Need a clarification of the definition $tilde d(p_1,p_2)=inftilde L(gamma):gamma(0)=p_1, gamma(1)=p_2$
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I cannot understand the definition of $tilde d(p_1,p_2)$ here? Can anyone please explain it clearly?
calculus real-analysis integration differential-geometry metric-spaces
edited Aug 15 at 3:32
user557902851
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asked Dec 21 '14 at 4:34
Extremal
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I cannot understand the definition of $tilde d(p_1,p_2)$ here? Can anyone please explain it clearly?
calculus real-analysis integration differential-geometry metric-spaces
edited Aug 15 at 3:32
user557902851
407117
asked Dec 21 '14 at 4:34
Extremal
3,1181626
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I cannot understand the definition of $tilde d(p_1,p_2)$ here? Can anyone please explain it clearly?
calculus real-analysis integration differential-geometry metric-spaces
edited Aug 15 at 3:32
user557902851
407117
asked Dec 21 '14 at 4:34
Extremal
3,1181626
I cannot understand the definition of $tilde d(p_1,p_2)$ here? Can anyone please explain it clearly?
calculus real-analysis integration differential-geometry metric-spaces
edited Aug 15 at 3:32
user557902851
407117
asked Dec 21 '14 at 4:34
Extremal
3,1181626
edited Aug 15 at 3:32
user557902851
407117
edited Aug 15 at 3:32
user557902851
407117
edited Aug 15 at 3:32
user557902851
407117
407117
asked Dec 21 '14 at 4:34
Extremal
3,1181626
asked Dec 21 '14 at 4:34
Extremal
3,1181626
asked Dec 21 '14 at 4:34
Extremal
3,1181626
3,1181626
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
down vote
As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $Bbb R^2-0$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
add a comment |Â
up vote
1
down vote
The definition given is $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$.
The condition $gamma(0) = p_1$, $gamma(1) = p_2$ means the curve $gamma(t)$ starts at point $p_1$ and ends at point $p_2$.
Since $barL(gamma)$ is the "length" of the curve $gamma$, $barL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the set of all possible "lengths" of curves which start at $p_0$ and end at $p_1$.
Therefore, $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the infimum (greatest lower bound) of the "lengths" of all curves which start at $p_1$ and end at $p_2$.
Note that the "length" of a curve is defined as $barL(gamma) = displaystyleint_gamma|x|,ds = int_0^1|x(t)|sqrtx'(t)^2+y'(t)^2,dt$ instead of the usual definition $L(gamma) = displaystyleint_gamma,ds = int_0^1sqrtx'(t)^2+y'(t)^2,dt$. With this definition, the length of a small segment of a curve is weighted by the absolute value of its $x$-coordinate.
Hence, the distance between two points $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is no longer the usual straight line Euclidean distance that you are used to seeing.
To compute $bard((1,1),(-1,-2))$, you'll need to figure our the infimum of the "lengths" of curves from $(1,1)$ to $(-1,-2)$. If you can come up with a curve from $(1,1)$ to $(-1,-2)$ and show that its "length" is less than the "length" of any other curve from $(1,1)$ to $(-1,-2)$, you'll have your answer. Hint: What is the "length" of any straight line segment that lies entirely on line $x = 0$? Use this fact to your advantage when deciding on what might be the curve from $(1,1)$ to $(-1,-2)$ with the least length.
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $Bbb R^2-0$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
add a comment |Â
up vote
2
down vote
As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $Bbb R^2-0$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $Bbb R^2-0$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
As in differential geometry, we are defining the "distance" between two points to be the inf of the "lengths" of all paths joining the points. (For example, with the usual notion of lengths of paths in $Bbb R^2-0$, what is the distance from $(-1,0)$ to $(1,0)$? Note that you cannot use the usual line segment joining them.)
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
answered Dec 21 '14 at 4:40
Ted Shifrin
60.1k44387
60.1k44387
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
add a comment |Â
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
But the distance is 2, right?
