Pas directement. Mais, en supposant que Maxima arrive à trouver les
racines du polynomial, il est facile à faire. Attention! La méthode
démontrée ne prend pas compte des eventuels polynomiaux où Maxima ne
trouve pas les racines explicites.
poly: x^4-x-1$
sols: solve(poly,x);
[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)),
x =
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))
-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 =
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))]
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(283)/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*sqrt(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 <[EMAIL PROTECTED]> 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: $(x-\sqrt3)^2$.
-------------------------------------------------------------------------
SF.Net email is sponsored by:
Check out the new SourceForge.net Marketplace.
It's the best place to buy or sell services
for just about anything Open Source.
http://ad.doubleclick.net/clk;164216239;13503038;w?http://sf.net/marketplace
_______________________________________________
Maxima-lang-fr mailing list
[email protected]
https://lists.sourceforge.net/lists/listinfo/maxima-lang-fr
