Set Theory in Javascript

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Vortex Solution to Navier Stokes Equations

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This is a solution I found to the Navier Stokes equations . It uses the flow lines of the Hopf Fibration (which is basically just a way of covering the entire 3D space in circles). The flow lines can be seen in this image below: You can see my paper here. Solution to Navier-Stokes equation . Unfortunately I won't get a million pounds for winning the Clay Millennium prize because this is a compressible not incompressible solution. :( Oh, well. Nice try! I think it's pretty nice anyway. :) Also it uses the nice identity: which paramaterizes the Pythagorean quadruples!
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When do two ellipses overlap/intersect

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See also here A general rotated ellipse is \(f(x,y) = a x^2+2 b x y+c y^2+d x+e y - 1\) = X.M.X Nearest point (x,y).(df/dx, df/dy) = 0. Let this point be (X1,Y1) and (X2,Y2). X1^2+Y1^2 = R1^2  Therefore a circle radius between R1 and R2 intersects. A circle of radius >R1 overlaps. We need to find the relationship between roots of two quadratic simultaneous equations and their coefficients. It intersects a circle of radius 1/C if (x1^2+y1^2-1/C)(x2^2+y2^1-1/C)<0 or: (9*e^4+18*d^2*e^2-24*c*e^2+24*a*e^2-96*b*d*e+9*d^4+24*c*d^2-24*a*d^2+16*c^2-32*a*c+64*b^2+16*a^2)* C^2 + (-6*a*e^4+12*b*d*e^3-6*c*d^2*e^2-6*a*d^2*e^2+8*a*c*e^2-16*b^2*e^2-8*a^2*e^2+12*b*d^3*e+16*b*c*d*e+16*a*b*d*e-6*c*d^4-8*c^2*d^2+8*a*c*d^2-16*b^2*d^2)* C +a^2*e^4-4*a*b*d*e^3+2*a*c*d^2*e^2+4*b^2*d^2*e^2-4*b*c*d^3*e+c^2*d^4    <   0 (which is a 6th degree polynomial) To convert this to the case where we have two ellipses we need to transform: a*x^2+2*b*x*y+c*y^...
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Existence of Solutions to Systems of Polynomial Inequalities

Existence of Solutions to Systems of Polynomial Inequalities I was wondering whether there is a general method to deciding if a system of polynomial inequalities has a solution. If there was such an algorithm this could be used to solve questions such as the sphere packing question. (The kissing number). Simply find all the inequalities for each sphere and determine if a solution can be found. To skip to the end it turns out that all systems of polynomial inequalities can be put in the following form: {A nm x n x m +B n x n +C  ≥ 0} or using the notation X ={x1,x2,x3,...,1} {Q nm X n X m   ≥ 0} Thus all conditions are built up from matrices, A , vectors B and constants C. We must deal with all the special cases where various components are zero. For the sphere packing problem we can set all the { B =0} which makes it much simpler. Simplest Case The simplest case is a single linear equation A x +B ≥ 0 This has no solution when A=0 & B<0 Star...
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