Mon premier Notebook Jupyter.ipynb

{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "def suite1(n):        # Pour Python cette suite u s'appelle suite1. On utilise la méthode def\n",
    "    u = n**2 + 4*n    # On utilise une variable u pour stocker la valeur de chaque terme de la suite\n",
    "    return u          # la méthode return renvoie le contenu de la variable u"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "45"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "suite1(5)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def suite3(n):              # on utilise def et on n'oublie pas les deux points\":\"\n",
    "    w = 2                   # on initialise une variable w à 0 qui est le premier terme de la suite w\n",
    "    for i in range(n):      # on a besoin de calculer tous les termes jusquà w_n pour trouver w_n+1: on utilise une boucle \"for\"\n",
    "        w = 2*w - (i+1)     # ici, à chaque rang, le terme u_n+1 dépend de w_n et aussi directement de n\n",
    "    return w                # on remarque ue le returne est aligné avec le \"for\" et aps avec le \"w\" précédent"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "7"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "suite3(5)                   # la suite démarre avec w_0 donc w_5 est le 6ème terme mais le compteur i démarre bien à 0 !"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "[Fraction(-9, 40), Fraction(-1, 4), Fraction(1, 1), Fraction(7, 8)]"
      ]
     },
     "execution_count": 5,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from fractions import * \n",
    "\n",
    "def f(x):\n",
    "    z = (2*x + 3)/(x**2 + 4)\n",
    "    return z\n",
    "\n",
    "L=[-6,-4,1,2]\n",
    "Frac=[Fraction (i) for i in L]\n",
    "M=[Fraction(f(x)) for x in Frac]\n",
    "M\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hello Anaconda\n"
     ]
    }
   ],
   "source": [
    "print(\"Hello Anaconda\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Hello Michel\n"
     ]
    }
   ],
   "source": [
    "print(\"Hello Michel\")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "23\n"
     ]
    }
   ],
   "source": [
    "a=10\n",
    "b=13\n",
    "print(a+b)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "False"
      ]
     },
     "execution_count": 9,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a=10\n",
    "b=31\n",
    "a==b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "True"
      ]
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a=10\n",
    "b=13\n",
    "a<b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "False"
      ]
     },
     "execution_count": 11,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a>b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "0.5833333333333333"
      ]
     },
     "execution_count": 12,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "1/3+1/4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.5142857142857142"
      ]
     },
     "execution_count": 13,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "5/7+9/5"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.0"
      ]
     },
     "execution_count": 14,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from math import *\n",
    "sqrt(4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# I. Mes premiers pas avec un Notebook"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## I.1.Mes premiers essais de titre"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### premier essai de boucle \"for\""
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "120\n"
     ]
    }
   ],
   "source": [
    "a=0\n",
    "for i in range(16) :\n",
    "    a=a+i\n",
    "print (a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### premier essai de fonction"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [],
   "source": [
    "def factorielle(n) :\n",
    "    fact = 1\n",
    "    for i in range(2,n+1) :\n",
    "        fact=fact*i\n",
    "    return fact\n",
    "    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "24"
      ]
     },
     "execution_count": 17,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "factorielle(4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "3628800"
      ]
     },
     "execution_count": 18,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "factorielle(10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000"
      ]
     },
     "execution_count": 19,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "factorielle(100)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {},
   "outputs": [],
   "source": [
