Mixing
What is the surface area of the 3-dimensional elliptope?
$begingroup$
The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-elliptope $$Gamma = {(x,y,z)in[-1,1]^3 : x^2+y^2+z^2leq 1+2xyz}.$$
The volume of $Gamma$ was considered in an earlier question (What is the volume of the $3$-dimensional elliptope?) and shown to be $pi^2/2$. However, what I'm interested in presently is the subset of singular 3-by-3 correlation matrices. This corresponds precisely to the boundary of the above set: $$partial Gamma = {(x,y,z)in [-1,1]^3: x^2+y^2+z^2=1+2xyz}.$$
With that in mind, what I want to know is the the surface area of $partial Gamma.$ Formally, this is not so hard: The surface can be expressed as the union of the surfaces $z=f_{pm}(x,y)=x ypm sqrt{1-x^2}sqrt{1-y^2}$, and the bottom surface is the mirror of the top. Hence their areas are the same, so the total area is given the double integral $$S=2int_{-1}^1int_{-1}^1 sqrt{1+left(frac{partial f_+}{partial x}right)^2+left(frac{partial f_+}{partial y}right)^2},dx,dy,$$ where
$$frac{partial f_+}{partial x} = y-xsqrt{frac{1-y^2}{1-x^2}},quad frac{partial f_+}{partial y}=x-ysqrt{frac{1-x^2}{1-y^2}}.$$ If one substitutes $x=cosalpha,y=cosbeta$ over the ranges $0leq alpha,betaleq pi$, then the result may be placed in the form
$$S = 2int_{0}^piint_0^pi sqrt{sin^2(alpha) sin^2(alpha -beta )+sin^2(beta ) sin^2(alpha -beta )+sin^2(alpha) sin^2(beta) };dalpha ,dbeta.$$
Alas, while this integral is intriguing it has defied my attempts at analytical solution (as well as those of Mathematica). Numerically, however, the integral seems to be exactly $5pi$. Can anyone show that this result is correct?
calculus multivariable-calculus definite-integrals analytic-geometry closed-form
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
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add a comment |
$begingroup$
The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-elliptope $$Gamma = {(x,y,z)in[-1,1]^3 : x^2+y^2+z^2leq 1+2xyz}.$$
The volume of $Gamma$ was considered in an earlier question (What is the volume of the $3$-dimensional elliptope?) and shown to be $pi^2/2$. However, what I'm interested in presently is the subset of singular 3-by-3 correlation matrices. This corresponds precisely to the boundary of the above set: $$partial Gamma = {(x,y,z)in [-1,1]^3: x^2+y^2+z^2=1+2xyz}.$$
With that in mind, what I want to know is the the surface area of $partial Gamma.$ Formally, this is not so hard: The surface can be expressed as the union of the surfaces $z=f_{pm}(x,y)=x ypm sqrt{1-x^2}sqrt{1-y^2}$, and the bottom surface is the mirror of the top. Hence their areas are the same, so the total area is given the double integral $$S=2int_{-1}^1int_{-1}^1 sqrt{1+left(frac{partial f_+}{partial x}right)^2+left(frac{partial f_+}{partial y}right)^2},dx,dy,$$ where
$$frac{partial f_+}{partial x} = y-xsqrt{frac{1-y^2}{1-x^2}},quad frac{partial f_+}{partial y}=x-ysqrt{frac{1-x^2}{1-y^2}}.$$ If one substitutes $x=cosalpha,y=cosbeta$ over the ranges $0leq alpha,betaleq pi$, then the result may be placed in the form
$$S = 2int_{0}^piint_0^pi sqrt{sin^2(alpha) sin^2(alpha -beta )+sin^2(beta ) sin^2(alpha -beta )+sin^2(alpha) sin^2(beta) };dalpha ,dbeta.$$
Alas, while this integral is intriguing it has defied my attempts at analytical solution (as well as those of Mathematica). Numerically, however, the integral seems to be exactly $5pi$. Can anyone show that this result is correct?
