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Calculating the length of a curve $(x,x^2)$ for $xin [0,2]$ using the Hausdorff measure of the set $A=(x,x^2)inmathbbR^2mid xin[0,2] $?
Clash Royale CLAN TAG#URR8PPP
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Let $A$ be the set of points given by the graph of $f(x) = x^2$ on the interval $[0,1]$:
$$
A = (x,x^2) in mathbbR^2bigg.
$$
I want to calculate the length of this one dimensional curve in $mathbbR^2$ using the Hausdorff measure. Using the standard method of calculating this length via an integral with an arc-length parameterization I (meaning WolframAlpha) get:
$$
I(A) = int_0^2 sqrt1 + fracdydx^2dx = int_0^2 sqrt1 + (2x)^2 dx = sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
The Hausdorff measure is defined as follows. Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $B subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $B$ is defined as:
$$
H^m(B) = lim_delta to 0 inf bigg B subset bigcup_j S_j, textdiam(S_j) < delta bigg
$$
So in my case I want the one-dimensional Hausdorff measure of the set $A$ I defined earlier and the definition becomes:
$$
H^1(A) = lim_delta to 0 inf biggsum_j (textdiam(S_j)) bigg
$$
I believe this should agree with earlier result obtained with the arc-length integral and we should get:
$$
H^1(A) = I(A) sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
But I have no idea how we can evaluate this Hausdorff measure of $A$ from the definition. It seems extremely complicated, and this is about the simplest example I can think of!
How do we pick the sets $S_j$ whose union covers $A$ and that have diameter less than $delta$ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit. How can we evaluate it?
measure-theory lebesgue-measure arc-length hausdorff-measure
edited Jul 31 at 18:04
Did
242k23207441
asked Jul 31 at 17:55
csss
1,22811221
add a comment |Â
up vote
1
down vote
favorite
Let $A$ be the set of points given by the graph of $f(x) = x^2$ on the interval $[0,1]$:
$$
A = (x,x^2) in mathbbR^2bigg.
$$
I want to calculate the length of this one dimensional curve in $mathbbR^2$ using the Hausdorff measure. Using the standard method of calculating this length via an integral with an arc-length parameterization I (meaning WolframAlpha) get:
$$
I(A) = int_0^2 sqrt1 + fracdydx^2dx = int_0^2 sqrt1 + (2x)^2 dx = sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
The Hausdorff measure is defined as follows. Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $B subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $B$ is defined as:
$$
H^m(B) = lim_delta to 0 inf bigg B subset bigcup_j S_j, textdiam(S_j) < delta bigg
$$
So in my case I want the one-dimensional Hausdorff measure of the set $A$ I defined earlier and the definition becomes:
$$
H^1(A) = lim_delta to 0 inf biggsum_j (textdiam(S_j)) bigg
$$
I believe this should agree with earlier result obtained with the arc-length integral and we should get:
$$
H^1(A) = I(A) sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
But I have no idea how we can evaluate this Hausdorff measure of $A$ from the definition. It seems extremely complicated, and this is about the simplest example I can think of!
How do we pick the sets $S_j$ whose union covers $A$ and that have diameter less than $delta$ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit. How can we evaluate it?
measure-theory lebesgue-measure arc-length hausdorff-measure
edited Jul 31 at 18:04
Did
242k23207441
asked Jul 31 at 17:55
csss
1,22811221
1
Why theBigg|s? (I corrected the one in the title.)
– Did
Jul 31 at 18:04
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $A$ be the set of points given by the graph of $f(x) = x^2$ on the interval $[0,1]$:
$$
A = (x,x^2) in mathbbR^2bigg.
$$
I want to calculate the length of this one dimensional curve in $mathbbR^2$ using the Hausdorff measure. Using the standard method of calculating this length via an integral with an arc-length parameterization I (meaning WolframAlpha) get:
$$
I(A) = int_0^2 sqrt1 + fracdydx^2dx = int_0^2 sqrt1 + (2x)^2 dx = sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
The Hausdorff measure is defined as follows. Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $B subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $B$ is defined as:
$$
H^m(B) = lim_delta to 0 inf bigg B subset bigcup_j S_j, textdiam(S_j) < delta bigg
$$
So in my case I want the one-dimensional Hausdorff measure of the set $A$ I defined earlier and the definition becomes:
$$
H^1(A) = lim_delta to 0 inf biggsum_j (textdiam(S_j)) bigg
$$
I believe this should agree with earlier result obtained with the arc-length integral and we should get:
$$
H^1(A) = I(A) sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
But I have no idea how we can evaluate this Hausdorff measure of $A$ from the definition. It seems extremely complicated, and this is about the simplest example I can think of!
