naturel, le logarithme (+ qqes trucs sur les séries entières)
En général, je crois savoir que l’on introduit le logarithme soit en le posant comme inverse de l’exponentielle, soit comme primitive de 1/x pour peu que la notion soit connue, puis on déroule ses propriétés. Quel dommage ! C’est se priver d’un joli frisson à peu de frais !
J’ai un souvenir des mathématiques scolaires absolument neutre, mais je me rappelle la grande illumination qui m’est tombée dessus au premier cours d’algèbre en première année de fac : un monde s’ouvrait à moi. Tout est dans ce petit exemple, je crois. À la lointaine époque où j’ai fréquenté le lycée, j’ai reçu les mathématiques comme une série d’outils, connectés en surface, mais à l’univers tellement décousu qu’il n’en avait que peu de substance. C’était passer à coté de toute la force, de tout le potentiel créatif qui y réside. Se dire simplement qu’on aimerait étudier les implications d’un énoncé comme “je décide d’appeler logarithme une fonction qui transforme le produit en somme”, c’est un tout autre paradigme qu’une description de l’outil et de ses fonctionnalités. Forcément, je retrouve mes dadas favoris, il s’agit peut-être exactement de reconnecter le sujet et l’objet.
(edit nov : j’ai retrouvé mes books de maths de terminale… et le chapître sur le logarithme commence par un TP plutôt pas mal fait sur l’étude de fonctions qui transforment le produit en somme… dont acte !)
On appelle logarithme une fonction qui transforme le produit en somme.
C’est tout. De là découle tout le reste.
Donc, on appelle logarithme une fonction l telle que l(x.y) = l(x) + l(y) (1), d’ailleurs c’est de là que vient le nom de logarithme : raison entre les nombres.
On pose z = y/x. de (1), on a l(z) = l(x) + l(z/x), soit l(z/x) = l(z) - l(x).
On pose x=y=1, de (1), on a l(1) = l(1) + l(1), d’où l(1) = 0.
De (1), on a aussi en posant x=y : l(x²) = 2 l(x).
On peut aussi par exemple établir sans fatigue : l(x.y.z) = l(x.y) + l(z) = l(x) + l(y) + l(z), ce qui nous donne, en posant x = x1/3.x1/3.x1/3, l(x1/3) = (1/3) l(x). De manière générale : l(xm/n) = (m/n)l(x).
Puis, s’il existe un réel a tel que l(a)=1, alors on a l(ax) = x, d’où il vient que l est la fonction inverse de la fonction exponentielle ax, l est un logarithme de base a.
En ce qui concerne le lien avec l’aire sous l’hyperbole y = 1/x : soient a et b quelconques positifs, on peut se convaincre (quitte à faire explicitement le changement de variables dans l’intégrale), que par exemple, l’aire sous la courbe entre 1 et a est la même que celle entre b et a.b (on dilate par b dans une direction, on comprime de 1/b dans l’autre) donc Aire (1->a.b) = Aire(1->b) + Aire (b->ab) = Aire (1->b) + Aire(1->a). Soit en posant l(a) = Aire(1->a), on a bien l(a.b) = l(a) + l(b) : c’est bien un logarithme.
On peut ensuite poser e comme le nombre tel que Aire (1->e) = 1.
Faisons un petit tour de plus afin de recoller d’autres bouts : retrouver le développement en série entière de ln (1+x) et celui de e.
Tout d’abord, le développement de 1/(1+x) en série géométrique : pour x tel que |x|<1 on pose 1/(1+x) = 1+δ (2) où δ est le reste de l’approximation par 1. On a alors : 1 = 1 + δ + x + x.δ , soit en négligeant le terme x.δ, δ est approximé par -x. L’expression (2) peut alors se réecrire 1/(1+x) = 1 -x + δ (un nouveau δ à approximer), en répétant le procédé ci-dessus, on trouve δ proche de x², puis par récurrence, la série géométrique : 1 - x + x² - x³ + x⁴ - x⁵…. (Viète 1593). En intégrant cette série terme à terme, on obtient bien le développement de ln(1+x) au voisinage de 0.
