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Orbital Analysis: Magnitude Sweep Functions
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Timothy Golden
May 1, 2021, 1:55:45 PM5/1/21
to
https://drive.google.com/file/d/1Vvqq2f_Ch6IozwNimJjcS4kw3tnVmtPd/view?usp=sharing
Though this math is done in P3 it is exactly the same result that the complex plane can yield. So why is that this graphic has not been presented by others? There are proofs by Fatou and/or Julia and those wayback that claim the space to be filled in completely, yet they could not discuss density. We have the computer power that they did not have. Can something be worked out on paper? Maybe, but this graphic is quite an accurate representation.
We see that some fractal behavior is occurring here. No escape criterion is given; though it is true that large values go large and so are off and will stay off of this graph. These are the orbits along a ray from the origin outward as documented in the graphic. The ray is there too. I can't quite pick it out though. m is simply swept from zero by dm along the clocking vector, of which this is but one plot. In this way a sort of survey alights of the orbitals. They do take to some n-ary process, but quite some additional dynamics take place as well.
Though this math is done in P3 it is exactly the same result that the complex plane can yield. So why is that this graphic has not been presented by others? There are proofs by Fatou and/or Julia and those wayback that claim the space to be filled in completely, yet they could not discuss density. We have the computer power that they did not have. Can something be worked out on paper? Maybe, but this graphic is quite an accurate representation.
We see that some fractal behavior is occurring here. No escape criterion is given; though it is true that large values go large and so are off and will stay off of this graph. These are the orbits along a ray from the origin outward as documented in the graphic. The ray is there too. I can't quite pick it out though. m is simply swept from zero by dm along the clocking vector, of which this is but one plot. In this way a sort of survey alights of the orbitals. They do take to some n-ary process, but quite some additional dynamics take place as well.
mitchr...@gmail.com
May 1, 2021, 2:09:42 PM5/1/21
to
Orbiting meteor showers will go to zero after
being swept away by the Earth.
Comets will be used up just the same.
The Eternal future has limits on
the quantities of these objects.
Or they don't collide...
Mitchell Raemsch
being swept away by the Earth.
Comets will be used up just the same.
The Eternal future has limits on
the quantities of these objects.
Or they don't collide...
Mitchell Raemsch
Brett Intriago
May 1, 2021, 3:45:09 PM5/1/21
to
Timothy Golden wrote:
> Though this math is done in P3 it is exactly the same result that the
> complex plane can yield. So why is that this graphic has not been
> presented by others? There are proofs by Fatou and/or Julia and those
This is so cool!!
> Though this math is done in P3 it is exactly the same result that the
> complex plane can yield. So why is that this graphic has not been
> presented by others? There are proofs by Fatou and/or Julia and those
Protesters clash with police, set up GUILLOTINE with effigy of PM in
Turin, Italy amid May Day protest https://on.rt.com/b7b2
but guillotine is too easy, I would suggest impaling. All too many people
killed already, too many children abused, too many families destroyed
into pieces by the "western" predatory capitalism.
Essentially they knew their fake money based wealth has to crash, already
2009. The numbers were showing the inevitable. Crashing the fake money
bring their illegal stolen, to are round *big_bang_zero* and worse.
Don't be fooled, thinking they are doing it because are stupid. It's
simple, fake money capitalism means literary *big_bang_zero* and worse.
Sad they bombed children in many countries, just to keep the lie go.
Brett Intriago
May 1, 2021, 3:51:57 PM5/1/21
to
Brett Intriago wrote:
> Timothy Golden wrote:
>
>> Though this math is done in P3 it is exactly the same result that the
>> complex plane can yield. So why is that this graphic has not been
>> presented by others? There are proofs by Fatou and/or Julia and those
>
> This is so cool!!
>
> Protesters clash with police, set up GUILLOTINE with effigy of PM in
> Turin, Italy amid May Day protest https://on.rt.com/b7b2
>
> but guillotine is too easy, I would suggest impaling. All too many
> people killed already, too many children abused, too many families
> destroyed into pieces by the "western" predatory capitalism.
>
> Essentially they knew their fake money based wealth has to crash,
> already 2009. The numbers were showing the inevitable. Crashing the fake
> money bring their illegal stolen, to are round *big_bang_zero* and
> worse.
stolen wealth
> Timothy Golden wrote:
>
>> Though this math is done in P3 it is exactly the same result that the
>> complex plane can yield. So why is that this graphic has not been
>> presented by others? There are proofs by Fatou and/or Julia and those
>
> This is so cool!!
>
> Protesters clash with police, set up GUILLOTINE with effigy of PM in
> Turin, Italy amid May Day protest https://on.rt.com/b7b2
>
> but guillotine is too easy, I would suggest impaling. All too many
> people killed already, too many children abused, too many families
> destroyed into pieces by the "western" predatory capitalism.
>
> Essentially they knew their fake money based wealth has to crash,
> already 2009. The numbers were showing the inevitable. Crashing the fake
> money bring their illegal stolen, to are round *big_bang_zero* and
> worse.
> Don't be fooled, thinking they are doing it because are stupid. It's
> simple, fake money capitalism means literary *big_bang_zero* and worse.
>
> Sad they bombed children in many countries, just to keep the lie go.
hosts, such as the value of their wealth preserves above the round
*big_bang_zero*.
Eugene Hattabaugh
May 2, 2021, 3:31:34 AM5/2/21
to
You can't fool me, there's a technically fundamental difference between a
*test* and a *generator*. You native americans are completely incompetent
in moon landings and "structural failures, because the fire is so
intense".
Timothy Golden
May 14, 2021, 7:52:38 AM5/14/21
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Timothy Golden
May 14, 2021, 7:55:10 AM5/14/21
to
michael Rodriguez
May 24, 2021, 8:29:00 PM5/24/21
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Tim, there is a other modulus, which is multiplicative for your p4 (quadruplisigned numbers)
in your format :
z = - a + b * c #d
in conventional format :
z = (a,b,c,d)
||z|| = ( sqrt( aa + bb + cc + dd - 2ac -2bd) )( - a + b - c + d )
where is multiplicative (guaranteed)
||w||·||z|| = ||wz||
It is not the distance function, and it is not spherical, rather hyperbolic.
In addition, this isotropic modulus, has signature, in analogy to split complex numbers".
It is a modulus not a an absolute value.
Read the first pages of "The Hyperbolic Number Plane" by Garret Sobczyk.
in your format :
z = - a + b * c #d
in conventional format :
z = (a,b,c,d)
||z|| = ( sqrt( aa + bb + cc + dd - 2ac -2bd) )( - a + b - c + d )
where is multiplicative (guaranteed)
||w||·||z|| = ||wz||
It is not the distance function, and it is not spherical, rather hyperbolic.
In addition, this isotropic modulus, has signature, in analogy to split complex numbers".
It is a modulus not a an absolute value.
Read the first pages of "The Hyperbolic Number Plane" by Garret Sobczyk.
michael Rodriguez
May 24, 2021, 8:30:42 PM5/24/21
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use the "wrong modulus" is like use the shoes swaped...
Timothy Golden
May 25, 2021, 10:19:00 AM5/25/21
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The ac bd split you've made is consistent with looking at P4 as RxC, where the real component is along
( 1, 0, 1, 0 ) to (0, 1, 0, 1 )
where you can see the squares do mimic R.
Still there is a symmetric distance function that as far as I can tell does hold just fine:
http://bandtechnology.com/PolySigned/DistanceFunction.html
Also the fact that this distance function forms this breakpoint
P1 P2 P3 | P4 P5 P6 P7 ...
allows a claim for support for emergent spacetime, so I don't see this breakpoint as a problem but as a benefit.
Still, I am interested in what you are discussing, though I don't really see what it means or how to use it.
The easy part: https://www.garretstar.com/secciones/publications/docs/HYP2.PDF
michael Rodriguez
May 28, 2021, 7:59:59 PM5/28/21
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z = - a + b * c # d = (+ b # d) @ ( -a * c) = (u) @ (w)
It seems that one could do a tiny manipulation, and if one thinks, in some way, that
hyperbolic moduli admit absolute values, in "some sense", then
|z|^2 = ||z|| = ( sqrt( aa + bb + cc + dd - 2ac -2bd) )( - a + b - c + d )
then
|z|^2 = sqrt( (aa + bb + cc + dd - 2ac -2bd)( ( - a + b - c + d )^2 ) )
then apply square root on both sides
|z| = qrt( (aa + bb + cc + dd - 2ac -2bd)( ( - a + b - c + d )^2 ) )
where qrt( ) is quartic root (the same as apply sqrt( ) two times )
The 'form of reference', this is, a curve of reference, or the surface of reference of
a 'number system' is like the "breathing" of such a system, sort of.
the versor( unitary p4 numbers ) possibly be :
versor( w ) = z / |z|
enough to get an pic of such shape, entering, for example, 10.000.000 random values.
What is take a walk in such surface ?
https://www.youtube.com/watch?v=wu6PUhqu3cs
It seems that one could do a tiny manipulation, and if one thinks, in some way, that
hyperbolic moduli admit absolute values, in "some sense", then
|z|^2 = ||z|| = ( sqrt( aa + bb + cc + dd - 2ac -2bd) )( - a + b - c + d )
then
|z|^2 = sqrt( (aa + bb + cc + dd - 2ac -2bd)( ( - a + b - c + d )^2 ) )
then apply square root on both sides
|z| = qrt( (aa + bb + cc + dd - 2ac -2bd)( ( - a + b - c + d )^2 ) )
where qrt( ) is quartic root (the same as apply sqrt( ) two times )
The 'form of reference', this is, a curve of reference, or the surface of reference of
a 'number system' is like the "breathing" of such a system, sort of.
the versor( unitary p4 numbers ) possibly be :
versor( w ) = z / |z|
enough to get an pic of such shape, entering, for example, 10.000.000 random values.
What is take a walk in such surface ?
https://www.youtube.com/watch?v=wu6PUhqu3cs
michael Rodriguez
May 28, 2021, 8:01:30 PM5/28/21
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..a typo
versor( z ) = z / |z|
versor( z ) = z / |z|
Timothy Golden
May 29, 2021, 7:42:09 AM5/29/21
to
- 1 + 1 * 1 # 1 = 0
means that a general value, say
- 1.1 + 2.2 * 3.3 # 4.4
can be reduced, in this case to
+ 1.1 * 2.2 # 3.3
because
- 1.1 + 1.1 * 1.1 # 1.1 = 0
Anyway the reduced form does expose something close to the dimensional information. That 3D space merely requires four directions: this is quite a gain from the usual six that are claimed.
The four rays of P4 are rays emanating from the center of a simplex out to its vertices. You can delete the frame of the tetrahedron and just look at these four rays as forming the vector space. The reduction above is tantamount to this geometry; step in each direction by some constant distance and you land back where you started. This is a ray based interpretation. In hindsight the ray is more fundamental than the line. The line is composed of two rays. Three rays are sufficient to form the plane and four rays our usual 3D space, and so on upward into general dimension.
The arithmetic product is well defined. It's rotational behavior can simply be viewed through iteration as
( - 1 ) ^ n
since in P4:
(-1) = - 1
(-1)(-1) = + 1
(-1)(-1)(-1) = * 1
(-1)(-1)(-1)(-1) = # 1
(-1)(-1)(-1)(-1)(-1) = - 1
etc.
FOIL type distribution gets the arithmetic product as is done here:
http://bandtechnology.com/PolySigned/FourSigned.html
So is this graphic what you are describing?
http://bandtechnology.com/PolySigned/Deformation/DeformationUnitSphereP4.html
Pretty sure this is the survey you are after.
Yes it is fine. For any z in P4 you can multiply by a scalar and you won't have any surprises. This means that studying the product of the unit shell is sufficient. Fortunately in P4 we are looking at a spherical shell which is fairly easy to perceive.
The identity axis and the complex plane both show up in that graphic.
I'm having a hard time swapping out my thinking above and going over to your hyperbolic viewpoint. The geometry is so simply laid out above. Of course I am studying operators and attempting to discover options for new operators and your thinking does seem nearby. To some degree I suspect that the creation of new operators does imply new number systems as well. This does bear out through the rationals where division is used in their construction. Arguably addition was used in the integers. Descartes lays out some claim that division should be taken up prior to products but this is simply grade school cancellation at some level.
Anyway, the mechanism of polysign is a modulo principal. It is already in use both in the magnitude of the real value and in the sign of the real value. The idea that more of the same awaits I find acceptable. This is clearly an area that modern mathematics has avoided. Otherwise the polysign numbers would have been discovered roughly four hundred years ago. That the modern curriculum ties students' hands rather than freeing them I find completely supported by the ignorance of so many. It is a fine time to make your break with a system that requires so much energy to be upheld on a false pedestal.
michael Rodriguez
Jun 4, 2021, 6:17:37 PM6/4/21
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> ...why orthogonality within the basis selection is a requirement...
In section http://polysign.org/PolySigned/NonOrthogonal/index.html
you make an observation regarding normals.
If it is the case, that in using parallel displacing in the business of naming
points, one is, subterraneanly using the quality of the normal (as "opposed" to
the parallel), then, should this, be an indication of searching alternatives,
since the knowledge built on top of this, would be "compromised" under your
perspective, poetically, as a "nail in the shoe that alters the totality of the
march", not denying possible alternatives, of course.
And in the case of the tangential and the normal, or the radial and the concentric
https://en.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing#Symbols
As in this or that framework, that makes easier to generate this or that curve.
This is, not using normals nor parallels. In that case, are you willing to start again,
all, one more time ?
In section http://polysign.org/PolySigned/NonOrthogonal/index.html
you make an observation regarding normals.
If it is the case, that in using parallel displacing in the business of naming
points, one is, subterraneanly using the quality of the normal (as "opposed" to
the parallel), then, should this, be an indication of searching alternatives,
since the knowledge built on top of this, would be "compromised" under your
perspective, poetically, as a "nail in the shoe that alters the totality of the
march", not denying possible alternatives, of course.
And in the case of the tangential and the normal, or the radial and the concentric
https://en.wikipedia.org/wiki/Geometric_dimensioning_and_tolerancing#Symbols
As in this or that framework, that makes easier to generate this or that curve.
This is, not using normals nor parallels. In that case, are you willing to start again,
all, one more time ?
michael Rodriguez
Jun 4, 2021, 6:21:37 PM6/4/21
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> So is this graphic what you are describing?
> http://bandtechnology.com/PolySigned/Deformation/DeformationUnitSphereP4.html
> Fortunately in P4 we are looking at a spherical shell which is fairly easy to perceive.
outside of the comfort zone. We will never know until try it.
take a brief look to hyperbolas, hyperbolic unit shell, squeeze mapping,
hyperbolic roots of unity or hyperbolic tilings
https://en.wikipedia.org/wiki/Squeeze_mapping
http://www.roguetemple.com/z/hyper/
https://en.wikipedia.org/wiki/Hyperbolic_orthogonality#/media/File:Orthogonality_and_rotation.svg
https://en.wikipedia.org/wiki/Hyperbolic_equilibrium_point#/media/File:Phase_Portrait_Sadle.svg
https://math.stackexchange.com/questions/1862340/what-are-the-hyperbolic-rotation-matrices-in-3-
and-4-dimensions
you can look at it as a goldesian version of directions,
rotations, "modulo behaviour" and turtle graphics
https://en.wikipedia.org/wiki/Six_degrees_of_freedom#/media/File:6DOF.svg
https://en.wikipedia.org/wiki/Ship_motions#/media/File:Rotations.png
https://www.geogebra.org/m/qarxbs9f (interactive)
> P1 P2 P3 | P4 P5 P6 P7 ...
or maybe P1 P2 P3 P4 P5 P6 P7 ... |
> ...though I don't really see what it means or how to use it.
maybe, although you may tinker now and then with it
> means that a general value, say
> - 1.1 + 2.2 * 3.3 # 4.4
> can be reduced, in this case to
> + 1.1 * 2.2 # 3.3
besides being multiplicative.
Well, supposing it is some hyperbolic-like type of surface
https://mathworld.wolfram.com/Hyperboloid.html
like an hyperboloid of one sheet + a hyperboloid of two sheets, then one can
do the following analogies.
