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5

0

$begingroup$

Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).

For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?

real-analysis calculus integration

share|cite|improve this question

asked Mar 21 at 0:08

QuokaQuoka

1,538316

$endgroup$

This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.

This question has not received enough attention.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a little too general for me... Maybe you could give some examples?
  $endgroup$
  – Yuriy S
  Mar 21 at 0:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
  $endgroup$
  – Quoka
  Mar 21 at 1:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
  $endgroup$
  – Christian Blatter
  yesterday
  

  
  
  
  

add a comment | 

5

0

$begingroup$

Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).

For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?

real-analysis calculus integration

share|cite|improve this question

asked Mar 21 at 0:08

QuokaQuoka

1,538316

$endgroup$

This question has an open bounty worth +50
reputation from rolandcyp ending ending at 2019-04-01 18:31:30Z">in 5 days.

This question has not received enough attention.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is a little too general for me... Maybe you could give some examples?
  $endgroup$
  – Yuriy S
  Mar 21 at 0:54
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @YuriyS I was thinking the double integral might be equal to something of the form $$C int_S f(x) dx$$ where $Ssubseteq mathbbR^3$ and $C>0$ is a constant. I'm not entirely sure if this is possible, but my intuition tells me it might work
  $endgroup$
  – Quoka
  Mar 21 at 1:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $xin S_r$ and $yin S_R$ you have no control about the point $x+y$ and the value of $f$ there. I don't think there is a simple formula without further assumptions on $f$.
  $endgroup$
  – Christian Blatter
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Suppose $f:mathbbR^3tomathbbR$ is an integrable function (I'm open to further restricting $f$, e.g. $f$ can be assumed to be smooth).

For some constants $0<rleq R$, is there a way to simplify the expression
$$
int_lvert y rvert = Rint_lvert xrvert =r f(x+y) ,mathrmdS(x)mathrmdS(y)
$$
to an expression not involving surface integrals?

real-analysis calculus integration

share|cite|improve this question

asked Mar 21 at 0:08

QuokaQuoka

1,538316

$endgroup$

share|cite|improve this question

asked Mar 21 at 0:08

QuokaQuoka

1,538316

share|cite|improve this question

asked Mar 21 at 0:08

QuokaQuoka

1,538316

asked Mar 21 at 0:08

QuokaQuoka

1,538316

asked Mar 21 at 0:08

QuokaQuoka

1,538316

2 Answers
2

active

oldest

votes

0

$begingroup$

Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.

In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Adam LatosińskiAdam Latosiński

3385

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
  $endgroup$
  – Adam Latosiński
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.

share|cite|improve this answer

answered 35 mins ago

user617446user617446

4443

$endgroup$

add a comment | 

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0

$begingroup$

Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.

In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Adam LatosińskiAdam Latosiński

3385

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
  $endgroup$
  – Adam Latosiński
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.

In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Adam LatosińskiAdam Latosiński

3385

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm having a hard time justifying your use of the Dirac-delta function since it is technically a distribution. Do you have any references explaining why I can use it this way?
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is a distribution, but as long as it's integrated with a continuous function, I don't believe there would be any problems with it. I think you can avoid using distributions and get the same result, although in a more complicated way.
  $endgroup$
  – Adam Latosiński
  yesterday
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I'm not sure how integration against a distribution is generally defined, could you elaborate on that a bit? Also, how did you obtain the equality $$int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = frac2pi R r thetabig(|z|-(R-r)big)thetabig(R+r-|z|big)?$$
  $endgroup$
  – Quoka
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Distributions are formally defined as linear functionals on some space of functions. Dirac's delta is defined on the space of continuous functions, in a following way: $$ langle delta_a, frangle = f(a) $$ Denoting $$ langle delta_a, frangle = int_-infty^infty rm dx, f(x)delta(x-a) $$ is just a useful notation, but it has all necessary properties of an integral.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As for how I obtained the equality, I've edited the answer, as there's little space in the comment.
  $endgroup$
  – Adam Latosiński
  20 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Let's start by writing the integral as
$$ I = int_mathbbR^3 int_mathbbR^3 f(x+y) delta(|x|-r) delta(|y|-R), rm d^3x rm d^3y$$
where $delta$ is Dirac's delta function. Changing the variables to $(z,y) = (x+y,y)$ we get
beginalign I &= int_mathbbR^3 int_mathbbR^3 f(z) delta(|z-y|-r) delta(|y|-R), rm d^3y rm d^3z \
&= int_mathbbR^3 f(z) Big(int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y Big) rm d^3zendalign
For $0le r le R$, it can be calculated that
To calculate $ int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y $ that let us use spherical coordinates for $y$ where angle $theta$ is measured from the direction of $z$, so that
$$|y| = rho $$
$$|z-y| = sqrt( = sqrtz$$
We have then
beginalign & int_mathbbR^3delta(|z-y|-r) delta(|y|-R), rm d^3y = \
&quad = int_0^infty rm drhoint_0^pirm dthetaint_0^2pirm dphi,rho^2sintheta delta(sqrtz-r) delta(rho-R) = \
&quad = 2pi R^2 int_0^pirm dtheta sintheta ;delta(sqrtz-r)=^u:=sqrtz\
&quad = 2pi R^2 int_^z rm du fracu delta(u-r)= \
&quad = frac2pi R int_-infty^infty rm du ,u ,theta((R+|z|)-u)theta(u - big|R-|z|big|)delta(u-r)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - big|R-|z|big|)=\
&quad = frac2pi R r ,theta((R+|z|)-r)theta(r - (R-|z|))theta(r - (|z|-R))=\
&quad = frac2pi R r ,theta(|z|+(R-r))theta(|z| - (R-r))theta(R+r - |z|) = ^textsince \
&quad = frac2pi R r
z thetabig(|z|-(R-r)big)thetabig(R+r-|z|big) endalign
where $theta$ is Heaviside's theta function.

In the end you get
$$ I = 2pi R r int_in[R-r, R+r] fracf(z) rm d^3z$$

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Adam LatosińskiAdam Latosiński

3385

$endgroup$

share|cite|improve this answer

edited 20 hours ago

answered yesterday

Adam LatosińskiAdam Latosiński

3385

answered yesterday

Adam LatosińskiAdam Latosiński

3385

answered yesterday

Adam LatosińskiAdam Latosiński

3385

0

$begingroup$

If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.

share|cite|improve this answer

answered 35 mins ago

user617446user617446

4443

$endgroup$

add a comment | 

0

$begingroup$

If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.

share|cite|improve this answer

answered 35 mins ago

user617446user617446

4443

$endgroup$

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$begingroup$

If you work in Fourier space you can write $f(x+y)=int e^ik(x+y)hatf_k dk$ and you can separate the integral to $int dk hatf_k int e^ikx dS(x) int e^ikydS(y) $.

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answered 35 mins ago

user617446user617446

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$endgroup$

share|cite|improve this answer

answered 35 mins ago

user617446user617446

4443

answered 35 mins ago

user617446user617446

4443

answered 35 mins ago

user617446user617446

4443