Consider the following system of equations:

\begin{align} \frac{1}{8} (\alpha +2 x)^2 \left((\alpha +2 x)^2-12\right)+\\ +\frac{4}{9} y^2 \left((\alpha +2 x)^2+2\right)+\frac{32 y^4}{81}&=0\\ \alpha ^3-6 \alpha +8 x^3-4 \alpha x^2-2 \left(\alpha ^2+2\right) x&=0 \end{align}

I would like to eliminate the $\alpha$ value in order to have an equation of the form $f(x,y)=0$.

Using Cardano's formula or Mathematica, I end up with imaginary values. However if I solve them numerically I obtain a nice shape $f(x,y)=0$

Due to the numerical plots I obtain I am thinking that there must be a nice (analytical) solution for these equations. How could I derive such an expression for $f(x,y)=0$? enter image description here

  • 2
    $\begingroup$ Apply Resultant get two polynomial factors, only one equal zero: link. $\endgroup$ Nov 16 '20 at 5:53
  • $\begingroup$ The Resultant works even faster than the GroebnerBasis, thanks! $\endgroup$
    – Matt
    Nov 16 '20 at 16:49

If you do have Mathematica, use the command

GroebnerBasis[{pol1, pol2}, {x,y},{a}]

where poli are the two expressions that equal $0$. This will eliminate the variable $a$.

On WolframAlpha I got a polynomial in $x$, $y$ with fairly large coefficients. Check this link here.
Hope there were no mistakes in the input.

$\bf{Added:}$. You will get the resultant of the two expressions considered as polynomials in $a$. The Groebner command is just more general, can eliminate some variables ( the ones in the second group) from a system of equations.


The second equation write $$\alpha^3-2x\alpha^2-2(3+2x^2)\alpha+4x(2 x-1)=0$$ The discriminant is $$\Delta=32 \left(128 x^4+18 x^2+27\right) > 0 \qquad \forall x$$ Using the trigonometric method for three real roots leads to $$\alpha_k=\frac{2}{3} \left(x+ \sqrt{2(8 x^2+9)} \cos \left(\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)\right)\right)$$ with $k=0,1,2$.

Replace in the first equation to have $f(x,y)=0$ for each value of $k$. This would lead to $$c_0+c_1\cos(A)+c_2\cos(2A)+c_3\cos(3A)+c_4\cos(4A)=0$$ with $$c_0=32 \left(70 x^4+3 x^2 \left(4 y^2+9\right)+y^4\right)+27 \left(8 y^2-9\right)$$ $$c_1=128 \sqrt{2} x \sqrt{8 x^2+9} \left(7 x^2+y^2\right)$$ $$c_2=2 \left(8 x^2+9\right) \left(112 x^2+8 y^2-9\right)$$ $$c_3=16 \sqrt{2} x \left(8 x^2+9\right)^{3/2}$$ $$c_4=\left(8 x^2+9\right)^2$$

$$A=\frac{1}{3} \left(2 \pi k-\cos ^{-1}\left(\frac{\sqrt{2} x \left(27-16 x^2\right)}{\left(8 x^2+9\right)^{3/2}}\right)\right)$$


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