- In regarding the study of deformation in the sphere, would be similar, with
the difference that if you draw a loxodrome or other kind of path over the
hyperboloid ( https://www.geogebra.org/m/e2mzbevc ), the multiplication of
any point with "all the other points" will yield the same hyperboloid.
And instead of a great circle, you may stumble upon a great hyperbola.
- taken as a 2d analogy, the split complex numbers, with the versor being
conjugated hyperbolas. you may add the asymtotes if you want.
https://en.wikipedia.org/wiki/Split-complex_number#/media/File:Drini-conjugatehyperbolas.svg
In the asymptotes are the idempotent elements, and the elements of one line
are annihilators with respect to the elements of the other line, which is
where the comparison with RxR is mentioned. Now, the observation here is that
doing a change of basis in the split-complex involves a corresponding change
in the modulus, where the shape of the new modulus is similar(conjugated hyperbolas)
but rotated 45 degrees.
|| (a,b) || = a^2 - b^2
|| (a,b) || = ab <-- with changed basis (slightly rotated hyperbola)
In the wikipedia article is mentioned the comparison RxR and the 'split-complex
with changed basis' :
"Such confusion may be perpetuated when the geometry of the split-complex plane
is not distinguished from that of RxR"
Supposing correct the p4 modulus as an hyperboloid, when doing a change of
basis (as in the split-complex numbers) will yield a surface called gabriel horn
https://mathcurve.com/surfaces.gb/trompette/gabriel.shtml
which is closer to what you refer as the "rod that convert into a flat disc".
Now, "rod that convert into a flat disc" is not the surface of study here,
rather, the hyperboloid.
in the thread "Treating Magnitude as Fundamental"
https://groups.google.com/g/sci.math/c/LiIG-10a9o0/m/YIsqi57MPYQJ
>> Good to hear that you want the usual meaning.
>> However, this means that hardly any Pn is algebraically closed,
>> esp. in RxC there is no solution of X^2 + (1,1) = (0,0) -- or
>> similarly
>> in P4 there is no solution of X^2 #1=0.
it is mentioned p4 does not have a square root of the inverse aditive of the #1
The observation here is that this feature make it naturally compatible with the
geometry of the proposed modulus.
Note that the analogy between split complex and p4 is not perfect, because
while one can use it as orthogonal coordinates, p4 has no idempotents, whence,
use it as a orthogonal basis will require an underlying correction factor in the
product. The 3d analogy of the asymptotes would be 'double cones'
michael Rodriguez
Jun 4, 2021, 6:36:07 PM6/4/21
to
1/(X @ Y) = 1/( (s)a @ (s)b )
with X being (s)a or (s)'a
with Y being (s)b or (s)'b
with a > b
and the ' symbol to save wrist's cramp
1/(X @ Y) = (#a^3 #'(a^2)b #a(b^2) #'b^3)/(#a^4 #'b^4)
a mini-reciprocal
with X being (s)a or (s)'a
with Y being (s)b or (s)'b
with a > b
and the ' symbol to save wrist's cramp
1/(X @ Y) = (#a^3 #'(a^2)b #a(b^2) #'b^3)/(#a^4 #'b^4)
a mini-reciprocal
Timothy Golden
Jun 9, 2021, 10:18:34 AM6/9/21
to
My own thinking is so simple-minded on polysign. I'm all for your advanced thinking, but I have a hard time following along. The hyperbolic stuff seems to be a theme here; leading out to the exponential form perhaps as the ultimate? Well, polysign has it's own extremely simple and proper form
( - 1 ) ^ n
which gets your turtle system working just fine.
It seems to me that the fundamental gains of polysign as a general dimensional system are going ignored. In terms of structural possibilities I hope that somebody works out some interdimensional aspects that gain us something new. That polysign develops geometry as well as algebra; the naming in this department have already been depleted with terms such as 'algebraic goemetry' where polysign will not fit in. Yet polysign will do polynomials in any dimension.
I've just been through quite a bit of fundamental mathematics over the last few months. Coming out the other end I feel like the various areas are too restrictive. I think we need to take freedoms like you do but down at a fundamental level. If there is something left down low to develop that is likely where the prize lays. If that something contradicts some of existing mathematics I really don't see that as a problem; mathematics has already loaded itself down with contradictions.
You are clearly having far more fun than I am working through this stuff. Keep Going! I guess I see that your mini-reciprocal method breaks a value in two, but I don't really understand the '(s)' notation. Your earlier work is so close to having it. I think maybe you should go back to that stage. It seems like some small variation on what you had will get it... assuming my code was giving reasonable results.
michael Rodriguez
Jun 18, 2021, 5:41:16 PM6/18/21
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Regarding your research of spatial path loops, you can diversify towards other
geometric shape, keeping more or less the primitive of the 'signon', let me explain.
Take for example, the vertices of the dodecahedron, they can be obtained as
https://en.wikipedia.org/wiki/Compound_of_five_tetrahedra
and since they are five tetrahedra, hence, five p4 signons.
Now, the issue with the dodecahedra's vertices is that can be imposed
five signons in two possible ways not just one way, look the handedness image in
https://mathworld.wolfram.com/Tetrahedron5-Compound.html
and when combined the two possibilities into one yield
https://en.wikipedia.org/wiki/Compound_of_ten_tetrahedra
This is, each vertex have a "p2 cancellation" with it antipodal vertex,
also have a "p4 cancellation" with respect other three points, and yet, another
"p4 cancellation" with respect yet another three points, more or less, each
vextex has seven cancellative partners. This may be applied in the research of
spatial loop paths.
Other nearby options in the area may be the cuboctahedron as compound, in the
way as the previous one, but using four noncoplanar triangles (also two handedness).
It can be considered like four "p3 signons" in two possible ways (spatial P3).
Each vertex with one antipodal partner, and two sets of p3 partners.
The icosidodecahedron the same but with six pentagons, also with a enantiomorph bother.
The cube can be considered also as compound of two tetrahedra, although it does
not has other brother in handedness.
Other topic regarding to your spatial path/loop
> https://drive.google.com/file/d/1mzcpItPfb7hEtKtiTzxtIGlOOx97OwpC/view
> https://drive.google.com/drive/folders/1xLjsTXOYvHeVau__OCKAHOBZIyps0cRh
You may upload the numbers associated to your loop to OEIS https://oeis.org/
To add a multiplicative structure to the above is possible too.
The word 'golden' infiltrated in https://en.wikipedia.org/wiki/Icosian
can not be randomly, it is sugestive, hyperbolic maybe...
For the polyhedra coordinates search http://eusebeia.dyndns.org/ in the section 4d platonic,
you may use the https://en.wikipedia.org/wiki/Compound_of_cube_and_octahedron to
adapt the coordinates, optionally.
Regading your sweep
> http://polysign.org/PolySigned/MagnitudeSweep/index.html
you may try a tiny trick while doing in quadruplisigned, this is,
in each iteration visually decompose a quadruplisigned number in -a*c and,+b#d,
like considering the -a*c slice and the +b#d slice visually separated, this may
produce the sensation that a spatial number can be "flattened" and work with that.
I guess for sextuplisigned can be separated into three flattened slices, or
perhaps, two 3d slices.Primordial soup machinery and primordial soup terminology.
> To some degree I suspect that the creation of new operators does imply new number systems as well
More a decision of a possibly angle to tackle the subject vs some arbitrary operations like
operationA(w,z)= (w @ z) / (1 @ (w~)z)
operationB(w,z)= (w @ z') / (1 @ (w~)z')
where z' is the aditive inverse of z, and w~ is the conjugate of w.
Or perhaps, for example you have motivation to make some non-comutative version
of your polysigned.
Conceptual slipage, rearranged :
1/(X@Y) = (X³ @'X²Y @XY² @'Y³)/(X⁴ @'Y⁴ )
X = (s1)m1
Y = (s2)m2
m1 > m2
In any case, I would say the general form is better, but both will yield the same result.
> leading out to the exponential form perhaps as the ultimate?
Possibly. I dont know if also have a 'direction cosine' analog.
Possibly you have in mind something as the seven Hamilton's operations.
geometric shape, keeping more or less the primitive of the 'signon', let me explain.
Take for example, the vertices of the dodecahedron, they can be obtained as
https://en.wikipedia.org/wiki/Compound_of_five_tetrahedra
and since they are five tetrahedra, hence, five p4 signons.
Now, the issue with the dodecahedra's vertices is that can be imposed
five signons in two possible ways not just one way, look the handedness image in
https://mathworld.wolfram.com/Tetrahedron5-Compound.html
and when combined the two possibilities into one yield
https://en.wikipedia.org/wiki/Compound_of_ten_tetrahedra
This is, each vertex have a "p2 cancellation" with it antipodal vertex,
also have a "p4 cancellation" with respect other three points, and yet, another
"p4 cancellation" with respect yet another three points, more or less, each
vextex has seven cancellative partners. This may be applied in the research of
spatial loop paths.
Other nearby options in the area may be the cuboctahedron as compound, in the
way as the previous one, but using four noncoplanar triangles (also two handedness).
It can be considered like four "p3 signons" in two possible ways (spatial P3).
Each vertex with one antipodal partner, and two sets of p3 partners.
The icosidodecahedron the same but with six pentagons, also with a enantiomorph bother.
The cube can be considered also as compound of two tetrahedra, although it does
not has other brother in handedness.
Other topic regarding to your spatial path/loop
> https://drive.google.com/file/d/1mzcpItPfb7hEtKtiTzxtIGlOOx97OwpC/view
> https://drive.google.com/drive/folders/1xLjsTXOYvHeVau__OCKAHOBZIyps0cRh
You may upload the numbers associated to your loop to OEIS https://oeis.org/
To add a multiplicative structure to the above is possible too.
The word 'golden' infiltrated in https://en.wikipedia.org/wiki/Icosian
can not be randomly, it is sugestive, hyperbolic maybe...
For the polyhedra coordinates search http://eusebeia.dyndns.org/ in the section 4d platonic,
you may use the https://en.wikipedia.org/wiki/Compound_of_cube_and_octahedron to
adapt the coordinates, optionally.
Regading your sweep
> http://polysign.org/PolySigned/MagnitudeSweep/index.html
you may try a tiny trick while doing in quadruplisigned, this is,
in each iteration visually decompose a quadruplisigned number in -a*c and,+b#d,
like considering the -a*c slice and the +b#d slice visually separated, this may
produce the sensation that a spatial number can be "flattened" and work with that.
I guess for sextuplisigned can be separated into three flattened slices, or
perhaps, two 3d slices.Primordial soup machinery and primordial soup terminology.
> To some degree I suspect that the creation of new operators does imply new number systems as well
operationA(w,z)= (w @ z) / (1 @ (w~)z)
operationB(w,z)= (w @ z') / (1 @ (w~)z')
where z' is the aditive inverse of z, and w~ is the conjugate of w.
Or perhaps, for example you have motivation to make some non-comutative version
of your polysigned.
Conceptual slipage, rearranged :
1/(X@Y) = (X³ @'X²Y @XY² @'Y³)/(X⁴ @'Y⁴ )
X = (s1)m1
Y = (s2)m2
m1 > m2
In any case, I would say the general form is better, but both will yield the same result.
> leading out to the exponential form perhaps as the ultimate?
Possibly you have in mind something as the seven Hamilton's operations.
Timothy Golden
Jun 19, 2021, 11:30:26 AM6/19/21
to
I still find it fascinating that their repetition develops a geometrical ring. Really I find this far more stimulating than the compounds you've introduced me to above. I say once again, congratulations to you for covering more and more ground. I on the other hand seem to be drilling a hole through the simplest of things and having trouble getting through. I really don't wish to give up on the idea that some guiding principle remains to be leveraged. If it exists then the notion of an emergent system can be a go. To what degree these compound shapes are chosen: I don't see how one could avoid compounding the faces in fours rather than in fives.
>
> To add a multiplicative structure to the above is possible too.
> The word 'golden' infiltrated in https://en.wikipedia.org/wiki/Icosian
> can not be randomly, it is sugestive, hyperbolic maybe...
>
> For the polyhedra coordinates search http://eusebeia.dyndns.org/ in the section 4d platonic,
> you may use the https://en.wikipedia.org/wiki/Compound_of_cube_and_octahedron to
> adapt the coordinates, optionally.
> Regading your sweep
>
>
>
> > http://polysign.org/PolySigned/MagnitudeSweep/index.html
> you may try a tiny trick while doing in quadruplisigned, this is,
> in each iteration visually decompose a quadruplisigned number in -a*c and,+b#d,
> like considering the -a*c slice and the +b#d slice visually separated, this may
> produce the sensation that a spatial number can be "flattened" and work with that.
Yes there is this detail, but I think looking at it as embedded P2 effects is better. Running a MU^n type of awareness we see that in P2 and in P4 there are redundant effects. This would be true of any factorables but this is the first encounter. In Pn we can establish an arbitrary path via factors of MU^n so lets say a sum:
> To add a multiplicative structure to the above is possible too.
> The word 'golden' infiltrated in https://en.wikipedia.org/wiki/Icosian
> can not be randomly, it is sugestive, hyperbolic maybe...
>
> For the polyhedra coordinates search http://eusebeia.dyndns.org/ in the section 4d platonic,
> you may use the https://en.wikipedia.org/wiki/Compound_of_cube_and_octahedron to
> adapt the coordinates, optionally.
> Regading your sweep
>
>
>
> > http://polysign.org/PolySigned/MagnitudeSweep/index.html
> you may try a tiny trick while doing in quadruplisigned, this is,
> in each iteration visually decompose a quadruplisigned number in -a*c and,+b#d,
> like considering the -a*c slice and the +b#d slice visually separated, this may
> produce the sensation that a spatial number can be "flattened" and work with that.
Sum over n ( -^n x[n] ) , which is in Pn and that -^n is simply cycling through the signs.
The only key piece of information in this path is the series x[n] which are simply a long list of hopefully small steps along the path, thus forming a smooth path. This same x[n] can be reused across another Pm, but what we see is that a P2 and a P3 could be combined and still get path freedom between the two (by alternating the values as needed), but a P2 and P4 combined path will not have this freedom because a step in minus in P2 will always be either a minus in P4 or a star in P4.
> I guess for sextuplisigned can be separated into three flattened slices, or
> perhaps, two 3d slices.Primordial soup machinery and primordial soup terminology.
> > To some degree I suspect that the creation of new operators does imply new number systems as well
> More a decision of a possibly angle to tackle the subject vs some arbitrary operations like
> operationA(w,z)= (w @ z) / (1 @ (w~)z)
> operationB(w,z)= (w @ z') / (1 @ (w~)z')
> where z' is the aditive inverse of z, and w~ is the conjugate of w.
> Or perhaps, for example you have motivation to make some non-comutative version
> of your polysigned.
>
> Conceptual slipage, rearranged :
> 1/(X@Y) = (X³ @'X²Y @XY² @'Y³)/(X⁴ @'Y⁴ )
> X = (s1)m1
> Y = (s2)m2
> m1 > m2
-^4 = @
+^4 = @
*^4 = @
#^4 = @
and this same will work out in Pn as s^n=@, so the course for a clean denominator looks good.
I guess in the notation it would be nice to know whether
'Y^4 == ' ( Y ^ 4 )
but I think I can conclude it does based on this analysis. And of course the prime is the inverse operator. I do remember trying to work some division out this way myself back in time though it is hazy now, and I don't believe that I ever settled into the inverse as you are doing. Perhaps a simple study of (z)('z) is in order. I know that in P4 the square of the sphere is a double cone; really a single cone with double coverage. Of course that can be interpreted as fitting in with RxC as well:
http://bandtechnology.com/PolySigned/S4SphereSquared.png
> In any case, I would say the general form is better, but both will yield the same result.
> > leading out to the exponential form perhaps as the ultimate?
> Possibly. I dont know if also have a 'direction cosine' analog.
> Possibly you have in mind something as the seven Hamilton's operations.
At least there is a touch of physics that can be gotten to through path mathematics, unless of course you believe that everything is tunneling all the time. I'm not buying it.
michael Rodriguez
Jul 1, 2021, 5:24:37 PM7/1/21
to
> I don't see how one could avoid compounding the faces in fours rather than in fives.