How do we pick the sets $S_j$ whose union covers $A$ and that have diameter less than $delta$ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit. How can we evaluate it?
measure-theory lebesgue-measure arc-length hausdorff-measure
edited Jul 31 at 18:04
Did
242k23207441
asked Jul 31 at 17:55
csss
1,22811221
Let $A$ be the set of points given by the graph of $f(x) = x^2$ on the interval $[0,1]$:
$$
A = (x,x^2) in mathbbR^2bigg.
$$
I want to calculate the length of this one dimensional curve in $mathbbR^2$ using the Hausdorff measure. Using the standard method of calculating this length via an integral with an arc-length parameterization I (meaning WolframAlpha) get:
$$
I(A) = int_0^2 sqrt1 + fracdydx^2dx = int_0^2 sqrt1 + (2x)^2 dx = sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
The Hausdorff measure is defined as follows. Let $alpha_m$ be the Lebesgue measure of the closed unt ball in $mathbbR^m$. For $B subset mathbbR^n$, the $m$-dimesnsional Hausdorff measure of $B$ is defined as:
$$
H^m(B) = lim_delta to 0 inf bigg B subset bigcup_j S_j, textdiam(S_j) < delta bigg
$$
So in my case I want the one-dimensional Hausdorff measure of the set $A$ I defined earlier and the definition becomes:
$$
H^1(A) = lim_delta to 0 inf biggsum_j (textdiam(S_j)) bigg
$$
I believe this should agree with earlier result obtained with the arc-length integral and we should get:
$$
H^1(A) = I(A) sqrt17 + frac14 sinh^-1(4) approx 4.64678.
$$
But I have no idea how we can evaluate this Hausdorff measure of $A$ from the definition. It seems extremely complicated, and this is about the simplest example I can think of!
How do we pick the sets $S_j$ whose union covers $A$ and that have diameter less than $delta$ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit. How can we evaluate it?
measure-theory lebesgue-measure arc-length hausdorff-measure
edited Jul 31 at 18:04
Did
242k23207441
asked Jul 31 at 17:55
csss
1,22811221
edited Jul 31 at 18:04
Did
242k23207441
edited Jul 31 at 18:04
Did
242k23207441
edited Jul 31 at 18:04
Did
242k23207441
242k23207441
asked Jul 31 at 17:55
csss
1,22811221
asked Jul 31 at 17:55
csss
1,22811221
asked Jul 31 at 17:55
csss
1,22811221
1,22811221
1
Why theBigg|s? (I corrected the one in the title.)
– Did
Jul 31 at 18:04
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52
add a comment |Â
1
Why theBigg|s? (I corrected the one in the title.)
– Did
Jul 31 at 18:04
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52
1
1
Why the
Bigg|s? (I corrected the one in the title.)– Did
Jul 31 at 18:04
Why the
Bigg|s? (I corrected the one in the title.)– Did
Jul 31 at 18:04
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Of course there is a general theorem saying that for a smooth arc $gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(gamma)$ of the arc
$$gamma:quad xmapsto(x,x^2)qquad(0leq xleq 2)$$
is defined as
$$L(gamma):=sup_cal Psum_k=1^N_cal Psqrt(x_k-x_k-1)^2+(x_k^2-x_k-1^2)^2 ,tag1$$
where the $sup$ is taken over all partitions $cal P$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(gamma)$, as defined in the question, is equal to the same integral.
