Monday, February 27, 2012

Since you've been so patient...

I've been busy with a lot of things, and making some interesting progress - though it seems like there's very little actual output at least where the Saga is concerned. But then we're at a very intricate point, and I have some messy timing problems to solve. It's especially uncanny since I've heard people are talking about colonizing the moon - and I mean here in the Real World, not in my subcreated one! It's a bit unsettling, since I sure didn't expect to be a prophet. I'd say more, but I don't want to spoil any surprises. Chesterton said there's a special place in Hell for people who do that: "To give away a secret when it should be kept is the worst of human crimes; and Dante was never more right than when he made the lowest circle in Hell the Circle of the Traitors. It is to destroy one human pleasure so that it can never be recovered..." [GKC ILN Nov 7 1908 CW28:210]

On the other hand, I have managed to get some great results in my little math puzzle, and I thought you might wish to hear about that. Let us always recall as Chesterton puts it:
You cannot evade the issue of God; whether you talk about pigs or the binomial theory, you are still talking about Him.
[GKC Daily News Dec 12 1903 quoted in Maycock, The Man Who Was Chesterton]
So, without further ado, let us explore some results on:

Finding the intersections of two hyperbolae

Consider the general hyperbola
Ax2 + 2Bxy + Cy2 + 2Dx + 2Ey + F = 0

The coefficient B can be eliminated by rotation through the angle a given by
if A=C
a is p/4
tan(2a) is 2B/(A–C)

That is, we make the following substitution:
x = x' cos a – y' sin a
y = x' sin a + y' cos a
resulting in
A'x' 2 + C'y' 2 + 2D'x' + 2E'y' + F' = 0

By letting s = sin a and c = cos a, these coefficients can be computed by:
A' = Ac2 + 2Bcs + Cs2
B' = 2cs(C–A)+2B(c2–s2)
C' = As2 – 2Bcs + Cc2
D' = Dc + Es
E' = Ec – Ds
F' = F
Note that the coefficient B' vanishes since 2cs = sin 2a, (c2–s2) = cos 2a, and tan 2a = 2B/(A–C).

Next, the coefficients D' and E' can be eliminated by translation by (Dx,Dy) where
Dx = – D'/A'
Dy = – E'/C'
which permits the following substitution:
x' = x'' + Dx
y' = y'' + Dy

The transformed equation of the original hyperbola thus reduces to:
A'x'' 2 + C'y'' 2 = N
where N = D' 2/A' + E' 2/C' – F'.

The above equation is called the "central equation" of the hyperbola since the center of the hyperbola is at the origin, and the foci are on one of the axes.

Next, we apply the same rotation and translation operation (a,(Dx,Dy)) to our second hyperbola, giving us a new equation in x'' and y'':

Px'' 2 + 2Qx''y'' + Ry'' 2 + 2Sx'' + 2Ty'' + U = 0

Note that the x''y'' term here does not vanish.

At this point we must verify to which of two cases the first hyperbola belongs. This "alignment" tells us whether this transformed "central" equation is oriented along the x or the y axis. If N/A' is less than zero, the foci are on the y-axis. This means a branch of the hyperbola can be considered as a function in the usual form y=f(x) and we can proceed straightforwardly. Otherwise, when N/A' is greater than zero, the foci are on the x-axis, and we interchange x and y (and all related coefficients) in what follows.

Solving the central equation of the first hyperbola for y'' we get:
y'' = ((N' – A'x'' 2)/C')1/2

Now, into that we substitute the above expression for y'' which gives a rather messy equation with a radical that can be reduced to the following quartic in x'':
Gx''4 + Hx''3 + Ix''2 + Jx'' + K = 0

G = CE2 + AQ2
H = 4ESC + 8AQT
I = 2EFC + 4(CS2 – NQ2 + AT2)
J = 4SFC – 8NQT
K = CF2 – 4NT2
E = P – AR/C
F = U + RN/C

The roots of the quartic are obtained by the usual method of solving the resolvent cubic. (See e.g. CRC Handbook of Mathematics 106.)

Up to four roots of the quartic will be found; these give the x values of the possible intersections, and the y value is found by substitution in the above equation for y''. The resulting coordinate is then transformed back to the original coordinate system, then checked to see whether they fall on the hyperbolae.


At 27 February, 2012 19:15, Anonymous some guy on the street said...

Five points define a Quadric, by a classic theorem of Blaise Pascal. Hence, four points determine a projective pencil of those Quadrics which meet these four points; a fifth point both determines a Quadric in the pencil, and lies at some position along that Quadric --- so, though two dimension determine the point, a one-dimensional variation does not alter the Quadric it determines.

Among the Quadrics meeting four given points, three are degenerate in that they correspond to quadratic equations of the form (a1 x + b1 y + c1) (a2 x + b2 y + c2) = 0 --- they are disjoint diagonal pairs for the complete hexagram on the four given points.

Two quadratic functions generically span a two-dimensional space of quadratic functions, and if their respective Quadrics meet (as generically) in four points, then all the quadratic functions in that span vanish at those four points; hence the Quadrics meeting those four points correspond to the quadratic functions in that span.

To locate the four points of intersection, it suffices to find those linear combinations of the two given quadratic functions which factor as above, and this in turn corsponds to finding those linear combinations such that the resultant quadratic vanishes at its critical point; to factor these degenerate quadratic functions; and to find the intersections of the given lines.

Of course, that is roughly what you already said, from resolvent cubic through the rest.


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