I forgot to say, each triangle occupying three out of six vertices of each hexagon ( lying on the corresponding great circle )
https://en.wikipedia.org/wiki/Cuboctahedron#/media/File:Cuboctahedron_equator.png
with your directions ( signations ?)
michael Rodriguez
Jul 13, 2021, 7:19:22 PM7/13/21
to
There is another way, very similar to find a "general" p4 conjugate (more or less the same)
1/w = (1/w)( w~/ w~ )(Wº/wº) = t/u
where w~ is the first p4 conjugate and multiplying by it one make disappear the signs - and *
the wº is the second p4 conjugate, and multiplying by it one make disappear the sign +
Now one end up with 't/u', and whatever is the composition of 't' it does not matter
the composition of 'u' can be #k, #'k or 0.
Being 0 means it is not invertible by this method
Being #k, it is ok, no problem
Being #'k, it is also ok, and although #'k is it a shorthand
#'k = -k+k*k
we have
(-k+k*k)(-1+1*1) = ?
1/w = (1/w)( w~/ w~ )(Wº/wº) = t/u
where w~ is the first p4 conjugate and multiplying by it one make disappear the signs - and *
the wº is the second p4 conjugate, and multiplying by it one make disappear the sign +
Now one end up with 't/u', and whatever is the composition of 't' it does not matter
the composition of 'u' can be #k, #'k or 0.
Being 0 means it is not invertible by this method
Being #k, it is ok, no problem
Being #'k, it is also ok, and although #'k is it a shorthand
#'k = -k+k*k
we have
(-k+k*k)(-1+1*1) = ?
Timothy Golden
Jul 17, 2021, 8:12:10 PM7/17/21
to
= + k * k # k * k # k - k # k - k + k
This is a good exercise to teach the sign mechanics. Triple foil is fun. Maybe that should be fomil.
= (-2 + 2 * 2 # 3) k
= # k
michael Rodriguez
Aug 4, 2021, 6:11:43 PM8/4/21
to
(@p-1$1)(@p+1&1)(@p*1#1) = 0
@p^3 '@p^2 '@2p @1 = 0
p =aprox= 1.80194
https://oeis.org/A160389
http://www.sacred-geometry.es/?q=en/content/golden-trisection ;0
Certainly put the ' in the middle in strange, in the left or the right side
@p^3 '@p^2 '@2p @1 = 0
p =aprox= 1.80194
https://oeis.org/A160389
http://www.sacred-geometry.es/?q=en/content/golden-trisection ;0
Certainly put the ' in the middle in strange, in the left or the right side
Timothy Golden
Aug 5, 2021, 10:37:19 AM8/5/21
to
I don't think you are serving anybody by taking so many identities. Anonymity is good for some I suppose and you must have some cause for need of it, yet to assume many identities is not really possible without some corroborative signals. Just the interest in polysign is enough probably. The trails of links as well.
That heptagon is an interesting figure as a shelter system with builtin bunks and couches at the sides possible. A few internal braces at six or eight feet will make really decent builtins upon tying in horizontals. Super triangulated with a shedding roof at a decent height. Underhung walls staying dry at ground level and still providing some outdoor storage space under the overhang. It's really quite good. Full standing headroom too. Tight certainly, but sometimes small spaces are good spaces. Everything is in reach. Everything must have a home or else it gets put outside.
As usual some interesting links. http://www.sacred-geometry.es/?q=en/content/metatrons-cube is quite good I think, but the simplex(signon)/rectilinear basis awareness could be better explained. Anyway metatron is some extensive analysis that I am not going to claim to have followed fully.
"We could define Golden Math, or Sacred Math, as the mathematics that naturally arise from the study of Unity."
That's a pretty sweet statement from the same site. I don't mean to go egotistical either. His usage of Golden is different than my name. This study and appreciation of unity as a guiding principal has some strange connections. No doubt these connections are strange due to our curricular habituation.
I guess generally I am wondering about the pacman product and why would an author present it without discussion?
Certainly it is their right to do so. Possibly it could lead to destructive bickering. If there is something there then I am for lighting it up; as in amplifying it rather than burning it down. I suspect it is degenerate out of a consideration for
commutative properties which are a sort of parallel or symmetrical concern, but as well if and when symmetry is broken and something exceptional has occurred then possibly this operator is relevant to some physical issue. Already we have a division problem that is not easily resolved. Rather if it is easily resolved it is not readily so by our habituation. This is the nature of all of polysign. This from a fully algebraic product including commutative behavior in general dimension. Division is a struggle in polysign. Why? I wish there was a simple answer.
I have been seeing some support for algebraic geometry on youtube with some very broad interpretations. As I recall I gave the wedge product a wedgie for good reason, yet I am prepared to review that again now that I am on different footing with mathematics. The interdimensional concern seems relevant and that wedge product as an intermediary comes to life with some stipulations. One of these I believe is a stipulation that a line of zero thickness does not exist. When the line takes a thickness then it as well does take an area. In this way we can achieve some interdimensional work which falls apart on the one hand, yet which carries a remedy on the other. Descartes trails off in Rules for the Direction of the Mind at a similar position. That mainstream mathematics has made a farce already of the Cartesian product is readily proven. And so all and any usage of such is worthy of review. As to whether a super-space is a Cartesian product of its own sub-spaces: no. This type of thinking is not necessary anymore. We have Pn and we must use it as Pn. The native form without any regard for standing mathematics could possibly lead the way through in a way that could take back the term 'algebraic geometry', and that would be entirely appropriate. Wouldn't that be something. Let's do it.
michael Rodriguez
Aug 5, 2021, 6:16:00 PM8/5/21
to
Goldenian conspiracy theory. For the division could exist some heuristic for the general case for polysigns, although require programming to do checkings. Presented with discussion. In the sense that some idealized properties may be attained at a high cost, however, tooo high, but for the sake of curiosity. I almost forgot your crusade against the cartesian product..
Timothy Golden
Aug 10, 2021, 9:46:58 AM8/10/21
to
On Thursday, August 5, 2021 at 6:16:00 PM UTC-4, michael Rodriguez wrote:
> Goldenian conspiracy theory. For the division could exist some heuristic for the general case for polysigns, although require programming to do checkings. Presented with discussion. In the sense that some idealized properties may be attained at a high cost, however, tooo high, but for the sake of curiosity. I almost forgot your crusade against the cartesian product..
Hah, well, the 'binary operator' misuse of the Cartesian product is deep down: https://en.wikipedia.org/wiki/Binary_operation
> Goldenian conspiracy theory. For the division could exist some heuristic for the general case for polysigns, although require programming to do checkings. Presented with discussion. In the sense that some idealized properties may be attained at a high cost, however, tooo high, but for the sake of curiosity. I almost forgot your crusade against the cartesian product..
Since the binary operator of computer science is flooding google search engine: https://mathworld.wolfram.com/BinaryOperation.html
I assure you that addition is not a binary operation by this definition.
That addition or summation or integration takes an n-ary form and that this n-ary form has no need of n-dimensional space I find great relief in.
A noob reads this and says, "But it's so simple. What's the big deal?"
And I say back to the noob: "Exactly."
Timothy Golden
Aug 20, 2021, 10:39:22 AM8/20/21
to
On Thursday, August 5, 2021 at 6:16:00 PM UTC-4, michael Rodriguez wrote:
> Goldenian conspiracy theory. For the division could exist some heuristic for the general case for polysigns, although require programming to do checkings. Presented with discussion. In the sense that some idealized properties may be attained at a high cost, however, tooo high, but for the sake of curiosity. I almost forgot your crusade against the cartesian product..
Yes: a division algorithm to some level of precision should be already done. It is not. Yet you have nearly solved the thing in Pn. I just have to admit that my own simple-mindedness and impedance stands the test of time. I would much prefer that you open the discussion of the division algorithm here as this is your work really. To open it up to others you may witness that your own works take some new joie de vivre. Really I think to post the work in its entirety here on USENET will form a better and more accessible record as well. We just simply call this working in the open, even if you wish to remain anonymous. I can respect that though it seems frustrating at times.
> Goldenian conspiracy theory. For the division could exist some heuristic for the general case for polysigns, although require programming to do checkings. Presented with discussion. In the sense that some idealized properties may be attained at a high cost, however, tooo high, but for the sake of curiosity. I almost forgot your crusade against the cartesian product..
Here a twelve year old kid could enter the discussion as well as an aged PhD and everything in between. In some ways we should be thankful that we are not swamped here. I guess I'm requesting that you take it from the top and maybe post it in a fresh thread. Possibly you'll get some heavy hitters to show up. That would be thrilling. Also though: this medium is not perfect and the ability of a thread to drop out here is real.
michael Rodriguez
Aug 31, 2021, 7:04:36 PM8/31/21
to
although in a hurry...
T = (s)m
( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
( (s1)m1° )( (s2)m2° ) = (s1@s2)m1m2°°
( (s1)m1° )( (s2)m2°° ) = (s1@s2)m1m2°°° = (s1@s2)m1m2
( (s1)m1 )( (s2)m2°° ) = (s1@s2)m1m2°°
usage of auxiliar Pn (auxPn)
since the search is in P6
we have that 6 = 2·3
two styles for p6
[6] = [2]x[3]
[6] = [3]x[2]
all auxPn numbers in use are not in the same Pn except by the rays that contain @ and @'
and if several are in use then all share the same @ and @' of the the main Pn.
a notation (s1,s2,,...,sr)m will not necessarily right since they have a small overlapping
this is not totally independent of the mainPn
Style [3]x[2]
(germe of generalization)
T @ T° @ T°° = 0
T° @ T°° = T'
T = T°°°
As you would say, it is a "mod behaviour inside a mod behaviour"
w = -11 $ +2 $ *13 $ #7 $ &3 $ $5
¶(w) = -11° $ +2°° $ *13 $ #7° $ &3°° $ $5
¶(¶(w)) = -11°° $ +2° $ *13 $ #7°° $ &3° $ $5
here one myay put attention to the distribution of signs
in what position
__________-11°_____+2°°____*13______#7°_____&3°°______@5__
______----------------------------------------------------
-11°°_|___+121_____*22°____#143°°___&77_____@33°_____-55°°
+2°___|___*22°°____#4______&26°_____@14°°___-6_______+10°
*13___|___#143°____&26°°___@169_____-91°____+39°°____*65
#7°°__|___&77______@14°____-91°°____+49_____*21°_____#35°°
&3°___|___@33°°___-6_______+39°_____*21°°___#9_______&15°
@5____|___-55°____+10°°____*65______#35°____&15°°____@25
(-55° $ -55°°) $ (-91°° $ -91° ) $ -6 $ -6
(+10°° $ +10° ) $ (+39° $ +39°° ) $ +49 $ +121
(*21°° $ *21° ) $ (*22°° $ *22° ) $ *65 $ *65
(#35° $ #35°°) $ (#143° $ #143°°) $ #4 $ #9
(&15°° $ &15° ) $ (&26°° $ &26° ) $ &77 $ &77
(@33°° $ @33° ) $ (@14° $ @14°° ) $ @169 $ @25
the auxPn extra-signs dissapear (no mor º symbol)
(-55') $ (-91') $ -6 $ -6 = (-146') $ -12 = (-134')
(+10') $ (+39') $ +49 $ +121 = (+49') $ +170 = +121
(*21') $ (*22') $ *65 $ *65 = (*43') $ *130 = *87
(#35') $ (#143') $ #4 $ #9 = (#178') $ #13 = (#165')
(&15') $ (&26') $ &77 $ &77 = (&41') $ &154 = &113
(@33') $ (@14') $ @169 $ @25 = (@47') $ @194 = @147
(-134') $ +121 $ *87 $ (#165') $ &113 $ @147
_______-134'_____+121______*87______#165'______&113_____@147
____------------------------------------------------------
-11_|__+1474'____*1331_____#957_____&1815'_____@1243____-1617
+2__|__*268'_____#242______&174_____@330'______-226_____+294
*13_|__#1742'____&1573_____@1131____-2145'_____+1469____*1911
#7__|__&938'_____@847______-609_____+1155'_____*791_____#1029
&3__|__@402'_____-363______+261_____*495'______#339_____&441
@5__|__-670'_____+605______*435_____#825'______&565_____@735
-670' $ -363 $ -609 $ -2145' $ -226 $ -1617 = -2815 $ -2817' = 0
+605 $ +261 $ +1155' $ +1469 $ +294 $ +1474' = -2629 $ -2629' = 0
*268' $ *1331 $ *435 $ *495' $ *791 $ *1911 = *4468 $ @763 = *3705
#825' $ #339 $ #1029 $ #1742' $ #242 $ #957 = #2567 $ #2547' = 0
&565 $ &441 $ &938' $ &1573 $ &174 $ &1815' = &2753 $ &2753' = 0
@735 $ @402' $ @847 $ @1131 $ @330' $ @1243 = @3956 $ @732' = @3224
only * and @
*3705 $ @3224
( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
_________-134'_______+121________*87_________#165'________&113________@147
------------------------------------------------------------------------
*3705__|_#496470'____&448305_____@322335_____-611325'_____+418665_____*544635
@3224'_|_-432016_____+390104'____*280488'____#531960______&364312'____@473928'
-432016 $ -611325' = -179309'
+390104' $ +418665 = +28561
*280488' $ *544635 = *264147
#531960 $ #496470' = #35490
&364312' $ &448305 = &83993
@473928' $ @322335 = @151593'
-179309' = +179309 $ *179309 $ #179309 $ &179309 $ @179309
@151593' = -151593 $ +151593 $ *151593 $ #151593 $ &151593
-151593 = -151593
+28561 $ +179309 $ +151593 = +359463
*264147 $ *179309 $ *151593 = *595049
#35490 $ #179309 $ #151593 = #366392
&83993 $ &179309 $ &151593 = &414895
@179309 = @179309
Ł(w) = *3705 $ @3224'
(1/w) = ( ( ( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) / ( (w)( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) )
-151593 $ +359463 $ *595049 $ #366392 $ &414895 $ @179309
q = @3332849
1/w = -(151593/q) $ +(359463/q) $ *(595049/q) $ #(366392/q) $ &(414895/q) $ @(179309/q)
Done ;)
The other alternative to get the reciprocal is doing [6] = [2]x[3]
w = -11 $ +2 $ *13 $ #7 $ &3 $ @5
€(w) = -'11 $ +2 $ *'13 $ #7 $ &'3 $ @5
(sandwiching the s and s' according position
after the first product, this time, only symbol + # @ are remain following the second method
T = (s)m
( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
( (s1)m1° )( (s2)m2° ) = (s1@s2)m1m2°°
( (s1)m1° )( (s2)m2°° ) = (s1@s2)m1m2°°° = (s1@s2)m1m2
( (s1)m1 )( (s2)m2°° ) = (s1@s2)m1m2°°
usage of auxiliar Pn (auxPn)
since the search is in P6
we have that 6 = 2·3
two styles for p6
[6] = [2]x[3]
[6] = [3]x[2]
all auxPn numbers in use are not in the same Pn except by the rays that contain @ and @'
and if several are in use then all share the same @ and @' of the the main Pn.
