User:Dr-Taher/Bezout Method

The '' Bézout Method" is a general method of solving algebraic equations. It was initiated and developed by Étienne Bézout in 1762.

This method tries to reduce the equation that we want to solve to other equations of lower degree. This tedious method certainly fails for equations of degree greater than or equal to five that have an unresolved Galois group. It has a concrete interest only for equations of degree 3.

Principle of the method
Consider an equation of degree $n$:

$$ \qquad a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 = 0$$

Let $r$ be an $n$ -th primary root of the unit.

We know that the $n$ roots $n$-ths of unity 1, $r$, $r 2$, ..., $r  n  - 1$ verify the relation:

$$ \qquad 1 + r + r^2 + \cdots + r^{n-1} = 0 $$

Bézout's method is looking for the roots of the studied equation in the form of linear combinations of the roots $'' n '$ -ths of the unit.

$$ \qquad x = b_0 + b_1r + b_2r^2 + \cdots + b_{n-1}r^{n-1} $$

For this, we start by eliminating r between the two relations:

$$ \qquad 1 + r + r^2 + \cdots + r^{n-1} = 0 $$

$$ \qquad x = b_0 + b_1r + b_2r^2 + \cdots + b_{n-1}r^{n-1} $$

Which gives us an equation of degree $n$ in $x$ whose coefficients are expressions depending on $b$0, $b$1, $b$2,...,$b_{n-1}$. By identifying the coefficients of this equation with the corresponding coefficients of the equation to be solved, we obtain a system of equations of unknowns $b$0, $b$1, $b$2,...,$b_{n-1}$ which after solving and reporting the different solutions in:

$$ \qquad x = b_0 + b_1r + b_2r^2 + \cdots + b_{n-1}r^{n-1} $$

will give us the solutions of the equation that we had set ourselves to solve.

Application to the resolution of cubic equations
We will expose the method on the following example:

$$ 6x^3 - 6x^2 + 12x + 7=0 ~$$

Let's ask: $$ \mathrm{j} = \mathrm{e}^{\frac{2\mathrm{i}\pi}{3}} ~$$

$j$ is one of the cubic roots of the unit and therefore verifies:

$$ \mathrm{j}^3 = 1 ~$$

Let's search for the roots in the form:

$$ x = a+b\mathrm{j}+c\mathrm{j}^2 $$

We will eliminate $j$ between the last two equations.

The last two equations are in the form:

$$ \left\{\begin{matrix} \mathrm{j}^3=1 \\ x-a-b\mathrm{j}=c\mathrm{j}^2 \end{matrix}\right. $$

By making successive member-to-member products and each time replacing those of the two equations whose degree with respect to $j$ is the highest by the result, we will gradually lower the degree of the equations with respect to $j$ until $j$ disappears from one of the equations.

A first member-to-member product gives us:

$$ \left\{\begin{matrix} b\mathrm{j}^2=\mathrm{j}x-a\mathrm{j}-c \\ x-a-b\mathrm{j}=c\mathrm{j}^2 \end{matrix}\right. $$

A second member-to-member product gives us:

$$ \left\{\begin{matrix} c\mathrm{j}x-ac\mathrm{j}+b^2\mathrm{j}=bx+c^2-ab \\ x-a-b\mathrm{j}=c\mathrm{j}^2 \end{matrix}\right. $$

A third member-to-member product gives us:

$$ \left\{\begin{matrix} c\mathrm{j}x-ac\mathrm{j}+b^2\mathrm{j}=bx+c^2-ab \\ ab^2-a^2c-b^2x+2acx-cx^2=2abc\mathrm{j}-b^3\mathrm{j}-c^3\mathrm{j}-2bc\mathrm{j}x \end{matrix}\right. $$

A last member-to-member product eliminates $j$ and gives us the equation:

$$ x^3 - 3ax^2 + (3a^2 - 3bc)x + 3abc - a^3 - b^3 - c^3 = 0 ~$$

By identifying the coefficients of this equation with the coefficients of the equation we need to solve, we obtain:

$$ \left\{\begin{matrix} -3a=-1 \\ 3a^2 - 3bc=2 \\ 3abc - a^3 - b^3 - c^3=\frac{7}{6} \end{matrix}\right. $$

From the first equation we deduce from this the value of $a$ that we report in the other equations, we obtain:

$$ \left\{\begin{matrix} a=\frac{1}{3} \\ bc=-\frac{5}{9} \\ bc - b^3 - c^3=\frac{65}{54} \end{matrix}\right. $$

Let's memorize the value of a and bring the product $bc$ in the third equation, we get:

$$ \left\{\begin{matrix} bc=-\frac{5}{9} \\ b^3 + c^3 = -\frac{95}{54} \end{matrix}\right. $$

By raising to the cube the two members of the first equation, we obtain:

$$ \left\{\begin{matrix} b^3c^3 = -\frac{125}{729} \\ b^3+c^3 = -\frac{95}{54} \end{matrix}\right. ~$$

$b 3$ and $c 3$ are therefore the roots of the equation:

$$ X^2 + \frac{95}{54}X -\frac{125}{729} = 0 ~$$

The two roots of this equation are:

$$ b^3 = \frac{5}{54}, \, c^3 = -\frac{50}{27} ~$$

The three pairs $( b, c )$ checking:

$$ bc = -\frac{5}{9} ~$$

thereby are :

$$b_1=\frac13\sqrt[3]{\frac52}\quad\text{و}\quad c_1=-\frac13\sqrt[3]{50}$$ ;

$$b_2=\frac{\mathrm{j}}3\sqrt[3]{\frac52}\quad\text{و}\quad c_2=-\frac{\mathrm{j}^2}3\sqrt[3]{50}$$ ;

$$b_3=\frac{\mathrm{j}^2}3\sqrt[3]{\frac52}\quad\text{و}\quad c_3=-\frac{\mathrm{j}}3\sqrt[3]{50}$$.

By reporting in $$ x = a+b\mathrm{j}+c\mathrm{j}^2 $$ the values ​​of $$a,b,c$$ found, we get

$$ x_1 = \frac{1}{3} + \frac{1}{3}\sqrt[3]{\frac{5}{2}}\mathrm{j} - \frac{1}{3}\sqrt[3]{50}\mathrm{j}^2$$,

$$ x_2 = \frac{1}{3} + \frac{\mathrm{j}}{3}\sqrt[3]{\frac{5}{2}}\mathrm{j} - \frac{\mathrm{j}^2}{3}\sqrt[3]{50}\mathrm{j}^2$$

$$ x_3 = \frac{1}{3} + \frac{\mathrm{j}^2}{3}\sqrt[3]{\frac{5}{2}}j- \frac{j}{3}\sqrt[3]{50}\mathrm{j}^2$$,

which, after simplification, gives

$$ x_1 = \frac{1}{3}\left(1 + \mathrm{j}\sqrt[3]{\frac{5}{2}} - \mathrm{j}^2\sqrt[3]{50}\right)$$,

$$ x_2 = \frac{1}{3}\left(1 + \mathrm{j}^2\sqrt[3]{\frac{5}{2}} - \mathrm{j}\sqrt[3]{50}\right)$$

$$ x_3 = \frac{1}{3}\left(1 + \sqrt[3]{\frac{5}{2}}- \sqrt[3]{50}\right)$$,

which are the three roots of the equation that we had to solve.

External link

 * Text Bézout (1764) on the resolution of algebraic equations, online and commented on BibNum

Category: Polynomial Equation