    "#le grand theoreme de Fermat dit qu'il n'y a pas de solutions pour ce problème.\n",
    "for a in range(1,200) :\n",
    "    for b in range(a+1,200) :\n",
    "        for c in range(b+1,200) :\n",
    "            if a**3 + b**3 == c**3 :\n",
    "                print(a,b,c)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
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     ]
    }
   ],
   "source": [
    "for a in range(1,500) :\n",
    "    for b in range(a+1,500) :\n",
    "        for c in range(b+1,500) :\n",
    "            for d in range(c+1,500) :\n",
    "                if a**3 + b**3 + c**3 == d**3 :\n",
    "                    print(a,b,c,d)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Deuxième essai de fonction\n",
    "#### Une fonction somme des n premiers entiers"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def somme_int(n) : \n",
    "    sumint=0\n",
    "    for i in range (1,n+1) :\n",
    "        sumint=sumint+i\n",
    "    return sumint"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "somme_int(5)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "somme_int(1254789)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Troisième et quatrième essais de fonction\n",
    "#### Une fonction somme des n premiers cubes et une fonction somme des n premiers entiers impairs"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def somme_cubes_int(n) : \n",
    "    sumcubint=0\n",
    "    for i in range (1,n+1) :\n",
    "        sumcubint=sumcubint+i**3\n",
    "    return sumcubint"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "somme_cubes_int(5)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "somme_cubes_int(2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def somme_odd_int(n) : \n",
    "    sumoddint=0\n",
    "    for i in range (n) :\n",
    "        sumoddint=sumoddint+2*i+1\n",
    "    return sumoddint"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "somme_odd_int(3)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Fonctions Géométriques \n",
    "- une fonction qui calcule le périmètre d'un triangle, l'aire d'un triangle\n",
    "- une fonction qui teste l'inégalité triangulaire ? qui <font color='green'>d'abord sait quelles sont les plus petites longueurs...</font> \n",
    "- une fonction de prof qui, pour une distance donnée, retourne les quatre coordonnées entières de deux points A et B \n",
    "\n",
    "\n",
    "### Fonctions Arithmétiques\n",
    "- une fonction pgcd ?  <font color='red'>avec une liste de restes de division euclidienne ?</font>\n",
    "- une fonction ppcm (qui utiliserait la fonction pgcd) ?\n",
    "- une fonction qui donne les coefficients de Bezout ? avec une liste de restes de division euclidienne\n",
    "- une fonction qui effectue la décomposition en facteurs premiers ?\n",
    "- une fonction qui donne la liste des diviseurs ?\n",
    "\n",
    "### Suites Numériques\n",
    "- Suite de Fibonacci  <font color='red'>OK DONE</font>\n",
    "\n",
    "### Autres\n",
    "- puissances de matrices? suites de matrices ? ancien programme de spé ?\n",
    "- nombres complexes ?\n",
    "- resolution de l'équation du second degré réelle ?\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Comment obtenir : \n",
    "\n",
    "- passer de la valeur exacte à la valeur décimale avec une précision donnée"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a=3\n",
    "b=2\n",
    "a/b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a=3\n",
    "b=2\n",
    "a//b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    ">>> from math import fabs, gcd, floor\n",
    "a=-3.5\n",
    "b=7\n",
    "c=9\n",
    "print(fabs(a))\n",
    "print(floor(a))\n",
    "print(gcd(b,c))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    ">>> from math import fabs\n",
    "def max(x,y) :\n",
    "    m=(fabs(x-y)+x+y)/2\n",
    "    return m\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "max(-3,15)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "max(42,5)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "max(-78,-96)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "1/3"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "13/17"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "%precision 20"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "1/3"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Problème d'erreur à partir de la 16ème décimale. Donc la commande \"%precision 20\" ne modifie que l'affichage"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    ">>> from math import *\n",
