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Uniform continuity of $cos(x^2)$
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I'm trying to figure out if the function $f(x)=cos(x^2)$ is uniformly continuous in $mathbbR^2$. Looking at it's graph It doesn't look like it is, as its graph oscillates more and more violently between $-1$ and $1$ as $xrightarrow infty$.
However I've failed to give an $epsilon-delta$ proof of this. I've also tried to pick two sequences $x_n,y_n$ such that $x_n-y_n rightarrow 0$ but $f(x_n)-f(y_n) notrightarrow 0$.
real-analysis uniform-continuity
edited Jul 30 at 18:45
Bernard
110k635102
asked Jul 30 at 18:40
Yagger
5101315
add a comment |Â
up vote
0
down vote
favorite
I'm trying to figure out if the function $f(x)=cos(x^2)$ is uniformly continuous in $mathbbR^2$. Looking at it's graph It doesn't look like it is, as its graph oscillates more and more violently between $-1$ and $1$ as $xrightarrow infty$.
However I've failed to give an $epsilon-delta$ proof of this. I've also tried to pick two sequences $x_n,y_n$ such that $x_n-y_n rightarrow 0$ but $f(x_n)-f(y_n) notrightarrow 0$.
real-analysis uniform-continuity
edited Jul 30 at 18:45
Bernard
110k635102
asked Jul 30 at 18:40
Yagger
5101315
4
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to figure out if the function $f(x)=cos(x^2)$ is uniformly continuous in $mathbbR^2$. Looking at it's graph It doesn't look like it is, as its graph oscillates more and more violently between $-1$ and $1$ as $xrightarrow infty$.
However I've failed to give an $epsilon-delta$ proof of this. I've also tried to pick two sequences $x_n,y_n$ such that $x_n-y_n rightarrow 0$ but $f(x_n)-f(y_n) notrightarrow 0$.
real-analysis uniform-continuity
edited Jul 30 at 18:45
Bernard
110k635102
asked Jul 30 at 18:40
Yagger
5101315
I'm trying to figure out if the function $f(x)=cos(x^2)$ is uniformly continuous in $mathbbR^2$. Looking at it's graph It doesn't look like it is, as its graph oscillates more and more violently between $-1$ and $1$ as $xrightarrow infty$.
However I've failed to give an $epsilon-delta$ proof of this. I've also tried to pick two sequences $x_n,y_n$ such that $x_n-y_n rightarrow 0$ but $f(x_n)-f(y_n) notrightarrow 0$.
real-analysis uniform-continuity
edited Jul 30 at 18:45
Bernard
110k635102
asked Jul 30 at 18:40
Yagger
5101315
edited Jul 30 at 18:45
Bernard
110k635102
edited Jul 30 at 18:45
Bernard
110k635102
edited Jul 30 at 18:45
Bernard
110k635102
110k635102
asked Jul 30 at 18:40
Yagger
5101315
asked Jul 30 at 18:40
Yagger
5101315
asked Jul 30 at 18:40
Yagger
5101315
5101315
4
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26
add a comment |Â
4
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26
4
4
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26
add a comment |Â
2 Answers
2
active
oldest
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up vote
0
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Try this out. Let $x=sqrt2kpi$. We show that if $$|cos(x^2)-cos((x+delta)^2)|<epsilon$$for some $epsilon>0$ therefore we can make $delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|cos(x^2)-cos((x+delta)^2)|=|1-cos(2xdelta+delta^2)|=2sin^2(xdelta+dfracdelta^22)<epsilon$$which means that $$-sqrtdfracepsilon2<sin(xdelta+dfracdelta^22)<sqrtdfracepsilon2$$or $$-2sin^-1sqrtdfracepsilon2<2xdelta+delta^2<2sin^-1sqrtdfracepsilon2$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2sin^-1sqrtdfracepsilon2<x^2+2xdelta+delta^2<x^2+2sin^-1sqrtdfracepsilon2$$which by rearranging means that $$delta<dfrac2sin^-1sqrtdfracepsilon2x+sqrtx^2+2sin^-1sqrtdfracepsilon2$$(the other bound is negative and not considered). So we can make $delta$ arbitrarily small by taking $x$ sufficiently large.
