Polynomials for Toratopes

Discussion of shapes with curves and holes in various dimensions.

Re: Polynomials for Toratopes

ICN5D wrote:It'll be very similar to the original f(xyz), but missing some terms. Just like the first shape I graphed, it was a lumpy torus with a hole. Compared to a smooth torus, the largest areas of deviation were focused in the 45 degree angles, between axes. All that was needed were the three oblique terms. By consolidating, you're adding unlike terms, and tossing out repeating terms from expanding the product of solutions.

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

which after fully expanding, lacks the 45 degree oblique region terms, 2x^2y^2 , 2x^2z^2 , and 2y^2z^2 .

Just like here, we're tossing out the 2x repeats of R^4 and r^4 , and combining everything into one.

What if you added some oblique terms? For example f(x,x,0) = 0.
Marek14
Pentonian

Posts: 1101
Joined: Sat Jul 16, 2005 6:40 pm

Re: Polynomials for Toratopes

Is that the symbolism for it? I guess the others would be f(x,0,x) and f(0,y,y). Well, adding in the obliques would make a full 3D torus. But, I haven't looked into a way to derive the terms manually, just extrapolate into 4D. The other night, I did end up simplifying the 4D oblique terms, for use with a consolidated equation from 1D intercepts:

2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) =

(x^2 + y^2)(z^2+w^2) + (x^2 + z^2)(y^2+w^2) + (x^2 + w^2)(y^2 + z^2)

which is a nice layout of a sum of three products, which alternate all possible 2D planes in 4D
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

Can we get a list of the order of the expanded polynomial corresponding to each toratope?

e.g.
(III) is order 2
((II)I) is order 4

(IIII) is order 2
((III)I) is order 4
((II)II) is order 4
((II)(II)) is order 8
(((II)I)I) is order 8

Can someone check this and do it for 5D? It'd be too hard to expand out the full polynomial, but maybe you can eyeball it to figure out the order. I suspect the order is 2 to the power of the number of bracket pairs.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

PWrong wrote:Can we get a list of the order of the expanded polynomial corresponding to each toratope?

e.g.
(III) is order 2
((II)I) is order 4

(IIII) is order 2
((III)I) is order 4
((II)II) is order 4
((II)(II)) is order 8
(((II)I)I) is order 8

Can someone check this and do it for 5D? It'd be too hard to expand out the full polynomial, but maybe you can eyeball it to figure out the order. I suspect the order is 2 to the power of the number of bracket pairs.

Basically, I'd say that when you can make a cut resulting in two lower-dimensional toratopes, the degree of your original equation is twice the degree for that toratope (since you're multiplying two od those together to create the cut).
Marek14
Pentonian

Posts: 1101
Joined: Sat Jul 16, 2005 6:40 pm

Re: Polynomials for Toratopes

We haven't actually proven that that works though, have we?

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

PWrong wrote:We haven't actually proven that that works though, have we?

I don't think so. I'd say that the degree of larger toratope must be AT LEAST twice one of the smaller polytope because the product must still be included somewhere in there.
Marek14
Pentonian

Posts: 1101
Joined: Sat Jul 16, 2005 6:40 pm

Re: Polynomials for Toratopes

Here's a proof of my conjecture. I'll write it mostly in Latex code.

Start with a toratope T = (T_1 T_2 ... T_n), where each T_j is a toratope. Then write
T_1 = (T_{11} T_{12} \ldots T_{1n_1})
T_2 = (T_{21} T_{22} \ldots T_{1n_2})
\vdots
T_n = (T_{n1} T_{n2} \ldots T_{1n_n})

Where all of the T_{ij} are toratopes. Then
T_{ij} = (T_{ij1} T_{ij2} \ldots T_{ijn_{ij}})
where the T_{ijk} are toratopes, and so on. At some point the T_{ijk...} will just be spheres.

The equation for this is a degree 2 polynomial of n functions:

P_2(\sqrt{T_1}, \sqrt{T_2}, \ldots, \sqrt{T_n}).
Expand this out and there will be a sum of n square roots, and some other stuff. Applying the ideas here, we get a polynomial of degree 2*2^n:

P_{2*2^n}(\sqrt{T_{11}}, \sqrt{T_{12}}, \ldots, \sqrt{T_{n n_n}).

Do this again, and we get a polynomial of degree 2 * 2^n * 2^{\sum_{i=1}^n n_i}.
P(\sqrt{T_{111}}, \sqrt{T_{112}}, \ldots, \sqrt{T_{n n_n n_{n_n}}).

If we keep doing this until all the T_{ijk...} are spheres, we eventually get a polynomial of degree 2^p, where
p = 1 + n + \sum_{i=1}^n n_i + \sum_{i=1}^n \sum_{j=1}^{n_i} n_{ij} + \ldots
which is precisely the number of bracket pairs.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Been moving lately, havent been able to respond. But, looks good with the theorem. Pretty much what I saw happening, too. Another thing I noticed, which ties in with Marek's observation was the number and type of intercepts. As we have seen, the intercepts can be expressed as a product of lower degree polynomials for a lower toratope. When factored out, we'll get multiple copies of a quadratic for a circle, or quartics for a torus, etc. Taking the (((II)I)(II)) for example, we get two octics for ditorus or tiger, four quartics for torus and eight quadratics for circle. All of which add up to degree 16. Same for (((II)I)((II)I)), we get two deg-16, four octics, eight quartics, and sixteen quadratics, which add up to deg-32. Even the empty cuts work this way, they're just all complex solutions.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

