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I need to find the critical points of the multivariate polynomial and types of critical points

$f(x_1,x_2,x_3,x_4)=x_2x_3+x_3x_4+x_1x_2+x_1x_2x_3x_4$

What are soft to find the critical points and how I can classification the critical points.

If taken $$\frac{\partial f}{\partial x_1}=0,\frac{\partial f}{\partial x_2}=0,\frac{\partial f}{\partial x_3}=0,\frac{\partial f}{\partial x_4}=0,$$ then we get $$x_2+x_2x_3x_4=0$$ $$x_3+x_1+x_1x_3x_4=0$$ $$x_2+x_4+x_1x_2x_4=0$$ $$x_3+x_1x_2x_3=0$$ 1. Do you use a program to solve equations or there is a better way?

  1. If there is a set of solutions. How do you critical points ?

thanks for the help.

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  • $\begingroup$ compute the partial derivatives $\endgroup$ – Dr. Sonnhard Graubner Mar 20 '17 at 17:29
  • $\begingroup$ @Dr.SonnhardGraubner See the new update $\endgroup$ – Emad kareem Mar 20 '17 at 17:55
  • $\begingroup$ Your partial derivatives are incorrect. For example, in computing $\frac{\partial f}{\partial x_1}$, what is the derivative of the first term $x_2x_3$ with respect to the variable $x_1$? $\endgroup$ – zipirovich Mar 20 '17 at 18:07
  • $\begingroup$ @zipirovich thanks ...See the new update... Why do later $\endgroup$ – Emad kareem Mar 20 '17 at 18:18
  • $\begingroup$ Looks like $(0,0,0,0)$ is the only critical point. Your first equation gives that either $x_2=0$ or $x_3x_4=-1$. The latter alternative together with the second equation gives $x_3=0$, which is absurd. Therefore $x_2=0$, and it follows that the rest of the vanish as well. The first three terms of $f$ give the Hessian at the origin, and you can study that with the usual technique. $\endgroup$ – Jyrki Lahtonen Mar 20 '17 at 18:36
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Look at the first equation, there are two possibilities: either $x_2 = 0$, or $1 + x_3 x_4 = 0$.

If $x_2 = 0$, then the remaining equations simplify to

$$\begin{cases} x_2 = 0 \\ x_3 + x_1 + x_3 x_1 x_4 = 0 \\ x_4 = 0 \\ x_3 = 0 ,\end{cases}$$

whence it follows that $x_1 = 0$ and so far we have obtained the solution $(0,0,0,0)$.

If $x_2 \neq 0$ then $x_3 x_4 = -1$, which means that the system simplifies to

$$\begin{cases} x_3 x_4 = -1 \\ x_3 = 0 \\ x_2 + x_4 + x_1 x_2 x_4 = 0 \\ x_3 + x_1 x_2 x_3 = 0 .\end{cases}$$

The first two equations cannot be satisfied simultaneously, therefore this system has no solution.

We have obtained, then, that $(0,0,0,0)$ is the only critical point of $f$.

The Hessian matrix of $f$ is

$$\begin{pmatrix} 0 & 1 + x_3 x_4 & x_2 x_4 & x_2 x_3 \\ 1 + x_3 x_4 & 0 & 1 + x_1 x_4 & x_1 x_3 \\ x_2 x_4 & 1 + x_1 x_4 & 0 & 1 + x_1 x_2 \\ x_2 x_3 & x_1 x_3 & 1 + x_1 x_2 & 0 \end{pmatrix}$$

which, upon evaluation in $(0,0,0,0)$ produces the matrix

$$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} .$$

Notice that the $1 \times 1$ determinant formed in the upper-left corner is $0$, therefore the matrix is not definite, which shows that $(0,0,0,0)$ is neither a maximum, nor a minimum. In fact, just noticing that $\dfrac {\partial ^2 f} {\partial x_1 ^2} (0,0,0,0,) = 0$ is enough to deduce this, it isn't really necessary to compute and evaluate the full Hessian.

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