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
add a comment |Â
up vote
-1
down vote
Hint: the derivative is continuous and its absolute value is not bounded from above.
answered Jul 30 at 18:48
A. Pongrácz
1,344115
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Try this out. Let $x=sqrt2kpi$. We show that if $$|cos(x^2)-cos((x+delta)^2)|<epsilon$$for some $epsilon>0$ therefore we can make $delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|cos(x^2)-cos((x+delta)^2)|=|1-cos(2xdelta+delta^2)|=2sin^2(xdelta+dfracdelta^22)<epsilon$$which means that $$-sqrtdfracepsilon2<sin(xdelta+dfracdelta^22)<sqrtdfracepsilon2$$or $$-2sin^-1sqrtdfracepsilon2<2xdelta+delta^2<2sin^-1sqrtdfracepsilon2$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2sin^-1sqrtdfracepsilon2<x^2+2xdelta+delta^2<x^2+2sin^-1sqrtdfracepsilon2$$which by rearranging means that $$delta<dfrac2sin^-1sqrtdfracepsilon2x+sqrtx^2+2sin^-1sqrtdfracepsilon2$$(the other bound is negative and not considered). So we can make $delta$ arbitrarily small by taking $x$ sufficiently large.
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
add a comment |Â
up vote
0
down vote
Try this out. Let $x=sqrt2kpi$. We show that if $$|cos(x^2)-cos((x+delta)^2)|<epsilon$$for some $epsilon>0$ therefore we can make $delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|cos(x^2)-cos((x+delta)^2)|=|1-cos(2xdelta+delta^2)|=2sin^2(xdelta+dfracdelta^22)<epsilon$$which means that $$-sqrtdfracepsilon2<sin(xdelta+dfracdelta^22)<sqrtdfracepsilon2$$or $$-2sin^-1sqrtdfracepsilon2<2xdelta+delta^2<2sin^-1sqrtdfracepsilon2$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2sin^-1sqrtdfracepsilon2<x^2+2xdelta+delta^2<x^2+2sin^-1sqrtdfracepsilon2$$which by rearranging means that $$delta<dfrac2sin^-1sqrtdfracepsilon2x+sqrtx^2+2sin^-1sqrtdfracepsilon2$$(the other bound is negative and not considered). So we can make $delta$ arbitrarily small by taking $x$ sufficiently large.
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Try this out. Let $x=sqrt2kpi$. We show that if $$|cos(x^2)-cos((x+delta)^2)|<epsilon$$for some $epsilon>0$ therefore we can make $delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|cos(x^2)-cos((x+delta)^2)|=|1-cos(2xdelta+delta^2)|=2sin^2(xdelta+dfracdelta^22)<epsilon$$which means that $$-sqrtdfracepsilon2<sin(xdelta+dfracdelta^22)<sqrtdfracepsilon2$$or $$-2sin^-1sqrtdfracepsilon2<2xdelta+delta^2<2sin^-1sqrtdfracepsilon2$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2sin^-1sqrtdfracepsilon2<x^2+2xdelta+delta^2<x^2+2sin^-1sqrtdfracepsilon2$$which by rearranging means that $$delta<dfrac2sin^-1sqrtdfracepsilon2x+sqrtx^2+2sin^-1sqrtdfracepsilon2$$(the other bound is negative and not considered). So we can make $delta$ arbitrarily small by taking $x$ sufficiently large.