    "floor(pi)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Triangle de Pascal"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "n=int(input(\"Entrez le nombre de lignes du triangle de Pascal: \"))\n",
    "x=[]\n",
    "for i in range(n):\n",
    "    x.append([])\n",
    "    x[i].append(1)\n",
    "    for j in range(1,i):\n",
    "        x[i].append(x[i-1][j-1]+x[i-1][j])\n",
    "    if(n!=0):\n",
    "        x[i].append(1)\n",
    "for i in range(n):\n",
    "    print(\"   \"*(n-i),end=\" \",sep=\" \")\n",
    "    for j in range(0,i+1):\n",
    "        print('{0:6}'.format(x[i][j]),end=\" \",sep=\" \")\n",
    "    print()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "n=int(input(\"Entrez le nombre de lignes du triangle de Pascal: \"))\n",
    "x=[]\n",
    "for i in range(n):\n",
    "    x.append([])\n",
    "    x[i].append(1)\n",
    "    for j in range(1,i):\n",
    "        x[i].append(x[i-1][j-1]+x[i-1][j])\n",
    "    if(n!=0):\n",
    "        x[i].append(1)\n",
    "for i in range(n):\n",
    "    print(\"   \"*(n-i))\n",
    "    for j in range(0,i+1):\n",
    "        print('{0:6}'.format(x[i][j]),end=\"\")\n",
    "    print()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    ">>> a = 0\n",
    ">>> while (a < 7):\n",
    "...    a = a + 1\n",
    "...    print(a)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fonctions Numériques\n",
    "## Des fonctions pour augmenter la précision des résultats \n",
    "On veut plus un plus grand nombre de \"décimales\" :\n",
    "- pour les fractions et ainsi, observer la périodicité\n",
    "- pour les irrationnels :\n",
    "    - racine carrée des entiers premiers\n",
    "    - Pi, e et gamma,..."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# on demande 1 000 décimales pour la fraction 1/7\n",
    "\n",
    "from decimal import *\n",
    "getcontext().prec = 1000\n",
    "f=Decimal(1)/Decimal(7)\n",
    "print(f)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# même chose avec la fraction 13/17\n",
    "\n",
    "from decimal import *\n",
    "getcontext().prec = 1000\n",
    "f=Decimal(13)/Decimal(17)\n",
    "print(f)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# même chose avec $\\sqrt{2}$\n",
    "\n",
    "from decimal import *\n",
    "from math import *\n",
    "getcontext().prec = 1000\n",
    "print(Decimal(2).sqrt())"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "sqrt(4)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Fonctions géométriques"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# la fonction qui permet de calculer les coordonnées du milieu d'un segment\n",
    "\n",
    "import math\n",
    "\n",
    "xA = int(input(\"Donnez moi xA = \"))\n",
    "yA = int(input(\"Donnez moi yB = \"))\n",
    "xB = int(input(\"Donnez moi xB = \"))\n",
    "yB = int(input(\"Donnez moi yB = \"))\n",
    "\n",
    "distance = math.sqrt((xB - xA)**2 + (yB - yA)**2)\n",
    "\n",
    "print(\"Bonjour, la distance AB vaut \",distance)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# La fonction qui permet de calculer la distance entre deux points\n",
    "\n",
    "import math\n",
    "L=[]\n",
    "\n",
    "xA = int(input(\"Donnez moi xA = \"))\n",
    "yA = int(input(\"Donnez moi yB = \"))\n",
    "xB = int(input(\"Donnez moi xB = \"))\n",
    "yB = int(input(\"Donnez moi yB = \"))\n",
    "\n",
    "xI=(xA+xB)/2\n",
    "yI=(yA+yB)/2\n",
    "L=[xI,yI]\n",
    "\n",
    "print(\"Bonjour, les coordonnées du milieu I du segment [AB] sont \", L)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Suites"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def u(n) :\n",
    "    u=1+((-1)**n)/n\n",
    "    return u"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "u(10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "n=20\n",
    "for i in range (1,n+1) :\n",
    "    print (u(i))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Suites récurrentes\n",
    "### Suites définies avec \"range\" comme fonction d'un indice\n",
    "#### Deux suites récurrentes qui convergent vers le Nombre d'Or"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def goldenratio_1(n) :\n",
    "    x=1\n",
    "    for i in range(n) :\n",
    "        x=1+1/x\n",
    "    print(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "goldenratio_1(40)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def goldenratio_2(n) :\n",
    "    x=1\n",
    "    for i in range(n) :\n",
    "        x=sqrt(1+x)\n",
    "    print(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "goldenratio_2(40)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Méthode de Héron pour $\\sqrt{2}$ et $\\sqrt{7}$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def racinededeux(n) :\n",
