A Root Finding Algorithm for Factoring Hypertoric Sections

Discussion of shapes with curves and holes in various dimensions.

A Root Finding Algorithm for Factoring Hypertoric Sections

Postby ICN5D » Mon Nov 23, 2015 11:34 pm

I discovered something pretty neat over the last couple of weeks. For a while, I knew these toratope equations could do this, but I hadn't sat down and actually tried it. I thought for sure it would be way more difficult, but it's actually very simple. And, come to find out, these things have some of the most amazing multivariate factoring properties. They are also naturally factorable. This should come as no surprise, though! All of those multiple objects in the intercept arrays are playing the role of a factored out root, in the exact solution for the plane equation. Plus, we're dealing with a type of symmetric polynomial, which should also lend its symmetries to other areas, such as algebraically factoring out roots.

During the process, I learned how factoring is the art of re-structuring the equation, while maintaining a mathematically true statement. The end result is a product of multiple copies of the same, smaller equation. This describes multiple copies of the same object, that differ in size and location to an exact amount. I have discovered 5 fundamental changes we can make to an equation, that step-by-step gets it into a factored form.

I call these re-structuring methods " Rules 1-5 "

The 5 Rules of Factoring a Multivariate Symmetric Polynomial of Toric Varieties

Rules 1,2,3,4 will factor: All Univariate real & complex, All Multivariate real & some complex
Rule 5 will factor: All Uni & Multivariate complex roots


  • Rule 1: Factor a Difference of Squares

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



  • Rule 2: Factor Square Root of the Square

    ((sqrt(a^2) -b)^2 +M) ----> ((a-b)^2 +M)*((a+b)^2 +M)

    A simplified version of sqrt(a^2) = |a| = -a , +a . The M stands for all other numbers/variables in the equation. The 'a' term can be a single number Re or Im, a variable, or a polynomial containing both.


  • Rule 3: Undo Square Root and Apply the Square

    (sqrt(a) - b-c)*(sqrt(a) - b+c) ----> (a - (b-c)^2)*(a - (b+c)^2)

    A simple process of collecting the b,c numbers/vars to the RH side, and squaring both sides. This Rule won't factor, but simplify the expression, and usually set up the equation for even further factoring.


  • Rule 4: Factor the Square and Apply Square Root

    a^2 +b -c ----> (a -sqrt(c-b))*(a +sqrt(c-b))

    A simple process of collecting the b,c numbers/vars to the RH side, and square root both sides.


  • Rule 5: Add Imaginary Variable, Factor as Complex Function f(x,i) using Rule 2

    a^2 +x^2 -b^2 = 0,

    a^2 = (sqrt(i^2)-a)^2 ,

    (sqrt(i^2)-a)^2 +x^2 -b^2 ----> ((i-a)^2 +x^2 -b^2)*((i+a)^2 +x^2 -b^2)

    This method will overcome an unsolvable complex solution. The problem is a loss of information, by cancelling the sqrt, and squaring the negative half of the number 'a'. By introducing an imaginary variable tied in with the Im number, we restore some of the factorization symmetry, and continue with Rule 2. A complex function will define the solution in a higher dimensional context, while using fewer dimensions to see it in the real numbers. It also factors into a much clearer picture of the nature of the roots.



They can be used on the original form of the cross section equation. The last thing you would want to do, is to expand all of the terms. We can factor directly from the original form, based on certain symmetries in the equation. But, this is a Root Finding Algorithm, not a direct root deriving one. We have 5 choices at our disposal. However, there are certain solutions that can use Rules 1,2,3, and 4 interchangeably, and seamlessly, in no particular order. Some other solutions have only one precise sequence, and even others that cannot factor with Rules 1,2,3, or 4, which have to use Rule 5.

But, how would we know what direction to take, or whether or not we arrived at a correctly factored root? Well, in this case, I'm cheating. Since Marek taught me the Cut Algorithm, I already know ahead of time what every single factored root of every single solution will be, out to the 10 billionth dimension, and beyond :) . Toratopic notation is a higher level abstraction of the polynomial symmetry, which we can interpret directly, to see what the exact solutions are in all coordinate planes. Basically, the 5 Rules are the direct algebraic steps of the Cut Algorithm. The idea is to get all of the variables organized into an equation for a toratope, while organizing all of the numbers together, which describe reflections of size and location.

Below are a rip of my notes, organized in a systematic fashion. I have solved the 1D, 2D roots of (((II)I)I) , 1D, 2D for ((II)(II)) , and a few 1D for (((II)(II))I). I need to refine those lists like the types below. Those will be further postings. My goal now is to rigorously prove every solution, using these steps. It's been a neat experience, seeing exact hypercomplex roots come from the cut ((()I)()), and things like that. The real solutions are very easy and straightforward, and usually use only Rules 1,2,3. What I'm more interested in are the complex solutions, which will use Rule 4 and 5, along with the rest. I've been using Desmos to prove in a direct way the factored roots, detailing the partially factored forms as well. I found that for certain complex roots, Desmos has a limitation, and won't graph the function. No matter, CalcPlot has proven to be most useful in all areas of graphing a tough function. I'm trusting CalcPlot more, since it accurately graphs a 9D torus.


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Finding Roots in the Degree-4 ((II)I) Polynomial for a torus

(sqrt(x11^2+x12^2) -R11^2)^2 +x2^2 = R1^2

(sqrt(x^2+y^2) -a)^2 +z^2 = b^2

((xy-a)z-b)

((xy)z)


The 2 Distinct Univariate Roots
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((x)) = ((y))
(()z)


The 2 Distinct Bi-Variate Roots
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((xy))
((x)z) = ((y)z)

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Solving for Univariate Roots. This equation has 2 distinct solutions when solving for a single axis. One is real, and one

is complex. All 1-plane solutions are 4 roots of a 0D point.



((x)) = ((y)) : 4 points in a row

Set yz=0 , solve for real roots on x axis


(sqrt(x^2+0^2) -a)^2 +0^2 = b^2

(sqrt(x^2) -a)^2 -b^2 = 0


Rule 1: Factor Diff of squares of x,a with b

(sqrt(x^2) -a)^2 -b^2 => (sqrt(x^2) - a-b)(sqrt(x^2) - a+b)

(sqrt(x^2) - a-b)(sqrt(x^2) - a+b) = 0


Rule 3: Undo Sqrt on x , Apply the Squaring to a,b

(x^2 - (a-b)^2)(x^2 - (a+b)^2) = 0


Rule 1: Factor Diff of squares of x with a,b

(x^2 - (a-b)^2) => (x -(a-b))(x +(a-b)) => (x-a+b)(x+a-b)
(x^2 - (a+b)^2) => (x -(a+b))(x +(a+b)) => (x-a-b)(x+a+b)


(x-a+b)(x+a-b)(x-a-b)(x+a+b) = 0




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(()z) : (±a),2 of 4 points

Set xy=0 , solve for complex roots on z axis


(sqrt(0^2+0^2) -a)^2 +z^2 = b^2

a^2 +z^2 -b^2 = 0


Rule 4: Factor the Squaring of z, Apply Sqrt to a,b

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


(z -sqrt(b^2 -a^2))(z +sqrt(b^2 -a^2)) = 0

(z -sqrt((b-a)(b+a)))(z +sqrt((b-a)(b+a))) = 0




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Solving for Bi-variate Roots. This equation has 2 distinct solutions when solving for a 2-plane. Both are real. All 2-

plane solutions are 2 roots of a 2D circle.


((xy)) : [R-2] of 2 circles

Set z=0 , solve for real roots on plane xy

(sqrt(x^2+y^2) -a)^2 +0^2 = b^2

(sqrt(x^2+y^2) -a)^2 -b^2 = 0


Rule 1: Factor Diff of Squares with x,y,a and b

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

(sqrt(x^2+y^2) - a-b)(sqrt(x^2+y^2) - a+b) = 0


Rule 3: Undo Sqrt on x,y , Apply the Squaring to a,b

(x^2+y^2 -(a-b)^2)(x^2+y^2 -(a+b)^2) = 0





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((x)z) = ((y)z) : 2x1 row of 2 circles


Set y=0 , solve for real roots on plane xz

(sqrt(x^2+0^2) -a)^2 +z^2 = b^2

(sqrt(x^2) -a)^2 +z^2 -b^2 = 0


Rule 2: Factor Sqrt of the Square for x,a

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


((x-a)^2 +z^2 -b^2)((x+a)^2 +z^2 -b^2) = 0




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Solving for Univariate Roots in the (((II)I)(II)) Equation

A 5 variable, degree-16 symmetric polynomial:

\left(\sqrt{\left(\sqrt{x_{111}^2 + x_{112}^2} -R_{111}\right)^2 + x_{12}^2} -R_{11}\right)^2 + \left(\sqrt{x_{21}^2 + x_

{22}^2} -R_{12}\right)^2 = R_{1}^2


This equation has 3 distinct solutions when solving for a 1-plane axis. Two are complex, and one is bi-complex. All 1D

solutions are 16 roots of a 0D point.

(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

The 5 Coordinate 1-plane Solutions
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(((x))()) = (((y))())
((()z)())
((())(w)) = ((())(v))




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(((x))()) : (±c),8


Set yzwv=0 , solve for complex roots on axis x

(sqrt((sqrt(x^2 + 0^2) -a)^2 + 0^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 + c^2 - d^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 = d^2 -c^2

Rule 4: Factor the Square of x,a,b terms , Apply Sqrt to d,c

sqrt((sqrt(x^2) -a)^2) - b-sqrt(d^2 -c^2)
sqrt((sqrt(x^2) -a)^2) - b+sqrt(d^2 -c^2)

Rule 3: Undo Sqrt to x,a terms , Apply Squaring to b,c,d terms

(sqrt(x^2) -a)^2 - (b-sqrt(d^2 -c^2))^2
(sqrt(x^2) -a)^2 - (b+sqrt(d^2 -c^2))^2

Rule 1: Factor Diff of Squares for x,a and b,c,d

(sqrt(x^2) -a)^2 - (b-sqrt(d^2 -c^2))^2 ==> (sqrt(x^2) - a-(b-sqrt(d^2 -c^2))) (sqrt(x^2) - a+(b-sqrt(d^2 -c^2)))
(sqrt(x^2) -a)^2 - (b+sqrt(d^2 -c^2))^2 ==> (sqrt(x^2) - a-(b+sqrt(d^2 -c^2))) (sqrt(x^2) - a+(b+sqrt(d^2 -c^2)))
------------------------------------------
(sqrt(x^2) - a-b+sqrt(d^2 -c^2))
(sqrt(x^2) - a+b-sqrt(d^2 -c^2))

(sqrt(x^2) - a-b-sqrt(d^2 -c^2))
(sqrt(x^2) - a+b+sqrt(d^2 -c^2))

Rule 2: Factor Sqrt of the Square of x, sqrt(x^2) = |x| = -x , +x

(sqrt(x^2) - a-b+sqrt(d^2 -c^2)) ==> (-x - a-b+sqrt(d^2 -c^2)) (x - a-b+sqrt(d^2 -c^2))
(sqrt(x^2) - a+b-sqrt(d^2 -c^2)) ==> (-x - a+b-sqrt(d^2 -c^2)) (x - a+b-sqrt(d^2 -c^2))

(sqrt(x^2) - a-b-sqrt(d^2 -c^2)) ==> (-x - a-b-sqrt(d^2 -c^2)) (x - a-b-sqrt(d^2 -c^2))
(sqrt(x^2) - a+b+sqrt(d^2 -c^2)) ==> (-x - a+b+sqrt(d^2 -c^2)) (x - a+b+sqrt(d^2 -c^2))

(-x - a-b+sqrt(d^2 -c^2)) => (x + a+b-sqrt(d^2 -c^2))
(x - a-b+sqrt(d^2 -c^2))
(-x - a+b-sqrt(d^2 -c^2)) => (x + a-b+sqrt(d^2 -c^2))
(x - a+b-sqrt(d^2 -c^2))
(-x - a-b-sqrt(d^2 -c^2)) => (x + a+b+sqrt(d^2 -c^2))
(x - a-b-sqrt(d^2 -c^2))
(-x - a+b+sqrt(d^2 -c^2)) => (x + a-b-sqrt(d^2 -c^2))
(x - a+b+sqrt(d^2 -c^2))
--------------------------------

(x + a+b-sqrt(d^2 -c^2))
(x - a-b+sqrt(d^2 -c^2))
(x + a-b+sqrt(d^2 -c^2))
(x - a+b-sqrt(d^2 -c^2))
(x + a+b+sqrt(d^2 -c^2))
(x - a-b-sqrt(d^2 -c^2))
(x + a-b-sqrt(d^2 -c^2))
(x - a+b+sqrt(d^2 -c^2))

The 16 complex roots of x


Sitting at a product of 2 points ±ic on the imaginary axis, are a product of 8 points ±a±b±d on the real axis


(sqrt((sqrt(x^2) -a)^2) -b)^2 + c^2 - d^2
expanded

-(2 b sqrt((a-sqrt(x^2))^2)+2 a sqrt(x^2))^2 + (b^2+c^2-d^2+x^2+a^2)^2

(x + a+b-sqrt(d^2 -c^2))(x - a-b+sqrt(d^2 -c^2))(x + a-b+sqrt(d^2 -c^2))(x - a+b-sqrt(d^2 -c^2))(x + a+b+sqrt(d^2 -c^2))(x

- a-b-sqrt(d^2 -c^2))(x + a-b-sqrt(d^2 -c^2))(x - a+b+sqrt(d^2 -c^2)) = 0




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((()z)()) : (±a)x(±c),4

Set xywv=0 , solve for bi-complex roots on axis z


(sqrt((sqrt(0^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

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

(sqrt(z^2 + a^2) -b)^2 + c^2 - d^2 = 0

The real numbers b,d are governed by two imaginary numbers a,c.

