97 lines
2.7 KiB
TeX
97 lines
2.7 KiB
TeX
% 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}
|