## maxima-lang-fr — French-language discussion of the Maxima computer algebra system

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 [Maxima-lang-fr] =?iso-8859-1?q?Trac=E9_de_courbes_param=E9trique?= =?iso-8859-1?q?s?= From: Guillaume MARTIN - 2007-12-12 23:26:09 Attachments: Message as HTML Bonjour, je souhaiterais tracer des courbes paramétriques du type : x = cos w y = sin w Pourriez vous m'indiquer quelle est la commande à utiliser dans maxima ? d'avance merci, Guillaume MARTIN PS : thanks you Mr DODIER : I've just tried the ilt function, it works very fine ! Guillaume --------------------------------- Ne gardez plus qu'une seule adresse mail ! Copiez vos mails vers Yahoo! Mail 
 Re: [Maxima-lang-fr] factorisation avec radicaux From: Stavros Macrakis - 2007-12-12 22:10:32 Pas directement. Mais, en supposant que Maxima arrive =E0 trouver les racines du polynomial, il est facile =E0 faire. Attention! La m=E9thode d=E9montr=E9e ne prend pas compte des eventuels polynomiaux o=F9 Maxima ne trouve pas les racines explicites. poly: x^4-x-1$sols: solve(poly,x); [x =3D -sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3= ^-(3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6)), x =3D sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^= -(3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6)), x =3D sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqr= t(283)/2+1/2)^(1/6)) -sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)/2= +1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+= 1/2)^(2/3)-4)^(1/4)), x =3D sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283= )/2+1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+1= /2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6))] fact: xreduce("*",makelist(x-rhs(q),q,sols)); (x-sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)/2= +1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/2+= 1/2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt(2= 83)/2+1/2)^(1/6))) *(x+sqrt(6*sqrt(3)*sqrt(3^-(3/2)*sqrt(283)/2+1/2)-(3*(3^-(3/2)*sqrt(283)= /2+1/2)^(2/3)-4)^(3/2)) /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) -sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) *(x-sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^-= (3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) *(x+sqrt((3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)^(3/2)+6*sqrt(3)*sqrt(3^-= (3/2)*sqrt(283)/2+1/2))*%i /(2*sqrt(3)*(3^-(3/2)*sqrt(283)/2+1/2)^(1/6)*(3*(3^-(3/2)*sqrt(283)/= 2+1/2)^(2/3)-4)^(1/4)) +sqrt(3*(3^-(3/2)*sqrt(283)/2+1/2)^(2/3)-4)/(2*sqrt(3)*(3^-(3/2)*sqrt= (283)/2+1/2)^(1/6))) expand(%); x^4+(3*(sqrt(283)/(6*sqrt(3))+1/2)^(2/3)-4)^(3/4)*sqrt((3*(sqrt(283)/(6*sqr= t(3))+1/2)^(2/3)-4)^(3/2) +6*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2))*%i*x /(12*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2)) -(sqrt(283)/(6*sqrt(3))+1/2)^(1/6)*sqrt((3*(sqrt(283)/(6*sqrt(3))+= 1/2)^(2/3)-4)^(3/2) +6*sqrt(3)*sqrt(sqrt(283)/(6*sqrt(3))+1/2))*%i*x etc. etc. radcan(%),algebraic:true; x^4-x-1 On Dec 9, 2007 2:14 PM, Valere Bonnet wrote: > j'aimerais savoir s'il est possible d'utiliser maxima pour factoriser des > expressions avec > des radicaux; par exemple:$x^2-2\sqrt3+3$; deviendrait:$(x-\sqrt3)^2$.   Re: [Maxima-lang-fr] Question integration From: Stavros Macrakis - 2007-12-12 14:22:47 Cyril, Je n'ai pas essay=E9 de reproduire ton probl=E8me moi-m=EAme, mais je crois que le changement suivant le r=E9soudra : > f(x):=3Dx$ > g(A):=3Dromberg(f(x),x,0,A)$> wxplot2d([g(A)], [A,-5,5])$ wxplot2d( '[g(A)], [A, -5,5] )$En effet, Maxima =E9value l'expression g(A) deux fois, d'abord pour =E9tablir l'expression =E0 traiter (qui peut =EAtre, par exemple, %o4, un r=E9sultat pr=E9c=E9dant) et ensuite =E0 chaque valeur de la variable. Mai= s Romberg ne peut pas s'appeler avec une variable symbolique. Pourquoi alors, pourrait-on se le demander, le premier cas, wxplot(f(x)...), marche? Parce que la valeur de f(x) pour x symbolique (pas num=E9rique) reste x. Par ailleurs, il est un peut dr=F4le (mais pas incorrect) d'utiliser systematiquement le m=EAme nom de variable en: f(x):=3Dx$ g(A):=3Dromberg(f(x),x,0,A)$wxplot2d('[g(A)], [A,-5,5])$ On peut aussi bien =E9crire: f(x):=3Dx$g(A):=3Dromberg(f(q),q,0,A)$ wxplot2d('[g(z)], [z,-5,5])$En =E9sp=E9rant t'avoir aid=E9, -s   [Maxima-lang-fr] Question integration From: - 2007-12-12 11:18:03 Bonjour, Je suis un nouvel utilisateur de MAXIMA. Est-ce que la liste presente est toujours d'actualite ? J'ai un probleme que j'aurais aime souhaiter soumettre aux membres de cette liste. Le voici. Pourquoi est-ce que le programme ci-dessous marche : kill(all)$ load(romberg)$f(x):=x$ g(A):=romberg(f(x),x,0,A)$wxplot2d([f(x)], [x,-5,5])$ g(-4); et pourquoi celui-ci ne marche pas ? kill(all)$load(romberg)$ f(x):=x$g(A):=romberg(f(x),x,0,A)$ wxplot2d([g(A)], [A,-5,5])\$ Je calcule numeriquement l'integrale de f de 0 a A, je stocke le resulte dans la fonction g et j'aimerais voir le graphe de la courbe g. mais je n'y arrive pas! J'ai essaye de regarder de l'aide mais je ne vois rien qui concerne vraiment mon probleme... Cordialement, Cyril Grunspan -- This message and any attachments (the "message") is intended solely for the addressees and is confidential. If you receive this message in error, please delete it and immediately notify the sender. Any use not in accord with its purpose, any dissemination or disclosure, either whole or partial, is prohibited except formal approval. The internet can not guarantee the integrity of this message. 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