a notation (s1,s2,,...,sr)m will not necessarily right since they have a small overlapping
this is not totally independent of the mainPn
Style [3]x[2]
(germe of generalization)
T @ T° @ T°° = 0
T° @ T°° = T'
T = T°°°
As you would say, it is a "mod behaviour inside a mod behaviour"
w = -11 $ +2 $ *13 $ #7 $ &3 $ $5
¶(w) = -11° $ +2°° $ *13 $ #7° $ &3°° $ $5
¶(¶(w)) = -11°° $ +2° $ *13 $ #7°° $ &3° $ $5
here one myay put attention to the distribution of signs
in what position
__________-11°_____+2°°____*13______#7°_____&3°°______@5__
______----------------------------------------------------
-11°°_|___+121_____*22°____#143°°___&77_____@33°_____-55°°
+2°___|___*22°°____#4______&26°_____@14°°___-6_______+10°
*13___|___#143°____&26°°___@169_____-91°____+39°°____*65
#7°°__|___&77______@14°____-91°°____+49_____*21°_____#35°°
&3°___|___@33°°___-6_______+39°_____*21°°___#9_______&15°
@5____|___-55°____+10°°____*65______#35°____&15°°____@25
(-55° $ -55°°) $ (-91°° $ -91° ) $ -6 $ -6
(+10°° $ +10° ) $ (+39° $ +39°° ) $ +49 $ +121
(*21°° $ *21° ) $ (*22°° $ *22° ) $ *65 $ *65
(#35° $ #35°°) $ (#143° $ #143°°) $ #4 $ #9
(&15°° $ &15° ) $ (&26°° $ &26° ) $ &77 $ &77
(@33°° $ @33° ) $ (@14° $ @14°° ) $ @169 $ @25
the auxPn extra-signs dissapear (no mor º symbol)
(-55') $ (-91') $ -6 $ -6 = (-146') $ -12 = (-134')
(+10') $ (+39') $ +49 $ +121 = (+49') $ +170 = +121
(*21') $ (*22') $ *65 $ *65 = (*43') $ *130 = *87
(#35') $ (#143') $ #4 $ #9 = (#178') $ #13 = (#165')
(&15') $ (&26') $ &77 $ &77 = (&41') $ &154 = &113
(@33') $ (@14') $ @169 $ @25 = (@47') $ @194 = @147
(-134') $ +121 $ *87 $ (#165') $ &113 $ @147
_______-134'_____+121______*87______#165'______&113_____@147
____------------------------------------------------------
-11_|__+1474'____*1331_____#957_____&1815'_____@1243____-1617
+2__|__*268'_____#242______&174_____@330'______-226_____+294
*13_|__#1742'____&1573_____@1131____-2145'_____+1469____*1911
#7__|__&938'_____@847______-609_____+1155'_____*791_____#1029
&3__|__@402'_____-363______+261_____*495'______#339_____&441
@5__|__-670'_____+605______*435_____#825'______&565_____@735
-670' $ -363 $ -609 $ -2145' $ -226 $ -1617 = -2815 $ -2817' = 0
+605 $ +261 $ +1155' $ +1469 $ +294 $ +1474' = -2629 $ -2629' = 0
*268' $ *1331 $ *435 $ *495' $ *791 $ *1911 = *4468 $ @763 = *3705
#825' $ #339 $ #1029 $ #1742' $ #242 $ #957 = #2567 $ #2547' = 0
&565 $ &441 $ &938' $ &1573 $ &174 $ &1815' = &2753 $ &2753' = 0
@735 $ @402' $ @847 $ @1131 $ @330' $ @1243 = @3956 $ @732' = @3224
only * and @
*3705 $ @3224
( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
_________-134'_______+121________*87_________#165'________&113________@147
------------------------------------------------------------------------
*3705__|_#496470'____&448305_____@322335_____-611325'_____+418665_____*544635
@3224'_|_-432016_____+390104'____*280488'____#531960______&364312'____@473928'
-432016 $ -611325' = -179309'
+390104' $ +418665 = +28561
*280488' $ *544635 = *264147
#531960 $ #496470' = #35490
&364312' $ &448305 = &83993
@473928' $ @322335 = @151593'
-179309' = +179309 $ *179309 $ #179309 $ &179309 $ @179309
@151593' = -151593 $ +151593 $ *151593 $ #151593 $ &151593
-151593 = -151593
+28561 $ +179309 $ +151593 = +359463
*264147 $ *179309 $ *151593 = *595049
#35490 $ #179309 $ #151593 = #366392
&83993 $ &179309 $ &151593 = &414895
@179309 = @179309
Ł(w) = *3705 $ @3224'
(1/w) = ( ( ( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) / ( (w)( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) )
-151593 $ +359463 $ *595049 $ #366392 $ &414895 $ @179309
q = @3332849
1/w = -(151593/q) $ +(359463/q) $ *(595049/q) $ #(366392/q) $ &(414895/q) $ @(179309/q)
Done ;)
The other alternative to get the reciprocal is doing [6] = [2]x[3]
w = -11 $ +2 $ *13 $ #7 $ &3 $ @5
€(w) = -'11 $ +2 $ *'13 $ #7 $ &'3 $ @5
(sandwiching the s and s' according position
after the first product, this time, only symbol + # @ are remain following the second method
Timothy Golden
Sep 2, 2021, 8:41:45 AM9/2/21
to
On Tuesday, August 31, 2021 at 7:04:36 PM UTC-4, michael Rodriguez wrote:
> although in a hurry...
>
> T = (s)m
>
> ( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
> ( (s1)m1° )( (s2)m2° ) = (s1@s2)m1m2°°
> ( (s1)m1° )( (s2)m2°° ) = (s1@s2)m1m2°°° = (s1@s2)m1m2
> ( (s1)m1 )( (s2)m2°° ) = (s1@s2)m1m2°°
>
> usage of auxiliar Pn (auxPn)
> since the search is in P6
> we have that 6 = 2·3
>
> two styles for p6
>
> [6] = [2]x[3]
>
> [6] = [3]x[2]
>
> all auxPn numbers in use are not in the same Pn except by the rays that contain @ and @'
> and if several are in use then all share the same @ and @' of the the main Pn.
> a notation (s1,s2,,...,sr)m will not necessarily right since they have a small overlapping
> this is not totally independent of the mainPn
I'm not so sure that this is a clean construction. Obviously thought if you can compute with it then the proof should be in the pudding.
> although in a hurry...
>
> T = (s)m
>
> ( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
> ( (s1)m1° )( (s2)m2° ) = (s1@s2)m1m2°°
> ( (s1)m1° )( (s2)m2°° ) = (s1@s2)m1m2°°° = (s1@s2)m1m2
> ( (s1)m1 )( (s2)m2°° ) = (s1@s2)m1m2°°
>
> usage of auxiliar Pn (auxPn)
> since the search is in P6
> we have that 6 = 2·3
>
> two styles for p6
>
> [6] = [2]x[3]
>
> [6] = [3]x[2]
>
> all auxPn numbers in use are not in the same Pn except by the rays that contain @ and @'
> and if several are in use then all share the same @ and @' of the the main Pn.
> a notation (s1,s2,,...,sr)m will not necessarily right since they have a small overlapping
> this is not totally independent of the mainPn
I barely follow. Trying though. If it's true then I am playing on a single tone instrument while you are running an orchestra. It sounds to me like you need mary and lary to get past nary. I think until you instantiate a Pm in a Pn then this is vague. This will turn into a lot of work in high sign so doing this as auxP4 with a P3 main... or would that be a main P4 with an auxP3? That may seem limited but there is still a rotational freedom within the P3 right? I guess I'm thinking geometrically of this.
>
>
> Style [3]x[2]
> (germe of generalization)
>
>
> T @ T° @ T°° = 0
>
> T° @ T°° = T'
>
> T = T°°°
>
> As you would say, it is a "mod behaviour inside a mod behaviour"
>
> w = -11 $ +2 $ *13 $ #7 $ &3 $ $5
> ¶(w) = -11° $ +2°° $ *13 $ #7° $ &3°° $ $5
> ¶(¶(w)) = -11°° $ +2° $ *13 $ #7°° $ &3° $ $5
>
> here one myay put attention to the distribution of signs
> in what position
>
> __________-11°_____+2°°____*13______#7°_____&3°°______@5__
> ______----------------------------------------------------
> -11°°_|___+121_____*22°____#143°°___&77_____@33°_____-55°°
> +2°___|___*22°°____#4______&26°_____@14°°___-6_______+10°
> *13___|___#143°____&26°°___@169_____-91°____+39°°____*65
> #7°°__|___&77______@14°____-91°°____+49_____*21°_____#35°°
> &3°___|___@33°°___-6_______+39°_____*21°°___#9_______&15°
> @5____|___-55°____+10°°____*65______#35°____&15°°____@25
>
>
> (-55° $ -55°°) $ (-91°° $ -91° ) $ -6 $ -6
> (+10°° $ +10° ) $ (+39° $ +39°° ) $ +49 $ +121
> (*21°° $ *21° ) $ (*22°° $ *22° ) $ *65 $ *65
> (#35° $ #35°°) $ (#143° $ #143°°) $ #4 $ #9
> (&15°° $ &15° ) $ (&26°° $ &26° ) $ &77 $ &77
> (@33°° $ @33° ) $ (@14° $ @14°° ) $ @169 $ @25
>
> the auxPn extra-signs dissapear (no mor º symbol)
>
>
> (-55') $ (-91') $ -6 $ -6 = (-146') $ -12 = (-134')
> (+10') $ (+39') $ +49 $ +121 = (+49') $ +170 = +121
> (*21') $ (*22') $ *65 $ *65 = (*43') $ *130 = *87
> (#35') $ (#143') $ #4 $ #9 = (#178') $ #13 = (#165')
> (&15') $ (&26') $ &77 $ &77 = (&41') $ &154 = &113
> (@33') $ (@14') $ @169 $ @25 = (@47') $ @194 = @147
>
OK, I see you are really doing some math here.
>
> (-55') $ (-91') $ -6 $ -6 = (-146') $ -12 = (-134')
> (+10') $ (+39') $ +49 $ +121 = (+49') $ +170 = +121
> (*21') $ (*22') $ *65 $ *65 = (*43') $ *130 = *87
> (#35') $ (#143') $ #4 $ #9 = (#178') $ #13 = (#165')
> (&15') $ (&26') $ &77 $ &77 = (&41') $ &154 = &113
> (@33') $ (@14') $ @169 $ @25 = (@47') $ @194 = @147
>
> (-134') $ +121 $ *87 $ (#165') $ &113 $ @147
>
>
> _______-134'_____+121______*87______#165'______&113_____@147
> ____------------------------------------------------------
> -11_|__+1474'____*1331_____#957_____&1815'_____@1243____-1617
> +2__|__*268'_____#242______&174_____@330'______-226_____+294
> *13_|__#1742'____&1573_____@1131____-2145'_____+1469____*1911
> #7__|__&938'_____@847______-609_____+1155'_____*791_____#1029
> &3__|__@402'_____-363______+261_____*495'______#339_____&441
> @5__|__-670'_____+605______*435_____#825'______&565_____@735
>
>
> -670' $ -363 $ -609 $ -2145' $ -226 $ -1617 = -2815 $ -2817' = 0
> +605 $ +261 $ +1155' $ +1469 $ +294 $ +1474' = -2629 $ -2629' = 0
> *268' $ *1331 $ *435 $ *495' $ *791 $ *1911 = *4468 $ @763 = *3705
*268' $ *1331 $ *435 $ *495' $ *791 $ *1911 = *4468 $ *763' = *3705
> *268' $ *1331 $ *435 $ *495' $ *791 $ *1911 = *4468 $ @763 = *3705
> #825' $ #339 $ #1029 $ #1742' $ #242 $ #957 = #2567 $ #2547' = 0
#825' $ #339 $ #1029 $ #1742' $ #242 $ #957 = #2567 $ #2567' = 0
> &565 $ &441 $ &938' $ &1573 $ &174 $ &1815' = &2753 $ &2753' = 0
> @735 $ @402' $ @847 $ @1131 $ @330' $ @1243 = @3956 $ @732' = @3224
>
> only * and @
>
> *3705 $ @3224
>
> ( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
Arggh. I know you are close here, but I'm not sure why you inverted the neutral part.
> @735 $ @402' $ @847 $ @1131 $ @330' $ @1243 = @3956 $ @732' = @3224
>
> only * and @
>
> *3705 $ @3224
>
> ( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
>
>
> _________-134'_______+121________*87_________#165'________&113________@147
> ------------------------------------------------------------------------
> *3705__|_#496470'____&448305_____@322335_____-611325'_____+418665_____*544635
> @3224'_|_-432016_____+390104'____*280488'____#531960______&364312'____@473928'
>
> -432016 $ -611325' = -179309'
-432016-179309 = -611325
>
> _________-134'_______+121________*87_________#165'________&113________@147
> ------------------------------------------------------------------------
> *3705__|_#496470'____&448305_____@322335_____-611325'_____+418665_____*544635
> @3224'_|_-432016_____+390104'____*280488'____#531960______&364312'____@473928'
>
> -432016 $ -611325' = -179309'
= 611325 check.
> +390104' $ +418665 = +28561
> *280488' $ *544635 = *264147
> #531960 $ #496470' = #35490
> &364312' $ &448305 = &83993
> @473928' $ @322335 = @151593'
>
> -179309' = +179309 $ *179309 $ #179309 $ &179309 $ @179309
> @151593' = -151593 $ +151593 $ *151593 $ #151593 $ &151593
>
> -151593 = -151593
> +28561 $ +179309 $ +151593 = +359463
> *264147 $ *179309 $ *151593 = *595049
> #35490 $ #179309 $ #151593 = #366392
> &83993 $ &179309 $ &151593 = &414895
> @179309 = @179309
>
> Ł(w) = *3705 $ @3224'
>
> (1/w) = ( ( ( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) / ( (w)( ¶(w) )( ¶(¶(w)) )( Ł(w) ) ) )
>
> -151593 $ +359463 $ *595049 $ #366392 $ &414895 $ @179309
>
> q = @3332849
>
> 1/w = -(151593/q) $ +(359463/q) $ *(595049/q) $ #(366392/q) $ &(414895/q) $ @(179309/q)
>
> Done ;)
>
> The other alternative to get the reciprocal is doing [6] = [2]x[3]
>
> w = -11 $ +2 $ *13 $ #7 $ &3 $ @5
> €(w) = -'11 $ +2 $ *'13 $ #7 $ &'3 $ @5
> (sandwiching the s and s' according position
>
> after the first product, this time, only symbol + # @ are remain following the second method
Do you have it? I'd have to review that algorithm I left off with to see where we were at there.
Done. I see you had the primes solved, so if this method can be done generally then you have done it.
Will it work in P4?
I guess it will, but for me the answer is not so trivial as it must be for you. You are nimble in polysign, sir.
And such good use of the inverse too. As you said: 'germe of generalization' and I believe it.
I'll try to have a go here but it's not a full hearted attempt so please forgive me if I trail off in digression...
T @ T° @ T°° = 0
P4: T @ T° = 0 , or should this be T @ T°° = 0 ?
I've honestly lost the meaning of this abstraction.
Could it be that expressing this beginning stage as a sum of two values is more specific?
z = - z1 + z2 : T @ T°° = 0 ?
z = - z1 * z2 : T @ T° = 0 ?
anyway, choose:
w = ( 0, 1, 2, 3 ) = @ 0 - 1 + 2 * 3
and I'm lost. Sorry. Maybe I should have run (0,1,2,3,4,5) first.
michael Rodriguez
Sep 3, 2021, 4:46:03 PM9/3/21
to
One may say that all Pn n>2 have a natural sub p2.
in p6
-1 $ +1 $ *1 $ #1 $ &1 = @1'
(-1 $ +1 $ *1 $ #1 $ &1)(-1 $ +1 $ *1 $ #1 $ &1) = @1
(@1)' = @1'
with @1' one may save even more space compared with (@1)'
> Hmmm....
I guess the (-55° $ -55°°) = -55' looks a bit odd, in the sense of leveraging the
degree symbol ° and pretend they well reflect a p3 dynamics
A notation (mainPnSign S1, auxPnSign S2 ) magnitude M
or (S1,S2)M or (-,-)55 $ (-,+)55
will not be precise because both spaces are not totally independent (in this
context, how they are used for the purposes of the thread)
perhaps another attempt of a more precise notation to express both spaces simultaneusly
(_n_,_o,_p,_q,_r,_s_)
(_g_,_h,_i,_j,_k,_m_)
(_a_,_b,_c,_d,_e,_f_)
→ - + * # & $ (mainP6)
↓ ° °° °°° (auxP3)
for example for p6 we have
( a,b,c,d,e,f) = -a @ +b @ *c @ #d @ &e @ $f
trying to maintain the style
(_0_,_0,_c,_0,_0,_f_)
(_0_,_b,_0,_0,_e,_0_) = -ºa @ +ººb @ *c @ #dº @ &eºº @ $f
(_a_,_0,_0,_d,_0,_0_)
(_0_,_0,_0,_0,_0,_0_)
(_a_,_b,_c,_d,_e,_f_) = -ºa @ +ºb @ *ºc @ #dº @ &eº @ $fº
(_0_,_0,_0,_0,_0,_0_)
(_0_,_a,_0,_0,_d,_0_)
(_0_,_b,_0,_0,_0,_0_) = +a @ +bºº @ +cº @ &d @ &eº
(_0_,_c,_0,_0,_e,_0_)
the auxP3(°) space and the mainPn space share the same subP2.
It may seems bizarre, but they do help in finding reciprocals.
