{"id":3189,"date":"2020-08-20T14:36:56","date_gmt":"2020-08-20T12:36:56","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?page_id=3189"},"modified":"2023-04-16T16:18:17","modified_gmt":"2023-04-16T14:18:17","slug":"encadrement-de-sqrt2-par-balayage-en-python","status":"publish","type":"page","link":"https:\/\/www.mathweb.fr\/euclide\/encadrement-de-sqrt2-par-balayage-en-python\/","title":{"rendered":"Encadrement de \\(\\sqrt2\\) par balayage en Python"},"content":{"rendered":"\n

En classe de Seconde, le programme stipule que l’on doit savoir obtenir un encadrement de \\(\\sqrt2\\) par balayage \u00e0 l’aide de Python.<\/p>\n\n\n\n

Nous allons voir sur cette page l’id\u00e9e qu’il y a derri\u00e8re cette op\u00e9ration et le script Python.<\/p>\n\n\n

\n
\"encadrement<\/figure>\n<\/div>\n\n\n

Le principe math\u00e9matique<\/h2>\n\n\n\n

On sait que si \\(0 < a < r < b\\) alors \\(0 < a^2 < r^2 < b^2\\).<\/p>\n\n\n\n

On cherche deux nombres a<\/em> et b<\/em> tels que:$$a < \\sqrt2 < b$$ donc tels que:$$a^2 < (\\sqrt2)^2 < b^2.$$<\/p>\n\n\n\n

De plus, on sait que $$1 < 2 < 3$$donc l’id\u00e9e est de partir de \\(a=\\sqrt1=1\\) et de lui ajouter un pas tr\u00e8s petit, par exemple \\(10^{-n}\\) o\u00f9 n<\/em> est un entier naturel, jusqu’\u00e0 obtenir:$$a^2 < 2 < (a+10^{-n})^2.$$<\/p>\n\n\n\n

Un exemple pas \u00e0 pas<\/h2>\n\n\n\n

Posons a<\/em> = 1 et b<\/em> = a + 0,1. On calcule ensuite a<\/em>\u00b2 et b<\/em>\u00b2 et on regarde si a<\/em>\u00b2 < 2 < b<\/em>\u00b2. On a a<\/em>\u00b2 = 1 et b<\/em>\u00b2 = 1,1\u00b2 = 1,21 donc 2 n’est pas compris entre a<\/em>\u00b2 et b<\/em>\u00b2.<\/p>\n\n\n\n

Dans ce cas, on pose a<\/em> = b<\/em> = 1,1 puis b<\/em> = a<\/em> + 0,1 = 1,2 et on calcule : a<\/em>\u00b2 = 1,21 et b<\/em>\u00b2 = 1,44. \u00ab\u00a02\u00a0\u00bb n’est pas compris entre a<\/em>\u00b2 et b<\/em>\u00b2 donc on continue.<\/p>\n\n\n\n

On pose a<\/em> = b<\/em> = 1,2 et b<\/em> = a<\/em> + 1 = 1,3… On r\u00e9sume cela dans un tableau:<\/p>\n\n\n\n

Valeurs de a<\/em><\/td>1<\/td>1,1<\/td>1,2<\/td>1,3<\/td>1,4<\/td><\/tr>
Valeurs de b<\/em><\/td>1,1<\/td>1,2<\/td>1,3<\/td>1,4<\/td>1,5<\/td><\/tr>
Valeurs de a<\/em>\u00b2<\/td>1<\/td>1,21<\/td>1,44<\/td>1,69<\/td>1,96<\/td><\/tr>
Valeurs de b<\/em>\u00b2<\/td>1,21<\/td>1,44<\/td>1,69<\/td>1,96<\/td>2,25<\/td><\/tr>
Est-ce que a<\/em>\u00b2 < 2 < b<\/em>\u00b2 ?<\/td>non<\/td>non<\/td>non<\/td>non<\/td>oui<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

On s’arr\u00eate donc lorsque a<\/em> = 1,4 et b<\/em> = 1,5, ce qui signifie que:$$1,4 < \\sqrt2 < 1,5.$$<\/p>\n\n\n\n