– Extremal
Dec 21 '14 at 5:13
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
Yes, even though there's no path of length $2$ joining those points, because the origin is deleted.
– Ted Shifrin
Dec 21 '14 at 12:39
add a comment |Â
up vote
1
down vote
The definition given is $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$.
The condition $gamma(0) = p_1$, $gamma(1) = p_2$ means the curve $gamma(t)$ starts at point $p_1$ and ends at point $p_2$.
Since $barL(gamma)$ is the "length" of the curve $gamma$, $barL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the set of all possible "lengths" of curves which start at $p_0$ and end at $p_1$.
Therefore, $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the infimum (greatest lower bound) of the "lengths" of all curves which start at $p_1$ and end at $p_2$.
Note that the "length" of a curve is defined as $barL(gamma) = displaystyleint_gamma|x|,ds = int_0^1|x(t)|sqrtx'(t)^2+y'(t)^2,dt$ instead of the usual definition $L(gamma) = displaystyleint_gamma,ds = int_0^1sqrtx'(t)^2+y'(t)^2,dt$. With this definition, the length of a small segment of a curve is weighted by the absolute value of its $x$-coordinate.
Hence, the distance between two points $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is no longer the usual straight line Euclidean distance that you are used to seeing.
To compute $bard((1,1),(-1,-2))$, you'll need to figure our the infimum of the "lengths" of curves from $(1,1)$ to $(-1,-2)$. If you can come up with a curve from $(1,1)$ to $(-1,-2)$ and show that its "length" is less than the "length" of any other curve from $(1,1)$ to $(-1,-2)$, you'll have your answer. Hint: What is the "length" of any straight line segment that lies entirely on line $x = 0$? Use this fact to your advantage when deciding on what might be the curve from $(1,1)$ to $(-1,-2)$ with the least length.
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
 |Â
show 6 more comments
up vote
1
down vote
The definition given is $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$.
The condition $gamma(0) = p_1$, $gamma(1) = p_2$ means the curve $gamma(t)$ starts at point $p_1$ and ends at point $p_2$.
Since $barL(gamma)$ is the "length" of the curve $gamma$, $barL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the set of all possible "lengths" of curves which start at $p_0$ and end at $p_1$.
Therefore, $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the infimum (greatest lower bound) of the "lengths" of all curves which start at $p_1$ and end at $p_2$.
Note that the "length" of a curve is defined as $barL(gamma) = displaystyleint_gamma|x|,ds = int_0^1|x(t)|sqrtx'(t)^2+y'(t)^2,dt$ instead of the usual definition $L(gamma) = displaystyleint_gamma,ds = int_0^1sqrtx'(t)^2+y'(t)^2,dt$. With this definition, the length of a small segment of a curve is weighted by the absolute value of its $x$-coordinate.
Hence, the distance between two points $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is no longer the usual straight line Euclidean distance that you are used to seeing.
To compute $bard((1,1),(-1,-2))$, you'll need to figure our the infimum of the "lengths" of curves from $(1,1)$ to $(-1,-2)$. If you can come up with a curve from $(1,1)$ to $(-1,-2)$ and show that its "length" is less than the "length" of any other curve from $(1,1)$ to $(-1,-2)$, you'll have your answer. Hint: What is the "length" of any straight line segment that lies entirely on line $x = 0$? Use this fact to your advantage when deciding on what might be the curve from $(1,1)$ to $(-1,-2)$ with the least length.
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
 |Â
show 6 more comments
up vote
1
down vote
up vote
1
down vote
The definition given is $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$.
The condition $gamma(0) = p_1$, $gamma(1) = p_2$ means the curve $gamma(t)$ starts at point $p_1$ and ends at point $p_2$.
Since $barL(gamma)$ is the "length" of the curve $gamma$, $barL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the set of all possible "lengths" of curves which start at $p_0$ and end at $p_1$.
Therefore, $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the infimum (greatest lower bound) of the "lengths" of all curves which start at $p_1$ and end at $p_2$.