For P4 ( 4 = 2 x 2 ) we will need subP2 (two times)
For P6 ( 6 = 2 x 3 ) we will need auxP3 and subP2
and presumably :
For P15 ( 15 = 3 x 5 ) we will need auxP3 and auxP5
For P30 ( 30 = 3 x 5 x 2 ) we will need auxP3, auxP5 and subP2
For P105 ( 105 = 3 x 7 x 5 ) we will need auxP3, auxP7 and auxP5
For P9 ( 9 = 3 x 3 ) we will need auxP3 (two times)
For P8 ( 8 = 2 x 2 x 2 ) we will need subP2 (three times)
Well you may rant a bit that if the only thing that have in common all the auxPn
in use and the mainPn is pair of rays (@ and @'), that will seem as a Real number
infiltration. To this I think that in some post you said that is not much the
real numbers by themselves, but to the degree that the way they are presented and
structured, do not allow looking into Pn (just citing from memory, may be vague)
> I think until you instantiate a Pm in a Pn then this is vague
Certainly the notation is odd, although temporally, until stumble upon a
better notation to indicate spaces with a shared elements.
Yes, rotational freedom although limited
I think an auxP3 is different compared to a subP3.
The main task, is to to ask is if I have a certain Pn number, what thing I have to
do to get @h or @k', that is like 90% of the procedure to the reciprocal
€ : eight
$ : six
if we have, we say, a p8 number, one may start with :
z = -5 € +11 € *7 € #23 € &3 € $19 € μ13 € @17
and using auxP2
þ(z) = -5' € +11 € *7' € #23 € &3' € $19 € μ13' € @17
or maybe using auxP4
þ(z) = -5° € +11°° € *7°°° € #23 € &3° € $19°° € μ13°°° € @17
T° @ T°° @ T°°° @ T = 0
T°°°° = T
(T°)(U°°) = (TU)°°°
I bet for just using auxPn primes, but you will have to check
then you compute ( z )( þ(z) ) = ...
It you can code a little calculator you may get to check such claims for a number
of cases. Pen, paper and hand calculator also have their limitations too.
And this is just may be one method to get reciprocals, you may find others.
>> the auxPn extra-signs dissapear (no mor º symbol)
> Hmmm....
disappear in the sense that the result of the product where ° are involved, is
structured in such a way that the result only yield terms with ' (in worst
case), but not terms with °, this is the result is exclusively in the mainPn space.
>> ( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
ouch, a typo.
( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ ( (@3224)^2 )' = @3332849
> Arggh. I know you are close here, but I'm not sure why you inverted the neutral part.
in conventional arithmetic, we have (A + B)(A - B) = (A^2 - B^2)
this is, difference of squares, translated to your notation is :
(A @ B)(A @ B') = A^2 @ (B^2)'
>> -432016 $ -611325' = -179309'
> -432016-179309 = -611325
> = 611325 check.
*40 $ (*50)' = (*10)'
*50 $ (*30)' = *20
> I guess it will, but for me the answer is not so trivial as it must be for you
>Do you have it?
nopes
> Will it work in P4?
þ(z) = -1' @ +2 @ *3' @ #0
_________-1_____+2_____*3
_____|___________________
-1'__|___+1'____*2'____#3'
+2___|___*2_____#4_____-6
*3'__|___#3'____-6'____+9'
-6 @ -6' = 0
+1' @ +9' = +10'
*2 @ *2' = 0
#3' @ #4 @ #3' = #6' @ #4 = #2'
+10' @ #2'
Now, notice that the effect of
using difference of squares (A @ B)(A @ B') = A^2 @ (B^2)'
is that the + symbol is converted to #
+ @ + = #
also notice that (A')' = A
( +10' @ #2' )( +10' @ #2 ) = #100 @ #4' = #96
It is like flipping some of the subP2 signs
sumarizing, it is almost like squaring, but fliping some of the subP2 signs
( -1 @ +2 @ *3 )( -1' @ +2 @ *3' ) = +10' @ #2'
and, with the result, we do it again
( +10' @ #2' )( +10' @ #2 ) = #100 @ #4' = #96
other notation could be :
z = -1 @ +2 @ *3 @ #0
( z )( flippedSubP2(z) )( flippedSubP2( ( z )( flippedSubP2(z) ) ) ) = #96
we just may say
flip(z) = flippedSubP2(z)
flop(z) = flippedSubP2( ( z )( flippedSubP2(z) ) )
( z )( flip(z) )( flop(z) ) = #96
resuming
(#1)/z = ( (#1)( flip(z) )( flop(z) ) ) / ( ( z )( flip(z) )( flop(z) ) )
ok, we know ( z )( flip(z) )( flop(z) ), and proceed
to compute (#1)( flip(z) )( flop(z) ), this is calculate the numerator
_________-1'_____+2_____*3'
______|____________________
+10'__|__*10_____#20'___-30
#2____|__-2'_____+4_____*6'
-2' @ -30 = -28
+4 = +4
*10 @ *6' = *4
#20' = #20'
-28 @ +4 @ *4 @ #20'
Near the end of the process, we can undo the terms with ' notation
#20' = -20 @ +20 @ *20
-28 @ +4 @ *4 @ #20' = -28 @ +4 @ *4 @ -20 @ +20 @ *20 = -48 @ +24 @ *24
this is (#1)( flip(z) )( flop(z) ) = -48 @ +24 @ *24
resuming
(#1)/z = ( (#1)( flip(z) )( flop(z) ) ) / #96 )
(#1)/z = ( -48 @ +24 @ *24 ) / #96
(here you may notice that, if were the case and instead of #96 you got #96', symply multiply by (@1'/@1')
squashing #96 into the terms of the numerator
(#1)/z = -(48/96) @ +(24/96) @ *(24/96)
it is possible to reduce a bit
(#1)/z = -(1/2) @ +(1/4) @ *(1/4) = w
w : reciprocal of z
optionally
(#1)/z = -0.5 @ +0.25 @ 0.25
Ok, it is check time
_____________-1__________+2__________*3____
________|__________________________________
-(1/2)__|____+(1/2)______*(2/2)______#(3/2)
+(1/4)__|____*(1/4)______#(2/4)______-(3/4)
*(1/4)__|____#(1/4)______-(2/4)______+(3/4)
-(2/4) @ -(3/4) = -(5/4)
+(1/2) @ +(3/4) = +(5/4)
*(2/2) @ *(1/4) = *(5/4)
#(3/2) @ #(2/4) @ *(1/4) = #(9/4)
-(5/4) @ +(5/4) @ *(5/4) @ #(9/4) = #(4/4) = #1 ;)
It could exist more optimal ways, I do not know.
Some p4 are not invertible by this method like +3 @ #3, if you do a
difference of square, this number gets zeroed.
If p4 has a hyperbolic behaviour, or some other associated shape, you can
research the "consequences" of the geometry on the arithmetics.
Right now it has bugs, but, more or less. For low Pn you may try
if works in low composite numbers. If works you may try the reciprocal in p105 for a random p105 number
Perhaps, it would possible to leverage with a tiny bit of number-theoretic and matricial tools (
some family of matrices or other artifacts https://en.wikipedia.org/wiki/Permanent_(mathematics) )
For p15 one may start with :
z = -1 @ +2 @ *3 @ #4 @ &5 @ $6 @ μ7 @ €8 @ ŋ9 @ ð10 @ ¢11 @ ¶12 @ ß13 @ þ14 @ ħ15
z» = -1° @ +2°° @ *3°°° @ #4°°°° @ &5 @ $6° @ μ7°° @ €8°°° @ ŋ9°°°° @ ð10 @ ¢11° @ ¶12°°
@ ß13°°° @ þ14°°°° @ ħ15
z»» = -1°° @ +2°°°° @ *3° @ #4°°° @ &5 @ $6°° @ μ7°°°° @ €8° @ ŋ9°°° @ ð10 @ ¢11°° @
¶12°°°° @ ß13° @ þ14°°° @ ħ15
z»»» = -1°°° @ +2° @ *3°°°° @ #4°° @ &5 @ $6°°° @ μ7° @ €8°°°° @ ŋ9°° @ ð10 @ ¢11°°° @
¶12° @ ß13°°°° @ þ14°° @ ħ15
z»»»» = -1°°°° @ +2°°° @ *3°° @ #4° @ &5 @ $6°°°° @ μ7°°° @ €8°° @ ŋ9° @ ð10 @ ¢11°°°° @
¶12°°° @ ß13°° @ þ14° @ ħ15
T @ T° @ T°° @ T°°° @ T°°°° = 0
(A°)(B°°°) = AB°°°°
(z)(z»)(z»»)(z»»»)(z»»»») = ...
(@1/w) = (@1)(P)/(w)(P) = (@1)(P)(Q)/(w)(P)(Q) = ...etc
in p6
-1 $ +1 $ *1 $ #1 $ &1 = @1'
(-1 $ +1 $ *1 $ #1 $ &1)(-1 $ +1 $ *1 $ #1 $ &1) = @1
(@1)' = @1'
with @1' one may save even more space compared with (@1)'
> Hmmm....
I guess the (-55° $ -55°°) = -55' looks a bit odd, in the sense of leveraging the
degree symbol ° and pretend they well reflect a p3 dynamics
A notation (mainPnSign S1, auxPnSign S2 ) magnitude M
or (S1,S2)M or (-,-)55 $ (-,+)55
will not be precise because both spaces are not totally independent (in this
context, how they are used for the purposes of the thread)
perhaps another attempt of a more precise notation to express both spaces simultaneusly
(_n_,_o,_p,_q,_r,_s_)
(_g_,_h,_i,_j,_k,_m_)
(_a_,_b,_c,_d,_e,_f_)
→ - + * # & $ (mainP6)
↓ ° °° °°° (auxP3)
for example for p6 we have
( a,b,c,d,e,f) = -a @ +b @ *c @ #d @ &e @ $f
trying to maintain the style
(_0_,_0,_c,_0,_0,_f_)
(_0_,_b,_0,_0,_e,_0_) = -ºa @ +ººb @ *c @ #dº @ &eºº @ $f
(_a_,_0,_0,_d,_0,_0_)
(_0_,_0,_0,_0,_0,_0_)
(_a_,_b,_c,_d,_e,_f_) = -ºa @ +ºb @ *ºc @ #dº @ &eº @ $fº
(_0_,_0,_0,_0,_0,_0_)
(_0_,_a,_0,_0,_d,_0_)
(_0_,_b,_0,_0,_0,_0_) = +a @ +bºº @ +cº @ &d @ &eº
(_0_,_c,_0,_0,_e,_0_)
the auxP3(°) space and the mainPn space share the same subP2.
It may seems bizarre, but they do help in finding reciprocals.
For P4 ( 4 = 2 x 2 ) we will need subP2 (two times)
For P6 ( 6 = 2 x 3 ) we will need auxP3 and subP2
and presumably :
For P15 ( 15 = 3 x 5 ) we will need auxP3 and auxP5
For P30 ( 30 = 3 x 5 x 2 ) we will need auxP3, auxP5 and subP2
For P105 ( 105 = 3 x 7 x 5 ) we will need auxP3, auxP7 and auxP5
For P9 ( 9 = 3 x 3 ) we will need auxP3 (two times)
For P8 ( 8 = 2 x 2 x 2 ) we will need subP2 (three times)
Well you may rant a bit that if the only thing that have in common all the auxPn
in use and the mainPn is pair of rays (@ and @'), that will seem as a Real number
infiltration. To this I think that in some post you said that is not much the
real numbers by themselves, but to the degree that the way they are presented and
structured, do not allow looking into Pn (just citing from memory, may be vague)
> I think until you instantiate a Pm in a Pn then this is vague
better notation to indicate spaces with a shared elements.
Yes, rotational freedom although limited
I think an auxP3 is different compared to a subP3.
The main task, is to to ask is if I have a certain Pn number, what thing I have to
do to get @h or @k', that is like 90% of the procedure to the reciprocal
€ : eight
$ : six
if we have, we say, a p8 number, one may start with :
z = -5 € +11 € *7 € #23 € &3 € $19 € μ13 € @17
and using auxP2
þ(z) = -5' € +11 € *7' € #23 € &3' € $19 € μ13' € @17
or maybe using auxP4
þ(z) = -5° € +11°° € *7°°° € #23 € &3° € $19°° € μ13°°° € @17
T° @ T°° @ T°°° @ T = 0
T°°°° = T
(T°)(U°°) = (TU)°°°
I bet for just using auxPn primes, but you will have to check
then you compute ( z )( þ(z) ) = ...
It you can code a little calculator you may get to check such claims for a number
of cases. Pen, paper and hand calculator also have their limitations too.
And this is just may be one method to get reciprocals, you may find others.
>> the auxPn extra-signs dissapear (no mor º symbol)
> Hmmm....
structured in such a way that the result only yield terms with ' (in worst
case), but not terms with °, this is the result is exclusively in the mainPn space.
>> ( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ (@3224')^2 = @3332849
( *3705 $ @3224 )( *3705 $ @3224' ) = (*3705)^2 $ ( (@3224)^2 )' = @3332849
> Arggh. I know you are close here, but I'm not sure why you inverted the neutral part.
this is, difference of squares, translated to your notation is :
(A @ B)(A @ B') = A^2 @ (B^2)'
>> -432016 $ -611325' = -179309'
> -432016-179309 = -611325
> = 611325 check.
*50 $ (*30)' = *20
> I guess it will, but for me the answer is not so trivial as it must be for you
> Sadly google has adopted non fixed width font so we users of g suffer.
The character space accepted make things a bit challenging.
>Do you have it?
nopes
> Will it work in P4?
> anyway, choose:
> w = ( 0, 1, 2, 3 ) = @ 0 - 1 + 2 * 3
z = -1 @ +2 @ *3 @ #0
> w = ( 0, 1, 2, 3 ) = @ 0 - 1 + 2 * 3
þ(z) = -1' @ +2 @ *3' @ #0
_________-1_____+2_____*3
_____|___________________
-1'__|___+1'____*2'____#3'
+2___|___*2_____#4_____-6
*3'__|___#3'____-6'____+9'
-6 @ -6' = 0
+1' @ +9' = +10'
*2 @ *2' = 0
#3' @ #4 @ #3' = #6' @ #4 = #2'
+10' @ #2'
Now, notice that the effect of
using difference of squares (A @ B)(A @ B') = A^2 @ (B^2)'
is that the + symbol is converted to #
+ @ + = #
also notice that (A')' = A
( +10' @ #2' )( +10' @ #2 ) = #100 @ #4' = #96
It is like flipping some of the subP2 signs
sumarizing, it is almost like squaring, but fliping some of the subP2 signs
( -1 @ +2 @ *3 )( -1' @ +2 @ *3' ) = +10' @ #2'
and, with the result, we do it again
( +10' @ #2' )( +10' @ #2 ) = #100 @ #4' = #96
other notation could be :
z = -1 @ +2 @ *3 @ #0
( z )( flippedSubP2(z) )( flippedSubP2( ( z )( flippedSubP2(z) ) ) ) = #96
we just may say
flip(z) = flippedSubP2(z)
flop(z) = flippedSubP2( ( z )( flippedSubP2(z) ) )
( z )( flip(z) )( flop(z) ) = #96
resuming
(#1)/z = ( (#1)( flip(z) )( flop(z) ) ) / ( ( z )( flip(z) )( flop(z) ) )
ok, we know ( z )( flip(z) )( flop(z) ), and proceed
to compute (#1)( flip(z) )( flop(z) ), this is calculate the numerator
_________-1'_____+2_____*3'
______|____________________
+10'__|__*10_____#20'___-30
#2____|__-2'_____+4_____*6'
-2' @ -30 = -28
+4 = +4
*10 @ *6' = *4
#20' = #20'
-28 @ +4 @ *4 @ #20'
Near the end of the process, we can undo the terms with ' notation
#20' = -20 @ +20 @ *20
-28 @ +4 @ *4 @ #20' = -28 @ +4 @ *4 @ -20 @ +20 @ *20 = -48 @ +24 @ *24
this is (#1)( flip(z) )( flop(z) ) = -48 @ +24 @ *24
resuming
(#1)/z = ( (#1)( flip(z) )( flop(z) ) ) / #96 )
(#1)/z = ( -48 @ +24 @ *24 ) / #96
(here you may notice that, if were the case and instead of #96 you got #96', symply multiply by (@1'/@1')
squashing #96 into the terms of the numerator
(#1)/z = -(48/96) @ +(24/96) @ *(24/96)
it is possible to reduce a bit
(#1)/z = -(1/2) @ +(1/4) @ *(1/4) = w
w : reciprocal of z
optionally
(#1)/z = -0.5 @ +0.25 @ 0.25
Ok, it is check time
_____________-1__________+2__________*3____
________|__________________________________
-(1/2)__|____+(1/2)______*(2/2)______#(3/2)
+(1/4)__|____*(1/4)______#(2/4)______-(3/4)
*(1/4)__|____#(1/4)______-(2/4)______+(3/4)
-(2/4) @ -(3/4) = -(5/4)
+(1/2) @ +(3/4) = +(5/4)
*(2/2) @ *(1/4) = *(5/4)
#(3/2) @ #(2/4) @ *(1/4) = #(9/4)
-(5/4) @ +(5/4) @ *(5/4) @ #(9/4) = #(4/4) = #1 ;)
It could exist more optimal ways, I do not know.