(sqrt(z^2 + a^2) -b)^2 -d^2 +c^2 = 0

It is factorable by 4 levels in a 5-step process. There cannot be a sqrt elimination, only a switch-over, which allows for

further factoring with a and z:

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

Undo the squaring, and factor

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

sqrt elim, and rewrite

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

factor as z^2 = -z , z

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

(z -sqrt((b-sqrt(d^2-c^2))^2 -a^2))*(z +sqrt((b-sqrt(d^2-c^2))^2 -a^2))*(z -sqrt((b+sqrt(d^2-c^2))^2 -a^2))*(z +sqrt((b

+sqrt(d^2-c^2))^2 -a^2)) = 0


Let A = (b-sqrt(d^2-c^2))^2 -a^2  ,  B = (b+sqrt(d^2-c^2))^2 -a^2

(z -sqrt(A))*(z +sqrt(A))*(z -sqrt(B))*(z +sqrt(B)) = 0


Factor (b±sqrt(d^2-c^2))^2 -a^2

(b-sqrt(d^2-c^2))^2 -a^2 ==> (b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2)) = A
(b+sqrt(d^2-c^2))^2 -a^2 ==> (b-a+sqrt(d^2-c^2))(b+a+sqrt(d^2-c^2)) = B

(z -sqrt((b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2))))*
(z +sqrt((b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2))))*
(z -sqrt((b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2))))*
(z +sqrt((b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2)))) = 0


Factor d^2-c^2 ==> (d-c)(d+c)

(b-a-sqrt(d^2-c^2))(b+a-sqrt(d^2-c^2)) ==> (b-a-sqrt((d-c)(d+c)))(b+a-sqrt((d-c)(d+c))) = A
(b-a+sqrt(d^2-c^2))(b+a+sqrt(d^2-c^2)) ==> (b-a+sqrt((d-c)(d+c)))(b+a+sqrt((d-c)(d+c))) = B


(z -sqrt((b-a-sqrt((d-c)(d+c)))(b+a-sqrt((d-c)(d+c)))))*
(z +sqrt((b-a-sqrt((d-c)(d+c)))(b+a-sqrt((d-c)(d+c)))))*
(z -sqrt((b-a+sqrt((d-c)(d+c)))(b+a+sqrt((d-c)(d+c)))))*
(z +sqrt((b-a+sqrt((d-c)(d+c)))(b+a+sqrt((d-c)(d+c))))) = 0

\left(z -\sqrt{\left(b-a-\sqrt{(d-c)(d+c)}\right)\left(b+a-\sqrt{(d-c)(d+c)}\right)}\right)\cdot\left(z +\sqrt{\left(b-a-

\sqrt{(d-c)(d+c)}\right)\left(b+a-\sqrt{(d-c)(d+c)}\right)}\right)\cdot\left(z -\sqrt{\left(b-a+\sqrt{(d-c)(d

+c)}\right)\left(b+a+\sqrt{(d-c)(d+c)}\right)}\right)\cdot\left(z +\sqrt{\left(b-a+\sqrt{(d-c)(d+c)}\right)\left(b+a+

\sqrt{(d-c)(d+c)}\right)}\right) = 0

(z ±sqrt((b±a±sqrt((d±c))))) = 0

This is the correct, completely factored equation. The exact bi-complex solution of a degree-16 symmetric polynomial.

Describes a product of 4 roots on the real number line z = ±b±d , sitting at a product of four places in a square, in a 2D

imaginary plane i = ±a , j = ±c .

Solve individually for real and imaginary parts

Set a,c=0 , solve for real roots of z

(sqrt(z^2 + 0^2) -b)^2 + 0^2 - d^2 = 0

(sqrt(z^2) -b)^2 - d^2 = 0

(z±b)^2 - d^2 = 0

(z±b±d) = 0

z = ±b±d


Set b,d=0 , solve for imaginary roots of z

(sqrt(z^2 + a^2) -0)^2 + c^2 - 0^2 = 0

(sqrt(z^2 + a^2))^2 + c^2 = 0

z^2 + a^2 + c^2 = 0

z = ±sqrt(-a^2 -c^2) , A bi-complex number

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

(z - sqrt(-(a-ic)(a+ic)))(z + sqrt(-(a-ic)(a+ic))))


(z - sqrt((-a+ic)(a+ic)))(z + sqrt((a-ic)(-a-ic))))

and equal to

(z - sqrt((-c+ia)(c+ia)))(z + sqrt((c-ia)(-c-ia))))

z = ±ia±jc


(z -sqrt((b-a-sqrt((d-c)(d+c)))(b+a-sqrt((d-c)(d+c))))) (z +sqrt((b-a-sqrt((d-c)(d+c)))(b+a-sqrt((d-c)(d+c))))) (z -sqrt

((b-a+sqrt((d-c)(d+c)))(b+a+sqrt((d-c)(d+c))))) (z +sqrt((b-a+sqrt((d-c)(d+c)))(b+a+sqrt((d-c)(d+c))))) = 0

set b,d=0

(z -sqrt((-a-sqrt((-c)(c)))(a-sqrt((-c)(c)))))
(z +sqrt((-a-sqrt((-c)(c)))(a-sqrt((-c)(c)))))
(z -sqrt((-a+sqrt((-c)(c)))(a+sqrt((-c)(c)))))
(z +sqrt((-a+sqrt((-c)(c)))(a+sqrt((-c)(c)))))

(z -sqrt((-a-sqrt(-c^2))(a-sqrt(-c^2)))) => (z-sqrt(-a^2-c^2))
(z +sqrt((-a-sqrt(-c^2))(a-sqrt(-c^2)))) => (z+sqrt(-a^2-c^2))
(z -sqrt((-a+sqrt(-c^2))(a+sqrt(-c^2)))) => (z-sqrt(-a^2-c^2))
(z +sqrt((-a+sqrt(-c^2))(a+sqrt(-c^2)))) => (z+sqrt(-a^2-c^2))

(z-sqrt(-a^2-c^2))(z+sqrt(-a^2-c^2))(z-sqrt(-a^2-c^2))(z+sqrt(-a^2-c^2)) => (z-sqrt(-a^2-c^2))(z+sqrt(-a^2-c^2))

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

set a,c=0

(z -sqrt((b-sqrt(d^2))(b-sqrt(d^2)))) => (z -sqrt((b-d)(b-d))) => (z -(b-d)) => (z-b+d)
(z +sqrt((b-sqrt(d^2))(b-sqrt(d^2)))) => (z +sqrt((b-d)(b-d))) => (z +(b-d)) => (z+b+d)
(z -sqrt((b+sqrt(d^2))(b+sqrt(d^2)))) => (z -sqrt((b+d)(b+d))) => (z -(b+d)) => (z-b-d)
(z +sqrt((b+sqrt(d^2))(b+sqrt(d^2)))) => (z +sqrt((b+d)(b+d))) => (z +(b+d)) => (z+b+d)

(z-b+d)(z+b+d)(z-b-d)(z+b+d)


(z -sqrt((b-sqrt(d^2))(b-sqrt(d^2))))*(z +sqrt((b-sqrt(d^2))(b-sqrt(d^2)))) = -b^2+2 b sqrt(d^2)-d^2+z^2
(z -sqrt((b+sqrt(d^2))(b+sqrt(d^2))))*(z +sqrt((b+sqrt(d^2))(b+sqrt(d^2)))) = -b (b+2 sqrt(d^2))-d^2+z^2

(-b^2+2 b sqrt(d^2)-d^2+z^2)*(-b (b+2 sqrt(d^2))-d^2+z^2) = (-b+d+z) (-b-d+z) (b-d+z) (b+d+z)

A bi-complex number solution. Describes a product of 16 points in a 3D complex plane R^1 x C^2 . Sitting at a product of 4

points in a square in C^2 at ±ia and ±jc, exist a product of 4 points in R^1 at ±b ±d . When one of the four groups of 4

points sit in R^1 , there are three more groups of 4, spaced along 2 imaginary axes in a square.

It is not obvious which direction to go first. So, how about factoring in a higher dimension, like R^3? This will get down

to a product of 4 tori, which can then have two variables canceled out, leading to something interesting. Perhaps an

effective way would be to include the imaginary axes in the equation, and factor it as if it were in R^3. Use a 3D

coordinate system, with one real as z, and 2 imaginary as i,j in (z,i,j)

(((i)z)(j)) = (sqrt((sqrt(i^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(j^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(i^2) -a)^2 + z^2) -b)^2 + (sqrt(j^2) -c)^2 = d^2

(sqrt((sqrt(i^2) -a)^2 + z^2) -b)^2 + (sqrt(j^2) -c)^2 - d^2

Factor into product of 4 in a square array

((sqrt((i-a)^2 + z^2) -b)^2 + (j-c)^2 - d^2)
((sqrt((i-a)^2 + z^2) -b)^2 + (j+c)^2 - d^2)
((sqrt((i+a)^2 + z^2) -b)^2 + (j-c)^2 - d^2)
((sqrt((i+a)^2 + z^2) -b)^2 + (j+c)^2 - d^2) = 0

((sqrt((i-a)^2 + z^2) -b)^2 + (j-c)^2 - d^2)((sqrt((i-a)^2 + z^2) -b)^2 + (j+c)^2 - d^2)((sqrt((i+a)^2 + z^2) -b)^2 + (j-

c)^2 - d^2)((sqrt((i+a)^2 + z^2) -b)^2 + (j+c)^2 - d^2) = 0

(sqrt((i±a)^2 + z^2) -b)^2 + (j±c)^2 - d^2 = 0

Where, cancelling b,d leads to cancelling real variable z:

(sqrt((i±a)^2 + z^2) -0)^2 + (j±c)^2 - 0^2 = 0
(sqrt((i±a)^2 + z^2))^2 + (j±c)^2 = 0
(i±a)^2 + z^2 + (j±c)^2 = 0
z^2 + (i±a)^2 + (j±c)^2 = 0
0^2 + (i±a)^2 + (j±c)^2 = 0
((i-a)^2 + (j-c)^2) ((i-a)^2 + (j+c)^2) ((i+a)^2 + (j-c)^2) ((i+a)^2 + (j+c)^2) = 0
(i±a)^2 + (j±c)^2 = 0, four points in vertices of a square on plane ij at (0,±a,±c)

Cancelling a,c leads to cancelling imaginary variable i,j:

((sqrt(i^2 + z^2) -b)^2 + j^2 - d^2) = 0
((sqrt(0^2 + z^2) -b)^2 + 0^2 - d^2) = 0
(sqrt(z^2) -b)^2 - d^2 = 0
(sqrt(z^2) -b±d) = 0
(z±b±d) = 0 , four points in a row on axis z at (±b±d,0,0)


Using the factored product of four complex roots, we can set i=a , j=c to move across the imaginary plane, and place the

real number line z at one of its special positions. There are 4 exact locations on ij where the imaginary components in

one of the complex roots will cancel out, allowing it to factor further into 4 real roots.

((sqrt((i-a-a)^2 + z^2) -b)^2 + (j-c-c)^2 - d^2)
((sqrt((i-a-a)^2 + z^2) -b)^2 + (j-c+c)^2 - d^2)
((sqrt((i-a+a)^2 + z^2) -b)^2 + (j-c-c)^2 - d^2)
((sqrt((i-a+a)^2 + z^2) -b)^2 + (j-c+c)^2 - d^2)

((sqrt((i-2a)^2 + z^2) -b)^2 + (j-2c)^2 - d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 + (j)^2 - d^2)
((sqrt((i)^2 + z^2) -b)^2 + (j-2c)^2 - d^2)
((sqrt((i)^2 + z^2) -b)^2 + (j)^2 - d^2)

When we get (i)^2 or (j)^2 , we can cancel the imaginary variable entirely. Each of the 4 real groups exist at 4 precise

points in a square array, displaced away from origin along two imaginary axes. When one of the 4 real groups are

translated along ij to origin, the imaginary values will cancel out in one of the complex roots, leaving behind only the

real part. In order to algebraically proceed any further with factoring the real roots, one must remove the imaginary

variable.