Deriving the Degree-4 Polynomial of 4D Torisphere ((III)I) S^1 x S^2

Implicit Definition
----------------------
(sqrt(x^2 + y^2 + z^2) - a)^2 + w^2 - b^2 = 0

• Defining by axis intercepts in 1D

1D Intercepts
------------------
(sqrt(x^2) - a)^2 - b^2 --> x±b±a = (x-b-a)(x+b-a)(x-b+a)(x+b+a)

(sqrt(y^2) - a)^2 - b^2 --> y±b±a = (y-b-a)(y+b-a)(y-b+a)(y+b+a)

(sqrt(z^2) - a)^2 - b^2 --> z±b±a = (z-b-a)(z+b-a)(z-b+a)(z+b+a)

(- a)^2 + w^2 - b^2 --> w±b±ai = (w-b-ai)(w+b-ai)(w-b+ai)(w+b+ai)

Polynome for Torisphere
--------------------------------
(x±b±a)^4 + (y±b±a)^4 + (z±b±a)^4 + (w±b±ai)^4 + Q4 - Cc = 0

Theoretical Oblique Compliment in 4D , Q4
------------------------------------------------------
2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2)

--- This will be the first test for Q4, with the Torisphere, then Spheritorus ((II)II)

• Deriving the Consolidation Compliment, Cc

We know from the torus equation that (x±b±a)^4 = (x-b-a)(x+b-a)(x-b+a)(x+b+a) expands into x^4 + b^4 + a^4 - 2(bx)^2 - 2(ax)^2 - 2(ab)^2

Extrapolate X to Y, Z, W :
----------------------------------

x^4 + b^4 + a^4 - 2(bx)^2 - 2(ax)^2 - 2(ab)^2

y^4 + b^4 + a^4 - 2(by)^2 - 2(ay)^2 - 2(ab)^2

z^4 + b^4 + a^4 - 2(bz)^2 - 2(az)^2 - 2(ab)^2

w^4 + b^4 + ai^4 - 2(bw)^2 - 2(aiw)^2 - 2(aib)^2

Simplify Imaginary values for W, i^2 = -1 , i^4 = 1
--------------------------------------------------------------

w^4 + b^4 + ai^4 - 2(bw)^2 - 2(aiw)^2 - 2(aib)^2

simplifies into

w^4 + b^4 + a^4 - 2(bw)^2 + 2(aw)^2 + 2(ab)^2

Consolidate into one full 4D equation
-----------------------------------------------

x^4 + b^4 + a^4 - 2(bx)^2 - 2(ax)^2 - 2(ab)^2

y^4 + b^4 + a^4 - 2(by)^2 - 2(ay)^2 - 2(ab)^2

z^4 + b^4 + a^4 - 2(bz)^2 - 2(az)^2 - 2(ab)^2

w^4 + b^4 + a^4 - 2(bw)^2 + 2(aw)^2 + 2(ab)^2

sums to

x^4 + y^4 + z^4 + w^4 + b^4 + a^4 - 2((bx)^2 + (ax)^2 + (by)^2 + (ay)^2 + (bz)^2 + (az)^2 + (bw)^2 - (aw)^2 + (ab)^2)

Terms that repeat
-----------------------

+ a^4 --> 3x
+ b^4 --> 3x
- 2(ab)^2 --> 1x , after Z and W cancel out

Consolidation Compliment Cc for 4D Torisphere
-------------------------------------------------------------

Cc = 3(a^4 + b^4) - 2(ab)^2

• Polynome now becomes
(x±b±a)^4 + (y±b±a)^4 + (z±b±a)^4 + (w±b±ai)^4 + Q4 - 3(a^4 + b^4) + 2(ab)^2 = 0

Apply theoretical Q4
----------------------------
(x±b±a)^4 + (y±b±a)^4 + (z±b±a)^4 + (w±b±ai)^4 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(a^4 + b^4) + 2(ab)^2 = 0

Expand into

(x-b-a)(x+b-a)(x-b+a)(x+b+a) + (y-b-a)(y+b-a)(y-b+a)(y+b+a) + (z-b-a)(z+b-a)(z-b+a)(z+b+a) + (w-b-ai)(w+b-ai)(w-b+ai)(w+b+ai) + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(a^4 + b^4) + 2(ab)^2

Set diameter values a=3 , b=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-1-3)(z+1-3)(z-1+3)(z+1+3) + (w-1-3i)(w+1-3i)(w-1+3i)(w+1+3i) + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(3^4 + 1^4) + 2(3)^2

• Make Cuts into 3D , establish cut dimension as parameter a :

((IIa)I)
----------
(x-1-3)(x+1-3)(x-1+3)(x+1+3) + (y-1-3)(y+1-3)(y-1+3)(y+1+3) + (a-1-3)(a+1-3)(a-1+3)(a+1+3) + (z-1-3i)(z+1-3i)(z-1+3i)(z+1+3i) + 2((xy)^2 + (xa)^2 + (xz)^2 + (ya)^2 + (yz)^2 + (az)^2) - 3(3^4 + 1^4) + 2(3)^2

((III)a)
----------
(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-1-3)(z+1-3)(z-1+3)(z+1+3) + (a-1-3i)(a+1-3i)(a-1+3i)(a+1+3i) + 2((xy)^2 + (xz)^2 + (xa)^2 + (yz)^2 + (ya)^2 + (za)^2) - 3(3^4 + 1^4) + 2(3)^2

