Mixing
Symmetries of the roots of this polynomial?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I have a polynomial equation in $x$ and $y$,
$$
(a-b)(xy+1) + (ab+1)(y-x) = 0.
$$
What transformation can act on $x$ and $y$ so that any point that satisfies this equation is mapped to another point that satisfies it? If there is a general way to find these transformations given a polynomial equation, that would also be nice to know.
group-theory polynomials roots transformation
edited 19 mins ago
Batominovski
29.7k23187
asked 1 hour ago
Display Name
1606
add a comment |Â
up vote
2
down vote
favorite
I have a polynomial equation in $x$ and $y$,
$$
(a-b)(xy+1) + (ab+1)(y-x) = 0.
$$
What transformation can act on $x$ and $y$ so that any point that satisfies this equation is mapped to another point that satisfies it? If there is a general way to find these transformations given a polynomial equation, that would also be nice to know.
group-theory polynomials roots transformation
edited 19 mins ago
Batominovski
29.7k23187
asked 1 hour ago
Display Name
1606
not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
1
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a polynomial equation in $x$ and $y$,
$$
(a-b)(xy+1) + (ab+1)(y-x) = 0.
$$
What transformation can act on $x$ and $y$ so that any point that satisfies this equation is mapped to another point that satisfies it? If there is a general way to find these transformations given a polynomial equation, that would also be nice to know.
group-theory polynomials roots transformation
edited 19 mins ago
Batominovski
29.7k23187
asked 1 hour ago
Display Name
1606
I have a polynomial equation in $x$ and $y$,
$$
(a-b)(xy+1) + (ab+1)(y-x) = 0.
$$
What transformation can act on $x$ and $y$ so that any point that satisfies this equation is mapped to another point that satisfies it? If there is a general way to find these transformations given a polynomial equation, that would also be nice to know.
group-theory polynomials roots transformation
group-theory polynomials roots transformation
edited 19 mins ago
Batominovski
29.7k23187
asked 1 hour ago
Display Name
1606
edited 19 mins ago
Batominovski
29.7k23187
asked 1 hour ago
Display Name
1606
edited 19 mins ago
Batominovski
29.7k23187
edited 19 mins ago
Batominovski
29.7k23187
edited 19 mins ago
Batominovski
29.7k23187
29.7k23187
asked 1 hour ago
Display Name
1606
asked 1 hour ago
Display Name
1606
asked 1 hour ago
Display Name
1606
1606
not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
1
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago
add a comment |Â
not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
1
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago
not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
1
1
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.
One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$
answered 52 mins ago
Will Jagy
99.7k597198
How did you find these substitutions?
– Display Name
51 mins ago
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
add a comment |Â
up vote
2
down vote
You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.
Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Now, consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.
In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.
answered 45 mins ago
Batominovski
29.7k23187
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.
One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$
answered 52 mins ago
Will Jagy
99.7k597198
How did you find these substitutions?
– Display Name
51 mins ago
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
add a comment |Â
up vote
3
down vote
the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.
One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$
answered 52 mins ago
Will Jagy
99.7k597198
How did you find these substitutions?
– Display Name
51 mins ago
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.
One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$
answered 52 mins ago
Will Jagy
99.7k597198
the substitutions
$$ x = u + fracab+1b-a ; , ; $$
$$ y = v - fracab+1b-a ; , ; $$
take us to the hyperbola
$$ uv = C $$
where $C = C(a,b)$ is a constant.
One motion, for real $t neq 0,$ is
$$ (u,v) mapsto left(tu, fracvt right) $$
answered 52 mins ago
Will Jagy
99.7k597198
answered 52 mins ago
Will Jagy
99.7k597198
answered 52 mins ago
Will Jagy
99.7k597198
answered 52 mins ago
Will Jagy
99.7k597198
99.7k597198
How did you find these substitutions?
– Display Name
51 mins ago
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
add a comment |Â
How did you find these substitutions?
– Display Name
51 mins ago
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
How did you find these substitutions?
– Display Name
51 mins ago
How did you find these substitutions?
– Display Name
51 mins ago
2
2
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
@DisplayName the center of a hyperbola is the point where the gradient of the defining polynomial is zero. This is a pure translation. Your question falls under projective geometry, projective transformations take conic sections to conic sections.
– Will Jagy
45 mins ago
add a comment |Â
up vote
2
down vote
You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.
Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Now, consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.
In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.
answered 45 mins ago
Batominovski
29.7k23187
add a comment |Â
up vote
2
down vote
You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.
Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Now, consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.
In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.
answered 45 mins ago
Batominovski
29.7k23187
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.
Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Now, consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.
In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.
answered 45 mins ago
Batominovski
29.7k23187
You have $a,b,x,yinmathbbR$ ($mathbbR$ can be replaced by $mathbbC$ in every occurrence in this answer) such that
$$fraca-bab+1=fracx-yxy+1.tag*$$
Let $c:=arctan(a)$, $d:=arctan(b)$, $u:=arctan(x)$, and $v:=arctan(y)$. Consider $c$, $d$, $u$, and $v$ as elements of $mathbbR/pimathbbZ$. Then, we have
$$tan(c-d)=tan(u-v),.$$
Consequently, $c-d=u-v$.
Now, any transformation $T:mathbbR/pimathbbZtomathbbR/pimathbbZ$ of the form $$T(t)=t+qtext for all tinmathbbR/pimathbbZ,,$$ for some fixed constant $qinmathbbR/pimathbbZ$ satisfies
$$T(t_1)-T(t_2)=t_1-t_2,$$
for all $t_1,t_2inmathbbR/pimathbbZ$. Now, consider the function $f:mathbbRtomathbbR$ sending $$smapsto tanbig(arctan(s)+qbig)=fracs+p1-sptext for all sinmathbbR,,$$
where $qinmathbbR$ is a fixed constant and $p:=tan(q)$. Then, if $a,b,x,yinmathbbR$ satisfy (*), then
$$fraca-bab+1=fracf(x)-f(y)f(x),f(y)+1$$ too.
In fact, only functions $f$ of the given form satisfy the requirement. Thus, all solutions $(x,y)inmathbbRtimesmathbbR$ to (*) is of the form
$$(x,y)=left(fraca+p1-ap,fracb+p1-bpright)$$
for some $pinmathbbR$ such that $pneqdfrac1a$ and $pneq dfrac1b$.
answered 45 mins ago
Batominovski
29.7k23187
edited 20 mins ago
answered 45 mins ago
Batominovski
29.7k23187
answered 45 mins ago
Batominovski
29.7k23187
answered 45 mins ago
Batominovski
29.7k23187
29.7k23187
add a comment |Â
add a comment |Â
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not sure if this will work, but perhaps in equation $f(x,y) = g(x,y)$ you can transform $f$ to $g$ and vice versa?
– gt6989b
1 hour ago
1
Interesting...why the group theory tag?
– Rushabh Mehta
1 hour ago
@RushabhMehta I think the transformations I'm looking for should form a continuous group, and I know there are a lot of methods for finding those. If I had seen a "groups" tag without a "-theory" modifier I would have used it.
– Display Name
1 hour ago