    "    x=1\n",
    "    for i in range(n) :\n",
    "        x=(1/2)*(x+2/x)\n",
    "        print(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "racinededeux(20)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def racinedesept(n) :\n",
    "    x=1\n",
    "    for i in range(n) :\n",
    "        x=(1/2)*(x+7/x)\n",
    "        print(x)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "racinedesept(20)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Fonctions Arithmétiques : autour des triplets Pythagoriciens\n",
    "Un triplet (a,b,c) d'entiers naturels non nuls est dit Pythagoricien si il satisfait aux deux conditions $a\\leq b \\leq c$ et $a^2+b^2=c^2$"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Triplets Pythagoriciens\n",
    "\n",
    "for i in range (1,101) :\n",
    "    for j in range(i,101) :\n",
    "        for k in range (j,101) :\n",
    "            if i**2 + j**2 == k**2 :\n",
    "                print (i,j,k)\n",
    "                                            # définir une liste puis :\n",
    "                                                # retirer de la liste tous les triplets \"multiples\"\n",
    "                                                # présenter le résultat sous la forme d'un tableau de 6 colonnes\n",
    "                                                # ranger par ordre croissant la suite des \"k\"\n",
    "                                            # cette fonction permet aussi de trouver des couples de points A et B \n",
    "                                            # à coordonnées entières tels que AB soit égale à un entier donné"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# un algorithme qui permet de déterminer les triplets pythagoriciens de la forme (n, n+1, k)\n",
    "\n",
    "from math import *\n",
    "for k in range (2,100000) :\n",
    "    x=(-1+sqrt(2*k**2-1))/2\n",
    "    if x-floor(x)==0 :\n",
    "        print(floor(x),floor(x)+1,k)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# un algorithme qui permet de déterminer les triplets pythagoriciens de la forme (k, n, n+1)\n",
    "\n",
    "from math import *\n",
    "for k in range (1,10000) :\n",
    "    x=(sqrt(2*k+1))\n",
    "    if x-floor(x)==0 :\n",
    "        print(floor(x),k,k+1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# une fonction qui donne l'écriture binaire d'un nombre entier donné\n",
    "\n",
    "def ecriturebinairede(n):\n",
    "    L=[]\n",
    "    L.append(n%2)\n",
    "    nombre=n\n",
    "    if n==0:\n",
    "        L=[0]\n",
    "    else :\n",
    "        N=int(1+floor(log2(n)))\n",
    "        while nombre!=0 and len(L)<=N-1:\n",
    "            quotient=nombre//2\n",
    "            reste=quotient%2\n",
    "            L.append(reste)\n",
    "            nombre=quotient\n",
    "    L.reverse()\n",
    "    return (L)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(9)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(11)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(2)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(100)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(0)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(1)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "ecriturebinairede(1024)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "bin(256)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "bin(1024)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "bin(11)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def test_primalité (n):\n",
    "    testprime=True\n",
    "    for k in range (2,n):\n",
    "        if n%k==0:\n",
    "            testprime=False\n",
    "    return(testprime)\n",
    "\n",
    "n=int(input(\"afficher l'écriture binaire de tous les nombres premiers inférieurs à:\"))\n",
    "M=[]\n",
    "L=[i for i in range (2,n+1) if test_primalité(i)==True]\n",
    "\n",
    "print(\"la liste des écritures binaires de tous les nombres premiers inférieurs à\",n, \"est :\", L)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "def test_primalité (n):\n",
    "    testprime=True\n",
    "    for k in range (2,n):\n",
    "        if n%k==0:\n",
    "            testprime=False\n",
    "    return(testprime)\n",
    "\n",
    "n=int(input(\"afficher l'écriture binaire de tous les nombres premiers inférieurs à:\"))\n",
    "M=[]\n",
    "L=[i for i in range (2,n+1) if test_primalité(i)==True]\n",
    "\n",
    "for i in L :\n",
    "    print(bin(i))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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