After cancellation, we get one out of the four groups come into the real number line, and factor into 4 roots. While this

real solution exists, there are three more groups, spaced along the imaginary 2-plane ij at (2a,2c) (2a,0) and (0,2c):

((sqrt((i-2a)^2 + z^2) -b)^2 + (j-2c)^2 - d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 - d^2)
((sqrt(z^2) -b)^2 + (j-2c)^2 - d^2)
((sqrt(z^2) -b)^2 - d^2)

the 4 real roots factor into

((sqrt(z^2) -b)^2 - d^2) ==> ((z-b)^2 - d^2)((z+b)^2 - d^2) ==> (z-b-d)(z-b+d)(z+b-d)(z+b+d)

Leaving a product of 4 real roots on z, and 12 complex on (z,i,j)

((sqrt((i-2a)^2 + z^2) -b)^2 + (j-2c)^2 - d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 - d^2)
((sqrt(z^2) -b)^2 + (j-2c)^2 - d^2)
(z-b-d)(z-b+d)(z+b-d)(z+b+d)



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


(z - sqrt(-a^2+b^2-2 b sqrt(d^2-c^2)-c^2+d^2))*(sqrt(-a^2+b^2-2 b sqrt(d^2-c^2)-c^2+d^2)+z) =
a^2-b^2+2 b sqrt(d^2-c^2)+c^2-d^2+z^2

(z-sqrt(-a^2+b^2+2 b sqrt(d^2-c^2)-c^2+d^2))*(z + sqrt(-a^2+b^2+2 b sqrt(d^2-c^2)-c^2+d^2)) =
a^2-b^2-2 b sqrt(d^2-c^2)+c^2-d^2+z^2

(a^2-b^2+2 b sqrt(d^2-c^2)+c^2-d^2+z^2)*(a^2-b^2-2 b sqrt(d^2-c^2)+c^2-d^2+z^2) = 0

a^4-2 a^2 b^2+2 a^2 c^2-2 a^2 d^2+2 a^2 z^2+b^4+2 b^2 c^2-2 b^2 d^2-2 b^2 z^2+c^4-2 c^2 d^2+2 c^2 z^2+d^4-2 d^2 z^2+z^4

(sqrt((sqrt(0^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

a^4-2 a^2 b^2+2 a^2 c^2-2 a^2 d^2+2 a^2 z^2+b^4+2 b^2 c^2-2 b^2 d^2-2 b^2 z^2+c^4-2 c^2 d^2+2 c^2 z^2+d^4-2 d^2 z^2+z^4
a^4-2 a^2 b^2+2 a^2 c^2-2 a^2 d^2+2 a^2 z^2+b^4+2 b^2 c^2-2 b^2 d^2-2 b^2 z^2+c^4-2 c^2 d^2+2 c^2 z^2+d^4-2 d^2 z^2+z^4




##########################################################################################################################




((())(w)) : (±a±b),4

Set xyzv=0 , solve for complex roots on axis w


(sqrt((sqrt(0^2 + 0^2) -a)^2 + 0^2) -b)^2 + (sqrt(w^2 + 0^2) -c)^2 = d^2

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

(sqrt(a^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2

Rule 2: Factor sqrt(a^2) and sqrt(w^2),

((a-b)^2 + (w-c)^2 - d^2)
((a-b)^2 + (w+c)^2 - d^2)
((a+b)^2 + (w-c)^2 - d^2)
((a+b)^2 + (w+c)^2 - d^2)

Rule 1: Factor Diff of Squares of a,b with d

(a-b)^2 -d^2 => (a-b-d)(a-b+d)
(a+b)^2 -d^2 => (a+b-d)(a+b+d)
------------------------------
((w-c)^2 + (a-b-d)(a-b+d))
((w+c)^2 + (a-b-d)(a-b+d))
((w-c)^2 + (a+b-d)(a+b+d))
((w+c)^2 + (a+b-d)(a+b+d))

Rule 4: Factor the Square of w,c and Apply Sqrt to a,b,d

((w-c)^2 + (a-b-d)(a-b+d)) => (w - c-sqrt(-(a-b-d)(a-b+d)))(w - c+sqrt(-(a-b-d)(a-b+d)))

((w+c)^2 + (a-b-d)(a-b+d)) => (w + c-sqrt(-(a-b-d)(a-b+d)))(w + c+sqrt(-(a-b-d)(a-b+d)))

((w-c)^2 + (a+b-d)(a+b+d)) => (w - c-sqrt(-(a+b-d)(a+b+d)))(w - c+sqrt(-(a+b-d)(a+b+d)))

((w+c)^2 + (a+b-d)(a+b+d)) => (w + c-sqrt(-(a+b-d)(a+b+d)))(w + c+sqrt(-(a+b-d)(a+b+d)))
------------------------------------------------------------
(w - c-sqrt(-(a-b-d)(a-b+d)))(w - c+sqrt(-(a-b-d)(a-b+d)))
(w + c-sqrt(-(a-b-d)(a-b+d)))(w + c+sqrt(-(a-b-d)(a-b+d)))
(w - c-sqrt(-(a+b-d)(a+b+d)))(w - c+sqrt(-(a+b-d)(a+b+d)))
(w + c-sqrt(-(a+b-d)(a+b+d)))(w + c+sqrt(-(a+b-d)(a+b+d)))

Make the sign changes in the sqrt,

(w - c-sqrt((-a+b+d)(a-b+d)))(w - c+sqrt((-a+b+d)(a-b+d)))
(w + c-sqrt((-a+b+d)(a-b+d)))(w + c+sqrt((-a+b+d)(a-b+d)))
(w - c-sqrt((-a-b+d)(a+b+d)))(w - c+sqrt((-a-b+d)(a+b+d)))
(w + c-sqrt((-a-b+d)(a+b+d)))(w + c+sqrt((-a-b+d)(a+b+d)))

The factored 16 complex roots for w:

(w - c-sqrt((-a+b+d)(a-b+d)))
(w - c+sqrt((-a+b+d)(a-b+d)))
(w + c-sqrt((-a+b+d)(a-b+d)))
(w + c+sqrt((-a+b+d)(a-b+d)))
(w - c-sqrt((-a-b+d)(a+b+d)))
(w - c+sqrt((-a-b+d)(a+b+d)))
(w + c-sqrt((-a-b+d)(a+b+d)))
(w + c+sqrt((-a-b+d)(a+b+d)))


\left(w - c-\sqrt{(-a+b+d)(a-b+d)}\right)\cdot\left(w - c+\sqrt{(-a+b+d)(a-b+d)}\right)\cdot\left(w + c-\sqrt{(-a+b+d)(a-

b+d)}\right)\cdot\left(w + c+\sqrt{(-a+b+d)(a-b+d)}\right)\cdot\left(w - c-\sqrt{(-a-b+d)(a+b+d)}\right)\cdot\left(w - c+

\sqrt{(-a-b+d)(a+b+d)}\right)\cdot\left(w + c-\sqrt{(-a-b+d)(a+b+d)}\right)\cdot\left(w + c+\sqrt{(-a-b+d)(a+b+d)}\right)

= 0




##########################################################################################################################
##########################################################################################################################
##########################################################################################################################




Solving for Bi-variate roots of the degree-16 (((II)I)(II)) polynomial. This equation has 5 distinct solutions when

solving for a 2-plane. One is real, and four are complex. All 2-plane solutions are 8 roots of a 2D circle.

(sqrt((sqrt(x111^2 + x112^2) -R111)^2 + x12^2) -R11)^2 + (sqrt(x21^2 + x22^2) -R12)^2 = R1^2

(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

(((xy-a)z-b)(wv-c)-d)

(((xy)z)(wv))

The 10 coordinate 2-plane solutions
------------------------------------
(((xy))())
(((x)z)()) = (((y)z)())
(((x))(w)) = (((y))(w)) = (((x))(v)) = (((y))(v))
((()z)(w)) = ((()z)(v))
((())(wv))




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(((xy))()) : (±c),[R-4] of 8 circles

set zwv=0 , solve for roots on xy

(sqrt((sqrt(x^2 + y^2) -a)^2 + 0^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2 + y^2) -a)^2) -b)^2 -d^2 +c^2 = 0

Rule 3: Factor Sqrt of Square for x,y,a term

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

((sqrt(x^2 + y^2) -a-b)^2 -d^2 +c^2) ((sqrt(x^2 + y^2) -a+b)^2 -d^2 +c^2) = 0

Rule 4: Factor by Eliminate Square of x,y,a,b and Apply ±Sqrt of c,d

(sqrt(x^2 + y^2) -a-b)^2 -d^2 +c^2 => (sqrt(x^2 + y^2) -a-b-sqrt(d^2-c^2))*(sqrt(x^2 + y^2) -a-b+sqrt(d^2-c^2))
(sqrt(x^2 + y^2) -a+b)^2 -d^2 +c^2 => (sqrt(x^2 + y^2) -a+b-sqrt(d^2-c^2))*(sqrt(x^2 + y^2) -a+b+sqrt(d^2-c^2))

(sqrt(x^2 + y^2) -a-b-sqrt(d^2-c^2))*
(sqrt(x^2 + y^2) -a-b+sqrt(d^2-c^2))*
(sqrt(x^2 + y^2) -a+b-sqrt(d^2-c^2))*
(sqrt(x^2 + y^2) -a+b+sqrt(d^2-c^2)) = 0

Rule 3: Eliminate Sqrt of xy , Make Squaring to a,b,c,d

(sqrt(x^2+y^2) -a-b-sqrt(d^2-c^2)) => (x^2+y^2 -(a-b-sqrt(d^2-c^2))^2)
(sqrt(x^2+y^2) -a-b+sqrt(d^2-c^2)) => (x^2+y^2 -(a-b+sqrt(d^2-c^2))^2)
(sqrt(x^2+y^2) -a+b-sqrt(d^2-c^2)) => (x^2+y^2 -(a+b-sqrt(d^2-c^2))^2)
(sqrt(x^2+y^2) -a+b+sqrt(d^2-c^2)) => (x^2+y^2 -(a+b+sqrt(d^2-c^2))^2)

(x^2+y^2 -(a-b-sqrt(d^2-c^2))^2)*
(x^2+y^2 -(a-b+sqrt(d^2-c^2))^2)*
(x^2+y^2 -(a+b-sqrt(d^2-c^2))^2)*
(x^2+y^2 -(a+b+sqrt(d^2-c^2))^2) = 0

(x^2+y^2-(a-b-sqrt(d^2-c^2))^2) (x^2+y^2-(a-b+sqrt(d^2-c^2))^2) (x^2+y^2-(a+b-sqrt(d^2-c^2))^2) (x^2+y^2-(a+b+sqrt(d^2-

c^2))^2) = 0




##########################################################################################################################




(((x)z)()) = (((y)z)()) : (±c),2x1,[R-2] of 8 circles

Set ywv = 0 , solve for complex roots on plane xz

(sqrt((sqrt(x^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2) -a)^2 + z^2) -b)^2 +c^2 -d^2 = 0


Rule 2: Factor Sqrt of Square of x

(sqrt(x^2) -a) => (x-a) , (x+a)

((sqrt((x-a)^2 + z^2) -b)^2 +c^2 -d^2) ((sqrt((x+a)^2 + z^2) -b)^2 +c^2 -d^2) = 0


Rule 4: Factor by Eliminate Square of x,a,z,b and Apply ±Sqrt of c,d

(sqrt((x-a)^2 + z^2) -b)^2 +c^2 -d^2 => (sqrt((x-a)^2+z^2) - b-sqrt(d^2-c^2)) (sqrt((x-a)^2+z^2) - b+sqrt(d^2-c^2))
(sqrt((x+a)^2 + z^2) -b)^2 +c^2 -d^2 => (sqrt((x+a)^2+z^2) - b-sqrt(d^2-c^2)) (sqrt((x+a)^2+z^2) - b+sqrt(d^2-c^2))
----------------------------------------
(sqrt((x-a)^2+z^2) - b-sqrt(d^2-c^2))*
(sqrt((x-a)^2+z^2) - b+sqrt(d^2-c^2))*
(sqrt((x+a)^2+z^2) - b-sqrt(d^2-c^2))*
(sqrt((x+a)^2+z^2) - b+sqrt(d^2-c^2)) = 0


Rule 3: Eliminate Sqrt of x,a,z Apply Squaring of b,c,d

(sqrt((x-a)^2+z^2) - b-sqrt(d^2-c^2)) => ((x-a)^2+z^2 - (b-sqrt(d^2-c^2))^2)
(sqrt((x-a)^2+z^2) - b+sqrt(d^2-c^2)) => ((x-a)^2+z^2 - (b+sqrt(d^2-c^2))^2)
(sqrt((x+a)^2+z^2) - b-sqrt(d^2-c^2)) => ((x+a)^2+z^2 - (b-sqrt(d^2-c^2))^2)
(sqrt((x+a)^2+z^2) - b+sqrt(d^2-c^2)) => ((x+a)^2+z^2 - (b+sqrt(d^2-c^2))^2)
-------------------------------------

((x-a)^2+z^2 - (b-sqrt(d^2-c^2))^2)*
((x-a)^2+z^2 - (b+sqrt(d^2-c^2))^2)*
((x+a)^2+z^2 - (b-sqrt(d^2-c^2))^2)*
((x+a)^2+z^2 - (b+sqrt(d^2-c^2))^2) = 0


((x-a)^2+z^2 -(b-sqrt(d^2-c^2))^2)((x-a)^2+z^2 -(b+sqrt(d^2-c^2))^2)((x+a)^2+z^2 -(b-sqrt(d^2-c^2))^2)((x+a)^2+z^2 -(b

+sqrt(d^2-c^2))^2) = 0

((x-a)^2+y^2 -(b-\sqrt{d^2-c^2})^2)((x-a)^2+y^2 -(b+\sqrt{d^2-c^2})^2)((x+a)^2+y^2 -(b-\sqrt{d^2-c^2})^2)((x+a)^2+y^2 -(b

+\sqrt{d^2-c^2})^2) = 0




##########################################################################################################################