Cela étant dit, il est possible de retrouver uniquement avec des considérations de surface la valeur de l’intégrale d’un monôme xα pour α>0 entre 0 et un réel B : (Fermat 1636) on choisit un θ < 1 mais proche de 1.On va approximer l'aire sous la courbe xα par la somme des aires des rectangles :
1er rectangle : θ.B -> B de hauteur Bα; d’aire (B - θ.B).Bα = B ( 1 - θ) Bα
2nd rectangle : θ².B -> θ.B de hauteur (θ.B)α d’aire (θ.B - θ².B).(θ.B)α = B ( 1 - θ) θα+1Bα;
3ième rectangle : θ³.B -> θ².B de hauteur (θ².B)α; d’aire (θ².B - θ³.B).(θ².B)α = B ( 1 - θ) θ2α+2Bα; etc.
En additionnant tout, on obtient : Bα+1 ( 1 - θ) ( 1 + θα+1 + θ2 α+2 ….) soit la somme d’une série géométrique de raison θα+1, on a alors l’expression Bα+1 ( 1 - θ)/( 1 - θα+1) (*). Pour 1-θ = ε proche de 0, on a ε/(1 - (1-ε)α+1) équivalent à 1/α+1 (en développant par le binôme de Newton (1-ε)α+1 et éliminant les termes supérieurs au degré 1). Notre surface est alors en valeur approchée par le haut : Bα+1 / (α+1). On réitère le procédé en approximant par en dessous avec les rectangles de hauteur (θ.B), (θ².B) et on retombe sur la même expression, c’est donc une valeur exacte.
Ainsi, on calcule l’aire “sous” la série géométrique entre 0 et x, et on obtient bien le développement en série de ln (1+x) = x - x²/2 + x³/3 - x⁴/4 + x⁵/5 ….. (Mercator 1668)
Pour terminer, on peut retourner vers ex. Calculons ln(1+x/N)N à l’aide de la série de Mercator : ln(1+x/N)N = N . ln( 1+x/N) = N.(x/N -x²/2N …) qui tend vers x lorsque N tend vers ∞, or on définit justement ex comme la limite ( 1+x/N)N… (**)
(j’ai bien conscience que ces paragraphes nécessitent papier stylo)
Voilà, on a le logarithme néperien, logarithme naturel.
Tout ceci n’étant qu’une version édulcorée de ce que l’on peut trouver dans le super exquis et inépuisable “l’analyse au fil de l’histoire” de E. Hairer et G. Wanner, chez Springer. (un aperçu sur google books)
(franchement, je ne pensais pas que je serais encore capable à mon âge de m’émerveiller sur le logarithme)
* au cas où : retrouver la somme d’une série géométrique Sn= 1 + q + q² + … + qn : on multiplie par (1-q) de chaque coté, les termes s’annulent 2 à 2 sauf le premier et le dernier : Sn ( 1- q ) = 1 - qn+1.
** Rappelons que le nombre d’Euler; e, est obtenu comme la limite quand N tend vers ∞ de ( 1+1/N)N. Puis d’une part, en développant (1+x/N)N par le binôme de Newton on a la série 1 + x + x²/2! + x³/3! … D’autre part, en posant M = x.N (avec x rationnel, on peut ne considérer que les N et M entiers), on a ( 1+x/N)N = ( 1+1/M)Mx soit pour M tendant vers ∞ e à la puissance x, ex.
October 4th, 2010 at 10:12 am
alors là je ne comprends pas qu’un exposé aussi brillant, lumineux fasse si peu d’audimat…tu déchires le PWF avec ce billet, c’est simple je suis sans voix, spechless mais la prochaine écris-le en Ourdou ou en Yiddish pour ajouter à ce chef d’oeuvre une dimension exotique qui lui revient…En revanche je vois que les crèpes font recette…
” Voilà, on a le logarithme néperien, logarithme naturel.” C’est d’une beauté, je ne m’en lasse pas….
October 4th, 2010 at 10:30 am
sans ironie aucune, on peut s’arrêter (ne serait-ce pour des élèves de terminale par exemple) avant l’aire sous l’hyperbole, avec le log de base a. Et rien que ça, oui, je trouve que ça déchire grave.
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