Note that the "length" of a curve is defined as $barL(gamma) = displaystyleint_gamma|x|,ds = int_0^1|x(t)|sqrtx'(t)^2+y'(t)^2,dt$ instead of the usual definition $L(gamma) = displaystyleint_gamma,ds = int_0^1sqrtx'(t)^2+y'(t)^2,dt$. With this definition, the length of a small segment of a curve is weighted by the absolute value of its $x$-coordinate.
Hence, the distance between two points $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is no longer the usual straight line Euclidean distance that you are used to seeing.
To compute $bard((1,1),(-1,-2))$, you'll need to figure our the infimum of the "lengths" of curves from $(1,1)$ to $(-1,-2)$. If you can come up with a curve from $(1,1)$ to $(-1,-2)$ and show that its "length" is less than the "length" of any other curve from $(1,1)$ to $(-1,-2)$, you'll have your answer. Hint: What is the "length" of any straight line segment that lies entirely on line $x = 0$? Use this fact to your advantage when deciding on what might be the curve from $(1,1)$ to $(-1,-2)$ with the least length.
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
The definition given is $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$.
The condition $gamma(0) = p_1$, $gamma(1) = p_2$ means the curve $gamma(t)$ starts at point $p_1$ and ends at point $p_2$.
Since $barL(gamma)$ is the "length" of the curve $gamma$, $barL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the set of all possible "lengths" of curves which start at $p_0$ and end at $p_1$.
Therefore, $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is the infimum (greatest lower bound) of the "lengths" of all curves which start at $p_1$ and end at $p_2$.
Note that the "length" of a curve is defined as $barL(gamma) = displaystyleint_gamma|x|,ds = int_0^1|x(t)|sqrtx'(t)^2+y'(t)^2,dt$ instead of the usual definition $L(gamma) = displaystyleint_gamma,ds = int_0^1sqrtx'(t)^2+y'(t)^2,dt$. With this definition, the length of a small segment of a curve is weighted by the absolute value of its $x$-coordinate.
Hence, the distance between two points $bard(p_1,p_2) = infbarL(gamma) : gamma(0) = p_1, gamma(1) = p_2$ is no longer the usual straight line Euclidean distance that you are used to seeing.
To compute $bard((1,1),(-1,-2))$, you'll need to figure our the infimum of the "lengths" of curves from $(1,1)$ to $(-1,-2)$. If you can come up with a curve from $(1,1)$ to $(-1,-2)$ and show that its "length" is less than the "length" of any other curve from $(1,1)$ to $(-1,-2)$, you'll have your answer. Hint: What is the "length" of any straight line segment that lies entirely on line $x = 0$? Use this fact to your advantage when deciding on what might be the curve from $(1,1)$ to $(-1,-2)$ with the least length.
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
edited Dec 21 '14 at 23:58
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
answered Dec 21 '14 at 4:41
JimmyK4542
39.5k245105
39.5k245105
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
 |Â
show 6 more comments
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
1
1
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
I think this answer is excellent. Just wanted to tell @Mathi to prove the existence of "inf" (so you know the definition is, in fact, well-defined) by using the hypothesis of a smooth curve.
– CABJ
Dec 21 '14 at 4:48
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
@JimmyK4542: So, is $tilde d ((1,1),(-1,-2))= sqrt(1-(-1))^2+(1-(-2))^2$?
– Extremal
Dec 21 '14 at 5:10
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
No, it's quite different. For starters, what's the "distance" from $(0,-2)$ to $(0,3)$?
– Ted Shifrin
Dec 21 '14 at 12:38
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Why is it wrong? So, please tell me how to find $tilde d((1,1),(−1,−2))$?
– Extremal
Dec 21 '14 at 23:13
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
Ok so, as I understand here $displaystyle tilde d ((1,1),(-1,-2))=int_-1^1 sqrt13x ;dx=0$. Am I correct?
– Extremal
Dec 22 '14 at 0:22
 |Â
show 6 more comments
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