(((x))(w)) = (((y))(w)) = (((x))(v)) = (((y))(v)) : 4x2 rectangle of 8 circles

set yzv=0 , solve for real roots in plane xw

(sqrt((sqrt(x^2 + 0^2) -a)^2 + 0^2) -b)^2 + (sqrt(w^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2


Rule 2: Factor by Elim Sqrt of the Square for x,a

(sqrt(x^2) -a) => (x-a) , (x+a)
---------------------------------
((sqrt((x-a)^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2)
((sqrt((x+a)^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2) = 0


Rule 2: Factor by Elim Sqrt of the Square for (x±a),b

(sqrt((x-a)^2) -b) => (-(x-a)-b) , ((x-a)-b) => (x-a+b) , (x-a-b)
(sqrt((x+a)^2) -b) => (-(x+a)-b) , ((x+a)-b) => (x+a+b) , (x+a-b)
-------------------------------------
((x-a+b)^2 + (sqrt(w^2) -c)^2 - d^2)
((x-a-b)^2 + (sqrt(w^2) -c)^2 - d^2)
((x+a+b)^2 + (sqrt(w^2) -c)^2 - d^2)
((x+a-b)^2 + (sqrt(w^2) -c)^2 - d^2) = 0


Rule 2: Factor by Elim Sqrt of the Square for w,c

(sqrt(w^2) -c) => (w-c) , (w+c)
-------------------------------
((x-a+b)^2 + (w-c)^2 - d^2)
((x-a+b)^2 + (w+c)^2 - d^2)

((x-a-b)^2 + (w-c)^2 - d^2)
((x-a-b)^2 + (w+c)^2 - d^2)

((x+a+b)^2 + (w-c)^2 - d^2)
((x+a+b)^2 + (w+c)^2 - d^2)

((x+a-b)^2 + (w-c)^2 - d^2)
((x+a-b)^2 + (w+c)^2 - d^2) = 0


((x±a±b)^2 + (w±c)^2 - d^2) = 0




##########################################################################################################################




((()z)(w)) = ((()z)(v)) : (±a),2x2 of 8 circles

Set xyv = 0 , solve for complex roots on plane zw

(sqrt((sqrt(0^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + 0^2) -c)^2 = d^2

(sqrt(a^2 + z^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2 = 0


Rule 2: Factor by Elim Sqrt of the Square for w,c

(sqrt(w^2) -c) => (w-c) , (w+c)

((sqrt(a^2 + z^2) -b)^2 + (w-c)^2 -d^2)*
((sqrt(a^2 + z^2) -b)^2 + (w+c)^2 -d^2) = 0

Furthest Factored Equation that works. Somehow unable to undo the squaring of a,z,b terms, and make the sqrt of w,c,d

terms while maintaining correct equation. Maybe this is the true, lowest factored form, and trying to do anything more

with it would be 'fudging the numbers' . So, I just have to understand this as a fundamental root, while Rule 5 can offer

more factoring symmetry, and an even clearer description.


Rule 4: Factor by Eliminate Square of z,a,b and Apply ±Sqrt of w,c,d

((sqrt(a^2+z^2) -b)^2 + (w-c)^2 -d^2) => (sqrt(a^2+z^2) -b -sqrt(d^2-(w-c)^2)) (sqrt(a^2+z^2) -b +sqrt(d^2-(w-c)^2))
((sqrt(a^2+z^2) -b)^2 + (w+c)^2 -d^2) => (sqrt(a^2+z^2) -b -sqrt(d^2-(w+c)^2)) (sqrt(a^2+z^2) -b +sqrt(d^2-(w+c)^2))

(sqrt(a^2+z^2) -b -sqrt(d^2-(w-c)^2))*
(sqrt(a^2+z^2) -b +sqrt(d^2-(w-c)^2))*
(sqrt(a^2+z^2) -b -sqrt(d^2-(w+c)^2))*
(sqrt(a^2+z^2) -b +sqrt(d^2-(w+c)^2)) = 0

This function won't graph in Desmos, but will in CalcPlot

Trying different technique with Rule 4 then Rule 2:

(sqrt(a^2 + z^2) -b-sqrt(d^2-(sqrt(w^2)-c)^2))(sqrt(a^2 + z^2) -b+sqrt(d^2-(sqrt(w^2)-c)^2)) = 0

Works,
(\sqrt{a^2 + x^2} -b-\sqrt{d^2-(\sqrt{y^2}-c)^2})(\sqrt{a^2 + x^2} -b+\sqrt{d^2-(\sqrt{y^2}-c)^2}) = 0

Rule 2: Factor Sqrt of Square for w,c

(sqrt(w^2)-c) => (w-c) , (w+c)

(sqrt(a^2+z^2) -b-sqrt(d^2-(sqrt(w^2)-c)^2)) => (sqrt(a^2+z^2) -b-sqrt(d^2-(w-c)^2))(sqrt(a^2+z^2) -b-sqrt(d^2-(w+c)^2))
(sqrt(a^2+z^2) -b+sqrt(d^2-(sqrt(w^2)-c)^2)) => (sqrt(a^2+z^2) -b+sqrt(d^2-(w-c)^2))(sqrt(a^2+z^2) -b+sqrt(d^2-(w+c)^2))

(sqrt(a^2+z^2) -b-sqrt(d^2-(w-c)^2))
(sqrt(a^2+z^2) -b-sqrt(d^2-(w+c)^2))
(sqrt(a^2+z^2) -b+sqrt(d^2-(w-c)^2))
(sqrt(a^2+z^2) -b+sqrt(d^2-(w+c)^2)) = 0

(\sqrt{a^2+x^2} -b-\sqrt{d^2-(y-c)^2})(\sqrt{a^2+x^2} -b-\sqrt{d^2-(y+c)^2})(\sqrt{a^2+x^2} -b+\sqrt{d^2-(y-c)^2})(\sqrt

{a^2+x^2} -b+\sqrt{d^2-(y+c)^2}) = 0

Will not work in Desmos, but functions perfectly in CalcPlot.

This is where Rule 5 comes into play. We can restore some of the factorization symmetry, by expressing as f(x,y,i) and

setting a = (sqrt(i^2)-a) , and then factoring with Rule 2 from a high dimension context.

(sqrt((sqrt(i^2)-a)^2 + z^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2 = 0

Rule 2: Factor Sqrt of Square for i,a

(sqrt(i^2)-a) => (i-a) , (i+a)

((sqrt((i-a)^2 + z^2) -b)^2 + (sqrt(w^2) -c)^2 -d^2) ((sqrt((i+a)^2 + z^2) -b)^2 + (sqrt(w^2) -c)^2 -d^2) = 0

Rule 2: Factor Sqrt of Square for w,c

(sqrt(w^2)-c) => (w-c) , (w+c)

((sqrt((i-a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)
((sqrt((i+a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i+a)^2 + z^2) -b)^2 + (w+c)^2 -d^2) = 0

((\sqrt{(i-a)^2 + x^2} -b)^2 + (y-c)^2 -d^2)((\sqrt{(i-a)^2 + x^2} -b)^2 + (y+c)^2 -d^2)((\sqrt{(i+a)^2 + x^2} -b)^2 +

(y-c)^2 -d^2)((\sqrt{(i+a)^2 + x^2} -b)^2 + (y+c)^2 -d^2) = 0

With these complex roots, we can shift by i=a, and center one of the complex roots at origin. This will offset one

imaginary value by 2a , and cancel the other, leading to a factorable real solution.

i=a,

((sqrt((i-a-a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-a-a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)
((sqrt((i-a+a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-a+a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)

((sqrt((i-2a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)
((sqrt((i)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i)^2 + z^2) -b)^2 + (w+c)^2 -d^2)

((sqrt((i-2a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)
((sqrt(z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt(z^2) -b)^2 + (w+c)^2 -d^2)

Rule 2: Factor Sqrt of Square for z,b

(sqrt(z^2) -b) => (z-b) , (z+b)

((sqrt((i-2a)^2 + z^2) -b)^2 + (w-c)^2 -d^2)
((sqrt((i-2a)^2 + z^2) -b)^2 + (w+c)^2 -d^2)
((z-b)^2 + (w-c)^2 -d^2)
((z+b)^2 + (w-c)^2 -d^2)
((z-b)^2 + (w+c)^2 -d^2)
((z+b)^2 + (w+c)^2 -d^2)

When i=a or i=-a , the equation will factor further into 4 real roots of a 2x2 square array of circles. There is an

additional location, spaced at (0,0,2a) on xyi, where this real root factoring can also happen.



(sqrt(a^2+x^2) -b-sqrt(d^2-(y-c)^2))(sqrt(a^2+x^2) -b-sqrt(d^2-(y+c)^2))(sqrt(a^2+x^2) -b+sqrt(d^2-(y-c)^2))(sqrt(a^2+x^2)

-b+sqrt(d^2-(y+c)^2)) = 0

(\sqrt{(\sqrt{i^2}-a)^2 + z^2} -b)^2 + (\sqrt{w^2} -c)^2 - d^2 = 0



##########################################################################################################################



((())(wv)) : (±a±b),[R-2] of 8 circles

(sqrt((sqrt(0^2 + 0^2) -a)^2 + 0^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

(sqrt(a^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 -d^2 = 0


Rule 4: Factor Square terms w,v,c Apply Sqrt to a,b,d

(sqrt(w^2 + v^2) -c)^2 = d^2 -(sqrt(a^2) -b)^2
-----------------------------------------------
sqrt(w^2 + v^2) -c = -sqrt(d^2 -(sqrt(a^2) -b)^2)
sqrt(w^2 + v^2) -c = sqrt(d^2 -(sqrt(a^2) -b)^2)
------------------------------------------------
sqrt(w^2 + v^2) - c-sqrt(d^2 -(sqrt(a^2) -b)^2)
sqrt(w^2 + v^2) - c+sqrt(d^2 -(sqrt(a^2) -b)^2)


Rule 3: Undo Sqrt of w,v terms, Apply Square to a,b,c,d terms

w^2 + v^2 - (c-sqrt(d^2 -(sqrt(a^2) -b)^2))^2
w^2 + v^2 - (c+sqrt(d^2 -(sqrt(a^2) -b)^2))^2


Rule 3: Factor (sqrt(a^2) -b) => (a-b) , (a+b)

w^2 + v^2 - (c-sqrt(d^2 -(a-b)^2))^2
w^2 + v^2 - (c-sqrt(d^2 -(a+b)^2))^2

w^2 + v^2 - (c+sqrt(d^2 -(a-b)^2))^2
w^2 + v^2 - (c+sqrt(d^2 -(a+b)^2))^2


(x^2 + y^2 - (c-sqrt(d^2 -(a-b)^2))^2)
(x^2 + y^2 - (c-sqrt(d^2 -(a+b)^2))^2)
(x^2 + y^2 - (c+sqrt(d^2 -(a-b)^2))^2)
(x^2 + y^2 - (c+sqrt(d^2 -(a+b)^2))^2) = 0

This is the correct complex solution to ((())(wv)) , verified using CalcPlot. Desmos, which plots nothing, does actually

have some kind of limitation that calcplot doesn't. I'm trusting calcplot more, since it can accurately compute a 9D

torus.




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Solving for Tri-variate roots of the degree-16 (((II)I)(II)) polynomial. This equation has 5 distinct solutions when

solving for a 3-plane. Three are real, and two are complex. All 3-plane solutions are 4 roots of a 3D torus.

(((xy)z)(wv)) : (sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

The 10 Coordinate 3-plane Solutions
------------------------------------
(((xy)z)())
(((xy))(w)) = (((xy))(v))
(((x)z)(w)) = (((y)z)(w)) = (((x)z)(v)) = (((y)z)(v))
(((x))(wv)) = (((y))(wv))
((()z)(wv))



##########################################################################################################################



(((xy)z)()) : (±c),[Rm-2] of 4 tori

Set wv=0, solve for complex roots on plane xyz

(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 + (sqrt(0^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 + c^2 - d^2 = 0


Rule 4: Factor Square of x,y,z,b and Apply Sqrt to c,d

(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b)^2 = d^2 - c^2
--------------------------------------------------------
sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b) = -sqrt(d^2 - c^2)

sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) -b) = sqrt(d^2 - c^2)
----------------------------------------------------------
(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) - b-sqrt(d^2 - c^2))
(sqrt((sqrt(x^2 + y^2) -a)^2 + z^2) - b+sqrt(d^2 - c^2))


Rule 3: Undo Sqrt of x,y,a,z and Apply Squaring to b,c,d

((sqrt(x^2 + y^2) -a)^2 + z^2 - (b-sqrt(d^2 - c^2))^2)
((sqrt(x^2 + y^2) -a)^2 + z^2 - (b+sqrt(d^2 - c^2))^2)


((sqrt(x^2+y^2)-a)^2 +z^2 -(b-sqrt(d^2-c^2))^2)((sqrt(x^2+y^2)-a)^2 +z^2 -(b+sqrt(d^2-c^2))^2) = 0




##########################################################################################################################




(((xy))(w)) : 1x1x2,[R1-2] vertical column of 4 tori

Set zv=0 , solve for real roots on plane xyw


(sqrt((sqrt(x^2 + y^2) -a)^2 + 0^2) -b)^2 + (sqrt(w^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2 + y^2) -a)^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2 = 0