Result 1: Graphing both intercepts makes correct figure, Q4 confirmed as correct 4D oblique angle definition

((IIZ)z)
----------
(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*sin(a))-1-3)((z*sin(a))+1-3)((z*sin(a))-1+3)((z*sin(a))+1+3) + ((z*cos(a))-1-3i)((z*cos(a))+1-3i)((z*cos(a))-1+3i)((z*cos(a))+1+3i) + 2((xy)^2 + (x(z*sin(a)))^2 + (x(z*cos(a)))^2 + (y(z*sin(a)))^2 + (y(z*cos(a)))^2 + ((z*sin(a))(z*cos(a)))^2) - 3(3^4 + 1^4) + 2(3)^2

Result 2: Rotate equation functions perfectly, from Z to z, by parameter a

• Now for Spheritorus ((II)II)

Using results from Torisphere, the Q4 will most likely work for Spheritorus. The task is to derive the Consolidate Compliment, Cc

(x±b±a)^4 + (y±b±a)^4 + (z±b±ai)^4 + (w±b±ai)^4 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - Cc = 0

Extrapolate X to Y, Z, W :
----------------------------------

x^4 + b^4 + a^4 - 2(bx)^2 - 2(ax)^2 - 2(ab)^2

y^4 + b^4 + a^4 - 2(by)^2 - 2(ay)^2 - 2(ab)^2

z^4 + b^4 + ai^4 - 2(bz)^2 - 2(aiz)^2 - 2(aib)^2

w^4 + b^4 + ai^4 - 2(bw)^2 - 2(aiw)^2 - 2(aib)^2

Simplify Imaginary values for Z and W, i^2 = -1 , i^4 = 1
----------------------------------------------------------------------

z^4 + b^4 + ai^4 - 2(bz)^2 - 2(aiz)^2 - 2(aib)^2
w^4 + b^4 + ai^4 - 2(bw)^2 - 2(aiw)^2 - 2(aib)^2

simplifies into

z^4 + b^4 + a^4 - 2(bz)^2 + 2(az)^2 + 2(ab)^2
w^4 + b^4 + a^4 - 2(bw)^2 + 2(aw)^2 + 2(ab)^2

Consolidate into one equation
---------------------------------------

x^4 + b^4 + a^4 - 2(bx)^2 - 2(ax)^2 - 2(ab)^2

y^4 + b^4 + a^4 - 2(by)^2 - 2(ay)^2 - 2(ab)^2

z^4 + b^4 + a^4 - 2(bz)^2 + 2(az)^2 + 2(ab)^2

w^4 + b^4 + a^4 - 2(bw)^2 + 2(aw)^2 + 2(ab)^2

Consolidated 1D Intercept equation

x^4 + y^4 + z^4 + w^4 + b^4 + a^4 - 2((bx)^2 + (ax)^2 + (by)^2 + (ay)^2 + (bz)^2 - (az)^2 + (bw)^2 - (aw)^2)

Terms that repeat
-------------------------

+ a^4 --> 3x
+ b^4 --> 3x
- 2(ab)^2 --> Z and W cancel out with X and Y

Consolidation Compliment Cc for 4D Spheritorus
-------------------------------------------------------------

Cc = 3(a^4 + b^4)

((II)II) Polynome
----------------------

(x±b±a)^4 + (y±b±a)^4 + (z±b±ai)^4 + (w±b±ai)^4 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(a^4 + b^4) = 0

So far I've made three polynomials of this form, using 1D intercepts and imaginary numbers to define the hole. The next task is ditorus and tiger. I've been slowly working on them.

Torus ((II)I)
(x±b±a)^4 + (y±b±a)^4 + (z±b±ai)^4 + 2((xy)^2 + (xz)^2 + (yz)^2) - 2(a^4 + b^4) = 0

Torisphere ((III)I)
(x±b±a)^4 + (y±b±a)^4 + (z±b±a)^4 + (w±b±ai)^4 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(a^4 + b^4) + 2(ab)^2 = 0

Spheritorus ((II)II)
(x±b±a)^4 + (y±b±a)^4 + (z±b±ai)^4 + (w±b±ai)^4 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - 3(a^4 + b^4) = 0

Proposed Polynomials for ditorus and tiger:

(((II)I)I)
(x±c±b±a)^8 + (y±c±b±a)^8 + (z±c±b±ai)^8 + (w±c±bi±ai)^8 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - Cc

((II)(II))
(x±c±bi±a)^8 + (y±c±bi±a)^8 + (z±c±b±ai)^8 + (w±c±b±ai)^8 + 2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2) - Cc

I'll need to expand, consolidate and graph these two to test my theory. I fear it won't be this easy, but there's only one way to find out!
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

This is fantastic, good luck with the tiger and ditorus. Do you think the (xy)^2 terms will be enough though? Given that these are octic surfaces I thought you'd need to add something more complicated.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

By the way, you could extend the S_n notation from http://www.mathpages.com/home/kmath111/kmath111.htm as a convenient shorthand. This might make the long equations easier to handle.