answered Jul 31 at 19:06
Christian Blatter
163k7107305
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Of course there is a general theorem saying that for a smooth arc $gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(gamma)$ of the arc
$$gamma:quad xmapsto(x,x^2)qquad(0leq xleq 2)$$
is defined as
$$L(gamma):=sup_cal Psum_k=1^N_cal Psqrt(x_k-x_k-1)^2+(x_k^2-x_k-1^2)^2 ,tag1$$
where the $sup$ is taken over all partitions $cal P$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(gamma)$, as defined in the question, is equal to the same integral.
answered Jul 31 at 19:06
Christian Blatter
163k7107305
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
add a comment |Â
up vote
1
down vote
Of course there is a general theorem saying that for a smooth arc $gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(gamma)$ of the arc
$$gamma:quad xmapsto(x,x^2)qquad(0leq xleq 2)$$
is defined as
$$L(gamma):=sup_cal Psum_k=1^N_cal Psqrt(x_k-x_k-1)^2+(x_k^2-x_k-1^2)^2 ,tag1$$
where the $sup$ is taken over all partitions $cal P$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(gamma)$, as defined in the question, is equal to the same integral.
answered Jul 31 at 19:06
Christian Blatter
163k7107305
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Of course there is a general theorem saying that for a smooth arc $gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(gamma)$ of the arc
$$gamma:quad xmapsto(x,x^2)qquad(0leq xleq 2)$$
is defined as
$$L(gamma):=sup_cal Psum_k=1^N_cal Psqrt(x_k-x_k-1)^2+(x_k^2-x_k-1^2)^2 ,tag1$$
where the $sup$ is taken over all partitions $cal P$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(gamma)$, as defined in the question, is equal to the same integral.
answered Jul 31 at 19:06
Christian Blatter
163k7107305
Of course there is a general theorem saying that for a smooth arc $gamma$ the one-dimensional Hausdorff measure of this arc is equal to its length computed via an integral.
In the case at hand the length $L(gamma)$ of the arc
$$gamma:quad xmapsto(x,x^2)qquad(0leq xleq 2)$$
is defined as
$$L(gamma):=sup_cal Psum_k=1^N_cal Psqrt(x_k-x_k-1)^2+(x_k^2-x_k-1^2)^2 ,tag1$$
where the $sup$ is taken over all partitions $cal P$ of the $x$-interval $[0,2]$. It is proven in calculus 102 that this $L(gamma)$ is equal to the integral you have calculated.
On the other hand arguments similar to those used in the proof of $(1)$, but "the other way around", show that $H^1(gamma)$, as defined in the question, is equal to the same integral.
answered Jul 31 at 19:06
Christian Blatter
163k7107305
answered Jul 31 at 19:06
Christian Blatter
163k7107305
answered Jul 31 at 19:06
Christian Blatter
163k7107305
answered Jul 31 at 19:06
Christian Blatter
163k7107305
163k7107305
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
add a comment |Â
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
Ok thanks! I am now left wondering what 'good' is the Hausdorff measure (repeating my comment above) ..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:15
add a comment |Â
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1
Why the
Bigg|s? (I corrected the one in the title.)– Did
Jul 31 at 18:04
"How do we pick the sets Sj whose union covers A and that have diameter less than δ - there is an infinite number of possibilities here..and then we have to take an infinum, followed by taking a limit" Hmmm, yes all these are true -- which is why nobody would compute the length of the arc using this method. :-)
– Did
Jul 31 at 18:08
@Did Oh I used the big vertical line out of habit!
– csss
Aug 1 at 13:10
@Did Ok, I see from Christian's answer that he mentions there is a theorem that relates the one-dimensional Hausdorff measure to a length integral which I assume is what you are hinting at in your second comment. With that in mind, what 'good' is the Hausdorff measure..when we want to evaluate the measure of some $m$ dimensional set in $n$ dimensions would we ever use the Hausdorff measure or is there always an alternative related approach as in the one-dimensional case? And if there is always another easier/better approach, why do we care about the Hausdorff measure at all?
– csss
Aug 1 at 13:13
We do care, because there are subsets of the plane that are much more complicated to "measure" than the graphs of regular functions.
– Did
Aug 1 at 13:52