Mixing
How to draw this set in $mathbb{R}^{2}$?
$$S = {(a,b)in mathbb{R}^{2}:ax_1+bx_2leq 1, forall (x_1,x_2)in mathbb{R}^{2}text{ such that }|x_1|+|x_2|leq 1}$$
I want to sketch the solutions of this set in $mathbb{R}^{2},$ but I am not sure what it looks like. I tried taking specific vectors $(x_1,x_2)$ but this does not help either. Any ideas will be much appreciated.
Edit:
If I take the vector $(pm 1,0)$ then $|a|leq 1$ and similarily $(0,pm1)$ implies that $|b|leq 1.$ Can I still conclude that $S$ is a square such that $|a|leq 1$ and $|b|leq 1.$
graphing-functions
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
add a comment |
$$S = {(a,b)in mathbb{R}^{2}:ax_1+bx_2leq 1, forall (x_1,x_2)in mathbb{R}^{2}text{ such that }|x_1|+|x_2|leq 1}$$
I want to sketch the solutions of this set in $mathbb{R}^{2},$ but I am not sure what it looks like. I tried taking specific vectors $(x_1,x_2)$ but this does not help either. Any ideas will be much appreciated.
Edit:
If I take the vector $(pm 1,0)$ then $|a|leq 1$ and similarily $(0,pm1)$ implies that $|b|leq 1.$ Can I still conclude that $S$ is a square such that $|a|leq 1$ and $|b|leq 1.$
graphing-functions
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15
add a comment |
$$S = {(a,b)in mathbb{R}^{2}:ax_1+bx_2leq 1, forall (x_1,x_2)in mathbb{R}^{2}text{ such that }|x_1|+|x_2|leq 1}$$
I want to sketch the solutions of this set in $mathbb{R}^{2},$ but I am not sure what it looks like. I tried taking specific vectors $(x_1,x_2)$ but this does not help either. Any ideas will be much appreciated.
Edit:
If I take the vector $(pm 1,0)$ then $|a|leq 1$ and similarily $(0,pm1)$ implies that $|b|leq 1.$ Can I still conclude that $S$ is a square such that $|a|leq 1$ and $|b|leq 1.$
graphing-functions
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
$$S = {(a,b)in mathbb{R}^{2}:ax_1+bx_2leq 1, forall (x_1,x_2)in mathbb{R}^{2}text{ such that }|x_1|+|x_2|leq 1}$$
I want to sketch the solutions of this set in $mathbb{R}^{2},$ but I am not sure what it looks like. I tried taking specific vectors $(x_1,x_2)$ but this does not help either. Any ideas will be much appreciated.
Edit:
If I take the vector $(pm 1,0)$ then $|a|leq 1$ and similarily $(0,pm1)$ implies that $|b|leq 1.$ Can I still conclude that $S$ is a square such that $|a|leq 1$ and $|b|leq 1.$
graphing-functions
graphing-functions
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
edited Nov 21 '18 at 0:15
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
asked Nov 20 '18 at 23:49
Hello_World
3,90821630
3,90821630
The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15
add a comment |
The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15
The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15
add a comment |
1 Answer
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$$|x_1|+|x_2|leq 1implies begin{cases}x_1+x_2leq1 & x_1,x_2geq0\-x_1+x_2leq1 &x_1<0,:x_2geq0\x_1-x_2leq1 &x_1geq0,:x_2<0\-x_1-x_2leq1 &x_1,x_2<0end{cases}$$
Thus $$ax_1+bx_2leq 1: text{such that}: |x_1|+|x_2|leq 1implies begin{cases}ax_1+bx_2leq1 & x_1,x_2geq0\-ax_1+bx_2leq1 &x_1<0,:x_2geq0\ax_1-bx_2leq1 &x_1geq0,:x_2<0\-ax_1-bx_2leq1 &x_1,x_2<0end{cases}$$
(*You shall end up with some form of a rectangle in the plane $mathbb{R}^2$).
