{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Syntaxe Latex"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 1"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{align}\n",
    "\\dot{x} & = \\sigma(y-x) \\\\\n",
    "\\dot{y} & = \\rho x - y - xz \\\\\n",
    "\\dot{z} & = -\\beta z + xy\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{align}\n",
    "\\dot{x} & = \\sigma(y-x) \\\\\n",
    "\\dot{y} & = \\rho x - y - xz \\\\\n",
    "\\dot{z} & = -\\beta z + xy\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 2"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\left( \\sum_{k=1}^n a_k b_k \\right)^2 \\leq \\left( \\sum_{k=1}^n a_k^2 \\right) \\left( \\sum_{k=1}^n b_k^2 \\right)\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\left( \\sum_{k=1}^n a_k b_k \\right)^2 \\leq \\left( \\sum_{k=1}^n a_k^2 \\right) \\left( \\sum_{k=1}^n b_k^2 \\right)\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\mathbf{V}_1 \\times \\mathbf{V}_2 =  \\begin{vmatrix}\n",
    "\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n",
    "\\frac{\\partial X}{\\partial u} &  \\frac{\\partial Y}{\\partial u} & 0 \\\\\n",
    "\\frac{\\partial X}{\\partial v} &  \\frac{\\partial Y}{\\partial v} & 0\n",
    "\\end{vmatrix}\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\mathbf{V}_1 \\times \\mathbf{V}_2 =  \\begin{vmatrix}\n",
    "\\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\\n",
    "\\frac{\\partial X}{\\partial u} &  \\frac{\\partial Y}{\\partial u} & 0 \\\\\n",
    "\\frac{\\partial X}{\\partial v} &  \\frac{\\partial Y}{\\partial v} & 0\n",
    "\\end{vmatrix}\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 4"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "P(E)   = {n \\choose k} p^k (1-p)^{ n-k}\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "P(E)   = {n \\choose k} p^k (1-p)^{ n-k}\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\frac{1}{\\Bigl(\\sqrt{\\phi \\sqrt{5}}-\\phi\\Bigr) e^{\\frac25 \\pi}} =\n",
    "1+\\frac{e^{-2\\pi}} {1+\\frac{e^{-4\\pi}} {1+\\frac{e^{-6\\pi}}\n",
    "{1+\\frac{e^{-8\\pi}} {1+\\ldots} } } }\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "\\frac{1}{\\Bigl(\\sqrt{\\phi \\sqrt{5}}-\\phi\\Bigr) e^{\\frac25 \\pi}} =\n",
    "1+\\frac{e^{-2\\pi}} {1+\\frac{e^{-4\\pi}} {1+\\frac{e^{-6\\pi}}\n",
    "{1+\\frac{e^{-8\\pi}} {1+\\ldots} } } }\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 6"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "1 +  \\frac{q^2}{(1-q)}+\\frac{q^6}{(1-q)(1-q^2)}+\\cdots =\n",
    "\\prod_{j=0}^{\\infty}\\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},\n",
    "\\quad\\quad \\text{for $|q|<1$}.\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{equation*}\n",
    "1 +  \\frac{q^2}{(1-q)}+\\frac{q^6}{(1-q)(1-q^2)}+\\cdots =\n",
    "\\prod_{j=0}^{\\infty}\\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},\n",
    "\\quad\\quad \\text{for $|q|<1$}.\n",
    "\\end{equation*}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 7"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "\\begin{align}\n",
    "\\nabla \\times \\vec{\\mathbf{B}} -\\, \\frac1c\\, \\frac{\\partial\\vec{\\mathbf{E}}}{\\partial t} & = \\frac{4\\pi}{c}\\vec{\\mathbf{j}} \\\\   \\nabla \\cdot \\vec{\\mathbf{E}} & = 4 \\pi \\rho \\\\\n",
    "\\nabla \\times \\vec{\\mathbf{E}}\\, +\\, \\frac1c\\, \\frac{\\partial\\vec{\\mathbf{B}}}{\\partial t} & = \\vec{\\mathbf{0}} \\\\\n",
    "\\nabla \\cdot \\vec{\\mathbf{B}} & = 0\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\\begin{align}\n",
    "\\nabla \\times \\vec{\\mathbf{B}} -\\, \\frac1c\\, \\frac{\\partial\\vec{\\mathbf{E}}}{\\partial t} & = \\frac{4\\pi}{c}\\vec{\\mathbf{j}} \\\\   \\nabla \\cdot \\vec{\\mathbf{E}} & = 4 \\pi \\rho \\\\\n",
    "\\nabla \\times \\vec{\\mathbf{E}}\\, +\\, \\frac1c\\, \\frac{\\partial\\vec{\\mathbf{B}}}{\\partial t} & = \\vec{\\mathbf{0}} \\\\\n",
    "\\nabla \\cdot \\vec{\\mathbf{B}} & = 0\n",
    "\\end{align}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Exemple 8"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Code"
   ]
  },
  {
   "cell_type": "raw",
   "metadata": {},
   "source": [
    "This expression $\\sqrt{3x-1}+(1+x)^2$ is an example of a TeX inline equation in a [Markdown-formatted](https://daringfireball.net/projects/markdown/) sentence."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Résultat"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This expression $\\sqrt{3x-1}+(1+x)^2$ is an example of a TeX inline equation in a [Markdown-formatted](https://daringfireball.net/projects/markdown/) sentence."
   ]
  }
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