[math-fun] important contribution to Indian mathematics
The equation of the samosa! ParametricPlot3D[{Cos[t], Cos[u], Cos[u + t]}, {t, 0, 2 π}, {u, 0, π}] (Rounded regular tetrahedron, four vertices, no edges, but contains the line segments joining the vertices.) Implicit equation: acos(z)=acos(x)+acos(y) Volume pi^2/2 Area (empirical) 5 pi, but I can't get Mma to do the integral. This can't be new. Does this surface have a legitimate name? Can anyone prove the area? --rwg
This reminds me very much of a question I don't recall mentioning here before: Consider real random variables U, V, W having a joint distribution on R^3. Assume that the means E(U) = E(V) = E(W) = 0 and the variances E(U^2) = E(V^2) = E(W^2) = 1. Let rho denotes the Pearson correlation coefficient. Note that rho(U,V) = E(UV), rho(V,W) = E(VW), rho(W,X) = E(WX) PUZZLE: What is the set of possible triples of correlations T = {(x,y,z) in [-1,1]^3}, where x = rho(U,V), y = rho(V,W), z = rho(W,U). ??? Find a closed form polynomial inequality that describes the set T as a subset of [-1,1]^3. --Dan RWG wrote: << The equation of the samosa! ParametricPlot3D[{Cos[t], Cos[u], Cos[u + t]}, {t, 0, 2 π}, {u, 0, π}] (Rounded regular tetrahedron, four vertices, no edges, but contains the line segments joining the vertices.) Implicit equation: acos(z)=acos(x)+acos(y) Volume pi^2/2 Area (empirical) 5 pi, but I can't get Mma to do the integral. This can't be new. Does this surface have a legitimate name? Can anyone prove the area?
Um, I guess I may as well go right to the answer. (Pearson) correlations -- aka correlation coefficients -- can be interpreted as cosines of angles (between random variables). For three arbitrary random variables with a joint distribution, they can be thought of as the cosines of the angles between three arbitrary vectors in some Euclidean space. So 3-space gives all possibilities in full generality. The necessary and sufficient requirement that (x,y,z) be such a triple of cosines is that the matrix ( 1 x z ) ( ( x 1 y ) ( ( z y 1 ) of cosines be positive semidefinite. This is equivalent to requiring that every upper right square matrix have nonnegative determinant. We already know that 1 >= 0, and that 1 >= x^2 (since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that (*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0 whose locus fits RWG's samosa description to a T. (I named this shape the "tetrahedral pillow" a long time ago -- you could look it up -- but "samosa" has much greater elegance.) If you have a Mac you could visualize this in Grapher by setting the frame limits to be -1 <= x,y,z <= 1 and then graphing the implicit equation 2xyz + 1 - (x^2 + y^2 + z^2) = 0. --Dan On 2013-01-31, at 8:22 AM, Dan Asimov wrote:
This reminds me very much of a question I don't recall mentioning here before:
Consider real random variables U, V, W having a joint distribution on R^3.
Assume that the means E(U) = E(V) = E(W) = 0 and the variances E(U^2) = E(V^2) = E(W^2) = 1.
Let rho denotes the Pearson correlation coefficient.
Note that rho(U,V) = E(UV), rho(V,W) = E(VW), rho(W,X) = E(WX)
PUZZLE: What is the set of possible triples of correlations
T = {(x,y,z) in [-1,1]^3},
where x = rho(U,V), y = rho(V,W), z = rho(W,U). ???
Find a closed form polynomial inequality that describes the set T as a subset of [-1,1]^3.
--Dan
RWG wrote: << The equation of the samosa! ParametricPlot3D[{Cos[t], Cos[u], Cos[u + t]}, {t, 0, 2 π}, {u, 0, π}]
(Rounded regular tetrahedron, four vertices, no edges, but contains the line segments joining the vertices.) Implicit equation: acos(z)=acos(x)+acos(y) Volume pi^2/2 Area (empirical) 5 pi, but I can't get Mma to do the integral.
This can't be new. Does this surface have a legitimate name? Can anyone prove the area?
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Actually RWG's samosa, and the locus of (*) below in [-1,1]^3, should be identical, both corresponding to all triples of cosines of 3 angles that add up to 2pi. But it seems that peculiarities of the way acos() is interpreted might mess things up. This is the case in the Mac Grapher utility, where only one face of the pillow is displayed for the equation (**) acos(x) + acos(y) + acos(z) = 2pi. PUZZLE: Determine whether any planar geometric circle lies on the locus of (*). --Dan On 2013-02-01, at 2:25 AM, Dan Asimov wrote:
(since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that
(*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0
whose locus fits RWG's samosa description to a T.
participants (2)
-
Bill Gosper -
Dan Asimov