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'Completing the square' in higher degrees
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Let $p_2=ax^2+bxy+cy^2$, $a,b,c in mathbbR$ with $a neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain:
$p_2=a(x^2+fracbaxy+fraccay^2)=
a(x^2+2xfracb2ay+fracb^24a^2y^2-fracb^24a^2y^2+fraccay^2)=
a((x+fracb2ay)^2+(fracca-fracb^24a^2)y^2)=
a(x+fracb2ay)^2+afrac4ac-b^24a^2y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$ with $a neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+fracbax^2y+fraccaxy^2+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3
-3x(fracb3ay)^2+fraccaxy^2-(fracb3ay)^3+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3)+
a(-3xfracb^29a^2y^2+fraccaxy^2-fracb^327a^3y^3+fracday^3)=
a(x+fracb3ay)^3+
a(fracca-3fracb^29a^2)xy^2+a(fracda-fracb^327a^3)y^3=
a(x+fracb3ay)^3+
frac9ac-3b^29axy^2+frac27a^2d-b^327a^2y^3=
a(x+fracb3ay)^3+
epsilon xy^2+delta y^3$
where $epsilon:=frac9ac-3b^29a=frac3ac-b^23a$ and $delta:=frac27a^2d-b^327a^2$.
Now let $G: (x,y) mapsto (x-fracb3ay,y)$; $G$ is an (affine) automorphism of $mathbbR$. We get that: $G(p_3)=G(a(x+fracb3ay)^3+
epsilon xy^2+delta y^3)= aG(x+fracb3ay)^3+
epsilon G(xy^2)+delta G(y^3)=ax^3+
epsilon (x-fracb3ay)y^2+delta y^3=
ax^3+
epsilon xy^2+ (delta-epsilonfracb3a)y^3$
Therefore, $fracpartial (G(p_3))partial x=3ax^2+epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree.
Is there a similar trick for higher odd degrees $2n+1 in 5,7,ldots$? Namely, can one find an affine automorphism $G$ of $mathbbR[x,y]$ such that every monomial in $fracpartial G(p_2n+1)partial x$ has even $x$ degree and even $y$ degree. ?
The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then
$fracpartial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)partial x
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Jul 18 at 15:04
user237522
1,8141617
add a comment |Â
up vote
3
down vote
favorite
Let $p_2=ax^2+bxy+cy^2$, $a,b,c in mathbbR$ with $a neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain:
$p_2=a(x^2+fracbaxy+fraccay^2)=
a(x^2+2xfracb2ay+fracb^24a^2y^2-fracb^24a^2y^2+fraccay^2)=
a((x+fracb2ay)^2+(fracca-fracb^24a^2)y^2)=
a(x+fracb2ay)^2+afrac4ac-b^24a^2y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$ with $a neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+fracbax^2y+fraccaxy^2+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3
-3x(fracb3ay)^2+fraccaxy^2-(fracb3ay)^3+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3)+
a(-3xfracb^29a^2y^2+fraccaxy^2-fracb^327a^3y^3+fracday^3)=
a(x+fracb3ay)^3+
a(fracca-3fracb^29a^2)xy^2+a(fracda-fracb^327a^3)y^3=
a(x+fracb3ay)^3+
frac9ac-3b^29axy^2+frac27a^2d-b^327a^2y^3=
a(x+fracb3ay)^3+
epsilon xy^2+delta y^3$
where $epsilon:=frac9ac-3b^29a=frac3ac-b^23a$ and $delta:=frac27a^2d-b^327a^2$.
Now let $G: (x,y) mapsto (x-fracb3ay,y)$; $G$ is an (affine) automorphism of $mathbbR$. We get that: $G(p_3)=G(a(x+fracb3ay)^3+
epsilon xy^2+delta y^3)= aG(x+fracb3ay)^3+
epsilon G(xy^2)+delta G(y^3)=ax^3+
epsilon (x-fracb3ay)y^2+delta y^3=
ax^3+
epsilon xy^2+ (delta-epsilonfracb3a)y^3$
Therefore, $fracpartial (G(p_3))partial x=3ax^2+epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree.
Is there a similar trick for higher odd degrees $2n+1 in 5,7,ldots$? Namely, can one find an affine automorphism $G$ of $mathbbR[x,y]$ such that every monomial in $fracpartial G(p_2n+1)partial x$ has even $x$ degree and even $y$ degree. ?
