Today, I ran a little experiment on complex conjugates and expressing empty cuts.

Complex Conjugate Hypothesis:

Complex conjugates can be used to locate intercepts in a higher dimension, outside an empty cut. Using the empty 1D torus cut of (()I) as an example, we assign the major diameter R as an imaginary value, Ri. At this time, I believe the imaginary component goes with the diameter, and not the axis, as in xi.

The degree-4 equation of a torus. Expressed in terms of x, y, and z intercepts in 1D:

(x±r±R)^4 + (y±r±R)^4 + (z±r±Ri)^4 - A = 0

A = Consolidation Compliment to cancel out repeat terms

(x±r±R)^4

(y±r±R)^4

(z±r±Ri)^4

• Expands into

(x-r-R)(x+r-R)(x-r+R)(x+r+R)

(y-r-R)(y+r-R)(y-r+R)(y+r+R)

(z-r-Ri)(z+r-Ri)(z-r+Ri)(z+r+Ri)

• Multiplies into:

For

(x-r-R)(x+r-R)

- Code: Select all
` x -r -R`

x x^2 -rx -Rx

+r rx -r^2 -Rr

-R -Rx Rr R^2

x^2 - r^2 + R^2 + rx - rx -2Rx + Rr - Rr

simplifies into

x^2 - r^2 + R^2 -2Rx

For

(x-r+R)(x+r+R)

- Code: Select all
` x -r +R`

x x^2 -rx Rx

r rx -r^2 Rr

R Rx -Rr R^2

x^2 - r^2 + R^2 + rx - rx + 2Rx + Rr - Rr

simplifies into

x^2 - r^2 + R^2 + 2Rx

• Now take both quadratics:

(x^2 - r^2 + R^2 -2Rx)(x^2 - r^2 + R^2 + 2Rx)

- Code: Select all
` x^2 -r^2 R^2 -2Rx`

x^2 x^4 -r^2x^2 R^2x^2 -2Rx^3

-r^2 -r^2x^2 r^4 -R^2r^2 2Rr^2x

R^2 R^2x^2 -R^2r^2 R^4 -2R^3x

2Rx 2Rx^3 -2Rr^2x 2R^3x -4R^2x^2

x^4 + r^4 + R^4 + 2Rx^3 - 2Rx^3 - 2r^2x^2 + 2R^2x^2 + 2Rr^2x + 2R^3x - 2R^3x - 2Rr^2x - 2R^2r^2 - 2R^2r^2 - 4R^2x^2

simplifies into

x^4 + r^4 + R^4 - 2r^2x^2 - 2R^2x^2 - 2R^2r^2

• Using equation above, apply to axes Y and Z. Simply put in place of X, and swap R for Ri in Z :

x^4 + r^4 + R^4 - 2r^2x^2 - 2R^2x^2 - 2R^2r^2

y^4 + r^4 + R^4 - 2r^2y^2 - 2R^2y^2 - 2R^2r^2

z^4 + r^4 + Ri^4 - 2r^2z^2 - 2Ri^2z^2 - 2Ri^2r^2

• For the Z intercepts polynome, apply i^2 = -1 , i^4 = 1

z^4 + r^4 + R^4 - 2r^2z^2 + 2R^2z^2 + 2R^2r^2

-- Effectively, we get a sign change, depending on exponent. The imaginary component is what makes Z different.

• Consolidate all three into one, recording which terms repeat to derive the Consolidation Compliment of A :

x^4 + r^4 + R^4 - 2r^2x^2 - 2R^2x^2 - 2R^2r^2

y^4 + r^4 + R^4 - 2r^2y^2 - 2R^2y^2 - 2R^2r^2

z^4 + r^4 + R^4 - 2r^2z^2 + 2R^2z^2 + 2R^2r^2

x^4 + y^4 + z^4 + r^4 + R^4 - 2r^2x^2 - 2r^2y^2 - 2r^2z^2 - 2R^2x^2 - 2R^2y^2 + 2R^2z^2 - 2R^2r^2