S_1^1 = x + y + z + w
S_1^2 = x^2 + y^2 + z^2 + w^2
S_2^1 = (xy) + (xz) + (xw) + (yz) + (yw) + (zw)
S_2^2 = (xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2
S_3^1 = xyz + xyw + xzw + yzw

Or make it a function to change the arguments:

S_1^2(p,q,r) = p^2 + q^2 + r^2
S_1^4(x ± b ± a, y ± b ± a, z ± b ± ai, w ± b ± ai) = (x±b±a)^4 + (y±b±a)^4 + (z±b±ai)^4 + (w±b±ai)^4
e.t.c.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Hey I'm confused. I was under the impression that the x±b±a terms were derived from solving the equation with all variables except one set to zero. But for the torus, the solutions for z with x=0 and y=0 are
z = ±\sqrt{a^2 - b^2}

Have I misunderstood the process?

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

PWrong wrote:This is fantastic, good luck with the tiger and ditorus. Do you think the (xy)^2 terms will be enough though? Given that these are octic surfaces I thought you'd need to add something more complicated.

Thanks. I'm making an educated guess on the oblique terms. Those will be the first to test out. Since it has to define the intercepts of two degree-4 tori, maybe it does need more. I have a few ideas brewing on what else it could be. If it is different, it may work for both octics, which would be an interesting find. One thing to point out, is the similarity between the 3D and 4D terms used so far. One simply adds a few more 2-D planes, 2((xw)^2 + (yw)^2 + (zw)^2) for a 4D compliment. But, these may just be a degree-4 thing, that works for ((II)I), ((II)II) , and ((III)I).

By the way, you could extend the S_n notation from http://www.mathpages.com/home/kmath111/kmath111.htm as a convenient shorthand. This might make the long equations easier to handle.

Ah, yeah, I keep forgetting about that. I checked it out again, and it made a little more sense this time. The concept of symmetric subgroups has everything to do with toratopes. These shapes and their equations have amazingly high symmetry, which leads to complex patterns in the subgroups. I wonder if that's ever been explored before? I'm imagining some kind of root finding algorithm that lets you derive the intercepts algebraically. You'd end up with the factored out clusters of lower toratopes or something like that.

Hey I'm confused. I was under the impression that the x±b±a terms were derived from solving the equation with all variables except one set to zero. But for the torus, the solutions for z with x=0 and y=0 are
z = ±\sqrt{a^2 - b^2}

Have I misunderstood the process?

Well, the way I see it, is this:

((xy)z) : full torus
(sqrt(x^2 + y^2) - a)^2 + z^2 - b^2

(()z) : z-cut of torus, cancel out x and y

(sqrt() - a)^2 + z^2 - b^2

Which, from what I can see will cancel out the sqrt, too, into

(- a)^2 + z^2 - b^2

This cut is the Z axis going through the hole, which is the interior complex plane of the torus. The ring is real and the hole is imaginary, in terms of solutions. Think of a parabola that stands above the x-axis in +y space. Solving for x will make 2 complex solutions, with imaginary numbers as the x-intercept. The imaginary value is locating the parabola in a higher dimension, of y-space. Kind of like a "projective intercept value"

Same thing is going on here, with z. Radially outwards from the z-axis are the real solutions of the ring. We could move the torus by the value of " a " to get z = ± b, as we cut through a circle and get 2 points in a row. But, there is another circle as well, spaced by " 2a " in the complex plane. We have to account for the second circle intercept, even if we don't see it in the real plane. The value of " a " defines the size of the complex plane, and becomes " ai ".

And, since we can move the torus back and forth to both circles, making two points in a row in two locations, we assign the range as ± ai . Now we get 2 points ± b at the position ± ai , which simply becomes

z = ± b ± ai
or
z = -b-ai , -b+ai , +b-ai , +b+ai

as the 4 complex solutions when solving for z. Shifting the torus by the value of +ai will make two real solutions and two complex, becoming

z = - b , + b , -b+2ai , +b+2ai

In relating this concept to toratopic notation and the implicit definition, we get a clear layout of the solutions, in real and complex planes. For z-cut again,

(- a)^2 + z^2 - b^2

will equate and simplify to z = ± b ± ai

The (-a)^2 term will become ± ai, through a similar process to your equation.

z^2 = b^2 - (-a)^2

z = sqrt(b^2 - (-a)^2)

z = ± b ± ai

I don't think I've seen this relation anywhere before. As in, how sqrt(b^2 - (-a)^2) simplifies into ± b ± ai . It's very interesting, and toratopes tell it all.

The first test with complex numbers was the torus, and it worked out very well. It confirmed my suspicion with imaginary numbers and the empty cut through a hole. And from this basic principle, tested and verified, we can easily extrapolate to higher toratopes.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

I don't think I've seen this relation anywhere before. As in, how sqrt(b^2 - (-a)^2) simplifies into ± b ± ai .

It doesn't though. Try plugging in values for a and b.

I can see that the process is working somehow, but the justification is wrong.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Hmm. Well, then Im not completely clear on how it happens. We have the beginning and end result, with a process that makes a complex conjugate, where "a" becomes "ai". Am I right with thinking the sqrt cancels when X and Y are set to zero? The value A is left over, as " -a" , which is the only thing thats different than B. Maybe subtract the z^2, then do a sign change will put B to + , and A to - ?
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

I've been thinking more about this. All day, in fact! You're right, the process is working somehow. And, it seems that the justification is wrong. But, we still have this strange relation. The principle with z = ± b ± ai, and how the hole is encircled by the ring is clear. Which relates to how imaginary numbers locate a shape across empty (higher dimensional) space. All of these things are well-known, and make sense.

