (b.south);
\path[->] (b.south) edge [bend right] node [below] {$t_{moyen}$} (c.south);
\end{tikzpicture}
\end{minipage}
\end{methode}
\paragraph{Exemple}:~
Le chiffre d'affaire d'une entreprise est passé de 1 million d'euro à 1,5 millions d'euros en 2ans. Quel est le taux d'évolution annuel de son chiffre d'affaire?
\vspace{2cm}
\afaire{}
\end{document} \end{document}

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Tstmg/05_Suites_et_moyennes/2B_suites_moyennes.tex
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@@ -38,6 +38,7 @@
Soit $u$ une suite arithmétique telle que $u(5) = 10$ et $u(7) = 18$. Retrouver la valeur de $u(6)$. Soit $u$ une suite arithmétique telle que $u(5) = 10$ et $u(7) = 18$. Retrouver la valeur de $u(6)$.
\vspace{2cm}
\afaire{} \afaire{}
\begin{propriete}[Suite géométrique] \begin{propriete}[Suite géométrique]
@@ -54,8 +55,8 @@ Soit $u$ une suite arithmétique telle que $u(5) = 10$ et $u(7) = 18$. Retrouver
\node[roundnode, fill=green!30] (b) [right=1.5cm of a] {$u(n)$}; \node[roundnode, fill=green!30] (b) [right=1.5cm of a] {$u(n)$};
\node[roundnode] (c) [right=1.5cm of b] {$u(n+1)$}; \node[roundnode] (c) [right=1.5cm of b] {$u(n+1)$};
\path[->] (a.east) edge [bend left] node [above] {$+r$} (b.west); \path[->] (a.east) edge [bend left] node [above] {$\times q$} (b.west);
\path[->] (b.east) edge [bend left] node [above] {$+r$} (c.west); \path[->] (b.east) edge [bend left] node [above] {$\times q$} (c.west);
\end{tikzpicture} \end{tikzpicture}
\end{minipage} \end{minipage}
\end{propriete} \end{propriete}
@@ -64,6 +65,7 @@ Soit $u$ une suite arithmétique telle que $u(5) = 10$ et $u(7) = 18$. Retrouver
Soit $v$ une suite géométrique telle que $v(6) = 10$ et $u(8) = 40$. Retrouver la valeur de $u(7)$. Soit $v$ une suite géométrique telle que $v(6) = 10$ et $u(8) = 40$. Retrouver la valeur de $u(7)$.
\vspace{2cm}
\afaire{} \afaire{}
\end{document} \end{document}

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tools/style/notbeamer.sty
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@@ -105,3 +105,10 @@
}{ }{
\end{encadre} \end{encadre}
} }
\newenvironment{methode}[1][]
{
\begin{encadre}{Méthode: #1}
}{
\end{encadre}
}