Mixing
Let $f(x,y) = 2x+y-2$. Suppose $(a,b)$ satisfies $f(a,b) > 0$. Show that for $(c,d)$ in open ball centered...
$begingroup$
Prove that if the point $(a,b)$ in the plane satisfies $2a+b-2 >0$, and if $d$ ($=frac{|2a+b-2|}{sqrt{5}}$ can be found using a formula) is the shortest distance from the point $(a,b)$ to the line $2x+y-2=0$, that every element $(c,d)in B_{frac{d}{2}}((a,b))$ (the open ball centered at $(a,b)$ with radius $frac{d}{2}$) satisfies $2c+d-2>0$
I'm finishing up a proof for one of my classes, but I don't feel it's rigorous enough and need this fact to help solidify my understanding. I can't seen to find how small $2c+d-2$ could using my ball, but I know there is some lower bound. Also note that the radius doesn't need to be $frac{d}{2}$ just smaller than $d$.
real-analysis
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
$endgroup$
|
show 2 more comments
$begingroup$
Prove that if the point $(a,b)$ in the plane satisfies $2a+b-2 >0$, and if $d$ ($=frac{|2a+b-2|}{sqrt{5}}$ can be found using a formula) is the shortest distance from the point $(a,b)$ to the line $2x+y-2=0$, that every element $(c,d)in B_{frac{d}{2}}((a,b))$ (the open ball centered at $(a,b)$ with radius $frac{d}{2}$) satisfies $2c+d-2>0$
I'm finishing up a proof for one of my classes, but I don't feel it's rigorous enough and need this fact to help solidify my understanding. I can't seen to find how small $2c+d-2$ could using my ball, but I know there is some lower bound. Also note that the radius doesn't need to be $frac{d}{2}$ just smaller than $d$.
real-analysis
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
$endgroup$
$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51
|
show 2 more comments
$begingroup$
Prove that if the point $(a,b)$ in the plane satisfies $2a+b-2 >0$, and if $d$ ($=frac{|2a+b-2|}{sqrt{5}}$ can be found using a formula) is the shortest distance from the point $(a,b)$ to the line $2x+y-2=0$, that every element $(c,d)in B_{frac{d}{2}}((a,b))$ (the open ball centered at $(a,b)$ with radius $frac{d}{2}$) satisfies $2c+d-2>0$
I'm finishing up a proof for one of my classes, but I don't feel it's rigorous enough and need this fact to help solidify my understanding. I can't seen to find how small $2c+d-2$ could using my ball, but I know there is some lower bound. Also note that the radius doesn't need to be $frac{d}{2}$ just smaller than $d$.
real-analysis
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
$endgroup$
Prove that if the point $(a,b)$ in the plane satisfies $2a+b-2 >0$, and if $d$ ($=frac{|2a+b-2|}{sqrt{5}}$ can be found using a formula) is the shortest distance from the point $(a,b)$ to the line $2x+y-2=0$, that every element $(c,d)in B_{frac{d}{2}}((a,b))$ (the open ball centered at $(a,b)$ with radius $frac{d}{2}$) satisfies $2c+d-2>0$
I'm finishing up a proof for one of my classes, but I don't feel it's rigorous enough and need this fact to help solidify my understanding. I can't seen to find how small $2c+d-2$ could using my ball, but I know there is some lower bound. Also note that the radius doesn't need to be $frac{d}{2}$ just smaller than $d$.
real-analysis
real-analysis
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
asked Jan 17 at 3:07
TrigginometricTrigginometric
114
114
$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51
|
show 2 more comments
$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51
$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51
|
show 2 more comments
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$begingroup$
The jist should be f is continuous and the only points where f =0 are on the line 2x+y-2 = 0, so if f is positive at (a,b) and my neighborhood doesn't intersect that line, f is positive on every point in the neighborhood.
$endgroup$
– user113102
Jan 17 at 5:17
$begingroup$
That's actually exactly what my proof of the question that this proof is required for does. However, I don't feel it is sufficient enough and I'm trying to find a lower bound on $2c+d-2$ using the open ball. I spoke to my professor who told me there is a way, I just can't see it
$endgroup$
– Trigginometric
Jan 17 at 5:37
$begingroup$
You're trying to understand how small can $2x + y- 2$ be if $(x,y)$ is inside the circle centered in $(a,b)$ with radius $frac{d}{2}$? Correct?
$endgroup$
– Matteo
Jan 17 at 10:01
$begingroup$
Ah. Maybe think about it this way: the level sets of the function f(x,y)=2x+y-2 are the lines 2x+y-2 = t. As t varies, this shifts the line left (as t gets more positive) and right (as t gets more negative), but remains parallel to the line 2x+y-2 = 0. For your fixed radius, can you find where this line intersects your ball for the first time? Then that t is the minimum value of the function 2x+y-2 on your ball.
$endgroup$
– user113102
Jan 17 at 13:59
$begingroup$
@Matteo Yes I am.
$endgroup$
– Trigginometric
Jan 17 at 19:51