The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then
$fracpartial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)partial x
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Jul 18 at 15:04
user237522
1,8141617
I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $p_2=ax^2+bxy+cy^2$, $a,b,c in mathbbR$ with $a neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain:
$p_2=a(x^2+fracbaxy+fraccay^2)=
a(x^2+2xfracb2ay+fracb^24a^2y^2-fracb^24a^2y^2+fraccay^2)=
a((x+fracb2ay)^2+(fracca-fracb^24a^2)y^2)=
a(x+fracb2ay)^2+afrac4ac-b^24a^2y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$ with $a neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+fracbax^2y+fraccaxy^2+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3
-3x(fracb3ay)^2+fraccaxy^2-(fracb3ay)^3+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3)+
a(-3xfracb^29a^2y^2+fraccaxy^2-fracb^327a^3y^3+fracday^3)=
a(x+fracb3ay)^3+
a(fracca-3fracb^29a^2)xy^2+a(fracda-fracb^327a^3)y^3=
a(x+fracb3ay)^3+
frac9ac-3b^29axy^2+frac27a^2d-b^327a^2y^3=
a(x+fracb3ay)^3+
epsilon xy^2+delta y^3$
where $epsilon:=frac9ac-3b^29a=frac3ac-b^23a$ and $delta:=frac27a^2d-b^327a^2$.
Now let $G: (x,y) mapsto (x-fracb3ay,y)$; $G$ is an (affine) automorphism of $mathbbR$. We get that: $G(p_3)=G(a(x+fracb3ay)^3+
epsilon xy^2+delta y^3)= aG(x+fracb3ay)^3+
epsilon G(xy^2)+delta G(y^3)=ax^3+
epsilon (x-fracb3ay)y^2+delta y^3=
ax^3+
epsilon xy^2+ (delta-epsilonfracb3a)y^3$
Therefore, $fracpartial (G(p_3))partial x=3ax^2+epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree.
Is there a similar trick for higher odd degrees $2n+1 in 5,7,ldots$? Namely, can one find an affine automorphism $G$ of $mathbbR[x,y]$ such that every monomial in $fracpartial G(p_2n+1)partial x$ has even $x$ degree and even $y$ degree. ?
The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then
$fracpartial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)partial x
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Jul 18 at 15:04
user237522
1,8141617
Let $p_2=ax^2+bxy+cy^2$, $a,b,c in mathbbR$ with $a neq 0$;
$p_2$ is homogeneous of degree $2$.
By 'completing the square trick' we obtain:
$p_2=a(x^2+fracbaxy+fraccay^2)=
a(x^2+2xfracb2ay+fracb^24a^2y^2-fracb^24a^2y^2+fraccay^2)=
a((x+fracb2ay)^2+(fracca-fracb^24a^2)y^2)=
a(x+fracb2ay)^2+afrac4ac-b^24a^2y^2$
Now consider $p_3=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d in mathbbR$ with $a neq 0$; $p_3$ is homogeneous of degree $3$.
Similarly to the above trick we obtain:
$p_3=a(x^3+fracbax^2y+fraccaxy^2+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3
-3x(fracb3ay)^2+fraccaxy^2-(fracb3ay)^3+fracday^3)=
a(x^3+3x^2fracb3ay+3x(fracb3ay)^2+(fracb3ay)^3)+
a(-3xfracb^29a^2y^2+fraccaxy^2-fracb^327a^3y^3+fracday^3)=
a(x+fracb3ay)^3+
a(fracca-3fracb^29a^2)xy^2+a(fracda-fracb^327a^3)y^3=
a(x+fracb3ay)^3+
frac9ac-3b^29axy^2+frac27a^2d-b^327a^2y^3=
a(x+fracb3ay)^3+
epsilon xy^2+delta y^3$
where $epsilon:=frac9ac-3b^29a=frac3ac-b^23a$ and $delta:=frac27a^2d-b^327a^2$.
Now let $G: (x,y) mapsto (x-fracb3ay,y)$; $G$ is an (affine) automorphism of $mathbbR$. We get that: $G(p_3)=G(a(x+fracb3ay)^3+
epsilon xy^2+delta y^3)= aG(x+fracb3ay)^3+
epsilon G(xy^2)+delta G(y^3)=ax^3+
epsilon (x-fracb3ay)y^2+delta y^3=
ax^3+
epsilon xy^2+ (delta-epsilonfracb3a)y^3$
Therefore, $fracpartial (G(p_3))partial x=3ax^2+epsilon y^2$,
in which every monomial has even $x$ degree and even $y$ degree.
Is there a similar trick for higher odd degrees $2n+1 in 5,7,ldots$? Namely, can one find an affine automorphism $G$ of $mathbbR[x,y]$ such that every monomial in $fracpartial G(p_2n+1)partial x$ has even $x$ degree and even $y$ degree. ?