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
Try this out. Let $x=sqrt2kpi$. We show that if $$|cos(x^2)-cos((x+delta)^2)|<epsilon$$for some $epsilon>0$ therefore we can make $delta$ arbitrarily small by taking $x$ sufficiently large. We have $$|cos(x^2)-cos((x+delta)^2)|=|1-cos(2xdelta+delta^2)|=2sin^2(xdelta+dfracdelta^22)<epsilon$$which means that $$-sqrtdfracepsilon2<sin(xdelta+dfracdelta^22)<sqrtdfracepsilon2$$or $$-2sin^-1sqrtdfracepsilon2<2xdelta+delta^2<2sin^-1sqrtdfracepsilon2$$by adding up $x^2$ to the sides of the inequality we have$$x^2-2sin^-1sqrtdfracepsilon2<x^2+2xdelta+delta^2<x^2+2sin^-1sqrtdfracepsilon2$$which by rearranging means that $$delta<dfrac2sin^-1sqrtdfracepsilon2x+sqrtx^2+2sin^-1sqrtdfracepsilon2$$(the other bound is negative and not considered). So we can make $delta$ arbitrarily small by taking $x$ sufficiently large.
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
answered Jul 30 at 19:21
Mostafa Ayaz
8,4623630
8,4623630
add a comment |Â
add a comment |Â
up vote
-1
down vote
Hint: the derivative is continuous and its absolute value is not bounded from above.
answered Jul 30 at 18:48
A. Pongrácz
1,344115
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
add a comment |Â
up vote
-1
down vote
Hint: the derivative is continuous and its absolute value is not bounded from above.
answered Jul 30 at 18:48
A. Pongrácz
1,344115
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Hint: the derivative is continuous and its absolute value is not bounded from above.
answered Jul 30 at 18:48
A. Pongrácz
1,344115
Hint: the derivative is continuous and its absolute value is not bounded from above.
answered Jul 30 at 18:48
A. Pongrácz
1,344115
answered Jul 30 at 18:48
A. Pongrácz
1,344115
answered Jul 30 at 18:48
A. Pongrácz
1,344115
answered Jul 30 at 18:48
A. Pongrácz
1,344115
1,344115
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
add a comment |Â
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
3
3
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
I'm not sure if that's sufficient to disprove absolute continuity? For example, what if you had a function where over each interval of $pi$, you come closer and closer to $sqrt$ but replace the singularities at $npi$ with a smaller and smaller $C^1$ spline?
– Daniel Schepler
Jul 30 at 18:56
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
The question is about uniform continuity, but your comment is correct nevertheless. However, I did not say it is enough in general, and I did not think it was. But it helps finding the candidate points. If $x_i- y_i$ tends to zero and the difference $|f(x_i)-f(y_i)|$ must stay above a fixed value $varepsilon$, then the fraction $|fracf(x_i)-f(y_i)x_i-y_i|$ tends to infiity, so $|f'(z_i)|$ must tend to infinity for some sequence $z_i$ with $y_ileq z_ileq x_i$.
– A. Pongrácz
Jul 30 at 19:05
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
Are you saying hints are not allowed? I wanted to point the author to the right direction, rather than spoonfeed them the answer. There was neither any criticism, nor any need for clarification.
– A. Pongrácz
Jul 30 at 19:51
add a comment |Â
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4
What if you take $x_n$ to be the points where $f$ takes its maximum 1 and $y_n$ to be the points where $f$ takes its minimum $-1$?
– Daniel Schepler
Jul 30 at 18:44
@DanielSchepler then $f(x_n)-f(y_n) rightarrow 2 neq 0$. But how can we give them? The only sequences I can think of would be $sqrt2pi n$ and $sqrt2pi n + pi$ , but its difference does not converge to zero.
– Yagger
Jul 30 at 19:02
Why not? Hint: $x_n - y_n = fracx_n^2 - y_n^2x_n + y_n$.
– Daniel Schepler
Jul 30 at 19:04
@DanielSchepler Got it! Thanks for the help
– Yagger
Jul 30 at 20:26