Restriction of interval while proving an inequality

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In many books and online , the inequality $sin x<x$ is defined only over the interval $[0, fracpi2]$ . However it is easy to check that the inequality holds good over the interval $[fracpi2 , pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, fracpi2]$ ?

inequality

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edited Jul 16 at 16:55

ervx

9,39531336

asked Jul 16 at 16:53

Aditi

694314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
  – mechanodroid
  Jul 16 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
  – Aditi
  Jul 16 at 17:15
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

In many books and online , the inequality $sin x<x$ is defined only over the interval $[0, fracpi2]$ . However it is easy to check that the inequality holds good over the interval $[fracpi2 , pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, fracpi2]$ ?

inequality

share|cite|improve this question

edited Jul 16 at 16:55

ervx

9,39531336

asked Jul 16 at 16:53

Aditi

694314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
  – mechanodroid
  Jul 16 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
  – Aditi
  Jul 16 at 17:15
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

In many books and online , the inequality $sin x<x$ is defined only over the interval $[0, fracpi2]$ . However it is easy to check that the inequality holds good over the interval $[fracpi2 , pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, fracpi2]$ ?

inequality

share|cite|improve this question

edited Jul 16 at 16:55

ervx

9,39531336

asked Jul 16 at 16:53

Aditi

694314

In many books and online , the inequality $sin x<x$ is defined only over the interval $[0, fracpi2]$ . However it is easy to check that the inequality holds good over the interval $[fracpi2 , pi]$ as well. We can check it graphically . In fact it holds good for all positive numbers. Is there a special reason for proving this inequality over the restricted interval of $[0, fracpi2]$ ?

inequality

share|cite|improve this question

edited Jul 16 at 16:55

ervx

9,39531336

asked Jul 16 at 16:53

Aditi

694314

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 16:55

ervx

9,39531336

edited Jul 16 at 16:55

ervx

9,39531336

edited Jul 16 at 16:55

ervx

9,39531336

9,39531336

asked Jul 16 at 16:53

Aditi

694314

asked Jul 16 at 16:53

Aditi

694314

asked Jul 16 at 16:53

Aditi

694314

694314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
  – mechanodroid
  Jul 16 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
  – Aditi
  Jul 16 at 17:15
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
  – mechanodroid
  Jul 16 at 17:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
  – Aditi
  Jul 16 at 17:15
  

  
  
  
  

You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
– mechanodroid
Jul 16 at 17:08

You probably mean $left(0, fracpi2right]$. At $x = 0$ we have equality.
– mechanodroid
Jul 16 at 17:08

@mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
– Aditi
Jul 16 at 17:15

@mechanodroid right . I didn’t pay attention to that. Thanks for correcting me.
– Aditi
Jul 16 at 17:15

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $sin x<x$ on $[0,pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

share|cite|improve this answer

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright ! Thanks for your help :)
  – Aditi
  Jul 16 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your welcome :-)
  – ervx
  Jul 16 at 17:03
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $sin x<x$ on $[0,pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

share|cite|improve this answer

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright ! Thanks for your help :)
  – Aditi
  Jul 16 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your welcome :-)
  – ervx
  Jul 16 at 17:03
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $sin x<x$ on $[0,pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

share|cite|improve this answer

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright ! Thanks for your help :)
  – Aditi
  Jul 16 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your welcome :-)
  – ervx
  Jul 16 at 17:03
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $sin x<x$ on $[0,pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

share|cite|improve this answer

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

You are correct. The inequality holds for all positive $x$. Easiest way to see this is probably to compare derivative.

In regards to the restriction of the inequality: It is possible that in your case, you only need the fact that $sin x<x$ on $[0,pi/2]$, rather than on the whole positive real line.

Often times, the author will restrict to a smaller interval to emphasize that that is the only necessary place for the inequality to hold in the remainder of the argument.

share|cite|improve this answer

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 16:59

edited Jul 16 at 16:59

edited Jul 16 at 16:59

answered Jul 16 at 16:57

ervx

9,39531336

answered Jul 16 at 16:57

ervx

9,39531336

answered Jul 16 at 16:57

ervx

9,39531336

9,39531336

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright ! Thanks for your help :)
  – Aditi
  Jul 16 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your welcome :-)
  – ervx
  Jul 16 at 17:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright ! Thanks for your help :)
  – Aditi
  Jul 16 at 16:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your welcome :-)
  – ervx
  Jul 16 at 17:03
  

  
  
  
  

Alright ! Thanks for your help :)
– Aditi
Jul 16 at 16:58

Alright ! Thanks for your help :)
– Aditi
Jul 16 at 16:58

Your welcome :-)
– ervx
Jul 16 at 17:03

Your welcome :-)
– ervx
Jul 16 at 17:03

add a comment | 

 

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