Benutzer:Juergen Behrndt/Baustelle

schreib grad an wikipedia-artikel {{Inuse|<Juergen Behrndt>}}

Dieser Artikel ist im Entstehen und noch nicht Bestandteil der freien Enzyklopädie Wikipedia.

Solltest du über eine Suchmaschine darauf gestoVorlage:SSen sein, bedenke, dass der Text noch unvollständig sein und Fehler oder ungeprüfte Aussagen enthalten kann. Wenn du Fragen zum Thema hast, nimm Kontakt mit dem Autor Juergen Behrndt auf.

Resolvente (Algebra)

Vorversion in englisch

For Lagrange as for Vandermonde, the base idea is to consider the quantity $x+ay+a^{2}z$ where x,y,z are the solutions of the cubic in question and where a is a cube root of unity(a?1).
Let t denote this quantity. Lagrange observes that t has 6 values depending in the order in which the roots x y z are taken. These 6 six values are the solution of a 6th degree equation $f(X)=(X-t_{1})(X-t_{2})(X-t_{3})(X-t_{4})(X-t_{5})(X-t_{6})$ whose coefficients being symmetric in the 6 values of f are symmetric in x,y,z
and are therefore known quantities expressible in terms of the coefficients of the given cubic.

Lagrange calls this the resolvent equation.

Although it is of higher degree of the original, it is solvable because it is in fact a quadratic equation in X^3
and can be solved by solving a quadratic and then taking a cube root.
The fact that the resolvent equation f(X) = 0 is quadratic in X^3 is easily seen
by observing that the values of t can be ordered so that
$t_{1}=x+ay+a^{2}z$
$t_{2}=z+ax+a^{2}y$
$t_{3}=a^{2}t_{1}$
$t_{4}=x+az+a^{2}y$
$t_{5}=at_{4}$
$t_{6}=a^{2}t_{4}$

which gives
$(X-t_{1})(X-t_{2})(X-t_{3})=(X-t_{1})(X-at_{1})(X-a^{2}t_{1})=X^{3}-t_{1}^{3}$

and f(X) = (X^3 - t_1^3)(X^3 - t_4^3) = X^6 – (t_1^3 + t_4^3) * X^3 + t_1^3 * t_4^3
In the notations above, let
$u=t_{1}^{3}$
$v=t_{4}^{3}$
And the coefficients of f(X) are precisely the quantities u+v and uv whose expressions
in terms of the coefficients of the original cubic gave the solution above.
Once the 6 values of t are obtained from the solution of the resolvent equation, the equations of the cubic are given by
$(x=1/3[(x+y+z)+t_{1}+t_{4}]$
$(y=1/3[(x+y+z)+a^{2}t_{1}+at_{4}]$
$(z=1/3[(x+y+z)+a_{1}^{t}+a^{2}t_{4}]$

and the only problem is to identify t_1 and t_4 among the 6 solutions of the resolvent equation.
Let t be any solution of the resolvent equation. Then the roots x,y,z can be reordered if necessary,

so that $t=x+ay+a^{2}z.$

Lagrange observes that (x+ay+a^2z) (x+a^2y+az) is symmetric in x,y,z and is therefore a known quantity, say w.
Once this observation has been made , the solution is simple.
The 3 roots of the cubic are
$1/3[(x+y+z)+t+w/t]$
$1/3[(x+y+z)+at+w/at]$
$1/3[(x+y+z)+a^{2}t+w/a^{2}t]$
,where t is any one of the 6 solutions of the resolvent equation.