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
add a comment |
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1 Answer
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1 Answer
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$$|x_1|+|x_2|leq 1implies begin{cases}x_1+x_2leq1 & x_1,x_2geq0\-x_1+x_2leq1 &x_1<0,:x_2geq0\x_1-x_2leq1 &x_1geq0,:x_2<0\-x_1-x_2leq1 &x_1,x_2<0end{cases}$$
Thus $$ax_1+bx_2leq 1: text{such that}: |x_1|+|x_2|leq 1implies begin{cases}ax_1+bx_2leq1 & x_1,x_2geq0\-ax_1+bx_2leq1 &x_1<0,:x_2geq0\ax_1-bx_2leq1 &x_1geq0,:x_2<0\-ax_1-bx_2leq1 &x_1,x_2<0end{cases}$$
(*You shall end up with some form of a rectangle in the plane $mathbb{R}^2$).
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
add a comment |
$$|x_1|+|x_2|leq 1implies begin{cases}x_1+x_2leq1 & x_1,x_2geq0\-x_1+x_2leq1 &x_1<0,:x_2geq0\x_1-x_2leq1 &x_1geq0,:x_2<0\-x_1-x_2leq1 &x_1,x_2<0end{cases}$$
Thus $$ax_1+bx_2leq 1: text{such that}: |x_1|+|x_2|leq 1implies begin{cases}ax_1+bx_2leq1 & x_1,x_2geq0\-ax_1+bx_2leq1 &x_1<0,:x_2geq0\ax_1-bx_2leq1 &x_1geq0,:x_2<0\-ax_1-bx_2leq1 &x_1,x_2<0end{cases}$$
(*You shall end up with some form of a rectangle in the plane $mathbb{R}^2$).
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
add a comment |
$$|x_1|+|x_2|leq 1implies begin{cases}x_1+x_2leq1 & x_1,x_2geq0\-x_1+x_2leq1 &x_1<0,:x_2geq0\x_1-x_2leq1 &x_1geq0,:x_2<0\-x_1-x_2leq1 &x_1,x_2<0end{cases}$$
Thus $$ax_1+bx_2leq 1: text{such that}: |x_1|+|x_2|leq 1implies begin{cases}ax_1+bx_2leq1 & x_1,x_2geq0\-ax_1+bx_2leq1 &x_1<0,:x_2geq0\ax_1-bx_2leq1 &x_1geq0,:x_2<0\-ax_1-bx_2leq1 &x_1,x_2<0end{cases}$$
(*You shall end up with some form of a rectangle in the plane $mathbb{R}^2$).
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
$$|x_1|+|x_2|leq 1implies begin{cases}x_1+x_2leq1 & x_1,x_2geq0\-x_1+x_2leq1 &x_1<0,:x_2geq0\x_1-x_2leq1 &x_1geq0,:x_2<0\-x_1-x_2leq1 &x_1,x_2<0end{cases}$$
Thus $$ax_1+bx_2leq 1: text{such that}: |x_1|+|x_2|leq 1implies begin{cases}ax_1+bx_2leq1 & x_1,x_2geq0\-ax_1+bx_2leq1 &x_1<0,:x_2geq0\ax_1-bx_2leq1 &x_1geq0,:x_2<0\-ax_1-bx_2leq1 &x_1,x_2<0end{cases}$$
(*You shall end up with some form of a rectangle in the plane $mathbb{R}^2$).
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
answered Nov 20 '18 at 23:59
Yadati Kiran
1,693619
1,693619
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
add a comment |
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
Hey could you give feedback on the edit I made?
– Hello_World
Nov 21 '18 at 0:19
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
$S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $frac{1}{2}x_1+x_2leq1 $ subject to the same above constraints. Will the $S$ still be a square?
– Yadati Kiran
Nov 21 '18 at 0:22
add a comment |
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The first relation is a half-space. The second is a square. The intersection have some sort of convex with straight sides shape.
– Will M.
Nov 21 '18 at 0:04
What do you mean by half-space?
– Hello_World
Nov 21 '18 at 0:07
I mean one side of a hyperplane. If you do not know linear algebra, you can ignore my comments.
– Will M.
Nov 21 '18 at 0:08
Oh, I understand now what you mean.
– Hello_World
Nov 21 '18 at 0:10
@WillM. Does the edit make sense?
– Hello_World
Nov 21 '18 at 0:15