Mixing
Verify if $n$ dimensional spheres intersect
$begingroup$
Consider $x_1,dots,x_n$ vectors lying in $mathbb{R}^n$. Also, consider positive numbers $alpha_1,dots,alpha_n$. Consider $n$ spheres with center as $x_i$ and radii $alpha_i$. How do I check the set of solutions to the set of equations
begin{align}
||x-x_i||_2 = alpha_i~,~ forall i
end{align}
is non-empty. Is this problem studied in literature?
real-analysis linear-algebra geometry
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
$endgroup$
add a comment |
$begingroup$
Consider $x_1,dots,x_n$ vectors lying in $mathbb{R}^n$. Also, consider positive numbers $alpha_1,dots,alpha_n$. Consider $n$ spheres with center as $x_i$ and radii $alpha_i$. How do I check the set of solutions to the set of equations
begin{align}
||x-x_i||_2 = alpha_i~,~ forall i
end{align}
is non-empty. Is this problem studied in literature?
real-analysis linear-algebra geometry
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
$endgroup$
2
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
1
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49
add a comment |
$begingroup$
Consider $x_1,dots,x_n$ vectors lying in $mathbb{R}^n$. Also, consider positive numbers $alpha_1,dots,alpha_n$. Consider $n$ spheres with center as $x_i$ and radii $alpha_i$. How do I check the set of solutions to the set of equations
begin{align}
||x-x_i||_2 = alpha_i~,~ forall i
end{align}
is non-empty. Is this problem studied in literature?
real-analysis linear-algebra geometry
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
$endgroup$
Consider $x_1,dots,x_n$ vectors lying in $mathbb{R}^n$. Also, consider positive numbers $alpha_1,dots,alpha_n$. Consider $n$ spheres with center as $x_i$ and radii $alpha_i$. How do I check the set of solutions to the set of equations
begin{align}
||x-x_i||_2 = alpha_i~,~ forall i
end{align}
is non-empty. Is this problem studied in literature?
real-analysis linear-algebra geometry
real-analysis linear-algebra geometry
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
edited Feb 2 at 12:53
dineshdileep
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
asked Feb 2 at 12:39
dineshdileepdineshdileep
6,03611935
6,03611935
2
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
1
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49
add a comment |
2
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
1
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49
2
2
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
1
1
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The question is well-studied. For example, you can read this.
One of the methods works like this:
Rewrite your equation as
$$
2x_ix = x^2 +x_i^2 - alpha_i^2.
$$
Find $x$ in form $x=ru+v$, where $r=x^2$, $u$ is the solution of $2x_iu=1$, $v$ is the solution of $2x_iv=x_i^2-alpha_i^2$. Then
$$
r=x^2=(ru+v)^2=r^2u^2+2uv r+v^2
$$
is a quadratic equation on $r$. If this equation has roots, then your $n$ spheres have a common point.
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
The question is well-studied. For example, you can read this.
One of the methods works like this:
Rewrite your equation as
$$
2x_ix = x^2 +x_i^2 - alpha_i^2.
$$
Find $x$ in form $x=ru+v$, where $r=x^2$, $u$ is the solution of $2x_iu=1$, $v$ is the solution of $2x_iv=x_i^2-alpha_i^2$. Then
$$
r=x^2=(ru+v)^2=r^2u^2+2uv r+v^2
$$
is a quadratic equation on $r$. If this equation has roots, then your $n$ spheres have a common point.
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
add a comment |
$begingroup$
The question is well-studied. For example, you can read this.
One of the methods works like this:
Rewrite your equation as
$$
2x_ix = x^2 +x_i^2 - alpha_i^2.
$$
Find $x$ in form $x=ru+v$, where $r=x^2$, $u$ is the solution of $2x_iu=1$, $v$ is the solution of $2x_iv=x_i^2-alpha_i^2$. Then
$$
r=x^2=(ru+v)^2=r^2u^2+2uv r+v^2
$$
is a quadratic equation on $r$. If this equation has roots, then your $n$ spheres have a common point.
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
add a comment |
$begingroup$
The question is well-studied. For example, you can read this.
One of the methods works like this:
Rewrite your equation as
$$
2x_ix = x^2 +x_i^2 - alpha_i^2.
$$
Find $x$ in form $x=ru+v$, where $r=x^2$, $u$ is the solution of $2x_iu=1$, $v$ is the solution of $2x_iv=x_i^2-alpha_i^2$. Then
$$
r=x^2=(ru+v)^2=r^2u^2+2uv r+v^2
$$
is a quadratic equation on $r$. If this equation has roots, then your $n$ spheres have a common point.
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
$endgroup$
The question is well-studied. For example, you can read this.
One of the methods works like this:
Rewrite your equation as
$$
2x_ix = x^2 +x_i^2 - alpha_i^2.
$$
Find $x$ in form $x=ru+v$, where $r=x^2$, $u$ is the solution of $2x_iu=1$, $v$ is the solution of $2x_iv=x_i^2-alpha_i^2$. Then
$$
r=x^2=(ru+v)^2=r^2u^2+2uv r+v^2
$$
is a quadratic equation on $r$. If this equation has roots, then your $n$ spheres have a common point.
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
answered Feb 2 at 13:14
Vasily MitchVasily Mitch
2,6791312
2,6791312
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
add a comment |
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
Does this work for vectors?
$endgroup$
– dineshdileep
Feb 3 at 5:06
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
$begingroup$
It is vector equations
$endgroup$
– Vasily Mitch
Feb 3 at 10:26
add a comment |
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2
$begingroup$
Why can't you take an arbitrary $x_0$ and use $alpha_i = |x_0-x_i|$?
$endgroup$
– Vasily Mitch
Feb 2 at 12:42
1
$begingroup$
Another even more brutal solution : let M= maximal distance $|x_i-x_j|$. Then take all $alpha_i=M$...
$endgroup$
– Jean Marie
Feb 2 at 12:45
$begingroup$
@VasilyMitch I see your point, thanks. This is not what I actually wanted. Let me rewrite the question
$endgroup$
– dineshdileep
Feb 2 at 12:49