Which brings us back to this little equation, and how to interpret it.

(-a)^2 + z^2 - b^2

where,

z = sqrt(b^2 - (-a)^2)

which somehow also means,

z = ± b ± ai

We know that sqrt(a^2) = a , or more precisely, ± a. The sqrt cancels out the ^2 , and leaves us with ± a. If we dropped out b, we get

sqrt(-(-a)^2)

which is the source of the mystery . It seems like,

1) the negative inside the brackets, (-a), gets turned into +a, when squared
2) the squaring gets canceled by the square root
3) since we have two negatives, canceling the inside minus will leave the outside minus left over. This should make an imaginary value:

1) sqrt(-(-a)^2)

2) sqrt(-a^2)

3) ± ai

This actually makes more sense to me now. It's a double negative that gets used in different ways. Interesting.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

If the torus is a ring torus then a > b, so the solutions for z are imaginary:
z =±sqrt(b^2 - a^2) =±i sqrt(a^2 - b^2)

If it's a spindle torus b>a, there is no hole and the solution is actually visible.

Don't forget in the expanded polynomial you had to add a bunch of extra stuff to make it work. Maybe it works but it doesn't have anything to do with complex roots. A torus is a cyclide, which is a quadric of x,y,z,r^2. The polynomials you're creating are also cyclides. It only takes a bit of fudging to turn one into the other.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

PWrong wrote: A torus is a cyclide, which is a quadric of x,y,z,r^2. The polynomials you're creating are also cyclides. It only takes a bit of fudging to turn one into the other.

Yes, this is true! It's neat to see how they're all related.

Don't forget in the expanded polynomial you had to add a bunch of extra stuff to make it work. Maybe it works but it doesn't have anything to do with complex roots.

The first shape I graphed was the weird lumpy torus, using only the 1D solution terms. From the start, using complex roots for Z, it made a hole. The extra stuff was only to smooth out the lumps, into a circular ring in all oblique angles. Where, the solution that was complex was for Z, which goes through the hole. Without the complex root, there would be no sign change for Z in the expanded polynome, and there would be no hole, but most likely a spindle torus. The X and Y intercepts will have four roots, and the Z will have two.

Using the 1D solution polynome, the other toric types may look something like this:

Q - extra terms to smooth out the ring, for 3D : 2((xy)^2 + (xz)^2 + (yz)^2)
Cc - removes extra terms when consolidating X, Y, Z into full equation

Ring Torus
(x ± b ± a)^4 + (y ± b ± a)^4 + (z ± b ± ai)^4 + Q - Cc

Horn Torus
(x ± b ± a)^4 + (y ± b ± a)^4 + z^4 + Q - Cc

Spindle Torus
(x ± b ± a)^4 + (y ± b ± a)^4 + (z ± a)^4 + Q - Cc

Hmm, makes me want to test this out. Might strengthen my case!

EDIT: Made changes to spindle torus, Z will only have two solutions, not 4. Also changed the polynome, too.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

I wrote some Mathematica code to check all the cyclide toratopes.

Consider ((II...)II...) with m I's in brackets followed by n I's. We'll call this a (m,n)-cyclide toratope. The expanded polynomial is:

(x_1 ± a ± b)^4 + ... + (x_m ± a ± b)^4 + (x_{m+1} ± a ± b i)^4 + ... + (x_{m+n} ± a ± b i)^4
+ S_2^2(x_1, ..., x_{m+n}
+ (m + n - 1)a^4 + (m - n + 1) 2 a^2 b^2 + (m + n - 1)b^4

I've checked this up to (9,9). Unfortunately with the way I programmed the toratope generator I can't go higher than that without a bit more work. But the pattern is pretty clear.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Here's a thing that might be worth noting:

(x ± a ± b)^4 = (x^2 - a^2 - b^2)^2 - 4 a^2 b^2

x ± a ± bi)^4 = (x^2 - a^2 + b^2)^2 + 4 a^2 b^2

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Whoa, awesome! You know, it might be possible to write a code that uses the implicit equation to naively derive the axis roots, and output in that generalized polynomial. That's what I use for the root product. That way it can be used with any toratope, no matter what combination. It seems like when using the 1-D roots for a polynomial, it reflects most of the combinatorics of the implicit, as opposed to 2D circles, or 3D tori.

Here's a thing that might be worth noting:

(x ± a ± b)^4 = (x^2 - a^2 - b^2)^2 - 4 a^2 b^2

(x ± a ± bi)^4 = (x^2 - a^2 + b^2)^2 + 4 a^2 b^2

Thanks, that might come in handy.

So, I've noticed that this strange polynomial has right around three distinct parts:

Root Product Sum + Oblique Compliment - Consolidate Compliment

-The root products are axis solutions, (x±a±b±c±d), real and/or complex. Respectively for each toratope, all axes have the same number of roots.
-The oblique terms will compliment the axis root definition, and smooth out the lumpy ring.
-The consolidate compliment will remove the extra terms that accumulate in the root product sum.

So far, though. I still have yet to expand (x±c±b±a)^8 , and test (((II)I)I) and ((II)(II)) .
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

I did a little bit with the tiger polynomial in Mathematica earlier. It's not anywhere near as simple as the quartic toratopes. The oblique complement will have to include many more terms and less symmetry.

What would the root product sum be for the tiger anyway? Since it apparently isn't really based on the roots of the equation, where do we put the i's?

