Integral of a determinant over a unit simplex
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Recalling the definition of the unit simplex
$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$
I would like to calculate this integral for all integers $ngeq 2$
$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$
where det is the usual determinant. I tried the triangular change of variable
$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$
but that does not really help...
Any guess would be welcome !
calculus integration determinant simplex
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
$endgroup$
add a comment |
$begingroup$
Recalling the definition of the unit simplex
$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$
I would like to calculate this integral for all integers $ngeq 2$
$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$
where det is the usual determinant. I tried the triangular change of variable
$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$
but that does not really help...
Any guess would be welcome !
calculus integration determinant simplex
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
$endgroup$
2
$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
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– Song
Dec 9 '18 at 23:57
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@Song Help me, please, to find my error for $J(2).$
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– Yuri Negometyanov
Dec 10 '18 at 8:59
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@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42
add a comment |
$begingroup$
Recalling the definition of the unit simplex
$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$
I would like to calculate this integral for all integers $ngeq 2$
$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$
where det is the usual determinant. I tried the triangular change of variable
$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$
but that does not really help...
Any guess would be welcome !
calculus integration determinant simplex
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
$endgroup$
Recalling the definition of the unit simplex
$$ Delta^{(n)}=lbrace (x_1,dots,x_n)in mathbb{R}_+^{n} ; , sum_{i=1}^n x_i=1 rbrace,$$
I would like to calculate this integral for all integers $ngeq 2$
$$I(n)=int_{Delta^{(n)}}mathrm{d}x_1dots mathrm{d}x_n , mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]} $$
where det is the usual determinant. I tried the triangular change of variable
$$
y_1=x_1 quad qquad \
y_2=x_1+x_2quad ;\
y_3=x_1+x_2+x_3\
dots\
dots\
y_n=sum_{i=1}^n x_i qquad quad
$$
but that does not really help...
Any guess would be welcome !
calculus integration determinant simplex
calculus integration determinant simplex
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
edited Dec 7 '18 at 14:33
Alexandre Krajenbrink
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
asked Dec 6 '18 at 10:34
Alexandre KrajenbrinkAlexandre Krajenbrink
9215
9215
2
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The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57
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@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59
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@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42
add a comment |
2
$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57
$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59
$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42
2
2
$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57
$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57
$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59
$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59
$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42
$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).
Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$
At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$
$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$
(see also Wolfram Alpha),
If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$
At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$
$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$
If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$
At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$
And further progress becames too hard.
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
$endgroup$
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
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– Christoph
Dec 9 '18 at 18:57
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@Christoph You are right. Thanks!
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– Yuri Negometyanov
Dec 9 '18 at 19:00
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@Christoph First Edition.
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– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
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– Christoph
Dec 10 '18 at 8:54
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@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
|
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$begingroup$
Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).
Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$
At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$
$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$
(see also Wolfram Alpha),
If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$
At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$
$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$
If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$
At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$
And further progress becames too hard.
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
$endgroup$
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
|
show 1 more comment
$begingroup$
Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).
Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$
At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$
$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$
(see also Wolfram Alpha),
If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$
At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$
$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$
If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$
At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$
And further progress becames too hard.
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
$endgroup$
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
|
show 1 more comment
$begingroup$
Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).
Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$
At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$
$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$
(see also Wolfram Alpha),
If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$
At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$
$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$
If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$
At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$
And further progress becames too hard.
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
$endgroup$
Let
$$I_1(a,b,c,x) = int e^{-ax^2-bx-c},mathrm dx = dfrac1{2sqrt a}left(sqrt π ,e^{b^2/(4a)-c},mathrm{erf}left(frac{2ax+b}{2sqrt a}right)right) + constanttag1$$
(see also Wolfram Alpha),
$$I_{0}(a,b,x) = int e^{-a(x-b)^2},mathrm dx = I_1(a,-2ab,ab^2,x) =sqrt{dfrac{pi}{4a}},mathrm{erf}left(sqrt a(x-b)right) + constant,tag2$$
$$I_{20}(a,b,x) = int mathrm{erf}(a(x-b)),mathrm dx = dfrac1{sqrtpi a}e^{-a^2(x-b)^2} + (x-b), mathrm{erf}(a(x-b)) + constanttag3$$
(see also Wolfram Alpha),
$$I_{21}(a,b,x) = int(x-b) mathrm{erf}(a(x-b)),mathrm dx = {small dfrac{sqrtpi(1-2a^2 (b^2-2bx+x^2)),mathrm{erf}(a(b-x))+2a(x-b)e^{-a^2(x-b)^2}}{4sqrtpi a^2}},tag4$$
(see also Wolfram Alpha).
