Convergent functions and integrals

Clash Royale CLAN TAG#URR8PPP

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Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]tomathbbR$ with norm

$||f||:=sup:xin[0,1]$.

Let $f_n:[0,1]tomathbbR$ be continuous for $ninmathbbN$ and let there be a function $fin V$ such that

$lim_ntoinfty ||f-f_n|| = 0$.

I have to show that $lim_ntoinfty int_0^1f_n(x)dx=int^1_0f(x)dx$.

I have already shown that $f$ is continous.

My strategy:

Proof that $lim_ntoinfty int_0^1f_n(x)dx -int^1_0f(x)dx = 0$

so $lim_ntoinfty int_0^1(f_n(x) - f(x))dx = 0$.

I then think I have to use the squeeze theorem:

$lim_ntoinftyinf_[0,1](f_n(x) - f(x)) leqlim_ntoinfty int_0^1(f_n(x) - f(x))dx leq lim_ntoinftysup_[0,1](f_n(x) - f(x))$

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

real-analysis convergence definite-integrals

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edited Jul 15 at 12:12

asked Jul 15 at 12:07

The Coding Wombat

1618

add a comment | 

up vote
2
down vote

favorite

Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]tomathbbR$ with norm

$||f||:=sup:xin[0,1]$.

Let $f_n:[0,1]tomathbbR$ be continuous for $ninmathbbN$ and let there be a function $fin V$ such that

$lim_ntoinfty ||f-f_n|| = 0$.

I have to show that $lim_ntoinfty int_0^1f_n(x)dx=int^1_0f(x)dx$.

I have already shown that $f$ is continous.

My strategy:

Proof that $lim_ntoinfty int_0^1f_n(x)dx -int^1_0f(x)dx = 0$

so $lim_ntoinfty int_0^1(f_n(x) - f(x))dx = 0$.

I then think I have to use the squeeze theorem:

$lim_ntoinftyinf_[0,1](f_n(x) - f(x)) leqlim_ntoinfty int_0^1(f_n(x) - f(x))dx leq lim_ntoinftysup_[0,1](f_n(x) - f(x))$

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

real-analysis convergence definite-integrals

share|cite|improve this question

edited Jul 15 at 12:12

asked Jul 15 at 12:07

The Coding Wombat

1618

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]tomathbbR$ with norm

$||f||:=sup:xin[0,1]$.

Let $f_n:[0,1]tomathbbR$ be continuous for $ninmathbbN$ and let there be a function $fin V$ such that

$lim_ntoinfty ||f-f_n|| = 0$.

I have to show that $lim_ntoinfty int_0^1f_n(x)dx=int^1_0f(x)dx$.

I have already shown that $f$ is continous.

My strategy:

Proof that $lim_ntoinfty int_0^1f_n(x)dx -int^1_0f(x)dx = 0$

so $lim_ntoinfty int_0^1(f_n(x) - f(x))dx = 0$.

I then think I have to use the squeeze theorem:

$lim_ntoinftyinf_[0,1](f_n(x) - f(x)) leqlim_ntoinfty int_0^1(f_n(x) - f(x))dx leq lim_ntoinftysup_[0,1](f_n(x) - f(x))$

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

real-analysis convergence definite-integrals

share|cite|improve this question

edited Jul 15 at 12:12

asked Jul 15 at 12:07

The Coding Wombat

1618

Let $V$ be real normed vector space of Riemann integrable functions $f : [0,1]tomathbbR$ with norm

$||f||:=sup:xin[0,1]$.

Let $f_n:[0,1]tomathbbR$ be continuous for $ninmathbbN$ and let there be a function $fin V$ such that

$lim_ntoinfty ||f-f_n|| = 0$.

I have to show that $lim_ntoinfty int_0^1f_n(x)dx=int^1_0f(x)dx$.

I have already shown that $f$ is continous.

My strategy:

Proof that $lim_ntoinfty int_0^1f_n(x)dx -int^1_0f(x)dx = 0$

so $lim_ntoinfty int_0^1(f_n(x) - f(x))dx = 0$.

I then think I have to use the squeeze theorem:

$lim_ntoinftyinf_[0,1](f_n(x) - f(x)) leqlim_ntoinfty int_0^1(f_n(x) - f(x))dx leq lim_ntoinftysup_[0,1](f_n(x) - f(x))$

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

real-analysis convergence definite-integrals

share|cite|improve this question

edited Jul 15 at 12:12

asked Jul 15 at 12:07

The Coding Wombat

1618

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 12:12

edited Jul 15 at 12:12

edited Jul 15 at 12:12

asked Jul 15 at 12:07

The Coding Wombat

1618

asked Jul 15 at 12:07

The Coding Wombat

1618

asked Jul 15 at 12:07

The Coding Wombat

1618

1618

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

share|cite|improve this answer

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does the limit fit into this?
  – The Coding Wombat
  Jul 15 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
  – Robert Z
  Jul 15 at 12:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
  – The Coding Wombat
  Jul 15 at 12:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I edited my P.S. Any further doubt?
  – Robert Z
  Jul 15 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
  – Robert Z
  Jul 15 at 13:05
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=max$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

share|cite|improve this answer

edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

add a comment | 

up vote
0
down vote

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$

share|cite|improve this answer

answered Jul 15 at 19:48

mechanodroid

22.3k52041

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

share|cite|improve this answer

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does the limit fit into this?
  – The Coding Wombat
  Jul 15 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
  – Robert Z
  Jul 15 at 12:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
  – The Coding Wombat
  Jul 15 at 12:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I edited my P.S. Any further doubt?
  – Robert Z
  Jul 15 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
  – Robert Z
  Jul 15 at 13:05
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote



accepted

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

share|cite|improve this answer

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does the limit fit into this?
  – The Coding Wombat
  Jul 15 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
  – Robert Z
  Jul 15 at 12:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
  – The Coding Wombat
  Jul 15 at 12:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I edited my P.S. Any further doubt?
  – Robert Z
  Jul 15 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
  – Robert Z
  Jul 15 at 13:05
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

share|cite|improve this answer

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

You are almost done, just modify the last line as
$$0leq left|int_0^1(f_n(x) - f(x))dxright|leq int_0^1left|f_n(x) - f(x)right|dx leq sup_[0,1]left|f_n(x) - f(x)right|$$
then use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$
and apply the squeeze theorem.