calculus multivariable-calculus definite-integrals analytic-geometry closed-form
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
$endgroup$
add a comment |
$begingroup$
The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-elliptope $$Gamma = {(x,y,z)in[-1,1]^3 : x^2+y^2+z^2leq 1+2xyz}.$$
The volume of $Gamma$ was considered in an earlier question (What is the volume of the $3$-dimensional elliptope?) and shown to be $pi^2/2$. However, what I'm interested in presently is the subset of singular 3-by-3 correlation matrices. This corresponds precisely to the boundary of the above set: $$partial Gamma = {(x,y,z)in [-1,1]^3: x^2+y^2+z^2=1+2xyz}.$$
With that in mind, what I want to know is the the surface area of $partial Gamma.$ Formally, this is not so hard: The surface can be expressed as the union of the surfaces $z=f_{pm}(x,y)=x ypm sqrt{1-x^2}sqrt{1-y^2}$, and the bottom surface is the mirror of the top. Hence their areas are the same, so the total area is given the double integral $$S=2int_{-1}^1int_{-1}^1 sqrt{1+left(frac{partial f_+}{partial x}right)^2+left(frac{partial f_+}{partial y}right)^2},dx,dy,$$ where
$$frac{partial f_+}{partial x} = y-xsqrt{frac{1-y^2}{1-x^2}},quad frac{partial f_+}{partial y}=x-ysqrt{frac{1-x^2}{1-y^2}}.$$ If one substitutes $x=cosalpha,y=cosbeta$ over the ranges $0leq alpha,betaleq pi$, then the result may be placed in the form
$$S = 2int_{0}^piint_0^pi sqrt{sin^2(alpha) sin^2(alpha -beta )+sin^2(beta ) sin^2(alpha -beta )+sin^2(alpha) sin^2(beta) };dalpha ,dbeta.$$
Alas, while this integral is intriguing it has defied my attempts at analytical solution (as well as those of Mathematica). Numerically, however, the integral seems to be exactly $5pi$. Can anyone show that this result is correct?
calculus multivariable-calculus definite-integrals analytic-geometry closed-form
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
$endgroup$
The $n$-elliptope is defined as the set of $n$-by-$n$ correlation matrices; that is, the set of $n$-by-$n$ symmetric positive-definite matrices with ones on the diagonal. Such matrices are parametrized by their $n(n-1)/2$ upper off-diagonal elements. In the case of $n=3$, this yields the 3-elliptope $$Gamma = {(x,y,z)in[-1,1]^3 : x^2+y^2+z^2leq 1+2xyz}.$$
The volume of $Gamma$ was considered in an earlier question (What is the volume of the $3$-dimensional elliptope?) and shown to be $pi^2/2$. However, what I'm interested in presently is the subset of singular 3-by-3 correlation matrices. This corresponds precisely to the boundary of the above set: $$partial Gamma = {(x,y,z)in [-1,1]^3: x^2+y^2+z^2=1+2xyz}.$$
With that in mind, what I want to know is the the surface area of $partial Gamma.$ Formally, this is not so hard: The surface can be expressed as the union of the surfaces $z=f_{pm}(x,y)=x ypm sqrt{1-x^2}sqrt{1-y^2}$, and the bottom surface is the mirror of the top. Hence their areas are the same, so the total area is given the double integral $$S=2int_{-1}^1int_{-1}^1 sqrt{1+left(frac{partial f_+}{partial x}right)^2+left(frac{partial f_+}{partial y}right)^2},dx,dy,$$ where
$$frac{partial f_+}{partial x} = y-xsqrt{frac{1-y^2}{1-x^2}},quad frac{partial f_+}{partial y}=x-ysqrt{frac{1-x^2}{1-y^2}}.$$ If one substitutes $x=cosalpha,y=cosbeta$ over the ranges $0leq alpha,betaleq pi$, then the result may be placed in the form
$$S = 2int_{0}^piint_0^pi sqrt{sin^2(alpha) sin^2(alpha -beta )+sin^2(beta ) sin^2(alpha -beta )+sin^2(alpha) sin^2(beta) };dalpha ,dbeta.$$
Alas, while this integral is intriguing it has defied my attempts at analytical solution (as well as those of Mathematica). Numerically, however, the integral seems to be exactly $5pi$. Can anyone show that this result is correct?
calculus multivariable-calculus definite-integrals analytic-geometry closed-form
calculus multivariable-calculus definite-integrals analytic-geometry closed-form
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
edited Jan 10 at 22:58
Semiclassical
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
asked Jan 10 at 14:59
SemiclassicalSemiclassical
11.1k32466
11.1k32466
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