The problem is that, for example, for degree $5$ the above trick only yields the existence of an automorphism $H$ such that:
$H(p_5)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$, and then
$fracpartial (Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5)partial x
= 5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains $2Dxy^3$.
Related questions are: i (former question of mine), ii (geometric interpretation) and iii (which is not exactly what I have asked; I did not ask for a sum of squares).
Thank you very much!
algebraic-geometry polynomials commutative-algebra
asked Jul 18 at 15:04
user237522
1,8141617
edited Jul 18 at 15:45
asked Jul 18 at 15:04
user237522
1,8141617
asked Jul 18 at 15:04
user237522
1,8141617
asked Jul 18 at 15:04
user237522
1,8141617
1,8141617
I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34
add a comment |Â
I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34
I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34
add a comment |Â
1 Answer
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By dehomogenizing the equation, this is equivalent to finding a substitution $xmapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $xmapsto x+c$ is a group action by $Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_2n+1x^2n+1+a_2n-1x^2n-1+cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
answered Jul 18 at 19:22
KReiser
7,54511230
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By dehomogenizing the equation, this is equivalent to finding a substitution $xmapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $xmapsto x+c$ is a group action by $Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_2n+1x^2n+1+a_2n-1x^2n-1+cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
answered Jul 18 at 19:22
KReiser
7,54511230
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
add a comment |Â
up vote
1
down vote
accepted
By dehomogenizing the equation, this is equivalent to finding a substitution $xmapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $xmapsto x+c$ is a group action by $Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_2n+1x^2n+1+a_2n-1x^2n-1+cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
answered Jul 18 at 19:22
KReiser
7,54511230
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By dehomogenizing the equation, this is equivalent to finding a substitution $xmapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $xmapsto x+c$ is a group action by $Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_2n+1x^2n+1+a_2n-1x^2n-1+cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
answered Jul 18 at 19:22
KReiser
7,54511230
By dehomogenizing the equation, this is equivalent to finding a substitution $xmapsto x+c$ so that after substituting, $p(x)$ has no terms of even degree. This means that the even symmetric polynomials in the roots should evaluate to $0$ after translating all roots by $c$. This is not in general possible: $c$ is uniquely determined by the top two coefficients, and the set of $c$ which work for each individual lower degree coefficients is a Zariski-closed proper subset of $Bbb R$, ie a finite collection of points. Since the polynomials are algebraically independent (by the fundamental theorem on symmetric polynomials), asking for $c$ to be in each of these subsets is a probability 0 condition.
Alternately, since the translation $xmapsto x+c$ is a group action by $Bbb R$ on the space of polynomials, it's enough to demonstrate that the orbit of $a_2n+1x^2n+1+a_2n-1x^2n-1+cdots+a_1x+a_0$ is of small dimension. The dimension of the space of degree exactly $2n+1$ polnyomials is $2n+2$, and the dimension of the requested orbit is $n+3$ - for $n>1$, this gives that the requested orbit has smaller dimension than the whole space, and therefore a polynomial is in this orbit with probability 0.
answered Jul 18 at 19:22
KReiser
7,54511230
edited Jul 18 at 20:57
answered Jul 18 at 19:22
KReiser
7,54511230
answered Jul 18 at 19:22
KReiser
7,54511230
answered Jul 18 at 19:22
KReiser
7,54511230
7,54511230
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
add a comment |Â
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
Thank you. So in degree $3$ we just had 'luck', which we will probably not have in higher degrees? Also, what if we assume that all the coefficients of $p(2n+1)$ belong to $1,-1$ and still work over $mathbbR$ (not over $mathbbZ$)?
– user237522
Jul 18 at 20:46
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
@user237522 In degree 3, the translation gets rid of the coefficient of $x^2$ and the derivative gets rid of the coefficient of $x^0$. For the restricted coefficient problem, you'd have to compute whether any of the given polynomials lie in the requested orbit. Probability 0 would suggest it's unlikely (although possible).
– KReiser
Jul 18 at 20:58
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
Thanks for the comment and for editing your answer.
– user237522
Jul 18 at 21:07
add a comment |Â
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I think one rough way about thinking about it is that completing the square is doing an affine change of co-ordinates to get a desired form.
– Andres Mejia
Jul 18 at 15:25
Thanks for the comment. Please, do you think that it is hopeless to find such $G$ for degrees $ geq 5$ (at least without making any further assumptions)? What if we allow $G$ to be any automorphism of $mathbbR[x,y]$, not just an affine automorphism? In this case, $G(p_2n+1)$ will not necessarily be homogeneous of degree $2n+1$, but we can still ask whether its partial derivative w.r.t. $x$ has monomials of even $x$ and $y$ degrees.
– user237522
Jul 18 at 15:34