with vector $mathbfx=[x_1,x_2,..,x_n]$, is $f_i(mathbfx)=x_i exp(-(x_1+x_2+…x_i))$ concave for vector $mathbfx$ ????

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It looks not that complicated but I'm stuck in the middle.

$mathbfx=[x_1,x_2, cdots ,x_n]$.

1. $g(mathbfx)=exp(-mathbfx)$ is a decreasing, and convex function.


2. $h(mathbfx)=x_1+x_2+x_3+dotsm;$ is a linear, increasing function, convex/concave.

So, $g(h(mathbfx))$ is a convex function.
Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(mathbfx))$.
$x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(mathbfx)=x_i g(h(mathbfx))$ is concave?

Thanks a lot.

convex-analysis

share|cite|improve this question

edited Jul 21 at 8:17

Bernard

110k635103

asked Jul 21 at 3:02

Eunhye Park

82

add a comment | 

up vote
0
down vote

favorite

2

It looks not that complicated but I'm stuck in the middle.

$mathbfx=[x_1,x_2, cdots ,x_n]$.

1. $g(mathbfx)=exp(-mathbfx)$ is a decreasing, and convex function.


2. $h(mathbfx)=x_1+x_2+x_3+dotsm;$ is a linear, increasing function, convex/concave.

So, $g(h(mathbfx))$ is a convex function.
Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(mathbfx))$.
$x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(mathbfx)=x_i g(h(mathbfx))$ is concave?

Thanks a lot.

convex-analysis

share|cite|improve this question

edited Jul 21 at 8:17

Bernard

110k635103

asked Jul 21 at 3:02

Eunhye Park

82

add a comment | 

up vote
0
down vote

favorite

2

up vote
0
down vote

favorite

2

2

It looks not that complicated but I'm stuck in the middle.

$mathbfx=[x_1,x_2, cdots ,x_n]$.

1. $g(mathbfx)=exp(-mathbfx)$ is a decreasing, and convex function.


2. $h(mathbfx)=x_1+x_2+x_3+dotsm;$ is a linear, increasing function, convex/concave.

So, $g(h(mathbfx))$ is a convex function.
Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(mathbfx))$.
$x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(mathbfx)=x_i g(h(mathbfx))$ is concave?

Thanks a lot.

convex-analysis

share|cite|improve this question

edited Jul 21 at 8:17

Bernard

110k635103

asked Jul 21 at 3:02

Eunhye Park

82

It looks not that complicated but I'm stuck in the middle.

$mathbfx=[x_1,x_2, cdots ,x_n]$.

1. $g(mathbfx)=exp(-mathbfx)$ is a decreasing, and convex function.


2. $h(mathbfx)=x_1+x_2+x_3+dotsm;$ is a linear, increasing function, convex/concave.

So, $g(h(mathbfx))$ is a convex function.
Am I right so far?

and the problem is that $x_i$ is multiplied to $g(h(mathbfx))$.
$x_i$'s are positive and between 0 to 1 real value.

How can I prove $f_i(mathbfx)=x_i g(h(mathbfx))$ is concave?

Thanks a lot.

convex-analysis

share|cite|improve this question

edited Jul 21 at 8:17

Bernard

110k635103

asked Jul 21 at 3:02

Eunhye Park

82

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 8:17

Bernard

110k635103

edited Jul 21 at 8:17

Bernard

110k635103

edited Jul 21 at 8:17

Bernard

110k635103

110k635103

asked Jul 21 at 3:02

Eunhye Park

82

asked Jul 21 at 3:02

Eunhye Park

82

asked Jul 21 at 3:02

Eunhye Park

82

82

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
0
down vote



accepted

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

share|cite|improve this answer

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
  – Eunhye Park
  Jul 21 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
  – seifedd
  Jul 21 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hope that this helps !!
  – seifedd
  Jul 21 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
  – Eunhye Park
  Jul 22 at 4:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
  – Eunhye Park
  Jul 22 at 4:08
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

Note that $f(x) = x e^-x$ is strictly convex for $x ge 2$ (just compute $f''(x)$).

share|cite|improve this answer

answered Jul 21 at 16:25

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
  – Eunhye Park
  Jul 22 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might want to add that to your question, otherwise this is an answer in the negative.
  – copper.hat
  Jul 22 at 4:39
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

share|cite|improve this answer

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
  – Eunhye Park
  Jul 21 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
  – seifedd
  Jul 21 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hope that this helps !!
  – seifedd
  Jul 21 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
  – Eunhye Park
  Jul 22 at 4:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
  – Eunhye Park
  Jul 22 at 4:08
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

share|cite|improve this answer

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
  – Eunhye Park
  Jul 21 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
  – seifedd
  Jul 21 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hope that this helps !!
  – seifedd
  Jul 21 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
  – Eunhye Park
  Jul 22 at 4:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
  – Eunhye Park
  Jul 22 at 4:08
  
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

share|cite|improve this answer

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

You can do the hessian H of $f_i(x)$ and prove that H is negative semidefinite (if x'Hx ≤ 0 for all x).

the Hessian of $f_i$ is positive semi definite in this case which means that $f_i$ is convex if $x_i$ >0.