Some p4 are not invertible by this method like +3 @ #3, if you do a
difference of square, this number gets zeroed.
If p4 has a hyperbolic behaviour, or some other associated shape, you can
research the "consequences" of the geometry on the arithmetics.
Right now it has bugs, but, more or less. For low Pn you may try
if works in low composite numbers. If works you may try the reciprocal in p105 for a random p105 number
Perhaps, it would possible to leverage with a tiny bit of number-theoretic and matricial tools (
some family of matrices or other artifacts https://en.wikipedia.org/wiki/Permanent_(mathematics) )
For p15 one may start with :
z = -1 @ +2 @ *3 @ #4 @ &5 @ $6 @ μ7 @ €8 @ ŋ9 @ ð10 @ ¢11 @ ¶12 @ ß13 @ þ14 @ ħ15
z» = -1° @ +2°° @ *3°°° @ #4°°°° @ &5 @ $6° @ μ7°° @ €8°°° @ ŋ9°°°° @ ð10 @ ¢11° @ ¶12°°
@ ß13°°° @ þ14°°°° @ ħ15
z»» = -1°° @ +2°°°° @ *3° @ #4°°° @ &5 @ $6°° @ μ7°°°° @ €8° @ ŋ9°°° @ ð10 @ ¢11°° @
¶12°°°° @ ß13° @ þ14°°° @ ħ15
z»»» = -1°°° @ +2° @ *3°°°° @ #4°° @ &5 @ $6°°° @ μ7° @ €8°°°° @ ŋ9°° @ ð10 @ ¢11°°° @
¶12° @ ß13°°°° @ þ14°° @ ħ15
z»»»» = -1°°°° @ +2°°° @ *3°° @ #4° @ &5 @ $6°°°° @ μ7°°° @ €8°° @ ŋ9° @ ð10 @ ¢11°°°° @
¶12°°° @ ß13°° @ þ14° @ ħ15
T @ T° @ T°° @ T°°° @ T°°°° = 0
(A°)(B°°°) = AB°°°°
(z)(z»)(z»»)(z»»»)(z»»»») = ...
(@1/w) = (@1)(P)/(w)(P) = (@1)(P)(Q)/(w)(P)(Q) = ...etc
Timothy Golden
Sep 3, 2021, 6:18:26 PM9/3/21
to
On Tuesday, August 31, 2021 at 7:04:36 PM UTC-4, michael Rodriguez wrote:
> although in a hurry...
Michael this is great news. I'm still digesting it though.
I still can't quite grasp the T° meaning though I do see the symmetry.
is there really no equivalent expression for
* 2 ° ?
Are these rules constructed exclusively for P6?
I'm pretty sure this is the case, or rather for the P3 factor of P6
It is a lovely intermediary as you say: "the auxPn extra-signs dissapear (no mor º symbol) "
leaving clean inverses... very nicely done.
I should not do so much thinking inline here, but I can't help it.
* 2 ° * 2 °° = * 2 ' = @ 2 - 2 + 2 # 2 & 2 [sum]
( * 2 ° )( * 2 °° ) = @ 4 [product]
where '&' is sign five.
In P3 let z = -1
z z = + 1 .
z' z' = ( + 1 * 1)( + 1 * 1 )
= - 1 + 1 + 1 * 1
= + 1 .
Hmmm. I see you are staying in the rudimentary notation as well.
Does it hold more generally?
I'll try a computer proof.
Not done yet.
What can you say about the form of the work that you are doing here?
The T° product and sum are their own thing?
Huh?
They are defined on a particular n though, right?
It's a T3 sum to zero as if the ° is a sign.
T3 products are behaving themselves as P3 too.
In P6 we have six components:
(a,b,c,d,e,f)
if you like; all magnitudes. ...
Well here I see you've posted, and I've tried to read all of your new post and am overwhelmed, but see that you are developing an interpretation that I think all will need who wish to understand this. I find it remarkable how productive the inverse is in all of this. I am also stupefied that such a simple product could lead to this much work in the reversal known as division. This is the conjugate, and the fanciness of the term is well deserved here in your work. I do appreciate all the work you've put into this next post. Notationally you've covered it enough different ways to be convincing even if poorly understood. The flip flop conclusion is really excellent. So cool that you can do the P4 problem too. Congratulations, Michael.
> although in a hurry...
Michael this is great news. I'm still digesting it though.
I still can't quite grasp the T° meaning though I do see the symmetry.
is there really no equivalent expression for
* 2 ° ?
Are these rules constructed exclusively for P6?
I'm pretty sure this is the case, or rather for the P3 factor of P6
It is a lovely intermediary as you say: "the auxPn extra-signs dissapear (no mor º symbol) "
leaving clean inverses... very nicely done.
I should not do so much thinking inline here, but I can't help it.
* 2 ° * 2 °° = * 2 ' = @ 2 - 2 + 2 # 2 & 2 [sum]
( * 2 ° )( * 2 °° ) = @ 4 [product]
where '&' is sign five.
>
> T = (s)m
>
> ( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
This one line above is outstanding all on its own.
> T = (s)m
>
> ( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
In P3 let z = -1
z z = + 1 .
z' z' = ( + 1 * 1)( + 1 * 1 )
= - 1 + 1 + 1 * 1
= + 1 .
Hmmm. I see you are staying in the rudimentary notation as well.
Does it hold more generally?
I'll try a computer proof.
Not done yet.
What can you say about the form of the work that you are doing here?
The T° product and sum are their own thing?
Huh?
They are defined on a particular n though, right?
It's a T3 sum to zero as if the ° is a sign.
T3 products are behaving themselves as P3 too.
In P6 we have six components:
(a,b,c,d,e,f)
if you like; all magnitudes. ...
Well here I see you've posted, and I've tried to read all of your new post and am overwhelmed, but see that you are developing an interpretation that I think all will need who wish to understand this. I find it remarkable how productive the inverse is in all of this. I am also stupefied that such a simple product could lead to this much work in the reversal known as division. This is the conjugate, and the fanciness of the term is well deserved here in your work. I do appreciate all the work you've put into this next post. Notationally you've covered it enough different ways to be convincing even if poorly understood. The flip flop conclusion is really excellent. So cool that you can do the P4 problem too. Congratulations, Michael.
Timothy Golden
Sep 4, 2021, 8:51:14 AM9/4/21
to
This was as well discovered and developed by Michael Rodriguez, presuming he stays with this identity.
The sorry looking code is mine, but the square of computation is his.
/*Wed 11 Nov 2020 10:12:35 AM ESTtpg- Lalo Torres has posted a quotient. Checking it here and todo: if it works put it in nsigned.C
*/
Poly LaloTorresReciprocal( Poly & z )
{
//cout << "\n in LaloTorresReciprocal( " << z << " ); \n";
Poly fac[z.n](z.n); // n polysigned factors compose the unsigned result
for( int j = 0; j < z.n; j ++ )
{
for( int i = 0; i < z.n; i++ )
{
fac[j][i*j] += z[i];
// the signs march upward as single, double, triple, ...
}
fac[j].Reduce();
}
/*/
for( int i = 0; i < z.n; i++ ) // print the factors for debug...
{
cout << "\nfac[" << i << "] = " << fac[i];
} cout << "\n";
/**/
Poly numerator(z.n);
numerator = fac[0];
for( int i = 2; i < z.n; i++ )
{ // does not include fac[1]
numerator = numerator * fac[i];
numerator.Reduce();
}
Poly denominator(z.n );
denominator = numerator * z;
denominator.Reduce();
Poly result( numerator );
result = result * ( 1.0 / denominator[0]);
int flag = 0;
for( int i = 1; i < z.n; i++ )
{ // verify denominator integrity
double d = 0.0001 * denominator[0]; // all other components should be relatively small if this code works
if( d < denominator[i] )
{ // failure detected.
// throw( denominator ) // this doesn't exist yet
flag = 1;
}
}
if( flag )
{
result.error.bits.Quotient = 1;
//cerr << "\n Error detected in " << __FILE__ << " line #" << __LINE__ ;
//cerr << " denominator : " << denominator;
}
result.Reduce();
cout << "\nLTR(z):"<< result; result.error.bits.Quotient = 0;//suppress error on deallocation
return result;
}
Timothy Golden
Sep 4, 2021, 10:07:31 AM9/4/21
to
On Saturday, September 4, 2021 at 8:51:14 AM UTC-4, Timothy Golden wrote:
> On Friday, September 3, 2021 at 6:18:26 PM UTC-4, Timothy Golden wrote:
> > On Tuesday, August 31, 2021 at 7:04:36 PM UTC-4, michael Rodriguez wrote:
> > > although in a hurry...
> > Michael this is great news. I'm still digesting it though.
snip
> On Friday, September 3, 2021 at 6:18:26 PM UTC-4, Timothy Golden wrote:
> > On Tuesday, August 31, 2021 at 7:04:36 PM UTC-4, michael Rodriguez wrote:
> > > although in a hurry...
> > Michael this is great news. I'm still digesting it though.
> > > ( (s1)m' )( (s2)m2' ) = (s1@s2)m1m2
> > This one line above is outstanding all on its own.
> > In P3 let z = -1
> > z z = + 1 .
> > z' z' = ( + 1 * 1)( + 1 * 1 )
> > = - 1 + 1 + 1 * 1
> > = + 1 .
> > Hmmm. I see you are staying in the rudimentary notation as well.
> > Does it hold more generally?
> > I'll try a computer proof.
> > Not done yet.
snip
> > This one line above is outstanding all on its own.
> > In P3 let z = -1
> > z z = + 1 .
> > z' z' = ( + 1 * 1)( + 1 * 1 )
> > = - 1 + 1 + 1 * 1
> > = + 1 .
> > Hmmm. I see you are staying in the rudimentary notation as well.
> > Does it hold more generally?
> > I'll try a computer proof.
> > Not done yet.
Looks like it's general; on all z in Pn:
z1' z2' = z1 z2.
How do we prove this?
It must be extremely simple.
I do not find this to be intuitive.
The simple code that shows it:
void InverseProductStudy()
{ // look into Michael Rodriguez inverse behavior (inv(s1x1))(inv(s2x2))=s1s2x1x2
for( int n = 1; n < 20; n++ )
{ Poly z1(n); z1.Random();
Poly z2(n); z2.Random();
Poly z3(n); z3 = (-z1)*(-z2);
Poly z4(n); z4 = z1 * z2;
Poly diff(n); diff = z4 - z3;
cout << "\n n: " << n << " diff: " << diff.Magnitude();
}
}
yielding:
tatrix@deb9wd2tb:~$ bin/inversestudy
n: 1 diff: 0
n: 2 diff: 0
n: 3 diff: 3.67172e-17
n: 4 diff: 5.31479e-17
n: 5 diff: 6.7987e-17
n: 6 diff: 4.11682e-17
n: 7 diff: 2.794e-16
n: 8 diff: 1.798e-17
n: 9 diff: 1.28651e-17
n: 10 diff: 2.15789e-16
n: 11 diff: 5.07062e-17
n: 12 diff: 9.85758e-17
n: 13 diff: 4.27742e-17
n: 14 diff: 6.71803e-16
n: 15 diff: 1.18223e-16
n: 16 diff: 5.34754e-16
n: 17 diff: 4.44089e-16
n: 18 diff: 1.21066e-17
n: 19 diff: 4.39463e-17
tatrix@deb9wd2tb:~$
and now more verbosely:
void InverseProductStudy()
{ // look into Michael Rodriguez inverse behavior (inv(s1x1))(inv(s2x2))=s1s2x1x2
cout.precision(2);
for( int n = 1; n < 10; n++ )
{ Poly z1(n); z1.Random();
Poly z2(n); z2.Random();
Poly z1i(n); z1i = - z1;
Poly z2i(n); z2i = - z2;
Poly z3(n); z3 = z1i * z2i;
Poly z4(n); z4 = z1 * z2;
Poly diff(n); diff = z4 - z3;
z3.Reduce();
z4.Reduce();
cout << "\n\n\nInverseProductStudy n = " << n << " diff: " << diff.Magnitude()
<< "\n\n z1:" << z1
<< "\n\n z1i:" << z1i
<< "\n\n z2:" << z2
<< "\n\n z2i:" << z2i
<< "\n\n z1z2:" << z4 void InverseProductStudy()
{ // look into Michael Rodriguez inverse behavior (inv(s1x1))(inv(s2x2))=s1s2x1x2
cout.precision(2);
for( int n = 1; n < 10; n++ )
{ Poly z1(n); z1.Random();
Poly z2(n); z2.Random();
Poly z1i(n); z1i = - z1;
Poly z2i(n); z2i = - z2;
Poly z3(n); z3 = z1i * z2i;
Poly z4(n); z4 = z1 * z2;
Poly diff(n); diff = z4 - z3;
z3.Reduce();
z4.Reduce();
cout << "\n\n\nInverseProductStudy n = " << n << " diff: " << diff.Magnitude()
<< "\n\n z1:" << z1
<< "\n\n z1i:" << z1i
<< "\n\n z2:" << z2
<< "\n\n z2i:" << z2i
<< "\n\n z1z2:" << z4
<< "\n\n z1iz2i:" << z3;
}
}
<< "\n\n z1iz2i:" << z3;
}
}
tatrix@deb9wd2tb:~$ bin/inversestudy
InverseProductStudy n = 1 diff: 0
z1:[P1 0 ]
z1i:[P1 -0 ]
z2:[P1 0 ]
z2i:[P1 -0 ]
z1z2:[P1 0 ]
z1iz2i:[P1 0 ]
InverseProductStudy n = 2 diff: 0
z1:[P2 0.42, 0 ]
z1i:[P2 0, 0.42 ]
z2:[P2 0.88, 0 ]
z2i:[P2 0, 0.88 ]
z1z2:[P2 0.37, 0 ]
z1iz2i:[P2 0.37, 0 ]
InverseProductStudy n = 3 diff: 1.3e-16
z1:[P3 0.65, 1.1, 0 ]
z1i:[P3 0.42, 0, 1.1 ]
z2:[P3 0, 0.56, 0.96 ]
z2i:[P3 0.96, 0.4, 0 ]
z1z2:[P3 0.66, 0, 0.86 ]
z1iz2i:[P3 0.66, 0, 0.86 ]
InverseProductStudy n = 4 diff: 0
z1:[P4 0.033, 0.026, 0.029, 0 ]
z1i:[P4 0, 0.0069, 0.0032, 0.033 ]
z2:[P4 0.42, 0, 0.87, 0.7 ]
z2i:[P4 0.45, 0.87, 0, 0.18 ]
z1z2:[P4 0.026, 0, 0.0095, 0.014 ]
z1iz2i:[P4 0.026, 0, 0.0095, 0.014 ]
InverseProductStudy n = 5 diff: 6.5e-18
z1:[P5 0.083, 0.074, 0.062, 0, 0.086 ]
z1i:[P5 0.0037, 0.013, 0.024, 0.086, 0 ]
z2:[P5 0.069, 0.28, 0, 0.26, 0.077 ]
void InverseProductStudy()
{ // look into Michael Rodriguez inverse behavior (inv(s1x1))(inv(s2x2))=s1s2x1x2
cout.precision(2);
for( int n = 1; n < 10; n++ )
{ Poly z1(n); z1.Random();void InverseProductStudy()
{ // look into Michael Rodriguez inverse behavior (inv(s1x1))(inv(s2x2))=s1s2x1x2
cout.precision(2);
for( int n = 1; n < 10; n++ )
{ Poly z1(n); z1.Random();
Poly z2(n); z2.Random();
Poly z1i(n); z1i = - z1;
Poly z2i(n); z2i = - z2;
Poly z3(n); z3 = z1i * z2i;
Poly z4(n); z4 = z1 * z2;
Poly diff(n); diff = z4 - z3;
z3.Reduce();
z4.Reduce();
cout << "\n\n\nInverseProductStudy n = " << n << " diff: " << diff.Magnitude()