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

The tiger is entirely imaginary ( sorry, Marek ). But, no, seriously, all axis roots are complex:

Tiger Implicit Equation ((xy)(zw))
(sqrt(x^2 + y^2) - a)^2 + (sqrt(z^2 + w^2) - b)^2 - c^2

Axial Roots X, Y, Z, and W

((x)())
(sqrt(x^2) - a)^2 + (- b)^2 - c^2
x = ± c ± bi ± a

((y)())
(sqrt(y^2) - a)^2 + (- b)^2 - c^2
y = ± c ± bi ± a

(()(z))
(- a)^2 + (sqrt(z^2) - b)^2 - c^2
z = ± c ± b ± ai

(()(w))
(- a)^2 + (sqrt(w^2) - b)^2 - c^2
w = ± c ± b ± ai

Which will become a root product sum of

(x±c±bi±a)^8 + (y±c±bi±a)^8 + (z±c±b±ai)^8 + (w±c±b±ai)^8

I'm pretty darn sure. Tiger has a fairly symmetric layout, which can extrapolate easily from the principles found in the Z intercept of torus equation. What I see in my mind is no matter which axis you solve for, moving across the complex plane, by the imaginary value ±Ai or ±Bi, will make 4 points of a torus intercept ±C±B or ±C±A. As for the oblique compliment, that'll be the fun part! Who knows what it could be. But, I'll most likely test and theorize, and ponder until I find it.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

PWrong wrote:Here's a thing that might be worth noting:

(x ± a ± b)^4 = (x^2 - a^2 - b^2)^2 - 4 a^2 b^2

x ± a ± bi)^4 = (x^2 - a^2 + b^2)^2 + 4 a^2 b^2

Hugh?
I don't know what you after here. But in a mere mathematical sense these equations cannot be true!
The left hand sides each would contain terms in x^3 and in x, while the right hand ones do not.

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Polynomials for Toratopes

The notation is a bit unclear. By (x ± a ± b)^4 we mean (x + a + b)(x + a - b)(x - a + b)(x - a - b). The cube and linear terms cancel out thanks to the "difference of two squares" rule.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

I've been thinking more about the oblique terms of a 4D octic surface. Now, I'm absolutely certain that 4D quartic terms aren't enough. I derive my inspiration from the 3D intercepts of ditorus and tiger. All cuts for both shapes, we get two tori, which are quartic surfaces. In order to get an equation for two intercepts of a torus, we multiply them together, as translated copies spaced apart by a diameter term. So, think about this for a sec. The 3D and 4D quartic surfaces either make a single torus, or two spheres, as 3D roots.

If we get two quartic surfaces as a solution of an octic, then the oblique terms for an octic surface may be a simple squaring of the whole quartic oblique term. The root product sum will take care of defining the points along x, y, z, w, for both tori. The consolidate compliment will remove repeat terms from the root product sum. This leaves the remaining terms to the oblique compliment, which probably just end up getting multiplied together, i.e. squared.

Oblique Compliment Terms, so far:

3D Quartic Surface

2((xy)^2 + (xz)^2 + (yz)^2)

4D Quartic Surfaces

2((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2)

Proposed 4D Octic Surfaces

4((xy)^2 + (xz)^2 + (xw)^2 + (yz)^2 + (yw)^2 + (zw)^2)^2

If this works, and I still have yet to test the octics, then we can easily generalize to higher orders and dimensions. As a theory, the principle seems sound. For a degree-16 in 5D, which makes 4x torus intercepts, we would simply raise the generic 5D quartic oblique terms to the power of four:

16((xy)^2 + (xz)^2 + (xw)^2 + (xv)^2 + (yz)^2 + (yw)^2 + (yv)^2 + (zw)^2 + (zv)^2 + (wv)^2)^4
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

Here's the polynomial for tiger:

Code: Select all
-64 b^2 r^2 w^2 x^2 - 64 b^2 r^2 w^2 y^2 - 64 b^2 r^2 x^2 z^2 - 64 b^2 r^2 y^2 z^2 + (a^4 - 2 a^2 b^2 + b^4 - 2 a^2 r^2 + 2 b^2 r^2 +    r^4 - 2 a^2 w^2 + 2 b^2 w^2 - 2 r^2 w^2 + w^4 - 2 a^2 x^2 -    2 b^2 x^2 + 2 r^2 x^2 + 2 w^2 x^2 + x^4 - 2 a^2 y^2 - 2 b^2 y^2 +    2 r^2 y^2 + 2 w^2 y^2 + 2 x^2 y^2 + y^4 - 2 a^2 z^2 + 2 b^2 z^2 -    2 r^2 z^2 + 2 w^2 z^2 + 2 x^2 z^2 + 2 y^2 z^2 + z^4)^2