• Terms that repeat :

r^4 twice

R^4 twice

- 2R^2r^2 will cancel out with the + 2R^2r^2 , leaving the other - 2R^2r^2 behind

A = 2(R^4 + r^4)So, referring to step one before expansion, the quartic equation for a torus may be expressible as:

(x±r±R)^4 + (y±r±R)^4 + (z±r±Ri)^4 - 2(R^4 + r^4) = 0

But before getting further into that, let's try graphing this equation, and see what we get.

x^4 + y^4 + z^4 + r^4 + R^4 - 2r^2x^2 - 2r^2y^2 - 2r^2z^2 - 2R^2x^2 - 2R^2y^2 + 2R^2z^2 - 2R^2r^2

simplifies into

x^4 + y^4 + z^4 + r^4 + R^4 - 2(rx)^2 - 2(ry)^2 - 2(rz)^2 - 2(Rx)^2 - 2(Ry)^2 + 2(Rz)^2 - 2(Rr)^2

R = 3

r = 1

Set diameter values:

x^4 + y^4 + z^4 + 1^4 + 3^4 - 2(1x)^2 - 2(1y)^2 - 2(1z)^2 - 2(3x)^2 - 2(3y)^2 + 2(3z)^2 - 2(3)^2

Graphs into:

Makes interesting surface, topologically identical to a torus, but oblate in the 45 degree angles. Neat to see how using 1D intercepts can get close. There's some missing terms needed to smooth out and define a fully circular diameter. Hmm. But, since the resulting shape has a hole along Z, the theory is sound. Complex conjugates can be used to define empty cuts.

Some more research is needed on this. I think cross-multiplying the x,y,z products in step one may yield the correct terms (xy,yz,xz) required. They seem to be the only ones missing. In studying the shape produced from the equation, you can see perfect circles sitting right on the axes. So, perhaps by cross multiplying, the oblique 1D intercepts are defined, well enough to smooth out the oblique blobs.

What I have:

x^4 + y^4 + z^4 + r^4 + R^4 - 2(rx)^2 - 2(ry)^2 - 2(rz)^2 - 2(Rx)^2 - 2(Ry)^2 + 2(Rz)^2 - 2(Rr)^2

What it should be:

x^4 + y^4 + z^4 + R^4 + r^4 +

2(xy)^2 + 2(xz)^2 + 2(yz)^2 - 2(rx)^2 - 2(ry)^2 - 2(rz)^2 - 2(Rx)^2 - 2(Ry)^2 + 2(Rz)^2 - 2(Rr)^2 = 0

So, somehow, the x,y,z terms need to get multiplied together at some point, no pun intended.

Now that I know the equation needs more terms, the first mentioned form at the top would need the

Oblique Compliment of 2[(xy)^2 + (xz)^2 + (yz)^2],

(x±r±R)^4 + (y±r±R)^4 + (z±r±Ri)^4 + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(R^4 + r^4) = 0

Which expands into

(x-r-R)(x+r-R)(x-r+R)(x+r+R) + (y-r-R)(y+r-R)(y-r+R)(y+r+R) + (z-r-Ri)(z+r-Ri)(z-r+Ri)(z+r+Ri) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(R^4 + r^4) = 0

R = 3

r = 1

(x-1-3)(x+1-3)(x-1+3)(x+1+3) + (y-1-3)(y+1-3)(y-1+3)(y+1+3) + (z-1-3i)(z+1-3i)(z-1+3i)(z+1+3i) + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(3^4 + 1^4) = 0

Which .....

Graphs correctly!

Wow. Extremely freaking cool. This was a very neat exercise in deriving an equation. I've never constructed one from scratch, using 1D slices. Also, in exploring other forms of polynomials.

By far the most important result was the use of complex conjugates to define empty cuts and holes. I'll need to look into the cross multiplying of x,y,z , and try to find a sound method to get the oblique compliment of 2((xy)^2 + (xz)^2 + (yz)^2) . Perhaps it's just as simple as assigning oblique 1D angles. I have a much better understanding of how this degree-4 equation works, and why the terms are the way they are.