Rule 3: Factor Sqrt of the Square for x,y,a

(sqrt((sqrt(x^2 + y^2) -a)^2) -b) => ((sqrt(x^2 + y^2) - a-b) , ((sqrt(x^2 + y^2) - a+b)
------------------------------------------------------------------------------------------
(((sqrt(x^2 + y^2) - a-b)^2 + (sqrt(w^2) -c)^2 - d^2)
(((sqrt(x^2 + y^2) - a+b)^2 + (sqrt(w^2) -c)^2 - d^2)


Rule 3: Factor Sqrt of the Square for w,c

(sqrt(w^2) -c) => (w-c) , (w+c)
--------------------------------

((sqrt(x^2 + y^2) - a-b)^2 + (w-c)^2 - d^2)
((sqrt(x^2 + y^2) - a-b)^2 + (w+c)^2 - d^2)
((sqrt(x^2 + y^2) - a+b)^2 + (w-c)^2 - d^2)
((sqrt(x^2 + y^2) - a+b)^2 + (w+c)^2 - d^2)



((sqrt(x^2+y^2) -a-b)^2 +(w-c)^2 -d^2)((sqrt(x^2+y^2) -a-b)^2 +(w+c)^2 -d^2)((sqrt(x^2+y^2) -a+b)^2 +(w-c)^2 -d^2)((sqrt

(x^2+y^2) -a+b)^2 +(w+c)^2 -d^2) = 0


((sqrt(x^2+y^2) -a-b)^2 +(z-c)^2 -d^2)((sqrt(x^2+y^2) -a-b)^2 +(z+c)^2 -d^2)((sqrt(x^2+y^2) -a+b)^2 +(z-c)^2 -d^2)((sqrt

(x^2+y^2) -a+b)^2 +(z+c)^2 -d^2) = 0




##########################################################################################################################




(((x)z)(w)) : 2x1x2 vertical square of 4 tori

Set yv=0 , solve for real roots in plane xzw

(sqrt((sqrt(x^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + 0^2) -c)^2 = d^2

(sqrt((sqrt(x^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2) -c)^2 - d^2 = 0

Rule 2: Factor Sqrt of the Square for x,a and w,c

(sqrt(x^2) -a) => (x-a) , (x+a)

(sqrt(w^2) -c) => (w-c) , (w+c)
---------------------------------

((sqrt((x-a)^2 + z^2) -b)^2 + (w-c)^2 - d^2)
((sqrt((x-a)^2 + z^2) -b)^2 + (w+c)^2 - d^2)
((sqrt((x+a)^2 + z^2) -b)^2 + (w-c)^2 - d^2)
((sqrt((x+a)^2 + z^2) -b)^2 + (w+c)^2 - d^2)


((sqrt((x-a)^2 +z^2) -b)^2 +(w-c)^2 -d^2)((sqrt((x-a)^2 +z^2) -b)^2 +(w+c)^2 -d^2)((sqrt((x+a)^2 +z^2) -b)^2 +(w-c)^2 -

d^2)((sqrt((x+a)^2 +z^2) -b)^2 +(w+c)^2 -d^2) = 0



((sqrt((x-a)^2 +z^2) -b)^2 +(y-c)^2 -d^2)((sqrt((x-a)^2 +z^2) -b)^2 +(y+c)^2 -d^2)((sqrt((x+a)^2 +z^2) -b)^2 +(y-c)^2 -

d^2)((sqrt((x+a)^2 +z^2) -b)^2 +(y+c)^2 -d^2) = 0





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(((x))(wv)) : 1x1x4 vertical column of 4 tori

Set yz=0 , solve for real roots on plane xwv

(sqrt((sqrt(x^2 + 0^2) -a)^2 + 0^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 - d^2 = 0


Rule 2: Two-fold Factor Sqrt of the Square for x,a,b

(sqrt((sqrt(x^2) -a)^2) -b) => (x-a-b) , (x-a+b) , (x+a-b) , (x+a+b)
---------------------------------------------------------------------
((x-a-b)^2 + (sqrt(w^2 + v^2) -c)^2 - d^2)
((x-a+b)^2 + (sqrt(w^2 + v^2) -c)^2 - d^2)
((x+a-b)^2 + (sqrt(w^2 + v^2) -c)^2 - d^2)
((x+a+b)^2 + (sqrt(w^2 + v^2) -c)^2 - d^2)



((sqrt(w^2+v^2) -c)^2 +(x-a-b)^2 -d^2)((sqrt(w^2+v^2) -c)^2 +(x-a+b)^2 -d^2)((sqrt(w^2+v^2) -c)^2 +(x+a-b)^2 -d^2)((sqrt

(w^2+v^2) -c)^2 +(x+a+b)^2 -d^2) = 0


((sqrt(y^2+z^2) -c)^2 +(x-a-b)^2 -d^2)((sqrt(y^2+z^2) -c)^2 +(x-a+b)^2 -d^2)((sqrt(y^2+z^2) -c)^2 +(x+a-b)^2 -d^2)((sqrt

(y^2+z^2) -c)^2 +(x+a+b)^2 -d^2) = 0




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((()z)(wv)) : (±a),1x1x2  of 4 tori

Set xy=0 , solve for complex roots on plane zwv

(sqrt((sqrt(0^2 + 0^2) -a)^2 + z^2) -b)^2 + (sqrt(w^2 + v^2) -c)^2 = d^2

(sqrt(a^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2 = 0

An unsolvable complex root with Rules 1,2,3. Rule 4 will only split into two factors, which do not tell a very clear

picture of the intercept.


Rule 5: Add Imaginary Variable i , Factor as f(x,y,z,i) with Rule 2

a^2 = (sqrt(i^2)-a)^2

(sqrt((sqrt(i^2)-a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2 = 0


Rule 2: Factor Sqrt of the Square for i,a

(sqrt(i^2)-a) => (i-a) , (i+a)
------------------------------
((sqrt((i-a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((sqrt((i+a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)

Set i=a to center one of the complex solutions at origin, and cancel its imaginary part.

((sqrt((i-a-a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((sqrt((i-a+a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
--------------------------------------------------------
((sqrt((i-2a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((sqrt((i)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
------------------------------------------------------
((sqrt((i-2a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((sqrt(z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)


Rule 2: Factor Sqrt of the Square for z,b in the real solution

(sqrt(z^2)-b) => (z-b) , (z+b)
-------------------------------

((sqrt((i-2a)^2+z^2)-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((z-b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)
((z+b)^2 + (sqrt(w^2+v^2) -c)^2 -d^2)




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Some Desmos Scripts for the bi-variate roots of (((II)I)(II)):

Factoring (((xy))())

Factoring ((()z)(w))

Factoring ((())(wv))
in search of combinatorial objects of finite extent
ICN5D
Pentonian
 
Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: A Root Finding Algorithm for Factoring Hypertoric Sectio

Postby ICN5D » Thu Nov 26, 2015 3:57 am

Finding Roots in the Degree-8 (((II)I)I) Polynomial for a 3-Torus

(sqrt((sqrt(x111^2+x112^2) -R111^2)^2+x12^2) -R11^2)^2 +x2^2 = R1^2

(sqrt((sqrt(x^2+y^2) -a)^2 +z^2) -b)^2 +w^2 = c^2

(((xy-a)z-b)w-c)

(((xy)z)w)


The 4 Coordinate 1-plane Solutions
-----------------------------------
(((x))) = (((y)))
((()z))
((())w)


The 6 Coordinate 2-plane Solutions
-----------------------------------
(((xy)))
(((x)z)) = (((y)z))
(((x))w) = (((y))w)
((()z)w)


The 4 Coordinate 3-plane Solutions
-----------------------------------
(((xy)z))
(((xy))w)
(((x)z)w) = (((y)z)w)


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Solving for roots of 1 variable. This equation has 3 distinct solutions when solving for a single axis. One is real, and two are complex. All 1-plane solutions are 8 roots of a 0D point.


(((x))) = (((y))) : 8 points in a row

Set yzw=0 , solve for real roots on x axis

(sqrt((sqrt(x^2+0^2) -a)^2 +0^2) -b)^2 +0^2 = c^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 -c^2 = 0

Rule 2: factor (sqrt(x^2)-a) ---> (x-a) , (x+a)

((sqrt((x-a)^2) -b)^2 -c^2)
((sqrt((x+a)^2) -b)^2 -c^2)


Rule 1: factor x,a,b with c

(sqrt((x-a)^2) - b-c)(sqrt((x-a)^2) - b+c)
(sqrt((x+a)^2) - b-c)(sqrt((x+a)^2) - b+c)


Rule 3: undo sqrt to x,a apply square to b,c

((x-a)^2 - (b-c)^2)
((x-a)^2 - (b+c)^2)
((x+a)^2 - (b-c)^2)
((x+a)^2 - (b+c)^2)


Rule 1: factor x,a with b,c

(x-a - (b-c))(x-a + (b-c)) --> (x-a-b+c)(x-a+b-c)
(x-a - (b+c))(x-a + (b+c)) --> (x-a-b-c)(x-a+b+c)
(x+a - (b-c))(x+a + (b-c)) --> (x+a-b+c)(x+a+b-c)
(x+a - (b+c))(x+a + (b+c)) --> (x+a-b-c)(x+a+b+c)


8 Real Roots of a 0D point spaced apart in a row

(x-a-b+c)(x-a+b-c)(x-a-b-c)(x-a+b+c)(x+a-b+c)(x+a+b-c)(x+a-b-c)(x+a+b+c) = 0


##########################################################################################################################


((()z)) : (2),4 of 8 points

Set xyw=0 , solve for complex roots on z axis

(sqrt((sqrt(0^2+0^2) -a)^2 +z^2) -b)^2 +0^2 = c^2

(sqrt(a^2 +z^2) -b)^2 -c^2 = 0


Rule 1: Factor a,z,b with c

((sqrt(a^2 +z^2) - b-c)((sqrt(a^2 +z^2) - b+c) = 0


Rule 3: Undo sqrt to a,z,b , apply square to b,c

(a^2 +z^2 - (b-c)^2)
(a^2 +z^2 - (b+c)^2) = 0

Move a,b,c to RH side,

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


Rule 1: Factor b,c with a

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


Rule 4: Factor square of z , apply sqrt to b,c,a

(z -sqrt((b-c-a)(b-c+a)))
(z +sqrt((b-c-a)(b-c+a)))
(z -sqrt((b+c-a)(b+c+a)))
(z +sqrt((b+c-a)(b+c+a)))


8 Complex Roots of a 0D point: 2 disjoint groups on the Im line, each group is a product of 4 points on the Re line

(z -sqrt((b-c-a)(b-c+a))) (z +sqrt((b-c-a)(b-c+a))) (z -sqrt((b+c-a)(b+c+a))) (z +sqrt((b+c-a)(b+c+a))) = 0


##########################################################################################################################


((())w) : (4),2 of 8 points

Set xyz=0 , solve for complex roots on w axis

(sqrt((sqrt(0^2+0^2) -a)^2 +0^2) -b)^2 +w^2 = c^2

w^2 +(sqrt(a^2)-b)^2 -c^2 = 0


Rule 2: Factor (sqrt(a^2)-b) --> (a-b) , (a+b)

(w^2 +(a-b)^2 -c^2)(w^2 +(a+b)^2 -c^2)


Rule 4: Factor the squaring of w, apply sqrt to a,b,c

(w -sqrt(c^2 -(a-b)^2))
(w +sqrt(c^2 -(a-b)^2))
(w -sqrt(c^2 -(a+b)^2))
(w +sqrt(c^2 -(a+b)^2))


Rule 1: Optional, factor c with a,b

(w -sqrt((c-(a-b))(c+(a-b))))
(w +sqrt((c-(a-b))(c+(a-b))))
(w -sqrt((c-(a+b))(c+(a+b))))
(w +sqrt((c-(a+b))(c+(a+b)))) ,
------------------------------
(w -sqrt((c-a+b)(c+a-b)))
(w +sqrt((c-a+b)(c+a-b)))
(w -sqrt((c-a-b)(c+a+b)))
(w +sqrt((c-a-b)(c+a+b)))


8 Complex Roots of a 0D point: 4 disjoint groups on the Im line, each group is a product of 2 points on the Re line.

(w -sqrt((c-a+b)(c+a-b))) (w +sqrt((c-a+b)(c+a-b))) (w -sqrt((c-a-b)(c+a+b))) (w +sqrt((c-a-b)(c+a+b))) = 0


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Solving for roots of 2 variables. This equation has 4 distinct solutions when solving for a 2-plane. Three are real, and one is complex. All 2-plane solutions are 4 roots of a 2D circle.


