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An affine $G$ such that $G(p)_x(mathbbR^2) geq 0$, $p in mathbbR[x,y]$ is homogeneous of odd degree
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Let $pin mathbbR[x,y]$ be a homogeneous polynomial of odd degree $d geq 1$.
Can one find an affine $mathbbR$-algebra automorphism $G$ of $mathbbR[x,y]$ such that
$G(p)_x(mathbbR^2) geq 0$,
where $G(p)_x$ is the partial derivative of $G(p) in mathbbR[x,y]$ with respect to $x$? Notice that $G(p)_x$ is homogeneous of even degree $d-1$.
I do not mind to further assume that the coefficients of $p$ belong to $mathbbR^geq 0=[0,infty)$, if this simplifies the answer.
Examples:
(1) $d=1$: Write $p=ax+by$, $a,b in mathbbR$.
If $a geq 0$ then for $G=1$ we have $G(p)=p=ax+by$, so $G(p)_x=a geq 0$. If $a < 0$ then for $G: (x,y) mapsto (-x,y)$ we have $G(p)=G(ax+by)=aG(x)+bG(y)=a(-x)+by=-ax+by$, so $G(p)_x=-a > 0$.
(2) $d=3$: Write $p=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$.
If $a neq 0$, then we can define $G: (x,y) mapsto (x-fracb3ay,y)$. Then a direct computation (if I do not have an error) yields $G(p)=ax^3+(frac-b^23a+c)xy^2+epsilon y^3$, for some $epsilon in mathbbR$.
If we further assume that $a,b,c,d in mathbbR^+$, then we are almost done; indeed, $G(p)_x=3ax^2+(frac-b^23a+c)y^2$, which is non-negative in case
$frac-b^23a+c geq 0$, so we need $3ac-b^2 geq 0$.
(Allowing $G$ to be an affine $mathbbC$-algebra automorphism of $mathbbC[x,y]$ still does not help; if we take $H: (x,y) mapsto (x-fracbi3ay,y)$ instead of our former $G$, then $delta xy$ will appear in $H(p)_x$, for some $delta in mathbbC$).
(3) $d=5$: Write $p=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$, $a,b,c,d,e,f in mathbbR^+$.
"Completing the square" trick, as far as I see, only guarantees that we can find an affine $G$ such that $G(p)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$ for some $A,C,D,E,F in mathbbR$, but this is not enough, because $G(p)_x=5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains the problematic term $2Dxy^3$.
Actually, one also has to be careful about the signs of $A,C,E$, which are dependent on the relations between the original coefficients $a,b,c,d,e,f$ (even if all of them are positive), as we have seen in the case $d=3$.
Any comments are welcome!
See also this question.
algebraic-geometry polynomials real-algebraic-geometry
asked Jul 17 at 12:40
user237522
1,8141617
add a comment |Â
up vote
1
down vote
favorite
Let $pin mathbbR[x,y]$ be a homogeneous polynomial of odd degree $d geq 1$.
Can one find an affine $mathbbR$-algebra automorphism $G$ of $mathbbR[x,y]$ such that
$G(p)_x(mathbbR^2) geq 0$,
where $G(p)_x$ is the partial derivative of $G(p) in mathbbR[x,y]$ with respect to $x$? Notice that $G(p)_x$ is homogeneous of even degree $d-1$.
I do not mind to further assume that the coefficients of $p$ belong to $mathbbR^geq 0=[0,infty)$, if this simplifies the answer.
Examples:
(1) $d=1$: Write $p=ax+by$, $a,b in mathbbR$.
If $a geq 0$ then for $G=1$ we have $G(p)=p=ax+by$, so $G(p)_x=a geq 0$. If $a < 0$ then for $G: (x,y) mapsto (-x,y)$ we have $G(p)=G(ax+by)=aG(x)+bG(y)=a(-x)+by=-ax+by$, so $G(p)_x=-a > 0$.
(2) $d=3$: Write $p=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$.
If $a neq 0$, then we can define $G: (x,y) mapsto (x-fracb3ay,y)$. Then a direct computation (if I do not have an error) yields $G(p)=ax^3+(frac-b^23a+c)xy^2+epsilon y^3$, for some $epsilon in mathbbR$.
If we further assume that $a,b,c,d in mathbbR^+$, then we are almost done; indeed, $G(p)_x=3ax^2+(frac-b^23a+c)y^2$, which is non-negative in case
$frac-b^23a+c geq 0$, so we need $3ac-b^2 geq 0$.
(Allowing $G$ to be an affine $mathbbC$-algebra automorphism of $mathbbC[x,y]$ still does not help; if we take $H: (x,y) mapsto (x-fracbi3ay,y)$ instead of our former $G$, then $delta xy$ will appear in $H(p)_x$, for some $delta in mathbbC$).
(3) $d=5$: Write $p=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$, $a,b,c,d,e,f in mathbbR^+$.
"Completing the square" trick, as far as I see, only guarantees that we can find an affine $G$ such that $G(p)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$ for some $A,C,D,E,F in mathbbR$, but this is not enough, because $G(p)_x=5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains the problematic term $2Dxy^3$.
Actually, one also has to be careful about the signs of $A,C,E$, which are dependent on the relations between the original coefficients $a,b,c,d,e,f$ (even if all of them are positive), as we have seen in the case $d=3$.
Any comments are welcome!
See also this question.
algebraic-geometry polynomials real-algebraic-geometry
asked Jul 17 at 12:40
user237522
1,8141617
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $pin mathbbR[x,y]$ be a homogeneous polynomial of odd degree $d geq 1$.
