Prove that, the map $xmapsto=[x]cos^2pi x$ is discontinuous at every integer points. [closed]

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I think that the map $xmapsto=[x]cos^2pi x$ may have jump discontinuities at every integer point. However I cannot establish that fact in a rigorous manner.

Can anybody assist me to find the proper way to solve my question?

Thanks for assistance in advance.

real-analysis limits functions trigonometry discontinuous-functions

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edited Jul 20 at 19:53

Biswarup Saha

2318

asked Jul 20 at 19:01

Arnab Roy

134

closed as off-topic by Adrian Keister, Batominovski, Shaun, amWhy, Xander Henderson Jul 21 at 2:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Shaun, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
  – Arnab Roy
  Jul 20 at 19:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
  – Biswarup Saha
  Jul 20 at 19:26
  

  
  
  
  

add a comment | 

up vote
-4
down vote

favorite

I think that the map $xmapsto=[x]cos^2pi x$ may have jump discontinuities at every integer point. However I cannot establish that fact in a rigorous manner.

Can anybody assist me to find the proper way to solve my question?

Thanks for assistance in advance.

real-analysis limits functions trigonometry discontinuous-functions

share|cite|improve this question

edited Jul 20 at 19:53

Biswarup Saha

2318

asked Jul 20 at 19:01

Arnab Roy

134

closed as off-topic by Adrian Keister, Batominovski, Shaun, amWhy, Xander Henderson Jul 21 at 2:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Shaun, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
  – Arnab Roy
  Jul 20 at 19:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
  – Biswarup Saha
  Jul 20 at 19:26
  

  
  
  
  

add a comment | 

up vote
-4
down vote

favorite

up vote
-4
down vote

favorite

I think that the map $xmapsto=[x]cos^2pi x$ may have jump discontinuities at every integer point. However I cannot establish that fact in a rigorous manner.

Can anybody assist me to find the proper way to solve my question?

Thanks for assistance in advance.

real-analysis limits functions trigonometry discontinuous-functions

share|cite|improve this question

edited Jul 20 at 19:53

Biswarup Saha

2318

asked Jul 20 at 19:01

Arnab Roy

134

I think that the map $xmapsto=[x]cos^2pi x$ may have jump discontinuities at every integer point. However I cannot establish that fact in a rigorous manner.

Can anybody assist me to find the proper way to solve my question?

Thanks for assistance in advance.

real-analysis limits functions trigonometry discontinuous-functions

share|cite|improve this question

edited Jul 20 at 19:53

Biswarup Saha

2318

asked Jul 20 at 19:01

Arnab Roy

134

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 19:53

Biswarup Saha

2318

edited Jul 20 at 19:53

Biswarup Saha

2318

edited Jul 20 at 19:53

Biswarup Saha

2318

2318

asked Jul 20 at 19:01

Arnab Roy

134

asked Jul 20 at 19:01

Arnab Roy

134

asked Jul 20 at 19:01

Arnab Roy

134

134

closed as off-topic by Adrian Keister, Batominovski, Shaun, amWhy, Xander Henderson Jul 21 at 2:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Shaun, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Adrian Keister, Batominovski, Shaun, amWhy, Xander Henderson Jul 21 at 2:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Adrian Keister, Shaun, amWhy

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
  – Arnab Roy
  Jul 20 at 19:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
  – Biswarup Saha
  Jul 20 at 19:26
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
  – Arnab Roy
  Jul 20 at 19:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
  – Biswarup Saha
  Jul 20 at 19:26
  

  
  
  
  

@Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
– Arnab Roy
Jul 20 at 19:22

@Shaun, I'll try to remember your words, sorry for the inconveniences caused to all of you. I'm a beginner, I'm new in this portal, that's my fault. But don't understand why people giving vote down to my post, they should remember they were also being a student once upon a time, they should encourage new leaners instead of giving down vote. Thank you,
– Arnab Roy
Jul 20 at 19:22

@ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
– Biswarup Saha
Jul 20 at 19:26

@ArnabRoy, it is really unexpected to see people are giving vote down to a new learner, beginner. Everybody should remember their past and should keep in mind that it's one's duty to guide new joiners in a proper way. But Arnab, I am advising you to ask more contextual problems here. Give a proper description of the problem. You are a beginner, hope you will be a good Mathematics lover in the future.
– Biswarup Saha
Jul 20 at 19:26

add a comment | 

1 Answer
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Let, $cinBbbZ$

Let, $g:BbbRtoBbbR$ is defined by $g(x)=[x]cos^2pi xquadforall xinBbbR$

I'll prove that $g$ has jump discontinuity at $c$.

Note that, $g(c)=[c]cos^2pi c=c$

My claim is $limlimits_xto c- g(x)=c-1$

Choose $varepsilon>0$

As the map $xmapstocos^2pi x$ is continuous on the whone real line(Why?), we have $limlimits_xto c+ cos^2pi x=limlimits_xto c- cos^2pi x=cos^2pi c=1$.

Now using $limlimits_xto c- cos^2pi x=1$, for $fracvarepsilon>0$, we can have $delta_varepsilon>0$ such that

$|cos^2pi x-1|<fracvarepsilonquadforall xin(c-delta_varepsilon, c)$

$implies|(c-1)cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$
$implies|[x]cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$ (because $xin(c-delta_varepsilon, c)implies[x]=c-1$)

$implies|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

Therefore, for any $epsilon>0,existsdelta_varepsilon>0$ such that $|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

$implieslimlimits_xto c- g(x)=c-1ne c=g(c)$

$implies g$ has (left)jump discontinuities at every integer points on $BbbR$.