<< "\n\n z1:" << z1
<< "\n\n z1i:" << z1i
<< "\n\n z2:" << z2
<< "\n\n z2i:" << z2i
<< "\n\n z1z2:" << z4
<< "\n\n z1iz2i:" << z3;
}
}
Poly z2(n); z2.Random();
Poly z1i(n); z1i = - z1;
Poly z2i(n); z2i = - z2;
Poly z3(n); z3 = z1i * z2i;
Poly z4(n); z4 = z1 * z2;
Poly diff(n); diff = z4 - z3;
z3.Reduce();
z4.Reduce();
cout << "\n\n\nInverseProductStudy n = " << n << " diff: " << diff.Magnitude()
<< "\n\n z1:" << z1
<< "\n\n z1i:" << z1i
<< "\n\n z2:" << z2
<< "\n\n z2i:" << z2i
<< "\n\n z1z2:" << z4
<< "\n\n z1iz2i:" << z3;
}
}
z2i:[P5 0.21, 0, 0.28, 0.017, 0.2 ]
z1z2:[P5 0.02, 0.0013, 0.016, 0.014, 0 ]
z1iz2i:[P5 0.02, 0.0013, 0.016, 0.014, 0 ]
InverseProductStudy n = 6 diff: 1e-16
z1:[P6 0.26, 0, 0.032, 0.34, 0.35, 0.14 ]
z1i:[P6 0.092, 0.35, 0.32, 0.011, 0, 0.21 ]
z2:[P6 0.33, 0.087, 0.19, 0.17, 0, 0.38 ]
z2i:[P6 0.051, 0.3, 0.19, 0.22, 0.38, 0 ]
z1z2:[P6 0.1, 0, 0.095, 0.17, 0.087, 0.13 ]
z1iz2i:[P6 0.1, 0, 0.095, 0.17, 0.087, 0.13 ]
InverseProductStudy n = 7 diff: 5e-16
z1:[P7 0.13, 0.44, 0.99, 0.87, 0, 0.79, 0.5 ]
z1i:[P7 0.86, 0.55, 0, 0.12, 0.99, 0.2, 0.49 ]
z2:[P7 0.9, 0.62, 0.072, 0.19, 0.47, 0.39, 0 ]
z2i:[P7 0, 0.28, 0.82, 0.7, 0.42, 0.51, 0.9 ]
z1z2:[P7 0.33, 0.051, 0.69, 1, 0, 0.27, 0.79 ]
z1iz2i:[P7 0.33, 0.051, 0.69, 1, 0, 0.27, 0.79 ]
InverseProductStudy n = 8 diff: 2.3e-16
z1:[P8 0.22, 0.26, 0.31, 0.26, 0.24, 0.3, 0.036, 0 ]
z1i:[P8 0.097, 0.052, 0, 0.054, 0.077, 0.014, 0.28, 0.31 ]
z2:[P8 0.36, 0.39, 0.66, 0.51, 0.96, 0, 0.91, 0.51 ]
z2i:[P8 0.6, 0.57, 0.3, 0.45, 0, 0.96, 0.045, 0.45 ]
z1z2:[P8 0.16, 0.14, 0, 0.15, 0.18, 0.06, 0.18, 0.19 ]
z1iz2i:[P8 0.16, 0.14, 0, 0.15, 0.18, 0.06, 0.18, 0.19 ]
InverseProductStudy n = 9 diff: 8.3e-17
z1:[P9 0.15, 0.2, 0.12, 0, 0.2, 0.034, 0.2, 0.23, 0.099 ]
z1i:[P9 0.076, 0.026, 0.1, 0.23, 0.028, 0.19, 0.029, 0, 0.13 ]
z2:[P9 0.12, 0.37, 0, 0.58, 0.17, 0.64, 0.37, 0.67, 0.52 ]
z2i:[P9 0.54, 0.29, 0.67, 0.089, 0.49, 0.023, 0.29, 0, 0.15 ]
z1z2:[P9 0.084, 0, 0.053, 0.09, 0.058, 0.17, 0.021, 0.12, 0.0028 ]
z1iz2i:[P9 0.084, 0, 0.053, 0.09, 0.058, 0.17, 0.021, 0.12, 0.0028 ]
tatrix@deb9wd2tb:~$
michael Rodriguez
Sep 9, 2021, 5:29:33 PM9/9/21
to
well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
Timothy Golden
Sep 19, 2021, 9:47:48 AM9/19/21
to
On Thursday, September 9, 2021 at 5:29:33 PM UTC-4, michael Rodriguez wrote:
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
I am trying to get back to your division algorithm and going through your previous posts in this thread regarding your new modulus which still feel foreign to me. I am sorry that I am not keeping up. I should be able to implement your versor in code and graph out the shape. It would be pretty cool if it does come out to Gabriel's horn. With the possibility of negative values creeping in with large a and c components I wonder if this is turning into a P2 modulus. Likewise back in your division algorithm and the new º operator I wonder if you are reaching into what Descartes called imaginary roots, yet you are coming at it from another angle. You've actually got them, whereas he did not. If a value z in P6 is given can we state zº? So I give you:
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
z = - 1 + 2 * 3 in P6
and I ask you what is zº?
I can say confidently that I have absorbed the rules
z @ zº @ zºº = 0, zººº = z .
So I can make a little bit of progress:
zº @ zºº = z'= - 2 + 1 # 3 & 3 @ 3 (where '&' indicates sign five)
and I guess this is as far as I feel confident going. Possibly you will come back with a firm value, but if you do not then to what degree have you entered into an imaginary form here? Descartes managed to do this even on his unsigned algebraic numbers while he was studying polynomials and their roots. Likewise he bumped into negative numbers though he did not allow them. By the term 'imaginary' here I don't mean to tie it directly to complex numbers, but more broadly as Descartes did.
I know you've already worked very hard to express this and I don't mean to dodge all that. I do believe in the work. I'll be reviewing it again. Maybe I'll try taking manual notes; trying to absorb it another way. Somehow it will have to translate into code.
Ross A. Finlayson
Sep 19, 2021, 10:06:24 PM9/19/21
to
This is pretty interesting, please carry on with this.
Timothy Golden
Sep 20, 2021, 11:48:27 AM9/20/21
to
On Thursday, September 9, 2021 at 5:29:33 PM UTC-4, michael Rodriguez wrote:
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
So I've applied your algorithm onto the progressive value
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
Totally following your algorithm with myself making several errors along the way.
Still, the computations are fairly clean and easy to double check.
I did cheat a bit and used my computer for some of the product computations.
Strange that this particular value is the offset for packing signons as well, but for the scaling factor.
There must be some reason for this but it is beyond me.
I think we bumped into this before too.
The progressive value is just a simple value that would seem to be arbitrary or at least this is why I use it.
It is easy to code in Pn.
http://bandtechnology.com/PolySigned/Lattice/Lattice.html
I'll have to check its generality.
Bravo, Michael !
I'm not claiming a full understanding still, but I'm starting to get the gist... maybe...
Timothy Golden
Sep 21, 2021, 10:43:40 AM9/21/21
to
On Monday, September 20, 2021 at 11:48:27 AM UTC-4, Timothy Golden wrote:
> On Thursday, September 9, 2021 at 5:29:33 PM UTC-4, michael Rodriguez wrote:
> > well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
> So I've applied your algorithm onto the progressive value
> z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
> 1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
It happens that the value
> On Thursday, September 9, 2021 at 5:29:33 PM UTC-4, michael Rodriguez wrote:
> > well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
> So I've applied your algorithm onto the progressive value
> z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
> 1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
MU @ NU'
is ( 0, 2, 1, 1, 1, ... )
where MU is minus unity (0,1,0,0,0...) and NU is neutral unity (1,0,0,0,...)
(z)(0,2,1,1,1...)
= - z @ z'
which seems sort of strange. It's quite lively yet its just a constant.
Sorry if this is a digression.
I was reviewing the previous quotient code and just had to work on this detail.
Could it be that sometimes finding a gold nugget is sheer luck?
This output shows generally in Pn(up to 9 anyways and I did just look out to 99):
z1:[P1 0 ]
z2:[P1 0 ]
z1z2:[P1 0 ]
z1:[P2 0, 1 ]
z2:[P2 0, 2 ]
z1z2:[P2 2, 0 ]
z1:[P3 0, 1, 2 ]
z2:[P3 0, 2, 1 ]
z1z2:[P3 3, 0, 0 ]
z1:[P4 0, 1, 2, 3 ]
z2:[P4 0, 2, 1, 1 ]
z1z2:[P4 4, 0, 0, 0 ]
z1:[P5 0, 1, 2, 3, 4 ]
z2:[P5 0, 2, 1, 1, 1 ]
z1z2:[P5 5, 0, 0, 0, 0 ]
z1:[P6 0, 1, 2, 3, 4, 5 ]
z2:[P6 0, 2, 1, 1, 1, 1 ]
z1z2:[P6 6, 0, 0, 0, 0, 0 ]
z1:[P7 0, 1, 2, 3, 4, 5, 6 ]
z2:[P7 0, 2, 1, 1, 1, 1, 1 ]
z1z2:[P7 7, 0, 0, 0, 0, 0, 0 ]
z1:[P8 0, 1, 2, 3, 4, 5, 6, 7 ]
z2:[P8 0, 2, 1, 1, 1, 1, 1, 1 ]
z1z2:[P8 8, 0, 0, 0, 0, 0, 0, 0 ]
z1:[P9 0, 1, 2, 3, 4, 5, 6, 7, 8 ]
z2:[P9 0, 2, 1, 1, 1, 1, 1, 1, 1 ]
z1z2:[P9 9, 0, 0, 0, 0, 0, 0, 0, 0 ]
michael Rodriguez
Sep 21, 2021, 3:39:04 PM9/21/21
to
You may think that the space that that contains an auxPm and mainPn is a bigger
space. Since both share the same identity element ( and also the additive
inverse of the identity element) you may call this bigger space a 'siames
space', where auxPm only surfaces its head in the reciprocal procedure.
It is correct when you apply º to the whole number, although, the ° is
intended to be distributed in some terms of a Pn number to generate
its 'conjugates'. Notice that these 'conjugates' are in the the 'siames space'.
For example in the case of a P15 number, the 15 (of p15) has factors three and
five. One may start with the three, this is, one builds the siames
space p15_auxP3, find the two 'conjugates' in p15_auxP3, multiply those
conjugates to the p15 number. Then, some number of terms of the p15 number
with certain signs should dissapear. After we build a siames space p15_auxp5,
find the four conjugates of the remaining p15 number. Notice that also one
could starts with p15_auxP5 and then p15_auxP3.
WIth more examples will be more easy. In p9, suppose we have the number
z = -3 @ +5 @ *1 @ #2 @ &1 @ $2 @ μ2 @ €1 @ ŋ3
the identity element is ŋ1 (the ŋ symbol is similar to the n of nine)
since 9 (in p9) has factors three and three, the order does not matter.
In the siames space p9_auxP3, we build the 'conjugates' of z
z» = -3° @ +5°° @ *1 @ #2° @ &1°° @ $2 @ μ2° @ €1°° @ ŋ3
z»» = -3°° @ +5° @ *1 @ #2°° @ &1° @ $2 @ μ2°° @ €1° @ ŋ3
Instead of z» and z»» you may call them flip_1(z) and flip_2(z). I wouldn't name
these 'conjugates' as z° and z°° because that will conflate notation. For
example, something analogous happens if you have a p4 number x = -a+b*c#d, if
you write *x, one could infer that *x is actually (*1)(x)
Ok, we have z, but we actually start with z in the denominator, we want to know
if ŋ1/z is equivalent to something useful that we can manipulate as any other
p9 number, let's see
ŋ1 / ( -3 @ +5 @ *1 @ #2 @ &1 @ $2 @ μ2 @ €1 @ ŋ3 ) = ŋ1 / (z)
We want to multiply the denominator with (z»)(z»»), this is, we want
(z)(z»)(z»»)
but usually, we can't multiply the denominator without compesating that in the
numerator also, this is, if you multiply by x in the denominator, you multiply
by x in the numerator also.
ŋ1 / z = (ŋ1)(z»)(z»») / (z)(z»)(z»») = (z»)(z»») / (z)(z»)(z»»)
Presumably, getting z multiplied by (z»)(z»») will get us a p9 number with less
terms.
Ok, we may start computing the result of (z»)(z»»)
________-3°___+5°°__*1____#2°___&1°°__$2____μ2°___€1°°__ŋ3____
_____|________________________________________________________
-3°°_|__+9____*15°__#3°°__&6____$3°___μ6°°__€6____ŋ3°___-9°°__
+5°__|__*15°°_#25___&5°___$10°°_μ5____€10°__ŋ10°°_-5____+15°__
*1___|__#3°___&5°°__$1____μ2°___€1°°__ŋ2____-2°___+1°°__*3____
#2°°_|__&6____$10°__μ2°°__€4____ŋ2°___-4°°__+4____*2°___#6°°__
&1°__|__$3°°__μ5____€1°___ŋ2°°__-1____+2°___*2°°__#1____&3°___
$2___|__μ6°___€10°°_ŋ2____-4°___+2°°__*4____#4°___&2°°__$6____
μ2°°_|__€6____ŋ10°__-2°°__+4____*2°___#4°°__&4____$2°___μ6°°__
€1°__|__ŋ3°°__-5____+1°___*2°°__#1____&2°___$2°°__μ1____€3°___
ŋ3___|__-9°___+15°°_*3____#6°___&3°°__$6____μ6°___€3°°__ŋ9____
grouping terms according primary sign (mainPn sign)
-9° @ -5 @ -2°° @ -4° @ -1 @ -4°° @ -2° @ -5 @ -9°°
+9 @ +15°° @ +1° @ +4 @ +2°° @ +2° @ +4 @ +1°° @ +15°
*15°° @ *15° @ *3 @ *2°° @ *2° @ *4 @ *2°° @ *2° @ *3
#3° @ #25 @ #3°° @ #6° @ #1 @ #4°° @ #4° @ #1 @ #6°°
&6 @ &5°° @ &5° @ &6 @ &3°° @ &2° @ &4 @ &2°° @ &3°
$3°° @ $10° @ $1 @ $10°° @ $3° @ $6 @ $2°° @ $2° @ $6
μ6° @ μ5 @ μ2°° @ μ2° @ μ5 @ μ6°° @ μ6° @ μ1 @ μ6°°
€6 @ €10°° @ €1° @ €4 @ €1°° @ €10° @ €6 @ €3°° @ €3°
ŋ3°° @ ŋ10° @ ŋ2 @ ŋ2°° @ ŋ2° @ ŋ2 @ ŋ10°° @ ŋ3° @ ŋ9
re-arranging terms
(-9° @ -9°°) @ (-2°° @ -2°) @ (-4° @ -4°°) @ (-5 @ -5 @ -1 )
(+15°° @ +15°) @ (+1° @ +1°°) @ (+2°° @ +2°) @ (+9 @ +4 @ +4)
(*15°° @ *15°) @ (*2°° @ *2°) @ (*2° @ *2°°) @ (*3 @ *3 @ *4)
(#3° @ #3°°) @ (#6° @ #6°°) @ (#4°° @ #4°) @ (#25 @ #1 @ #1)
(&5°° @ &5°) @ (&3°° @ &3°) @ (&2° @ &2°°) @ (&6 @ &6 @ &4)
($3°° @ $3°) @ ($10° @ $10°°) @ ($2°° @ $2°) @ ($1 @ $6 @ $6)
(μ6° @ μ6°°) @ (μ2°° @ μ2°) @ (μ6° @ μ6°°) @ (μ5 @ μ5 @ μ1)
(€10°° @ €10°) @ (€1° @ €1°°) @ (€3°° @ €3°) @ (€6 @ €6 @ €4)
(ŋ3°° @ ŋ3°) @ (ŋ10° @ ŋ10°°) @ (ŋ2°° @ ŋ2°) @ (ŋ2 @ ŋ2 @ ŋ9)
further re-arranging terms
-9' @ -2' @ -4' @ -11 = -15' @ -11 = -4'
+15' @ +1' @ +2' @ +17 = +18' @ +17 = +1'
*15' @ *2' @ *2' @ *10 = *19' @ *10 = *9'
#3' @ #6' @ #4' @ #27 = #13' @ #27 = #14
&5' @ &3' @ &2' @ &16 = &10' @ &16 = &6
$3' @ $10' @ $2' @ $13 = $15' @ $13 = $2'
μ6' @ μ2' @ μ6' @ μ11 = μ14' @ μ11 = μ3'
€10' @ €1' @ €3' @ €16 = €14' @ @ €16 = €2
ŋ3' @ ŋ10' @ ŋ2' @ ŋ13 = ŋ15' @ ŋ13 = ŋ2'
(z»)(z»») = -4' @ +1' @ *9' @ #14 @ &6 @ $2' @ μ3' @ €2 @ ŋ2'
so far we have
(z»)(z»») / (z)(z»)(z»») = (-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') / (z)(z»)(z»»)
(z)(z»)(z»») = (-3+5*1#2&1$2μ2€1ŋ3)(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') = ?