P.S. Following your approach, you need
$$-sup_[0,1]left|f_n(x) - f(x)right|leqinf_[0,1](f_n(x) - f(x)) leq int_0^1(f_n(x) - f(x))dx\ leq sup_[0,1](f_n(x) - f(x))leqsup_[0,1]left|f_n(x) - f(x)right|$$
and again use $sup_[0,1]left|f_n(x) - f(x)right|=|f-f_n|to 0$.

share|cite|improve this answer

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 12:41

edited Jul 15 at 12:41

edited Jul 15 at 12:41

answered Jul 15 at 12:12

Robert Z

84.2k955123

answered Jul 15 at 12:12

Robert Z

84.2k955123

answered Jul 15 at 12:12

Robert Z

84.2k955123

84.2k955123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does the limit fit into this?
  – The Coding Wombat
  Jul 15 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
  – Robert Z
  Jul 15 at 12:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
  – The Coding Wombat
  Jul 15 at 12:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I edited my P.S. Any further doubt?
  – Robert Z
  Jul 15 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
  – Robert Z
  Jul 15 at 13:05
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How does the limit fit into this?
  – The Coding Wombat
  Jul 15 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
  – Robert Z
  Jul 15 at 12:20
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
  – The Coding Wombat
  Jul 15 at 12:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  I edited my P.S. Any further doubt?
  – Robert Z
  Jul 15 at 12:42
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
  – Robert Z
  Jul 15 at 13:05
  

  
  
  
  

How does the limit fit into this?
– The Coding Wombat
Jul 15 at 12:17

How does the limit fit into this?
– The Coding Wombat
Jul 15 at 12:17

1

1

Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
– Robert Z
Jul 15 at 12:20

Take the limit of both sides and you are done. Recall then $sup_[0,1]left|f_n(x) - f(x)right|=||f-f_n||to 0$
– Robert Z
Jul 15 at 12:20

Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
– The Coding Wombat
Jul 15 at 12:22

Alright! And the absolute value bars? Where did they come from, how did you transform it into this?
– The Coding Wombat
Jul 15 at 12:22

1

1

I edited my P.S. Any further doubt?
– Robert Z
Jul 15 at 12:42

I edited my P.S. Any further doubt?
– Robert Z
Jul 15 at 12:42

1

1

After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
– Robert Z
Jul 15 at 13:05

After the squeeze theorem, I state that the LIMIT of the absolute value of the integral is equal to 0, and therefore the LIMIT of integral itself is equal to 0
– Robert Z
Jul 15 at 13:05

 | 
show 2 more comments

up vote
1
down vote

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=max$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

share|cite|improve this answer

edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

add a comment | 

up vote
1
down vote

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=max$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

share|cite|improve this answer

edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

add a comment | 

up vote
1
down vote

up vote
1
down vote

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=max$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

share|cite|improve this answer

edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

Is it then allowed to put the limit inside the supremum and infimum, which makes them both equal to 0?

You can't switch infimum/supremum and the limit in general.

Consider $f_n(x)=max$. Then $sup_xinmathbb R|f_n(x)|=1$ while $lim_ntoinfty |f_n(x)|=0$. Hence,
$$
lim_ntoinftysup_xinmathbb R|f_n(x)|=1neq 0=sup_xinmathbb Rlim_ntoinfty |f_n(x)|.
$$

But your case is a bit more special because $[0,1]$ is compact and $f_n$ and $f$ are continuous. Hence, the infimum and supremum are achieved at some points. Then you can apply the limit as your last step.

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edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

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share|cite|improve this answer

edited Jul 15 at 12:32

edited Jul 15 at 12:32

edited Jul 15 at 12:32

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

answered Jul 15 at 12:14

Mundron Schmidt

7,1162727

7,1162727

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down vote

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$

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answered Jul 15 at 19:48

mechanodroid

22.3k52041

add a comment | 

up vote
0
down vote

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$

share|cite|improve this answer

answered Jul 15 at 19:48

mechanodroid

22.3k52041

add a comment | 

up vote
0
down vote

up vote
0
down vote

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$

share|cite|improve this answer

answered Jul 15 at 19:48

mechanodroid

22.3k52041

The map $f mapsto int_0^1 f(x),dx$ is continuous w.r.t your norm $|cdot|$.

Indeed, it is linear and bounded:

$$left|int_0^1 f(x),dxright| le int_0^1 |f(x)|,dx le int_0^1 |f|,dx = |f|$$

Therefore $f_n xrightarrow f$ implies $ int_0^1 f_n(x),dx to int_0^1 f(x),dx$

share|cite|improve this answer

answered Jul 15 at 19:48

mechanodroid

22.3k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 19:48

mechanodroid

22.3k52041

answered Jul 15 at 19:48

mechanodroid

22.3k52041

answered Jul 15 at 19:48

mechanodroid

22.3k52041

22.3k52041

add a comment | 

add a comment | 

 

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