Here is the expansion of (x±r±bi±a)^8 + (y±r±bi±a)^8 + (z±r±b±ai)^8 + (w±r±b±ai)^8:
Code: Select all
4 a^8 + 16 a^6 b^2 + 24 a^4 b^4 + 16 a^2 b^6 + 4 b^8 + 24 a^4 r^4 - 16 a^2 b^2 r^4 + 24 b^4 r^4 + 4 r^8 + 4 a^6 w^2 + 4 a^4 b^2 w^2 - 4 a^2 b^4 w^2 - 4 b^6 w^2 + 4 a^4 r^2 w^2 + 40 a^2 b^2 r^2 w^2 + 4 b^4 r^2 w^2 - 4 a^2 r^4 w^2 + 4 b^2 r^4 w^2 - 4 r^6 w^2 + 6 a^4 w^4 - 4 a^2 b^2 w^4 + 6 b^4 w^4 - 4 a^2 r^2 w^4 + 4 b^2 r^2 w^4 + 6 r^4 w^4 + 4 a^2 w^6 - 4 b^2 w^6 - 4 r^2 w^6 + w^8 - 4 a^6 x^2 - 4 a^4 b^2 x^2 + 4 a^2 b^4 x^2 + 4 b^6 x^2 + 4 a^4 r^2 x^2 + 40 a^2 b^2 r^2 x^2 + 4 b^4 r^2 x^2 + 4 a^2 r^4 x^2 - 4 b^2 r^4 x^2 - 4 r^6 x^2 + 6 a^4 x^4 - 4 a^2 b^2 x^4 + 6 b^4 x^4 + 4 a^2 r^2 x^4 - 4 b^2 r^2 x^4 + 6 r^4 x^4 - 4 a^2 x^6 + 4 b^2 x^6 - 4 r^2 x^6 + x^8 - 4 a^6 y^2 - 4 a^4 b^2 y^2 + 4 a^2 b^4 y^2 + 4 b^6 y^2 + 4 a^4 r^2 y^2 + 40 a^2 b^2 r^2 y^2 + 4 b^4 r^2 y^2 + 4 a^2 r^4 y^2 - 4 b^2 r^4 y^2 - 4 r^6 y^2 + 6 a^4 y^4 - 4 a^2 b^2 y^4 + 6 b^4 y^4 + 4 a^2 r^2 y^4 - 4 b^2 r^2 y^4 + 6 r^4 y^4 - 4 a^2 y^6 + 4 b^2 y^6 - 4 r^2 y^6 + y^8 + 4 a^6 z^2 + 4 a^4 b^2 z^2 - 4 a^2 b^4 z^2 - 4 b^6 z^2 + 4 a^4 r^2 z^2 + 40 a^2 b^2 r^2 z^2 + 4 b^4 r^2 z^2 - 4 a^2 r^4 z^2 + 4 b^2 r^4 z^2 - 4 r^6 z^2 + 6 a^4 z^4 - 4 a^2 b^2 z^4 + 6 b^4 z^4 - 4 a^2 r^2 z^4 + 4 b^2 r^2 z^4 + 6 r^4 z^4 + 4 a^2 z^6 - 4 b^2 z^6 - 4 r^2 z^6 + z^8

