I was wondering if someone knows whether Mathematica allows one to plot another 'probability function over the unit 2 simplex (it can be expressed as a function of two arguments subject to certain contrainsts function .
Where I am taking the domain to be of the function to be over vectors in the 2- standard simplex itself as it were,subject to certain contrain'ts.
https://en.wikipedia.org/wiki/Ternary_plot#Plotting_the_points
Does it actually have a closed form expression as a function x,y coordinates. as a function of two arguments. I presume mathematica can allows you plot it .and optimize certain functions over tenary plot, ternary graph, triangle plot, simplex plot, G? Is that correct
If you plot is assigned to G, then type FullForm[G]. Pay attention to the head.If you see Graphics[stuff], then you can use the Show command to combine plots, Its you see Graph[stuff], it might still be possible, but is harder. If you code is not too long, I recommend attaching it to your question.
I was wondering whether these two variable maps are actually convex and bi-jective, and continuous, ie that no triples go missing when defining it as a barycentric map with absolute coordinates, voer an equilateral triangle with unit lengths in bary centres coordinates, and equal side length in cartesian coordinates.
, (in both coordinates) as a function two cartesian coordinates
they are defined over an equilateral, trianle (equilaterial I presume in both cartesian coordinate and barycentric, or probability coordinates) using two cartesian coordinate,s where generally the cartesian coordinates (x1,y1), (x2,y2),(x3,y3(of each of three vertices (1,0,0) (0,0,1) (1,0,1) are determine the cartesian function function value of the other coordinates. so for P ==p3,p2,p1>=0,
There is also a mapping given from cartesian coordinates to the barycentric coordinates as well. Generally these look non symmetric with the first two vertices being something like (1,-1) or (-1,1) and (1/2 sqr3/2) or something like that for third dependent variables if I can remembers
yes,
there some constraints that I have been typing up basically there cannot be an instance of a triple of events which sum to one, vertically x,y,z,x+y+z=1, where x,y,z, are not located at the same cartesian point(m,l) ie do not sit on the same triple or world , ie F(m,l)= F(m1,l1)=, F(m2,l2)=
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