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Consider $x \in \Bbb{R}^n$ and suppose $f_m: \Bbb{R}^n \to \Bbb{R}$ which is defined as: $$f_m(x)=x^TAx+\alpha_{1}x_1^3+\alpha_{2}x_1^2x_2+\dots+\alpha_{r_1}x_n^3+\alpha_{r_1+1}x_1^4+\alpha_{r_1+2}x_1^3x_2+\dots+\alpha_{r_2}x_n^4+\dots+\alpha_{c}x_n^m$$

where subscript in $x_i$ refers to the $i$th element of $x$. If we know that $A$ is positive definite, what is the necessary and sufficient condition for $\alpha_1,...,\alpha_c$ to result in a positive function $f_m(x)$ for all $\Vert x\Vert < \rho$ and all $m\in\{3,4,...,M\}$? Is the following statement correct?

$f_m(x)$ is positive for $\Vert x\Vert < \rho$ if and only if $x=0$ is the only solution of $f_m(x)=0$ in $\Vert x\Vert < \rho$ and $A$ is positive definite.

If it is correct, how to show $x=0$ is the only solution? In other words, how to choose $\alpha_i$ so that $x=0$ becomes the unique solution?


For example in $x\in\Bbb{R}^2$, up to and including the 3rd order, $f_3(x)$ can be showed as follows:

$$f_3(x)=A_{1,1}x_1^2+A_{2,2}x_2^2+\alpha_{1}x_1^3+\alpha_{2}x_1^2x_2+\alpha_{3}x_2^2x_1+\alpha_{4}x_2^3$$ $$A_{1,1}>0,A_{2,2}>0$$ Choose $\alpha_1,\dots,\alpha_4$ on the basis of $A_{1,1}$ and $A_{2,2}$ so that $f_3(x)>0$ for $\Vert x\Vert < \rho$. Then choose $\alpha_5,\dots,\alpha_9$ on the basis of $A_{1,1}$ and $A_{2,2}$ so that $f_4(x)=f_3(x)+\alpha_{5}x_1^4+\alpha_{5}x_1^3x_2+\dots+\alpha_{9}x_2^4>0$ for $\Vert x\Vert < \rho$.

And so on.

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  • $\begingroup$ What does positive function mean in this context? Under one interpretation, since your formula implies $f(0)=0$, which is not a positive number, no $f$ of your form is a positive function.. $\endgroup$ – kimchi lover Aug 11 '18 at 14:47
  • $\begingroup$ @kimchilover , $x=0$ is the only solution for $f(x)=0$ and $A$ is positive definite; it should be shown that $f(x)>0$ for all $x\neq 0$. $\endgroup$ – Mehdi Moghadasian Aug 11 '18 at 14:59
  • $\begingroup$ How can a cubic polynomial be positive definite? $\endgroup$ – Rodrigo de Azevedo Aug 12 '18 at 10:32
  • $\begingroup$ @RodrigodeAzevedo Exactly! but what about specifying a bounded $\Vert x\Vert <\rho$ in which positiveness satisfies? Even a cubic form can be positive in a specific radius. $\endgroup$ – Mehdi Moghadasian Aug 12 '18 at 11:17
  • $\begingroup$ No, I don't have Mathematica $\endgroup$ – Mehdi Moghadasian Aug 12 '18 at 12:05

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