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
-1
down vote



accepted

Let, $cinBbbZ$

Let, $g:BbbRtoBbbR$ is defined by $g(x)=[x]cos^2pi xquadforall xinBbbR$

I'll prove that $g$ has jump discontinuity at $c$.

Note that, $g(c)=[c]cos^2pi c=c$

My claim is $limlimits_xto c- g(x)=c-1$

Choose $varepsilon>0$

As the map $xmapstocos^2pi x$ is continuous on the whone real line(Why?), we have $limlimits_xto c+ cos^2pi x=limlimits_xto c- cos^2pi x=cos^2pi c=1$.

Now using $limlimits_xto c- cos^2pi x=1$, for $fracvarepsilon>0$, we can have $delta_varepsilon>0$ such that

$|cos^2pi x-1|<fracvarepsilonquadforall xin(c-delta_varepsilon, c)$

$implies|(c-1)cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$
$implies|[x]cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$ (because $xin(c-delta_varepsilon, c)implies[x]=c-1$)

$implies|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

Therefore, for any $epsilon>0,existsdelta_varepsilon>0$ such that $|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

$implieslimlimits_xto c- g(x)=c-1ne c=g(c)$

$implies g$ has (left)jump discontinuities at every integer points on $BbbR$.

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

add a comment | 

up vote
-1
down vote



accepted

Let, $cinBbbZ$

Let, $g:BbbRtoBbbR$ is defined by $g(x)=[x]cos^2pi xquadforall xinBbbR$

I'll prove that $g$ has jump discontinuity at $c$.

Note that, $g(c)=[c]cos^2pi c=c$

My claim is $limlimits_xto c- g(x)=c-1$

Choose $varepsilon>0$

As the map $xmapstocos^2pi x$ is continuous on the whone real line(Why?), we have $limlimits_xto c+ cos^2pi x=limlimits_xto c- cos^2pi x=cos^2pi c=1$.

Now using $limlimits_xto c- cos^2pi x=1$, for $fracvarepsilon>0$, we can have $delta_varepsilon>0$ such that

$|cos^2pi x-1|<fracvarepsilonquadforall xin(c-delta_varepsilon, c)$

$implies|(c-1)cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$
$implies|[x]cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$ (because $xin(c-delta_varepsilon, c)implies[x]=c-1$)

$implies|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

Therefore, for any $epsilon>0,existsdelta_varepsilon>0$ such that $|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

$implieslimlimits_xto c- g(x)=c-1ne c=g(c)$

$implies g$ has (left)jump discontinuities at every integer points on $BbbR$.

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

add a comment | 

up vote
-1
down vote



accepted

up vote
-1
down vote



accepted

Let, $cinBbbZ$

Let, $g:BbbRtoBbbR$ is defined by $g(x)=[x]cos^2pi xquadforall xinBbbR$

I'll prove that $g$ has jump discontinuity at $c$.

Note that, $g(c)=[c]cos^2pi c=c$

My claim is $limlimits_xto c- g(x)=c-1$

Choose $varepsilon>0$

As the map $xmapstocos^2pi x$ is continuous on the whone real line(Why?), we have $limlimits_xto c+ cos^2pi x=limlimits_xto c- cos^2pi x=cos^2pi c=1$.

Now using $limlimits_xto c- cos^2pi x=1$, for $fracvarepsilon>0$, we can have $delta_varepsilon>0$ such that

$|cos^2pi x-1|<fracvarepsilonquadforall xin(c-delta_varepsilon, c)$

$implies|(c-1)cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$
$implies|[x]cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$ (because $xin(c-delta_varepsilon, c)implies[x]=c-1$)

$implies|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

Therefore, for any $epsilon>0,existsdelta_varepsilon>0$ such that $|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

$implieslimlimits_xto c- g(x)=c-1ne c=g(c)$

$implies g$ has (left)jump discontinuities at every integer points on $BbbR$.

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

Let, $cinBbbZ$

Let, $g:BbbRtoBbbR$ is defined by $g(x)=[x]cos^2pi xquadforall xinBbbR$

I'll prove that $g$ has jump discontinuity at $c$.

Note that, $g(c)=[c]cos^2pi c=c$

My claim is $limlimits_xto c- g(x)=c-1$

Choose $varepsilon>0$

As the map $xmapstocos^2pi x$ is continuous on the whone real line(Why?), we have $limlimits_xto c+ cos^2pi x=limlimits_xto c- cos^2pi x=cos^2pi c=1$.

Now using $limlimits_xto c- cos^2pi x=1$, for $fracvarepsilon>0$, we can have $delta_varepsilon>0$ such that

$|cos^2pi x-1|<fracvarepsilonquadforall xin(c-delta_varepsilon, c)$

$implies|(c-1)cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$
$implies|[x]cos^2pi x-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$ (because $xin(c-delta_varepsilon, c)implies[x]=c-1$)

$implies|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

Therefore, for any $epsilon>0,existsdelta_varepsilon>0$ such that $|g(x)-(c-1)|<varepsilonquadforall xin(c-delta_varepsilon, c)$

$implieslimlimits_xto c- g(x)=c-1ne c=g(c)$

$implies g$ has (left)jump discontinuities at every integer points on $BbbR$.

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 19:02

Biswarup Saha

2318

answered Jul 20 at 19:02

Biswarup Saha

2318

answered Jul 20 at 19:02

Biswarup Saha

2318

2318

add a comment | 

add a comment |