(((xy))) : [R-4] of 4 circles

Set zw=0 , solve for real roots on plane xy

(sqrt((sqrt(x^2+y^2) -a)^2 +0^2) -b)^2 +0^2 = c^2

(sqrt((sqrt(x^2+y^2) -a)^2) -b)^2 -c^2 = 0


Rule 1: Factor x,y,a,b with c

(sqrt((sqrt(x^2+y^2) -a)^2) - b-c) (sqrt((sqrt(x^2+y^2) -a)^2) - b+c) = 0


Rule 3: Undo sqrt on x,y,a , apply squaring to b,c

((sqrt(x^2+y^2) -a)^2 - (b-c)^2)
((sqrt(x^2+y^2) -a)^2 - (b+c)^2)


Rule 1: Factor x,y,a with b,c

(sqrt(x^2+y^2) -a -(b-c)) (sqrt(x^2+y^2) -a +(b-c))
(sqrt(x^2+y^2) -a -(b+c)) (sqrt(x^2+y^2) -a +(b+c)) ,
---------------------------------------------------------
(sqrt(x^2+y^2) -a-b+c)
(sqrt(x^2+y^2) -a+b-c)
(sqrt(x^2+y^2) -a-b-c)
(sqrt(x^2+y^2) -a+b+c)


Rule 3: Undo sqrt to x,y , Apply Squaring to a,b,c

(x^2+y^2 -(a-b+c)^2)
(x^2+y^2 -(a+b-c)^2)
(x^2+y^2 -(a-b-c)^2)
(x^2+y^2 -(a+b+c)^2)


4 Real roots of a circle, centered at origin that differ in diameter by R = a±b±c

(x^2+y^2 -(a-b+c)^2) (x^2+y^2 -(a+b-c)^2) (x^2+y^2 -(a-b-c)^2) (x^2+y^2 -(a+b+c)^2) = 0


##########################################################################################################################


(((x)z)) = (((y)z)) : 2x1,[R-2] of 4 circles

Set yw=0 , solve for real roots on plane xz

(sqrt((sqrt(x^2+0^2) -a)^2 +z^2) -b)^2 +0^2 = c^2

(sqrt((sqrt(x^2) -a)^2 +z^2) -b)^2 -c^2 = 0


Rule 1: factor x,a,z,b with c

(sqrt((sqrt(x^2) -a)^2 +z^2) - b-c) (sqrt((sqrt(x^2) -a)^2 +z^2) - b+c) = 0


Rule 3: undo sqrt to x,a,z,b , Apply square to b,c

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


Rule 2: Factor (sqrt(x^2) -a) ---> (x-a) , (x+a)

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


4 real roots of a circle, as 2 disjoint groups centered at x = ±a that differ in diameter by R = b±c

((x-a)^2 +z^2 - (b-c)^2) ((x+a)^2 +z^2 - (b-c)^2) ((x-a)^2 +z^2 - (b+c)^2) ((x+a)^2 +z^2 - (b+c)^2) = 0


##########################################################################################################################


(((x))w) = (((y))w) : 4x1 row of 4 circles

Set yz=0 , solve for real roots on plane xw

(sqrt((sqrt(x^2+0^2) -a)^2 +0^2) -b)^2 +w^2 = c^2

(sqrt((sqrt(x^2) -a)^2) -b)^2 +w^2 -c^2 = 0


Rule 2: Factor (sqrt(x^2) -a) --> (x-a) , (x+a)

((sqrt((x-a)^2) -b)^2 +w^2 -c^2)
((sqrt((x+a)^2) -b)^2 +w^2 -c^2)


Rule 2: Factor (sqrt((x±a)^2) -b) --> (x±a-b) , (x±a+b)

((x-a-b)^2 +w^2 -c^2)((x-a+b)^2 +w^2 -c^2)
((x+a-b)^2 +w^2 -c^2)((x+a+b)^2 +w^2 -c^2)


4 real roots of a circle spaced along a line, centered at x = ±a±b

((x-a-b)^2 +w^2 -c^2) ((x-a+b)^2 +w^2 -c^2) ((x+a-b)^2 +w^2 -c^2) ((x+a+b)^2 +w^2 -c^2) = 0

##########################################################################################################################


((()z)w) : (2),2x1 of 4 circles

Set xy=0 , solve for complex roots on plane zw

(sqrt((sqrt(0^2+0^2) -a)^2 +z^2) -b)^2 +w^2 = c^2

(sqrt(a^2 +z^2) -b)^2 +w^2 -c^2 = 0


Rule 5: Add Im variable, a ---> (sqrt(i^2) -a)

(sqrt((sqrt(i^2) -a)^2 +z^2) -b)^2 +w^2 -c^2 = 0


Rule 2: Factor (sqrt(i^2) -a) --> (i-a) , (i+a)

((sqrt((i-a)^2 +z^2) -b)^2 +w^2 -c^2)
((sqrt((i+a)^2 +z^2) -b)^2 +w^2 -c^2)


Set i=a , cancel Im part in one of the complex solutions

((sqrt((i-a-a)^2 +z^2) -b)^2 +w^2 -c^2)
((sqrt((i-a+a)^2 +z^2) -b)^2 +w^2 -c^2) ,
----------------------------------------
((sqrt((i-2a)^2 +z^2) -b)^2 +w^2 -c^2)
((sqrt((i)^2 +z^2) -b)^2 +w^2 -c^2)
----------------------------------------
((sqrt((i-2a)^2 +z^2) -b)^2 +w^2 -c^2)
((sqrt(z^2) -b)^2 +w^2 -c^2)


Rule 2: Factor (sqrt(z^2) -b) ---> (z-b) , (z+b)

((sqrt((i-2a)^2 +z^2) -b)^2 +w^2 -c^2)
((z-b)^2 +w^2 -c^2)
((z+b)^2 +w^2 -c^2)


4 complex roots of a 2D circle, as 2 disjoint groups at i = ±a, each group is a product of 2 disjoint circles at z = ±b

((sqrt((i-a)^2 +z^2) -b)^2 +w^2 -c^2) ((sqrt((i+a)^2 +z^2) -b)^2 +w^2 -c^2) = 0


Translating along i by ±a will cancel the imaginary part of one of the groups, bringing 2 circles into the real plane zw

((sqrt((i-2a)^2 +z^2) -b)^2 +w^2 -c^2) ((z-b)^2 +w^2 -c^2) ((z+b)^2 +w^2 -c^2) = 0


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################


Solving for roots of 3 variables. This equation has 3 distinct solutions when solving for a 3-plane. Three are real, and none are complex. All 3-plane solutions are 2 roots of a torus.


(((xy)z)) : [Rm-2] of 2 tori

Set w=0 , solve for real roots on plane xyz

(sqrt((sqrt(x^2+y^2) -a)^2 +z^2) -b)^2 +0^2 = c^2

(sqrt((sqrt(x^2+y^2) -a)^2 +z^2) -b)^2 -c^2 = 0


Rule 1: Factor x,y,a,z,b,c with d

(sqrt((sqrt(x^2+y^2) -a)^2 +z^2) - b-c)
(sqrt((sqrt(x^2+y^2) -a)^2 +z^2) - b+c)


Rule 3: Undo sqrt to x,y,a,z , apply square to b,c

((sqrt(x^2+y^2) -a)^2 +z^2 - (b-c)^2)
((sqrt(x^2+y^2) -a)^2 +z^2 - (b+c)^2)


2 Roots of a 3D torus, centered at origin, that differ in the minor diameter by Rmin = b±c

((sqrt(x^2+y^2) -a)^2 +z^2 -(b-c)^2) ((sqrt(x^2+y^2) -a)^2 +z^2 -(b+c)^2) = 0


##########################################################################################################################


(((xy))w) : [R1-2] of 2 tori

Set z=0 , solve for real roots on plane xyw

(sqrt((sqrt(x^2+y^2) -a)^2 +0^2) -b)^2 +w^2 = c^2

(sqrt((sqrt(x^2+y^2) -a)^2) -b)^2 +w^2 -c^2 = 0


Rule 2: Factor (sqrt((sqrt(x^2+y^2) -a)^2) -b) --> (sqrt(x^2+y^2) - a-b) , (sqrt(x^2+y^2) - a+b)

((sqrt(x^2+y^2) - a-b)^2 +w^2 -c^2)
((sqrt(x^2+y^2) - a+b)^2 +w^2 -c^2)


2 Roots of a 3D torus, centered at origin, that differ in the major diameter by Rmaj = a±b

((sqrt(x^2+y^2) - a-b)^2 +w^2 -c^2) ((sqrt(x^2+y^2) - a+b)^2 +w^2 -c^2) = 0


##########################################################################################################################


(((x)z)w) = (((y)z)w) : 2x1 row of 2 tori

Set y=0 , solve for real roots on plane xzw

(sqrt((sqrt(x^2+0^2) -a)^2 +z^2) -b)^2 +w^2 = c^2

(sqrt((sqrt(x^2) -a)^2 +z^2) -b)^2 +w^2 -c^2 = 0


Rule 2: factor (sqrt(x^2) -a) ---> (x-a) , (x+a)

((sqrt((x-a)^2 +z^2) -b)^2 +w^2 -c^2)
((sqrt((x+a)^2 +z^2) -b)^2 +w^2 -c^2)


2 Roots of a 3D torus, centered at x = ±a

((sqrt((x-a)^2 +z^2) -b)^2 +w^2 -c^2) ((sqrt((x+a)^2 +z^2) -b)^2 +w^2 -c^2) = 0
Last edited by ICN5D on Fri Nov 27, 2015 11:48 pm, edited 1 time in total.
in search of combinatorial objects of finite extent
ICN5D
Pentonian
 
Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: A Root Finding Algorithm for Factoring Hypertoric Sectio

Postby ICN5D » Fri Nov 27, 2015 11:41 pm

Finding Roots in the Degree-8 ((II)(II)) Polynomial for a Tiger

(sqrt(x11^2 + x12^2) -R11)^2 + (sqrt(x21^2 + x22^2) -R12)^2 = R1^2

(sqrt(x^2+y^2) -a)^2 +(sqrt(z^2+w^2) -b)^2 = c^2

((xy-a)(zw-b)-c)

((xy)(zw))


The 4 Coordinate 1-plane Solutions
------------------------------------
((x)()) = ((y)()) = (()(z)) = (()(w))


The 6 Coordinate 2-plane Solutions
-------------------------------------
((xy)()) = (()(zw))
((x)(z)) = ((y)(z)) = ((x)(w)) = ((y)(w))


The 4 Coordinate 3-plane Solutions
------------------------------------
((xy)(z)) = ((xy)(w)) = ((x)(zw)) = ((y)(zw))


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################

Solving for roots of 1 variable. This equation has 1 distinct complex solution when solving for a single axis. All coordinate 1-plane solutions are 8 roots of a 0D point.


((x)()) = ((y)()) = (()(z)) = (()(w)) : (2),4 array 8 points

Set yzw=0 , solve for complex roots on x

(sqrt(x^2+0^2) -a)^2 +(sqrt(0^2+0^2) -b)^2 = c^2

(sqrt(x^2) -a)^2 +b^2 -c^2 = 0


Rule 2: Factor (sqrt(x^2)-a) --> (x-a) , (x+a)

((x-a)^2 +b^2 -c^2)
((x+a)^2 +b^2 -c^2)


Rule 4: Factor the squaring on x,a , apply sqrt to b,c

(x -a-sqrt(c^2-b^2))
(x -a+sqrt(c^2-b^2))
(x +a-sqrt(c^2-b^2))
(x +a+sqrt(c^2-b^2))


8 complex roots of a 0D point: 2 disjoint groups at x = ±ib, each group is a product of 4 points x = ±a±c

(x -a-sqrt(c^2-b^2)) (x -a+sqrt(c^2-b^2)) (x +a-sqrt(c^2-b^2)) (x +a+sqrt(c^2-b^2)) = 0


\left(x -a-\sqrt{c^2-b^2}\right)\cdot
\left(x -a+\sqrt{c^2-b^2}\right)\cdot
\left(x +a-\sqrt{c^2-b^2}\right)\cdot
\left(x +a+\sqrt{c^2-b^2}\right) = 0


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################

Solving for roots of 2 variables. This equation has 2 distinct solutions when solving for a 2-plane. One is real, and one is complex. All coordinate 2-plane solutions are 4 roots of a 2D circle.


((xy)()) = (()(zw)) : (2),[R-2] array of 4 circles

Set zw=0 , solve for complex roots on xy

(sqrt(x^2+y^2) -a)^2 +(sqrt(0^2+0^2) -b)^2 = c^2

(sqrt(x^2+y^2) -a)^2 +b^2 -c^2 = 0


Rule 4: Factor the squaring of x,y,a , apply sqrt to b,c

(sqrt(x^2+y^2) - a-sqrt(c^2-b^2))
(sqrt(x^2+y^2) - a+sqrt(c^2-b^2))


Rule 3: Undo sqrt to x,y , apply squaring to a,b,c

(x^2+y^2 - (a-sqrt(c^2-b^2))^2)
(x^2+y^2 - (a+sqrt(c^2-b^2))^2)


4 complex roots of a circle: 2 disjoint groups at ±ib , each group is a product of 2 circles of diamters R = a±c

(x^2+y^2 -(a-sqrt(c^2-b^2))^2) (x^2+y^2 -(a+sqrt(c^2-b^2))^2) = 0

\left(x^2+y^2 -\left(a-\sqrt{c^2-b^2}\right)^2\right)\cdot
\left(x^2+y^2 -\left(a+\sqrt{c^2-b^2}\right)^2\right) = 0


##########################################################################################################################


((x)(z)) = ((y)(z)) = ((x)(w)) = ((y)(w)) : 2x2 square array of 4 circles

Set yw=0 , solve for real roots on xz

(sqrt(x^2+0^2) -a)^2 +(sqrt(z^2+0^2) -b)^2 = c^2

(sqrt(x^2)-a)^2 +(sqrt(z^2)-b)^2 -c^2 = 0


Rule 2: Factor (sqrt(x^2)-a) --> (x-a) , (x+a)

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


Rule 2: Factor (sqrt(z^2)-b) --> (z-b) , (z+b)

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


4 real roots of a circle in a 2x2 square, centered at x = ±a , z = ±b, of diameter R = c

((x-a)^2 +(z-b)^2 -c^2) ((x-a)^2 +(z+b)^2 -c^2) ((x+a)^2 +(z-b)^2 -c^2) ((x+a)^2 +(z+b)^2 -c^2) = 0


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################

Solving for roots of 3 variables. This equation has 1 distinct real solution when solving for a 3-plane. All coordinate 3-plane solutions are 2 roots of a 3D torus.