Then
$$J(1) = intlimits_0^1 ,mathrm dx_1 = 1.$$
If $n=2,$ then the simplex perimeter consists of two intervals on the axices and one slope interval $y=1-x.$
At the same time, determinant equals to
$$d_2=begin{vmatrix}
1 & e^{-(x-y)^2}\
e^{-(x-y)^2} & 1\
end{vmatrix}
=1-e^{-2(x-y)^2},$$
$$J(2) = intlimits_0^1left(2-2e^{-2x^2}+1-e^{-2(2x-1)^2}right),mathrm dx
=3-(2I_0(2,0,x)+I_0(8,1/2,x))bigg|_0^1
=3-left(sqrt{dfracpi2},mathrm{erf}(sqrt2x)+dfrac14sqrt{dfracpi2},mathrm{erf}left(sqrt2(2x-1)right)right)bigg|_0^1 = 3-dfrac34sqrt{2pi},mathrm{erf}(sqrt2) approx 1.20557$$
(see also Wolfram Alpha),
If $n=3,$ then the simplex surface consists of three triangles on the axices' planes and one slope triangle at the plane $x_3=1-x_1-x_2.$
At the same time, determinant equals to
$$d_3=begin{vmatrix}
1 & e^{-(x_1-x_2)^2} & e^{-(x_1-x_3)^2} \
e^{-(x_1-x_2)^2} & 1 & e^{-(x_2-x_3)^2}\
e^{-(x_1-x_3)^2} & e^{-(x_2-x_3)^2} & 1\
end{vmatrix}
= 1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-left(e^{-2(x_1-x_2)^2} + e^{-2(x_2-x_3)^2} + e^{-2(x_3-x_1)^2}right),$$
$$J(3) = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-(x-y)^2-y^2-x^2} - e^{y^2-x^2} - e^{-(x-y)^2-x^2}-e^{-(x-y)^2-y^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x_1}bigg(1+2e^{-(x_1-x_2)^2-(x_2-x_3)^2-(x_3-x_1)^2}-e^{-2(x_1-x_2)^2} - e^{-2(x_2-x_3)^2} - e^{-2(x_3-x_1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-x^2}e^{-(y-x)^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$ + intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-(y-x)^2-(2y+x-1)^2-(y+2x-1)^2} -e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dx_2,mathrm dx_1 $$
$$ = 3intlimits_0^1intlimits_0^{1-x}left(1+2e^{-2y^2+2xy-2x^2} - e^{-x^2}e^{-y^2} - e^{-y^2+2xy-2x^2}-e^{-2y^2+2xy-x^2}right),mathrm dy,mathrm dx$$
$$+intlimits_0^1intlimits_0^{1-x}bigg(1+2e^{-6(y^2+(x-1)y+x^2-x)-2}-e^{-2(y-x)^2} - e^{-2(2y+x-1)^2} - e^{-2(y+2x-1)^2}bigg),mathrm dy,mathrm dx $$
$$ = 3J_{31} + J_{32} approx 0.0929729 + 0.00144917 = 0.09442207.$$
Integral 1 and Integral 2 can be calculated using $(1)-(3).$
If $n=4,$ then the simplex hypersurface consists of four triangular pyramids on the axices' hyperplanes and one slope triangular pyramid at the hyperplane $x_4=1-x_1-x_2-x_3.$
At the same time, determinant equals to
$$d_4=begin{vmatrix}
1 & e^{-(x-y)^2} & e^{-(x-z)^2} & e^{-(x-t)^2}\
e^{-(y-x)^2} & 1 & e^{-(y-z)^2} & e^{-(y-t)^2}\
e^{-(z-x)^2} & e^{-(z-y)^2} & 1 & e^{-(z-t)^2}\
e^{-(t-x)^2} & e^{-(t-y)^2} & e^{-(t-z)^2} & 1\
end{vmatrix}
= e^{-(t-x)^2-(x-t)^2-(t-y)^2-(x-y)^2-(y-t)^2-(y-x)^2-(t-z)^2-(x-z)^2-(y-z)^2-(z-t)^2-(z-x)^2-(z-y)^2}bigg(
e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
-e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
+e^{(t-x)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-y)^2}
-e^{(x-t)^2+(t-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-t)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(x-z)^2+(y-z)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(t-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-y)^2+(y-t)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(x-t)^2+(t-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-x)^2+(x-t)^2+(x-y)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
-e^{(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}
+e^{(t-x)^2+(x-t)^2+(t-y)^2+(x-y)^2+(y-t)^2+(y-x)^2+(t-z)^2+(x-z)^2+(y-z)^2+(z-t)^2+(z-x)^2+(z-y)^2}bigg).$$
And further progress becames too hard.
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
edited Dec 14 '18 at 19:05
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
answered Dec 9 '18 at 18:38
Yuri NegometyanovYuri Negometyanov
11.6k1728
11.6k1728
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
|
show 1 more comment
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
You seem to be integrating over the region in the positive orthant defined by $sum_i x_i le 1$. Note that OP (as stated) is integrating over the $(n-1)$-simplex defined by $sum_i x_i = 1$.
$endgroup$
– Christoph
Dec 9 '18 at 18:57
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph You are right. Thanks!
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 19:00
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
@Christoph First Edition.
$endgroup$
– Yuri Negometyanov
Dec 9 '18 at 21:05
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
Not sure why you are putting all this work into this when the integral asked for is just $0$ as pointed out by Song in the comments and OP doesn't care to clarify.
$endgroup$
– Christoph
Dec 10 '18 at 8:54
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
$begingroup$
@Christoph Thank you, I will check my work and his arguments
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:58
|
show 1 more comment
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2
$begingroup$
The problem as you stated has a trivial solution $ I(n)=0$ since$ Delta^{(n)}$ has ($n$-dimensional Lebesgue) measure $0$. For $I(n)$ to be meaningful, the integral should be defined in terms of $n-1$-dimensional Lebesgue meausre.
$endgroup$
– Song
Dec 9 '18 at 23:57
$begingroup$
@Song Help me, please, to find my error for $J(2).$
$endgroup$
– Yuri Negometyanov
Dec 10 '18 at 8:59
$begingroup$
@Song, thanks for your comment. Originally, the representation of the integral I'm interested in is $I(n)=int_{mathbb{R}_+^n}mathrm{d}x_1dots mathrm{d}x_n , delta(sum_{i=1}^n x_i-1)mathrm{det}big[exp big(-frac{(x_i-x_j)^2}{2}big)big]_{i,jin [1,n]}$ and I did interpret the $delta$ as a constraint leading to the simplex, perhaps it was badly formulated.
$endgroup$
– Alexandre Krajenbrink
Dec 10 '18 at 10:42