Obtenir un encadrement par balayage en Python : deux programmes<\/h2>\n\n\n\n
def approximation(n):\n    a = 1\n    while ((a+10**(-n))**2 < 2):\n        a = a + 10**(-n)\n        \n    return round(a,n) , round(a+10**(-n),n)\n\np , q = approximation(5)\nprint(f'{p} < racine(2) < {q}')\n<\/pre>\n\n\n\n
Expliquons ce programme.<\/p>\n\n\n\n
J’ai d\u00e9fini une fonction approximation<\/em> admettant un nombre en argument : n<\/em>. Ce nombre va d\u00e9signer l’amplitude de l’encadrement souhait\u00e9, c’est-\u00e0-dire la diff\u00e9rence entre les deux bornes de l’encadrement.<\/p>\n\n\n\n
Dans cette fonction, j’ai affect\u00e9 \u00e0 la variable a<\/em> la valeur 1 car on commence \u00e0 1 (comme dans l’exemple pr\u00e9c\u00e9dent). Je vais ajout\u00e9 aux diff\u00e9rentes valeurs de a<\/em> le nombre \\(10^{-n}\\), que l’on \u00e9crit en python : 10**(-n). Dans l’exemple pr\u00e9c\u00e9dent, j’ajoutais 0,1 qui correspond  \u00e0 \\(10^{-1}\\).<\/p>\n\n\n\n
Tant que (a<\/em> + \\(10^{-n}\\)) <\/em>\u00b2 est plus petit que 2, cela signifie que je n’ai pas encore obtenu mon encadrement, donc je continue \u00e0 ajouter \\(10^{-n}\\) \u00e0 a<\/em>.<\/p>\n\n\n\n
La boucle while<\/em> s’arr\u00eate quand (a<\/em> + \\(10^{-n}\\))\u00b2 > 2. Dans ce cas, la fonction approximation <\/em>retourne deux nombres arrondis (round<\/em>) : a<\/em> et (a<\/em> + \\(10^{-n}\\))\u00b2 qui sont les deux bornes de l’encadrement.<\/p>\n\n\n\n
Ensuite (ligne 8), j’affecte les deux valeurs retourn\u00e9es par la fonction aux variables p<\/em> et q<\/em>, pour ensuite les afficher \u00e0 la ligne 9. En lan\u00e7ant le programme, on obtient:<\/p>\n\n\n\n
1.41421 < racine(2) < 1.41422<\/pre>\n\n\n\n
Si je veux un encadrement \u00e0 \\(10^{-7}\\), il suffira de taper:<\/p>\n\n\n\n
>>> approximation(7)\n1.4142135 < racine(2) < 1.4142136<\/pre>\n\n\n\n
Mais attention : \u00e0 partir de n<\/em> = 7, \u00e7a commence \u00e0 \u00eatre tr\u00e8s long… Ce programme (comme tout programme de balayage) n’est pas du tout optimal pour les grandes valeurs de n<\/em> (essayez avec n<\/em> = 10… vous pourrez vous pr\u00e9parer un bon chocolat chaud en attendant tellement c’est long !).<\/p>\n\n\n\n
Autre solution<\/h3>\n\n\n\n
def balayage(p):\n    # encadrement de racine carr\u00e9e de 2 d'amplitude 10^(-p)\n    a, b = 1, 2\n    x = 2**0.5 # racine carr\u00e9e de 2\n    e = 0 # exposant de 10\n    while e > -p:\n        e = e - 1\n        # balayage de [a;b]\n        while x > a:\n            a = a + 10**e\n        \n        b = a\n        a = b - 10**e\n        \n    return round(a,abs(e)) , round(b,abs(e))<\/pre>\n\n\n\n
>>> balayage(10)\n(1.4142135623, 1.4142135624)<\/pre>\n\n\n\n
Ce programme est plus intuitif que le premier… et en plus, il est plus performant!<\/p>\n\n\n\n
N’oubliez pas que si vous rencontrez des difficult\u00e9s en math\u00e9matiques, je peux vous aider par webcam<\/a> !<\/p>\n\n\n\n
[Retourner aux ressources Python]<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"
En classe de Seconde, le programme stipule que l’on doit savoir obtenir un encadrement de \\(\\sqrt2\\) par balayage \u00e0 l’aide de Python. Nous allons voir sur cette page l’id\u00e9e qu’il y a derri\u00e8re cette op\u00e9ration et le script Python. Le principe math\u00e9matique On sait que si \\(0 < a < […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-3189","page","type-page","status-publish","hentry"],"yoast_head":"\nEncadrement de \\(\\sqrt2\\) par balayage en Python - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"En classe de Seconde, le programme stipule que l'on doit savoir obtenir un encadrement de \u221a2 par balayage \u00e0 l'aide de Python.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.mathweb.fr\/euclide\/encadrement-de-sqrt2-par-balayage-en-python\/\" \/>\n<meta property=\"og:locale\" content=\"fr_FR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Encadrement de \\(\\sqrt2\\) par balayage en Python - 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