((xy)(z)) = ((xy)(w)) = ((x)(zw)) = ((y)(zw)) : 1x1x2 vertical column of 2 tori

Set w=0 , solve for real roots on xyz

(sqrt(x^2+y^2) -a)^2 +(sqrt(z^2+0^2) -b)^2 = c^2

(sqrt(x^2+y^2) -a)^2 +(sqrt(z^2) -b)^2 -c^2 = 0


Rule 2: Factor (sqrt(z^2)-b) --> (z-b) , (z+b)

((sqrt(x^2+y^2) -a)^2 +(z-b)^2 -c^2)
((sqrt(x^2+y^2) -a)^2 +(z+b)^2 -c^2)


2 roots of a 3D torus, centered at z = ±b, of diameter Rmaj = a , Rmin = c

((sqrt(x^2+y^2) -a)^2 +(z-b)^2 -c^2) ((sqrt(x^2+y^2) -a)^2 +(z+b)^2 -c^2) = 0
in search of combinatorial objects of finite extent
ICN5D
Pentonian
 
Posts: 1047
Joined: Mon Jul 28, 2008 4:25 am
Location: Orlando, FL

Re: A Root Finding Algorithm for Factoring Hypertoric Sectio

Postby ICN5D » Fri Nov 27, 2015 11:47 pm

Finding Roots in the Degree-16 (((II)(II))I) Polynomial for a Toratiger, T^2 x C2

(sqrt((sqrt(x111^2+x112^2) -R111^2)^2 + (sqrt(x211^2+x212^2) -R211^2)^2) -R11^2)^2 +x2^2 = R1^2

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2) -c)^2 +v^2 = d^2

(((xy-a)(zw-b)-c)v-d)

(((xy)(zw))v)


The 5 Coordinate 1-plane Solutions
------------------------------------
(((x)())) = (((y)())) = ((()(z))) = ((()(w)))
((()())v)


The 10 Coordinate 2-plane Solutions
------------------------------------
(((xy)())) = ((()(zw)))
(((x)(z))) = (((y)(z))) = (((x)(w))) = (((y)(w)))
(((x)())v) = (((y)())v) = ((()(z))v) = ((()(w))v)


The 10 Coordinate 3-plane Solutions
------------------------------------
(((xy)(z))) = (((xy)(w))) = (((x)(zw))) = (((y)(zw)))
(((xy)())v) = ((()(zw))v)
((x)(z))v) = ((y)(z))v) = ((x)(w))v) = ((y)(w))v)


The 5 Coordinate 4-plane Solutions
------------------------------------
(((xy)(zw)))
(((x)(zw))v) = (((y)(zw))v) = (((xy)(z))v) = (((xy)(w))v)


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################


Solving for roots of 1 variable. This equation has 2 distinct solutions when solving for a single axis. One is complex, and one is bi-complex. All 1-plane solutions are 16 roots of a 0D point.


(((x)())) = (((y)())) = ((()(z))) = ((()(w))) : (2),8 array of 16 points

Set yzwv=0 , solve for complex roots on x

(sqrt((sqrt(x^2+0^2) -a)^2 + (sqrt(0^2+0^2) -b)^2) -c)^2 +0^2 = d^2

(sqrt((sqrt(x^2)-a)^2 +b^2) -c)^2 -d^2 = 0


Rule 1: Factor x,a,b,c with d

(sqrt((sqrt(x^2)-a)^2 +b^2) - c-d)
(sqrt((sqrt(x^2)-a)^2 +b^2) - c+d)


Rule 3: undo sqrt on x,a,b , apply squaring to c,d

((sqrt(x^2)-a)^2 +b^2 - (c-d)^2)
((sqrt(x^2)-a)^2 +b^2 - (c+d)^2)


Rule 2: Factor (sqrt(x^2)-a) --> (x-a) , (x+a)

((x-a)^2 +b^2 - (c-d)^2)
((x+a)^2 +b^2 - (c-d)^2)
((x-a)^2 +b^2 - (c+d)^2)
((x+a)^2 +b^2 - (c+d)^2)


Rule 4: Factor the squaring of x,a , apply sqrt to b,c,d

(x-a-sqrt((c-d)^2-b^2))
(x-a+sqrt((c-d)^2-b^2))
(x+a-sqrt((c-d)^2-b^2))
(x+a+sqrt((c-d)^2-b^2))
(x-a-sqrt((c+d)^2-b^2))
(x-a+sqrt((c+d)^2-b^2))
(x+a-sqrt((c+d)^2-b^2))
(x+a+sqrt((c+d)^2-b^2))


Optional: Rule 1, factor c,d with b

(x-a-sqrt((c-d-b)(c-d+b)))
(x-a+sqrt((c-d-b)(c-d+b)))
(x+a-sqrt((c-d-b)(c-d+b)))
(x+a+sqrt((c-d-b)(c-d+b)))
(x-a-sqrt((c+d-b)(c+d+b)))
(x-a+sqrt((c+d-b)(c+d+b)))
(x+a-sqrt((c+d-b)(c+d+b)))
(x+a+sqrt((c+d-b)(c+d+b)))


16 complex roots of a 0D point: 2 disjoint groups on x = ±ib , each group is a product of 8 points x = ±a±c±d

(x-a-sqrt((c-d-b)(c-d+b))) (x-a+sqrt((c-d-b)(c-d+b))) (x+a-sqrt((c-d-b)(c-d+b))) (x+a+sqrt((c-d-b)(c-d+b))) (x-a-sqrt((c+d-b)(c+d+b))) (x-a+sqrt((c+d-b)(c+d+b))) (x+a-sqrt((c+d-b)(c+d+b))) (x+a+sqrt((c+d-b)(c+d+b))) = 0


\left(x-a-\sqrt{(c-d)^2-b^2}\right)\cdot
\left(x-a+\sqrt{(c-d)^2-b^2}\right)\cdot
\left(x+a-\sqrt{(c-d)^2-b^2}\right)\cdot
\left(x+a+\sqrt{(c-d)^2-b^2}\right)\cdot
\left(x-a-\sqrt{(c+d)^2-b^2}\right)\cdot
\left(x-a+\sqrt{(c+d)^2-b^2}\right)\cdot
\left(x+a-\sqrt{(c+d)^2-b^2}\right)\cdot
\left(x+a+\sqrt{(c+d)^2-b^2}\right) = 0


##########################################################################################################################


((()())v) : ((2x2)-(II)),2 array of 16 points

Set xyzw=0 , solve for bi-complex roots on v

(sqrt((sqrt(0^2+0^2) -a)^2 + (sqrt(0^2+0^2) -b)^2) -c)^2 +v^2 = d^2

(sqrt(a^2 + b^2) -c)^2 +v^2 -d^2 = 0


Rule 4: Factor v^2 , apply sqrt to a,b,c,d

(v -sqrt(d^2 - (sqrt(a^2 + b^2) -c)^2))
(v +sqrt(d^2 - (sqrt(a^2 + b^2) -c)^2))


Optional Rule 1, factor d with a,b,c

d^2 - (sqrt(a^2 + b^2) -c)^2 ,
-------------------------------
(d - (sqrt(a^2 + b^2) -c))
(d+ (sqrt(a^2 + b^2) -c)) ,
--------------------------------
(d+c-sqrt(a^2 + b^2))(d-c+sqrt(a^2 + b^2)) ,


(v -sqrt((d+c-sqrt(a^2 + b^2))(d-c+sqrt(a^2 + b^2))))
(v +sqrt((d+c-sqrt(a^2 + b^2))(d-c+sqrt(a^2 + b^2))))


16 bi-complex roots of a 0D point: A 2x2 square array of 4 circles centered at i = ±a , j = ±c on 2D Im plane, when the surface of any circle is placed at origin, the Im part will cancel, and factor into 2 real roots v = ±d

(v -sqrt((d+c-sqrt(a^2+b^2))(d-c+sqrt(a^2+b^2)))) (v +sqrt((d+c-sqrt(a^2+b^2))(d-c+sqrt(a^2+b^2)))) = 0

\left(v -\sqrt{\left(d+c-\sqrt{a^2 + b^2}\right)\left(d-c+\sqrt{a^2 + b^2}\right)}\right)\cdot
\left(v +\sqrt{\left(d+c-\sqrt{a^2 + b^2}\right)\left(d-c+\sqrt{a^2 + b^2}\right)}\right) = 0

##########################################################################################################################
##########################################################################################################################
##########################################################################################################################


Solving for roots of 2 variables. This equation has 3 distinct solutions when solving for a 2-plane. One is real, and two are complex. All 2-plane solutions are 8 roots of a 2D circle.



(((xy)())) = ((()(zw))) : (2),[R-4] array of 8 circles

Set zwv=0 , solve for complex roots on xy

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(0^2+0^2) -b)^2) -c)^2 +0^2 = d^2

(sqrt((sqrt(x^2+y^2) -a)^2 +b^2) -c)^2 -d^2 = 0


Rule 1: factor x,y,a,b,c with d

(sqrt((sqrt(x^2+y^2) -a)^2 +b^2) - c-d)
(sqrt((sqrt(x^2+y^2) -a)^2 +b^2) - c+d)


Rule 3: undo sqrt to x,y,a,b , apply squaring to c,d

((sqrt(x^2+y^2) -a)^2 +b^2 - (c-d)^2)
((sqrt(x^2+y^2) -a)^2 +b^2 - (c+d)^2)


Rule 4: Factor squaring of x,y,a , apply sqrt to c,d,b

(sqrt(x^2+y^2) - a-sqrt((c-d)^2-b^2))
(sqrt(x^2+y^2) - a+sqrt((c-d)^2-b^2))
(sqrt(x^2+y^2) - a-sqrt((c+d)^2-b^2))
(sqrt(x^2+y^2) - a+sqrt((c+d)^2-b^2))


Rule 3: Undo sqrt to x,y , apply squaring to a,b,c,d

(x^2+y^2 - (a-sqrt((c-d)^2-b^2))^2)
(x^2+y^2 - (a+sqrt((c-d)^2-b^2))^2)
(x^2+y^2 - (a-sqrt((c+d)^2-b^2))^2)
(x^2+y^2 - (a+sqrt((c+d)^2-b^2))^2)


8 complex roots of a circle: 2 disjoint groups at ±ib, each group is 4 circles of diameter R = a±c±d

(x^2+y^2 -(a-sqrt((c-d)^2-b^2))^2) (x^2+y^2 -(a+sqrt((c-d)^2-b^2))^2) (x^2+y^2 -(a-sqrt((c+d)^2-b^2))^2) (x^2+y^2 -(a+sqrt((c+d)^2-b^2))^2) = 0


\left(x^2+y^2 - \left(a-\sqrt{(c-d)^2-b^2}\right)^2\right)\cdot
\left(x^2+y^2 - \left(a+\sqrt{(c-d)^2-b^2}\right)^2\right)\cdot
\left(x^2+y^2 - \left(a-\sqrt{(c+d)^2-b^2}\right)^2\right)\cdot
\left(x^2+y^2 - \left(a+\sqrt{(c+d)^2-b^2}\right)^2\right) = 0


##########################################################################################################################


(((x)(z))) = (((y)(z))) = (((x)(w))) = (((y)(w))) : 2x2,[R-2] square array of 8 circles

Set ywv=0 , solve for real roots on xz

(sqrt((sqrt(x^2+0^2) -a)^2 + (sqrt(z^2+0^2) -b)^2) -c)^2 +0^2 = d^2

(sqrt((sqrt(x^2)-a)^2 + (sqrt(z^2)-b)^2) -c)^2 -d^2 = 0


Rule 2: Factor (sqrt(x^2)-a) --> (x-a) , (x+a)

((sqrt((x-a)^2 + (sqrt(z^2)-b)^2) -c)^2 -d^2)
((sqrt((x+a)^2 + (sqrt(z^2)-b)^2) -c)^2 -d^2)


Rule 2: Factor (sqrt(z^2)-b) --> (z-b) , (z+b)

((sqrt((x-a)^2 + (z-b)^2) -c)^2 -d^2)
((sqrt((x-a)^2 + (z+b)^2) -c)^2 -d^2)
((sqrt((x+a)^2 + (z-b)^2) -c)^2 -d^2)
((sqrt((x+a)^2 + (z+b)^2) -c)^2 -d^2)


Rule 1: factor x,a,z,b,c with d

(sqrt((x-a)^2 + (z-b)^2) - c-d)(sqrt((x-a)^2 + (z-b)^2) - c+d)
(sqrt((x-a)^2 + (z+b)^2) - c-d)(sqrt((x-a)^2 + (z+b)^2) - c+d)
(sqrt((x+a)^2 + (z-b)^2) - c-d)(sqrt((x+a)^2 + (z-b)^2) - c+d)
(sqrt((x+a)^2 + (z+b)^2) - c-d)(sqrt((x+a)^2 + (z+b)^2) - c+d)


