Mixing
“Recentering” open balls in $mathbb{R}^n$
1
$begingroup$
Say I have an open ball around the origin with radius r: B(r,0), and I want to transform it so that some point p!= 0 is the center (want to keep radius the same). How can I do this?
Edit: I want p to be invariant under this map.
real-analysis
asked Jan 26 at 20:41
user549064user549064
64
$endgroup$
add a comment |
1
$begingroup$
Say I have an open ball around the origin with radius r: B(r,0), and I want to transform it so that some point p!= 0 is the center (want to keep radius the same). How can I do this?
Edit: I want p to be invariant under this map.
real-analysis
asked Jan 26 at 20:41
user549064user549064
64
$endgroup$
2
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44
add a comment |
1
1
1
0
$begingroup$
Say I have an open ball around the origin with radius r: B(r,0), and I want to transform it so that some point p!= 0 is the center (want to keep radius the same). How can I do this?
Edit: I want p to be invariant under this map.
real-analysis
asked Jan 26 at 20:41
user549064user549064
64
$endgroup$
Say I have an open ball around the origin with radius r: B(r,0), and I want to transform it so that some point p!= 0 is the center (want to keep radius the same). How can I do this?
Edit: I want p to be invariant under this map.
real-analysis
real-analysis
asked Jan 26 at 20:41
user549064user549064
64
asked Jan 26 at 20:41
user549064user549064
64
edited Jan 26 at 21:08
user549064
asked Jan 26 at 20:41
user549064user549064
64
asked Jan 26 at 20:41
user549064user549064
64
asked Jan 26 at 20:41
user549064user549064
64
64
2
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44
add a comment |
2
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44
2
2
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44
add a comment |
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2
$begingroup$
Consider the map $F(x) = x+p$ where $pinmathbb{R}^n$. Then $F(B(r,0))=B(r,p)$
$endgroup$
– Yanko
Jan 26 at 20:43
$begingroup$
@Yanko. When p is in the ball, p is not invariant under the map.
$endgroup$
– William Elliot
Jan 26 at 23:51
$begingroup$
@WilliamElliot Ah now I see his edit... well..
$endgroup$
– Yanko
Jan 27 at 10:44