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snippets/tpl_fonctions.tex

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+% vim:ft=tex:
+%
+\documentclass[12pt]{article}
+\usepackage[utf8x]{inputenc}
+\usepackage[francais]{babel}
+\usepackage[T1]{fontenc}
+\usepackage{amssymb}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+
+
+\title{
+    Snippets pour Opytex \\
+    Fonctions
+}
+\author{
+    Benjamin Bertrand
+}
+
+
+\begin{document}
+\maketitle
+
+\section{Calculer des images}
+\begin{enumerate}
+    %-set f = Expression.random("{a}*x^2 + {b}*x + {c}")
+    \item $\forall x \in \mathbb{R} \qquad f(x) = \Var{f}$
+
+        Solution:
+        \begin{align*}
+            f(0) &= \Var{f(0).explain() | join('=')} \\
+            f(1) &= \Var{f(1).explain() | join('=')} \\
+            f(2) &= \Var{f(2).explain() | join('=')} \\
+            f({10}) &= \Var{f(10).explain() | join('=')} \\
+            f({100}) &= \Var{f(100).explain() | join('=')} 
+        \end{align*}
+\end{enumerate}
+
+\section{Résolution d'équation du 2nd degré}
+%- macro solveEquation(P)
+
+    On commence par calculer le discriminant de $P(x) = \Var{P}$.
+    \begin{eqnarray*}
+        \Delta & = & b^2-4ac \\
+        \Var{P.delta.explain()|calculus(name="\\Delta")}
+    \end{eqnarray*}
+
+    \Block{if P.delta > 0}
+    comme $\Delta = \Var{P.delta} > 0$ donc $P$ a deux racines
+
+    \begin{eqnarray*}
+        x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} =  \frac{\Var{-P.b} - \sqrt{\Var{P.delta}}}{2 \times \Var{P.a}} = \Var{P.roots[0] } \\
+        x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} =  \frac{\Var{-P.b} + \sqrt{\Var{P.delta}}}{2 \times \Var{P.a}} = \Var{P.roots[1] }
+    \end{eqnarray*}
+
+    Les solutions de l'équation $\Var{P} = 0$ sont donc $\mathcal{S} = \left\{ \Var{P.roots[0]}; \Var{P.roots[1]} \right\}$
+
+    \Block{elif P.delta == 0}
+    Comme $\Delta = 0$ donc $P$ a une racine
+
+    \begin{eqnarray*}
+        x_1 = \frac{-b}{2a} = \frac{-\Var{P.b}}{2\times \Var{P.a}} = \Var{P.roots[0]} \\
+    \end{eqnarray*}
+
+    La solution de $\Var{P} = 0$ est donc $\mathcal{S} = \left\{ \Var{P.roots[0]}\right\}$
+
+    \Block{else}
+    Alors $\Delta = \Var{P.delta} < 0$ donc $P$ n'a pas de racine donc l'équation $\Var{P} = 0$ n'a pas de solution.
+
+    \Block{endif}
+%- endmacro
+
+\begin{enumerate}
+    %-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c>0"])
+    \item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
+
+        Solution:
+
+        \Var{solveEquation(P)}
+
+    %-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c==0"])
+    \item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
+
+        Solution:
+
+        \Var{solveEquation(P)}
+
+    %-set P = Expression.random("{a}*x^2 + {b}*x + {c}", ["b**2-4*a*c<0"])
+    \item Étude du polynôme $P$, $\forall x \in \mathbb{R} \quad P(x) = \Var{P}$
+
+        Solution:
+
+        \Var{solveEquation(P)}
+
+\end{enumerate}
+\end{document}

snippets/tpl_suite.tex

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+% vim:ft=tex:
+%
+\documentclass[12pt]{article}
+\usepackage[utf8x]{inputenc}
+\usepackage[francais]{babel}
+\usepackage[T1]{fontenc}
+\usepackage{amssymb}
+\usepackage{amsmath}
+\usepackage{amsfonts}
+
+
+\title{
+    Snippets pour Opytex \\
+    Suites
+}
+\author{
+    Benjamin Bertrand
+}
+
+\begin{document}
+\maketitle
+
+\section{Calculs de termes}
+\begin{enumerate}
+    \item Calculer les termes $u_0$, $u_1$, $u_2$, $u_{10}$ et $u_{100}$ pour les suites suivantes
+        \begin{enumerate}
+            %-set u = Expression.random("{a}*n+{b}")
+            \item $\forall n \in \mathbb{N} \qquad u_n = \Var{u}$
+
+                Solution:
+                \begin{align*}
+                    u_0 &= \Var{u(0).explain() | join('=')} \\
+                    u_1 &= \Var{u(1).explain() | join('=')} \\
+                    u_2 &= \Var{u(2).explain() | join('=')} \\
+                    u_{10} &= \Var{u(10).explain() | join('=')} \\
+                    u_{100} &= \Var{u(100).explain() | join('=')} 
+                \end{align*}
+
+                %-set v = Expression.random("({a}*n+{b})/{c}", ["c>1"])
+            \item $\forall n \in \mathbb{N} \qquad v_n = \Var{v|replace("frac","dfrac")}$
+
+                Solution:
+                \begin{align*}
+                    v_0 &= \Var{v(0).explain() | join('=')} \\
+                    v_1 &= \Var{v(1).explain() | join('=')} \\
+                    v_2 &= \Var{v(2).explain() | join('=')} \\
+                    v_{10} &= \Var{v(10).explain() | join('=')} \\
+                    v_{100} &= \Var{v(100).explain() | join('=')} 
+                \end{align*}
+
+                %-set v = Expression.random("({a}*n+{b})/{c}", ["c>1"])
+            \item $\forall n \in \mathbb{N} \qquad v_n = \Var{v}$
+
+                Solution:
+                \begin{align*}
+                    %- for j in [0, 1, 2, 10, 100]
+                    v_{\Var{j}} &= \Var{v(j).explain() | join('=')} \\
+                    %- endfor
+                \end{align*}
+
+                %-set f = Expression.random("{a}*x")
+                %-set v0 = randint(0, 10)
+            \item $\forall n \in \mathbb{N} \qquad v_{n+1} = \Var{f("v_n")} \mbox{ et } v_0 = \Var{v0}$
+
+                Solution:
+                \begin{align*}
+                    v_0 &= \Var{v0} \\
+                    %-set v = f(v0)
+                    v_1 &= \Var{v.explain() | join('=')} \\
+                    %-set v = f(v)
+                    v_2 &= \Var{v.explain() | join('=')} \\
+                \end{align*}
+                Pour le terme 10, il faut calculer tous les autres avant!
+                \begin{align*}
+                    %#- Trick to move around scoping rules
+                    %#- https://stackoverflow.com/a/49699589
+                    %- set v = namespace(val = v)
+                    %- for i in range(8)
+                    %- set v.val = f(v.val)
+                    v_{\Var{i+3}} &= \Var{v.val.explain() | join('=')} \\
+                    %- endfor
+                \end{align*}
+                
+        \end{enumerate}
+
+\end{enumerate}
+\end{document}