Rule 3: undo sqrt to x,a,z,b , apply squaring to c,d

((x-a)^2 + (z-b)^2 - (c-d)^2)
((x-a)^2 + (z-b)^2 - (c+d)^2)
((x-a)^2 + (z+b)^2 - (c-d)^2)
((x-a)^2 + (z+b)^2 - (c+d)^2)
((x+a)^2 + (z-b)^2 - (c-d)^2)
((x+a)^2 + (z-b)^2 - (c+d)^2)
((x+a)^2 + (z+b)^2 - (c-d)^2)
((x+a)^2 + (z+b)^2 - (c+d)^2)


8 real roots of a circle, centered at x = ±a , z = ±b , of diameters R = c±d

((x-a)^2+(z-b)^2-(c-d)^2) ((x-a)^2+(z-b)^2-(c+d)^2) ((x-a)^2+(z+b)^2-(c-d)^2) ((x-a)^2+(z+b)^2-(c+d)^2) ((x+a)^2+(z-b)^2-(c-d)^2) ((x+a)^2+(z-b)^2-(c+d)^2) ((x+a)^2+(z+b)^2-(c-d)^2) ((x+a)^2+(z+b)^2-(c+d)^2) = 0


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(((x)())v) = (((y)())v) = ((()(z))v) = ((()(w))v) : (2),4x1 array of 8 circles

Set yzw=0 , solve for complex roots on xv

(sqrt((sqrt(x^2+0^2) -a)^2 + (sqrt(0^2+0^2) -b)^2) -c)^2 +v^2 = d^2

(sqrt((sqrt(x^2)-a)^2 +b^2) -c)^2 +v^2 -d^2 = 0


Rule 2: Factor (sqrt(x^2)-a) --> (x-a) , (x+a)

((sqrt((x-a)^2 +b^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +b^2) -c)^2 +v^2 -d^2)

Unsolvable with Rule 1,2,3,4 beyond this point.
-----------------------------------------------


Rule 5: add Im variable, b --> (sqrt(i^2)-b)

((sqrt((x-a)^2 +(sqrt(i^2)-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(sqrt(i^2)-b)^2) -c)^2 +v^2 -d^2)


Rule 2: Factor (sqrt(i^2)-b) --> (i-b) , (i+b)

((sqrt((x-a)^2 +(i-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x-a)^2 +(i+b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i+b)^2) -c)^2 +v^2 -d^2)

Set i=b ,

((sqrt((x-a)^2 +(i-b-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-b-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x-a)^2 +(i-b+b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-b+b)^2) -c)^2 +v^2 -d^2) ,
----------------------------------------------
((sqrt((x-a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((x-a)^2 +(i)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i)^2) -c)^2 +v^2 -d^2) ,
--------------------------------------------
((sqrt((x-a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((x-a)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2) -c)^2 +v^2 -d^2)


Rule 2: Factor real part (sqrt((x±a)^2) -c) --> (x±a-c) , (x±a+c)

((sqrt((x-a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 +(i-2b)^2) -c)^2 +v^2 -d^2)
((x-a-c)^2 +v^2 -d^2)
((x-a+c)^2 +v^2 -d^2)
((x+a-c)^2 +v^2 -d^2)
((x+a+c)^2 +v^2 -d^2)


8 complex roots of a 2D circle: 2 disjoint groups at i = ±b, each group is 4 real roots of a circle of diameter R = d

((sqrt((x-a)^2 +(i-b)^2) -c)^2 +v^2 -d^2) ((sqrt((x-a)^2 +(i+b)^2) -c)^2 +v^2 -d^2) ((sqrt((x+a)^2 +(i-b)^2) -c)^2 +v^2 -d^2) ((sqrt((x+a)^2 +(i+b)^2) -c)^2 +v^2 -d^2) = 0


Translating by i = ±b will cancel the Im part of one of the groups, and bring 4 roots into the real plane xy as circles centered at x = ±a±c , of diameter R = d

Im group: ((sqrt((x-a)^2+(i-2b)^2)-c)^2 +v^2 -d^2) ((sqrt((x+a)^2+(i-2b)^2)-c)^2 +v^2 -d^2)

Re group: ((x-a-c)^2 +v^2 -d^2) ((x-a+c)^2 +v^2 -d^2) ((x+a-c)^2 +v^2 -d^2) ((x+a+c)^2 +v^2 -d^2)


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Solving for roots of 3 variables. This equation has 3 distinct solutions when solving for a 3-plane. Two are real, and one is complex. All 3-plane solutions are 4 roots of a 3D torus.


(((xy)(z))) = (((xy)(w))) = (((x)(zw))) = (((y)(zw))) : 1x1x2,[Rm-2] column of 4 tori

Set wv=0 , solve for real roots on xyz

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+0^2) -b)^2) -c)^2 +0^2 = d^2

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2)-b)^2) -c)^2 -d^2 = 0


Rule 2: Factor (sqrt(z^2)-b) --> (z-b) , (z+b)

((sqrt((sqrt(x^2+y^2)-a)^2 + (z-b)^2) -c)^2 -d^2)
((sqrt((sqrt(x^2+y^2)-a)^2 + (z+b)^2) -c)^2 -d^2)


Rule 1: Factor x,y,a,z,b,c with d

(sqrt((sqrt(x^2+y^2)-a)^2 + (z-b)^2) - c-d)
(sqrt((sqrt(x^2+y^2)-a)^2 + (z-b)^2) - c+d)
(sqrt((sqrt(x^2+y^2)-a)^2 + (z+b)^2) - c-d)
(sqrt((sqrt(x^2+y^2)-a)^2 + (z+b)^2) - c+d)


Rule 3: undo sqrt to x,y,a,z,b , apply squaring to c,d

((sqrt(x^2+y^2)-a)^2 + (z-b)^2 -(c-d)^2)
((sqrt(x^2+y^2)-a)^2 + (z-b)^2 -(c+d)^2)
((sqrt(x^2+y^2)-a)^2 + (z+b)^2 -(c-d)^2)
((sqrt(x^2+y^2)-a)^2 + (z+b)^2 -(c+d)^2)


4 real roots of a torus, centered at z = ±b , of diameters Rmaj = a , Rmin = c±d

((sqrt(x^2+y^2)-a)^2 + (z-b)^2 -(c-d)^2) ((sqrt(x^2+y^2)-a)^2 + (z-b)^2 -(c+d)^2) ((sqrt(x^2+y^2)-a)^2 + (z+b)^2 -(c-d)^2) ((sqrt(x^2+y^2)-a)^2 + (z+b)^2 -(c+d)^2) = 0


##########################################################################################################################


(((xy)())v) = ((()(zw))v) : (2),[R1-2] array of 4 tori

Set zw=0 , solve for complex roots on xyv

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(0^2+0^2) -b)^2) -c)^2 +v^2 = d^2

(sqrt((sqrt(x^2+y^2) -a)^2 +b^2) -c)^2 +v^2 -d^2 = 0

Unsolvable with Rules 1,2,3,4
------------------------------


Rule 5: add Im variable , b --> (sqrt(i^2)-b)

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(i^2)-b)^2) -c)^2 +v^2 -d^2 = 0


Rule 2: Factor (sqrt(i^2)-b) --> (i-b) , (i+b)

((sqrt((sqrt(x^2+y^2) -a)^2 + (i-b)^2) -c)^2 +v^2 -d^2)
((sqrt((sqrt(x^2+y^2) -a)^2 + (i+b)^2) -c)^2 +v^2 -d^2)

Optional, Set i=b,

((sqrt((sqrt(x^2+y^2) -a)^2 + (i-b-b)^2) -c)^2 +v^2 -d^2)
((sqrt((sqrt(x^2+y^2) -a)^2 + (i-b+b)^2) -c)^2 +v^2 -d^2) ,
----------------------------------------------------------
((sqrt((sqrt(x^2+y^2) -a)^2 + (i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((sqrt(x^2+y^2) -a)^2 + (i)^2) -c)^2 +v^2 -d^2) ,
-------------------------------------------------------
((sqrt((sqrt(x^2+y^2) -a)^2 + (i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt((sqrt(x^2+y^2) -a)^2) -c)^2 +v^2 -d^2)


Rule 2: factor (sqrt((sqrt(x^2+y^2) -a)^2) -c) --> (sqrt(x^2+y^2) -a-c) , (sqrt(x^2+y^2) -a+c)

((sqrt((sqrt(x^2+y^2) -a)^2 + (i-2b)^2) -c)^2 +v^2 -d^2)
((sqrt(x^2+y^2) -a-c)^2 +v^2 -d^2)
((sqrt(x^2+y^2) -a+c)^2 +v^2 -d^2)


4 complex roots of a 3D torus: 2 disjoint groups at i = ±b , each group is 2 real roots of a torus

((sqrt((sqrt(x^2+y^2)-a)^2 + (i-b)^2) -c)^2 +v^2 -d^2) ((sqrt((sqrt(x^2+y^2)-a)^2 + (i+b)^2) -c)^2 +v^2 -d^2) = 0


Translating along Im axis i = ±b will cancel the imaginary part of one of the groups, bringing 2 real roots of a torus into plane xyv, centered on origin, of diameters Rmaj = a±c , Rmin = d

Im group: ((sqrt((sqrt(x^2+y^2) -a)^2 + (i-2b)^2) -c)^2 +v^2 -d^2)

Re group: ((sqrt(x^2+y^2) -a-c)^2 +v^2 -d^2) ((sqrt(x^2+y^2) -a+c)^2 +v^2 -d^2)


##########################################################################################################################


((x)(z))v) = ((y)(z))v) = ((x)(w))v) = ((y)(w))v) : 2x2x1 square of 4 tori

Set yw=0 , solve for real roots on xzv

(sqrt((sqrt(x^2+0^2) -a)^2 + (sqrt(z^2+0^2) -b)^2) -c)^2 +v^2 = d^2

(sqrt((sqrt(x^2)-a)^2 + (sqrt(z^2)-b)^2) -c)^2 +v^2 -d^2 = 0


Rule 2: factor (sqrt(x^2)-a) ---> (x-a) , (x+a)

((sqrt((x-a)^2 + (sqrt(z^2)-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 + (sqrt(z^2)-b)^2) -c)^2 +v^2 -d^2)


Rule 2: factor (sqrt(z^2)-b) ---> (z-b) , (z+b)

((sqrt((x-a)^2 + (z-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x-a)^2 + (z+b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 + (z-b)^2) -c)^2 +v^2 -d^2)
((sqrt((x+a)^2 + (z+b)^2) -c)^2 +v^2 -d^2)


4 real roots of a torus centered at x = ±a , z = ±b , of diameters Rmaj = c , Rmin = d

((sqrt((x-a)^2 + (z-b)^2)-c)^2 +v^2 -d^2) ((sqrt((x-a)^2 + (z+b)^2)-c)^2 +v^2 -d^2) ((sqrt((x+a)^2 + (z-b)^2)-c)^2 +v^2 -d^2) ((sqrt((x+a)^2 + (z+b)^2)-c)^2 +v^2 -d^2) = 0


##########################################################################################################################
##########################################################################################################################
##########################################################################################################################

Solving for roots of 4 variables. This equation has 2 distinct, real solutions when solving for a 4-plane. All 4-plane solutions are either 2 roots of a 4D tiger, or 2 roots of a 4D 3-torus.


(((xy)(zw))) : [Rm-2] concentric pair of 2 tigers

Set v=0 , solve for real roots on xyzw

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2) -c)^2 +0^2 = d^2

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2) -c)^2 -d^2 = 0


Rule 1: factor x,y,a,z,w,b,c with d

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2) - c-d)
(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2) - c+d)


Rule 3: undo sqrt to x,y,a,z,w,b , apply squaring to c,d

((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2 - (c-d)^2)
((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+w^2) -b)^2 - (c+d)^2)


2 real roots of a 4D tiger, centered on origin, of diameters R1a = a , R1b = b , Rmin = c±d

((sqrt(x^2+y^2)-a)^2 + (sqrt(z^2+w^2)-b)^2 -(c-d)^2) ((sqrt(x^2+y^2)-a)^2 + (sqrt(z^2+w^2)-b)^2 -(c+d)^2) = 0


##########################################################################################################################


(((x)(zw))v) = (((y)(zw))v) = (((xy)(z))v) = (((xy)(w))v) : 1x1x2x1 row of 2 ditoruses (T^3)

Set w=0 , solve for real roots on xyzv

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2+0^2) -b)^2) -c)^2 +v^2 = d^2

(sqrt((sqrt(x^2+y^2) -a)^2 + (sqrt(z^2)-b)^2) -c)^2 +v^2 -d^2 = 0


Rule 2: factor (sqrt(z^2)-b) ---> (z-b) , (z+b)

((sqrt((sqrt(x^2+y^2) -a)^2 + (z-b)^2) -c)^2 +v^2 -d^2)
((sqrt((sqrt(x^2+y^2) -a)^2 + (z+b)^2) -c)^2 +v^2 -d^2)


2 real roots of a 3-torus, centered at z = ±b , of diameters R1 = a , R2 = c , Rmin = d

((sqrt((sqrt(x^2+y^2)-a)^2 + (z-b)^2)-c)^2 +v^2 -d^2) ((sqrt((sqrt(x^2+y^2)-a)^2 + (z+b)^2)-c)^2 +v^2 -d^2) = 0
in search of combinatorial objects of finite extent
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