Can one find an affine $mathbbR$-algebra automorphism $G$ of $mathbbR[x,y]$ such that
$G(p)_x(mathbbR^2) geq 0$,
where $G(p)_x$ is the partial derivative of $G(p) in mathbbR[x,y]$ with respect to $x$? Notice that $G(p)_x$ is homogeneous of even degree $d-1$.
I do not mind to further assume that the coefficients of $p$ belong to $mathbbR^geq 0=[0,infty)$, if this simplifies the answer.
Examples:
(1) $d=1$: Write $p=ax+by$, $a,b in mathbbR$.
If $a geq 0$ then for $G=1$ we have $G(p)=p=ax+by$, so $G(p)_x=a geq 0$. If $a < 0$ then for $G: (x,y) mapsto (-x,y)$ we have $G(p)=G(ax+by)=aG(x)+bG(y)=a(-x)+by=-ax+by$, so $G(p)_x=-a > 0$.
(2) $d=3$: Write $p=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$.
If $a neq 0$, then we can define $G: (x,y) mapsto (x-fracb3ay,y)$. Then a direct computation (if I do not have an error) yields $G(p)=ax^3+(frac-b^23a+c)xy^2+epsilon y^3$, for some $epsilon in mathbbR$.
If we further assume that $a,b,c,d in mathbbR^+$, then we are almost done; indeed, $G(p)_x=3ax^2+(frac-b^23a+c)y^2$, which is non-negative in case
$frac-b^23a+c geq 0$, so we need $3ac-b^2 geq 0$.
(Allowing $G$ to be an affine $mathbbC$-algebra automorphism of $mathbbC[x,y]$ still does not help; if we take $H: (x,y) mapsto (x-fracbi3ay,y)$ instead of our former $G$, then $delta xy$ will appear in $H(p)_x$, for some $delta in mathbbC$).
(3) $d=5$: Write $p=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$, $a,b,c,d,e,f in mathbbR^+$.
"Completing the square" trick, as far as I see, only guarantees that we can find an affine $G$ such that $G(p)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$ for some $A,C,D,E,F in mathbbR$, but this is not enough, because $G(p)_x=5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains the problematic term $2Dxy^3$.
Actually, one also has to be careful about the signs of $A,C,E$, which are dependent on the relations between the original coefficients $a,b,c,d,e,f$ (even if all of them are positive), as we have seen in the case $d=3$.
Any comments are welcome!
See also this question.
algebraic-geometry polynomials real-algebraic-geometry
asked Jul 17 at 12:40
user237522
1,8141617
Let $pin mathbbR[x,y]$ be a homogeneous polynomial of odd degree $d geq 1$.
Can one find an affine $mathbbR$-algebra automorphism $G$ of $mathbbR[x,y]$ such that
$G(p)_x(mathbbR^2) geq 0$,
where $G(p)_x$ is the partial derivative of $G(p) in mathbbR[x,y]$ with respect to $x$? Notice that $G(p)_x$ is homogeneous of even degree $d-1$.
I do not mind to further assume that the coefficients of $p$ belong to $mathbbR^geq 0=[0,infty)$, if this simplifies the answer.
Examples:
(1) $d=1$: Write $p=ax+by$, $a,b in mathbbR$.
If $a geq 0$ then for $G=1$ we have $G(p)=p=ax+by$, so $G(p)_x=a geq 0$. If $a < 0$ then for $G: (x,y) mapsto (-x,y)$ we have $G(p)=G(ax+by)=aG(x)+bG(y)=a(-x)+by=-ax+by$, so $G(p)_x=-a > 0$.
(2) $d=3$: Write $p=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$.
If $a neq 0$, then we can define $G: (x,y) mapsto (x-fracb3ay,y)$. Then a direct computation (if I do not have an error) yields $G(p)=ax^3+(frac-b^23a+c)xy^2+epsilon y^3$, for some $epsilon in mathbbR$.
If we further assume that $a,b,c,d in mathbbR^+$, then we are almost done; indeed, $G(p)_x=3ax^2+(frac-b^23a+c)y^2$, which is non-negative in case
$frac-b^23a+c geq 0$, so we need $3ac-b^2 geq 0$.
(Allowing $G$ to be an affine $mathbbC$-algebra automorphism of $mathbbC[x,y]$ still does not help; if we take $H: (x,y) mapsto (x-fracbi3ay,y)$ instead of our former $G$, then $delta xy$ will appear in $H(p)_x$, for some $delta in mathbbC$).
(3) $d=5$: Write $p=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$, $a,b,c,d,e,f in mathbbR^+$.
"Completing the square" trick, as far as I see, only guarantees that we can find an affine $G$ such that $G(p)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$ for some $A,C,D,E,F in mathbbR$, but this is not enough, because $G(p)_x=5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains the problematic term $2Dxy^3$.
Actually, one also has to be careful about the signs of $A,C,E$, which are dependent on the relations between the original coefficients $a,b,c,d,e,f$ (even if all of them are positive), as we have seen in the case $d=3$.
Any comments are welcome!
See also this question.
algebraic-geometry polynomials real-algebraic-geometry
asked Jul 17 at 12:40
user237522
1,8141617
edited Jul 17 at 19:00
asked Jul 17 at 12:40
user237522
1,8141617
asked Jul 17 at 12:40
user237522
1,8141617
asked Jul 17 at 12:40
user237522
1,8141617
1,8141617
add a comment |Â
add a comment |Â
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