Finally, here's the difference between them, expanded:
Code: Select all
3 a^8 + 20 a^6 b^2 + 18 a^4 b^4 + 20 a^2 b^6 + 3 b^8 + 4 a^6 r^2 - 12 a^4 b^2 r^2 + 12 a^2 b^4 r^2 - 4 b^6 r^2 + 18 a^4 r^4 - 4 a^2 b^2 r^4 + 18 b^4 r^4 + 4 a^2 r^6 - 4 b^2 r^6 + 3 r^8 + 8 a^6 w^2 - 8 a^4 b^2 w^2 + 8 a^2 b^4 w^2 - 8 b^6 w^2 + 48 a^2 b^2 r^2 w^2 - 8 a^2 r^4 w^2 + 8 b^2 r^4 w^2 + 8 a^2 b^2 w^4 - 8 a^2 r^2 w^4 + 8 b^2 r^2 w^4 + 8 a^2 w^6 - 8 b^2 w^6 - 8 a^4 b^2 x^2 + 8 b^6 x^2 - 8 a^4 r^2 x^2 + 48 a^2 b^2 r^2 x^2 + 8 b^4 r^2 x^2 + 16 a^2 r^4 x^2 - 8 b^2 r^4 x^2 - 8 r^6 x^2 - 12 a^4 w^2 x^2 + 8 a^2 b^2 w^2 x^2 + 4 b^4 w^2 x^2 + 8 a^2 r^2 w^2 x^2 + 40 b^2 r^2 w^2 x^2 + 4 r^4 w^2 x^2 + 12 a^2 w^4 x^2 - 4 b^2 w^4 x^2 + 4 r^2 w^4 x^2 - 4 w^6 x^2 - 8 a^2 b^2 x^4 + 16 a^2 r^2 x^4 + 12 a^2 w^2 x^4 + 4 b^2 w^2 x^4 - 4 r^2 w^2 x^4 - 6 w^4 x^4 + 8 b^2 x^6 - 8 r^2 x^6 - 4 w^2 x^6 - 8 a^4 b^2 y^2 + 8 b^6 y^2 - 8 a^4 r^2 y^2 + 48 a^2 b^2 r^2 y^2 + 8 b^4 r^2 y^2 + 16 a^2 r^4 y^2 - 8 b^2 r^4 y^2 - 8 r^6 y^2 - 12 a^4 w^2 y^2 + 8 a^2 b^2 w^2 y^2 + 4 b^4 w^2 y^2 + 8 a^2 r^2 w^2 y^2 + 40 b^2 r^2 w^2 y^2 + 4 r^4 w^2 y^2 + 12 a^2 w^4 y^2 - 4 b^2 w^4 y^2 + 4 r^2 w^4 y^2 - 4 w^6 y^2 - 12 a^4 x^2 y^2 - 8 a^2 b^2 x^2 y^2 - 12 b^4 x^2 y^2 + 24 a^2 r^2 x^2 y^2 + 8 b^2 r^2 x^2 y^2 - 12 r^4 x^2 y^2 + 24 a^2 w^2 x^2 y^2 + 8 b^2 w^2 x^2 y^2 - 8 r^2 w^2 x^2 y^2 - 12 w^4 x^2 y^2 + 12 a^2 x^4 y^2 + 12 b^2 x^4 y^2 - 12 r^2 x^4 y^2 - 12 w^2 x^4 y^2 - 4 x^6 y^2 - 8 a^2 b^2 y^4 + 16 a^2 r^2 y^4 + 12 a^2 w^2 y^4 + 4 b^2 w^2 y^4 - 4 r^2 w^2 y^4 - 6 w^4 y^4 + 12 a^2 x^2 y^4 + 12 b^2 x^2 y^4 - 12 r^2 x^2 y^4 - 12 w^2 x^2 y^4 - 6 x^4 y^4 + 8 b^2 y^6 - 8 r^2 y^6 - 4 w^2 y^6 - 4 x^2 y^6 + 8 a^6 z^2 - 8 a^4 b^2 z^2 + 8 a^2 b^4 z^2 - 8 b^6 z^2 + 48 a^2 b^2 r^2 z^2 - 8 a^2 r^4 z^2 + 8 b^2 r^4 z^2 - 12 a^4 w^2 z^2 + 24 a^2 b^2 w^2 z^2 - 12 b^4 w^2 z^2 - 8 a^2 r^2 w^2 z^2 + 8 b^2 r^2 w^2 z^2 - 12 r^4 w^2 z^2 + 12 a^2 w^4 z^2 - 12 b^2 w^4 z^2 + 12 r^2 w^4 z^2 - 4 w^6 z^2 - 12 a^4 x^2 z^2 + 8 a^2 b^2 x^2 z^2 + 4 b^4 x^2 z^2 + 8 a^2 r^2 x^2 z^2 + 40 b^2 r^2 x^2 z^2 + 4 r^4 x^2 z^2 + 24 a^2 w^2 x^2 z^2 - 8 b^2 w^2 x^2 z^2 + 8 r^2 w^2 x^2 z^2 - 12 w^4 x^2 z^2 + 12 a^2 x^4 z^2 + 4 b^2 x^4 z^2 - 4 r^2 x^4 z^2 - 12 w^2 x^4 z^2 - 4 x^6 z^2 - 12 a^4 y^2 z^2 + 8 a^2 b^2 y^2 z^2 + 4 b^4 y^2 z^2 + 8 a^2 r^2 y^2 z^2 + 40 b^2 r^2 y^2 z^2 + 4 r^4 y^2 z^2 + 24 a^2 w^2 y^2 z^2 - 8 b^2 w^2 y^2 z^2 + 8 r^2 w^2 y^2 z^2 - 12 w^4 y^2 z^2 + 24 a^2 x^2 y^2 z^2 + 8 b^2 x^2 y^2 z^2 - 8 r^2 x^2 y^2 z^2 - 24 w^2 x^2 y^2 z^2 - 12 x^4 y^2 z^2 + 12 a^2 y^4 z^2 + 4 b^2 y^4 z^2 - 4 r^2 y^4 z^2 - 12 w^2 y^4 z^2 - 12 x^2 y^4 z^2 - 4 y^6 z^2 + 8 a^2 b^2 z^4 - 8 a^2 r^2 z^4 + 8 b^2 r^2 z^4 + 12 a^2 w^2 z^4 - 12 b^2 w^2 z^4 + 12 r^2 w^2 z^4 - 6 w^4 z^4 + 12 a^2 x^2 z^4 - 4 b^2 x^2 z^4 + 4 r^2 x^2 z^4 - 12 w^2 x^2 z^4 - 6 x^4 z^4 + 12 a^2 y^2 z^4 - 4 b^2 y^2 z^4 + 4 r^2 y^2 z^4 - 12 w^2 y^2 z^4 - 12 x^2 y^2 z^4 - 6 y^4 z^4 + 8 a^2 z^6 - 8 b^2 z^6 - 4 w^2 z^6 - 4 x^2 z^6 - 4 y^2 z^6

If you're correct, then the difference should be just the oblique compliment and the constant terms. There shouldn't be any terms that combine the radii a,b,r with the variables x,y,z,w. So something is wrong.

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

Re: Polynomials for Toratopes

Holy moley, thats a lot of terms. Thanks for doing that for me. Hmm, Im going to dive into this a little later. Somewhere in there is what Im looking for, whether its easy or insanely difficult. Seems to be the latter! I like a good challenge, though.
in search of combinatorial objects of finite extent
ICN5D
Pentonian

Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: Polynomials for Toratopes

I just used the Mathematica code I wrote earlier. Here's a simplified form of the tiger polynomial.

-64 b^2 r^2 (x^2 + y^2) (z^2 + w^2)
+ ((-a^2 + b^2 + r^2)^2 + 2 (r^2 - b^2 ) (x^2 + y^2 - z^2 - w^2) - 2 a^2 (x^2 + y^2 + z^2 + w^2) + (x^2 + y^2 + z^2 + w^2)^2)^2 = 0

PWrong
Pentonian

Posts: 1599
Joined: Fri Jan 30, 2004 8:21 am
Location: Perth, Australia

PreviousNext