share|cite|improve this answer

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

share|cite|improve this answer

share|cite|improve this answer

edited Jul 21 at 4:44

edited Jul 21 at 4:44

edited Jul 21 at 4:44

answered Jul 21 at 3:56

seifedd

424

answered Jul 21 at 3:56

seifedd

424

answered Jul 21 at 3:56

seifedd

424

424

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
  – Eunhye Park
  Jul 21 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
  – seifedd
  Jul 21 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hope that this helps !!
  – seifedd
  Jul 21 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
  – Eunhye Park
  Jul 22 at 4:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
  – Eunhye Park
  Jul 22 at 4:08
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
  – Eunhye Park
  Jul 21 at 6:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
  – seifedd
  Jul 21 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hope that this helps !!
  – seifedd
  Jul 21 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
  – Eunhye Park
  Jul 22 at 4:01
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
  – Eunhye Park
  Jul 22 at 4:08
  
  

  
  
  
  

Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
– Eunhye Park
Jul 21 at 6:11

Thank you for your answer. I tried by deriving Hessian H of $f_i(mathbfx)$, but I failed to prove $mathbfv'Hmathbfvleq 0$ for all $mathbfv$. Is there other way to prove this? I guess this function will be concave in 0<x<1.. at least quasiconcave.
– Eunhye Park
Jul 21 at 6:11

Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
– seifedd
Jul 21 at 19:27

Try: $ f_i( lambda x +(1- lambda )y) leq lambda f_i(x) +(1-lambda) f_i(y)$ $g(x)=exp(-x)$ convex and you can prove that easily in $R^n$ So, $f_i( lambda x+(1-lambda) y) leq x_i*(lambda g_i(x)+ (1- lambda) g_i(y) )$
– seifedd
Jul 21 at 19:27

I hope that this helps !!
– seifedd
Jul 21 at 19:32

I hope that this helps !!
– seifedd
Jul 21 at 19:32

Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
– Eunhye Park
Jul 22 at 4:01

Thank you for your kindness!!, for $x,yin R^n$ and $f_i(x)=x_i g(x)$, $f_i(lambda x+ (1-lambda)y) = (lambda x_i +(1-lambda) y_i) g_i(lambda x + (1-lambda)y) leq (lambda x_i +(1-lambda) y_i) [lambda g_i(x) +(1-lambda)g_i(y)] $
– Eunhye Park
Jul 22 at 4:01

Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
– Eunhye Park
Jul 22 at 4:08

Then, i get $lambda^2 x_i g_i(x) +(1-lambda)^2 y_i g_i(y) +lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]=lambda^2 f_i(x) + (1-lambda)^2 f_i(y) + lambda (1-lambda)[y_i g_i(x) + x_i g_i(y)]$.. Here I stuck again..I don't know how the cross term $lambda(1-lambda)x_i f_i(y)$ can be dealt with. ;-(
– Eunhye Park
Jul 22 at 4:08

add a comment | 

up vote
0
down vote

Note that $f(x) = x e^-x$ is strictly convex for $x ge 2$ (just compute $f''(x)$).

share|cite|improve this answer

answered Jul 21 at 16:25

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
  – Eunhye Park
  Jul 22 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might want to add that to your question, otherwise this is an answer in the negative.
  – copper.hat
  Jul 22 at 4:39
  

  
  
  
  

add a comment | 

up vote
0
down vote

Note that $f(x) = x e^-x$ is strictly convex for $x ge 2$ (just compute $f''(x)$).

share|cite|improve this answer

answered Jul 21 at 16:25

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
  – Eunhye Park
  Jul 22 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might want to add that to your question, otherwise this is an answer in the negative.
  – copper.hat
  Jul 22 at 4:39
  

  
  
  
  

add a comment | 

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Note that $f(x) = x e^-x$ is strictly convex for $x ge 2$ (just compute $f''(x)$).

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answered Jul 21 at 16:25

copper.hat

122k557156

Note that $f(x) = x e^-x$ is strictly convex for $x ge 2$ (just compute $f''(x)$).

share|cite|improve this answer

answered Jul 21 at 16:25

copper.hat

122k557156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 16:25

copper.hat

122k557156

answered Jul 21 at 16:25

copper.hat

122k557156

answered Jul 21 at 16:25

copper.hat

122k557156

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
  – Eunhye Park
  Jul 22 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might want to add that to your question, otherwise this is an answer in the negative.
  – copper.hat
  Jul 22 at 4:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
  – Eunhye Park
  Jul 22 at 4:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You might want to add that to your question, otherwise this is an answer in the negative.
  – copper.hat
  Jul 22 at 4:39
  

  
  
  
  

you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
– Eunhye Park
Jul 22 at 4:34

you are right. $f''(x) = (2-x)e^-x$. My consideration is only in $xin[0,1]$ :) Thank you for your answer
– Eunhye Park
Jul 22 at 4:34

You might want to add that to your question, otherwise this is an answer in the negative.
– copper.hat
Jul 22 at 4:39

You might want to add that to your question, otherwise this is an answer in the negative.
– copper.hat
Jul 22 at 4:39

add a comment | 

 

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