_______-4'____+1'____*9'____#14____&6_____$2'____μ3'____€2_____ŋ2'____
____|_________________________________________________________________
-3__|__+12'___*3'____#27'___&42____$18____μ6'____€9'____ŋ6_____-6'____
+5__|__*20'___#5'____&45'___$70____μ30____€10'___ŋ15'___-10____+10'___
*1__|__#4'____&1'____$9'____μ14____€6_____ŋ2'____-3'____+2_____*2'____
#2__|__&8'____$2'____μ18'___€28____ŋ12____-4'____+6'____*4_____#4'____
&1__|__$4'____μ1'____€9'____ŋ14____-6_____+2'____*3'____#2_____&2'____
$2__|__μ8'____€2'____ŋ18'___-28____+12____*4'____#6'____&4_____$4'____
μ2__|__€8'____ŋ2'____-18'___+28____*12____#4'____&6'____$4_____μ4'____
€1__|__ŋ4'____-1'____+9'____*14____#6_____&2'____$3'____μ2_____€2'____
ŋ3__|__-12'___+3'____*27'___#42____&18____$6'____μ9'____€6_____ŋ6'____
-12' @ -1' @ -18' @ -28 @ -6 @ -4' @ -3' @ -10 @ -6'
+12' @ +3' @ +9' @ +28 @ +12 @ +2' @ +6' @ +2 @ +10'
*20' @ *3' @ *27' @ *14 @ *12 @ *4' @ *3' @ *4 @ *2'
#4' @ #5' @ #27' @ #42 @ #6 @ #4' @ #6' @ #2 @ #4'
&8' @ &1' @ &45' @ &42 @ &18 @ &2' @ &6' @ &4 @ &2'
$4' @ $2' @ $9' @ $70 @ $18 @ $6' @ $3' @ $4 @ $4'
μ8' @ μ1' @ μ18' @ μ14 @ μ30 @ μ6' @ μ9' @ μ2 @ μ4'
€8' @ €2' @ €9' @ €28 @ €6 @ €10' @ €9' @ €6 @ €2'
ŋ4' @ ŋ2' @ ŋ18' @ ŋ14 @ ŋ12 @ ŋ2' @ ŋ15' @ ŋ6 @ ŋ6'
(-12' @ -1' @ -18' @ -4' @ -3' @ -6') @ (-28 @ -6 @ -10) = -44' @ -44 = 0
(+12' @ +3' @ +9' @ +2' @ +6' @ +10') @ (+28 @ +12 @ +2) = +42' @ +42 = 0
(*20' @ *3' @ *27' @ *4' @ *3' @ *2') @ (*14 @ *12 @ *4) = *59' @ *30 = *29'
(#4' @ #5' @ #27' @ #4' @ #6' @ #4') @ (#42 @ #6 @ #2) = #50' @ 50 = 0
(&8' @ &1' @ &45' @ &2'@ &6' @ &2') @ (&42 @ &18 @ &4) = &64' @ &64 = 0
($4' @ $2' @ $9' @ $6' @ $3' @ $4') @ ($70 @ $18 @ $4) = $28' @ $92 = $64
(μ8' @ μ1' @ μ18' @ μ6' @ μ9' @ μ4') @ (μ14 @ μ30 @ μ2) = μ46' @ μ46 = 0
(€8' @ €2' @ €9' @ €10' @ €9' €2') @ (€28 @ €6 @ €6) = €40' @ €40 = 0
(ŋ4' @ ŋ2' @ ŋ18' @ ŋ2' @ ŋ15' @ ŋ6') @ (ŋ14 @ ŋ12 @ ŋ6) = ŋ47' @ ŋ32 = ŋ15'
(z)(z»)(z»») = *29' @ $64 @ ŋ15'
so far, we have
(z»)(z»») / (z)(z»)(z»») = (-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') / (*29' $64 ŋ15')
notice that in the denominator remain only terms with signs * $ ŋ
so far, we have finished the firt part of the procedure for this p9 number,
and we continue to the second part
we may rename (z)(z»)(z»») as w to save some chars
w = (*29' $64 ŋ15')
Since the second factor (of 9 in p9) is also three, we access again p9_auxP3
to get the two 'conjugates' of w
w~ = *29'° @ $64°° @ ŋ15'
w~~ = *29'°° @ $64° @ ŋ15'
we want to multiply the denominator with (w~)(w~~)
(z»)(z»») / w = (z»)(z»»)(w~)(w~~) / (w)(w~)(w~~)
ok, let's start computing (w~)(w~~)
(*29'° @ $64°° @ ŋ15')(*29'°° @ $64° @ ŋ15') = ??
___________*29'°______$64°°______ŋ15'_______
________|___________________________________
*29'°°__|__$841_______ŋ1856'°____*435°°_____
$64°____|__ŋ1856'°°___*4096______$960'°_____
ŋ15'____|__*435°______$960'°°____ŋ225_______
arranging a bit
*4096 @ (*435° @ *435°°) = *4096 @ *435' = *3661
$841 @ ($960'°° @ $960'°) = $841 @ $960 = $1801
ŋ225 @ (ŋ1856'°° @ ŋ1856'°) = ŋ225 @ ŋ1856 = ŋ2081
(w~)(w~~) = *3661 @ $1801 @ ŋ2081
now we can compute (w)(w~)(w~~)
(*29' $64 ŋ15')(*3661 @ $1801 @ ŋ2081) = ???
____________*3661______$1801______ŋ2081______
_________|___________________________________
*29'_____|__$106169'___ŋ52229'____*60349'____
$64______|__ŋ234304____*115264____$133184____
ŋ15'_____|__*54915'____$27015'____ŋ31215'____
arranging
(*54915' @ *60349') @ *115264 = *115264' @ *115264 = 0
($106169' @ $27015') @ $133184 = $133184' @ $$133184 = 0
(ŋ52229' @ ŋ31215') @ ŋ234304 = ŋ150860
(w)(w~)(w~~) = ŋ150860
so far we have
(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2')(*3661 @ $1801 @ ŋ2081) / (ŋ150860)
(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2')(*3661 @ $1801 @ ŋ2081) = ????
________*3661______$1801______ŋ2081______
_____|___________________________________
-4'__|__#14644'____μ7204'_____-8324'_____
+1'__|__&3661'_____€1801'_____+2081'_____
*9'__|__$32949'____ŋ16209'____*18729'____
#14__|__μ51254_____-25214_____#29134_____
&6___|__€21966_____+10806_____&12486_____
$2'__|__ŋ7322'_____*3602'_____$4162'_____
μ3'__|__-10983'____#5403'_____μ6243'_____
€2___|__+7322______&3602______€4162______
ŋ2'__|__*7322'_____$3602'_____ŋ4162'_____
(-10983' @ -8324') @ -25214 = -5907
+2081' @ (+10806 @ +7322) = +16047
(*7322' @ *3602' @ *18729') = *29653'
(#14644' @ #5403') @ #29134 = #9087
&3661' @ (&3602 @ &12486) = &12427
($32949' @ $3602' @ $4162') = $40713'
(μ7204' @ μ6243') @ μ51254 = μ37807
€1801' @ (€21966 @ €4162) = €24327
((ŋ7322' @ ŋ16209' @ ŋ4162') = ŋ27693'
the terms with ' have to be converted
*29653' = -29653 +29653 *0 #29653 &29653 $29653 μ29653 €29653 ŋ29653
$40713' = -40713 +40713 *40713 #40713 &40713 $0 μ40713 €40713 ŋ40713
ŋ27693' = -27693 +27693 *27693 #27693 &27693 $27693 μ27693 €27693 ŋ0
further arranging terms
-5907 @ -29653 @ -40713 @ -27693 = -103966
+16047 @ +29653 @ +40713 @ +27693 = +114106
*0 @ *40713 @ *27693 = *68406
#9087 @ #29653 @ #40713 @ #27693 = #107146
&12427 @ &29653 @ &40713 @ &27693 = &110486
$29653 @ $0 @ $27693 = $57346
μ37807 @ μ29653 @ μ40713 @ μ27693 = μ135866
€24327 @ €29653 @ €40713 @ €27693 = €122386
ŋ29653 @ ŋ40713 @ ŋ0 = ŋ70366
at last, we have that (z»)(z»»)(w~)(w~~) is equal to
-103966 +114106 *68406 #107146 &110486 $57346 μ135866 €122386 ŋ70366
ok, we have
(-103966 +114106 *68406 #107146 &110486 $57346 μ135866 €122386 ŋ70366 )/ŋ150860
if you use q = ŋ150860 then
ŋ1 / z = -103966/q @ +114106/q @ *68406/q @ #107146/q @ &110486/q @ $57346/q @ μ135866/q @ €122386/q @ ŋ70366/q
end of procedure
check time
(z)(ŋ1 / z) = ?????
due to lack of space the /q in each term is omitted, to prioritize compatibility
of format, or to prevent the tables get a space-time distorsione
___-103966__+114106__*68406__#107146__&110486__$57346__μ135866__€122386__ŋ70366_
__|_____________________________________________________________________________
-3|+311898__*342318__#205218_&321438__$331458__μ172038_€407598__ŋ367158__-211098
+5|*519830__#570530__&342030_$535730__μ552430__€286730_ŋ679330__-611930__+351830
*1|#103966__&114106__$68406__μ107146__€110486__ŋ57346__-135866__+122386__*70366_
#2|&207932__$228212__μ136812_€214292__ŋ220972__-114692_+271732__*244772__#140732
&1|$103966__μ114106__€68406__ŋ107146__-110486__+57346__*135866__#122386__&70366_
$2|μ207932__€228212__ŋ136812_-214292__+220972__*114692_#271732__&244772__$140732
μ2|€207932__ŋ228212__-136812_+214292__*220972__#114692_&271732__$244772__μ140732
€1|ŋ103966__-114106__+68406__*107146__#110486__&57346__$135866__μ122386__€70366_
ŋ3|-311898__+342318__*205218_#321438__&331458__$172038_μ407598__€367158__ŋ211098
arranging by sign
-311898 -114106 -136812 -214292 -110486 -114692 -135866 -611930 -211098 = -1961180
+311898 +342318 +68406 +214292 +220972 +57346 +271732 +122386 +351830 = +1961180
*519830 *342318 *205218 *107146 *220972 *114692 *135866 *244772 *70366 = *1961180
#103966 #570530 #205218 #321438 #110486 #114692 #271732 #122386 #140732 = #1961180
&207932 &114106 &342030 &321438 &331458 &57346 &271732 &244772 &70366 = &1961180
$103966 $228212 $68406 $535730 $331458 $172038 $135866 $244772 $140732 = $1961180
μ207932 μ114106 μ136812 μ107146 μ552430 μ172038 μ407598 μ122386 μ140732 = μ1961180
€207932 €228212 €68406 €214292 €110486 €286730 €407598 €367158 €70366 = €1961180
ŋ103966 ŋ228212 ŋ136812 ŋ107146 ŋ220972 ŋ57346 ŋ679330 ŋ367158 ŋ211098 = ŋ2112040
we get
-1961180 +1961180 *1961180 #1961180 &1961180 $1961180 μ1961180 €1961180 ŋ2112040
which can be reduced to ŋ150860
(z)(ŋ1/z) = (ŋ150860) / q = ŋ150860 / ŋ150860 = ŋ1 ;)
With a bit of luck this is not like those old telephone books, yet
Next time I going to make a spreadsheet.
using arrays may suit better, you may take a bit of inspiration in
https://en.wikipedia.org/wiki/De_Bruijn_torus#/media/File:2-2-4-4-de-Bruijn-torus.svg
even is just works for some, is an angle to approach.
> z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
> 1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
ok, check time
__|__-1__+2__*3__#4__&5__
__|______________________
-2|__+2__*4__#6__&8__$10_
+1|__*1__#2__&3__$4__-5__
*1|__#1__&2__$3__-4__+5__
#1|__&1__$2__-3__+4__*5__
&1|__$1__-2__+3__*4__#5__
-2 -3 -4 -5 = -14
+2 +3 +4 +5 = +14
*1 *4 *4 *5 = *14
#1 #2 #6 #5 = #14
&1 &2 &3 &8 = &14
$1 $2 $3 $4 $10 = $20
-14 +14 *14 #14 &14 $20 = $6 = @6
1/6.@6 = @1
exploring a bit the complete procedure for p4 numbers
the identity elements and the idempotents
identity element : #1
pancake idempotent ( local "identity element" just for pancakers)
#0,5 @ +0,5' = #1/2 +1/2' = #1 -1/2 *1/2 = id_C
pancake imaginary unit ( local imaginary unit just for pancakers)
-0,5 @ *0,5' = -1/2 @ *1/2' = im_C
rod idempotent ( local "identity element" just for roders)
#0,5 @ +0,5 = #1/2 @ +1/2 = id_R
hybrid format
w.id_R @ x.id_C @ y.im_C <--> ( w , x + iy )
also, jokingly, nill idempotent ( local "identity element" just for nillers)
-0,25 +0,25 *0,25 #0,25 = -1/4 @ +1/4 @ *1/4 @ #1/4 = id_0
you may say this is p1-like
To generate a number in the pancake, generate an arbitrary number, and after,
multiply by id_C. To get a number in the rod, generate an arbitray number,
and after, multiply by id_R. The meaning of multiply an arbitrary number by
some idempotent is that this idempotent leave intact the part of the number
that lies in the zone where the idempotent belongs, this is like multiply
arbitrary number expressed as (a,b)
( a , b )( id_C , 0 ) = ( a , 0 )
identity element
t = (t)(#1)
t = (t)(id_C @ id_R)
What happens when you stumble upon something else in the procedure ?
Now we start with z = - 6 + 1 # 5, which is a pancake element, let's check that
(- 6 + 1 # 5)(id_C) = ?
______-1/2__*1/2__#1___
___|___________________
-6_|__+3____#3____-6___
+1_|__*1/2__-1/2__+1___
#5_|__-5/2__*5/2__#5___
-5/2 -1/2 -6 = -9
+3 +1 = +4
*1/2 *5/2 = *3
#3 #5 = 8
-9 +4 *3 #8 = -6 +1 #5
(- 6 + 1 # 5)(id_C) = - 6 + 1 # 5
you may thought in multiplying by id_C as the pancake test
ok, z is a pancake element
Suppose we don't know that z is a pancake, and proceed according the
procedure to find its reciprocal
(- 6 + 1 # 5)(- 6' + 1 # 5) =
_______-6____+1____#5____
____|____________________
-6'_|__+36'__*6'___-30'__
+1__|__*6____#1____+5____
#5__|__-30___+5____#25___
-30 @ -30' = 0
+36' @ +5 @ +5 = +26'
*6 @ *6' = 0
#1 @ #25 = #26
(- 6 + 1 # 5)(- 6' + 1 # 5) = +26' @ #26
This time, since we don't know with what zone we will be working we present
the procedure as follows :
instead of (#1/z) we'll use (id_?/z), this is 'identity element unknown'
id_? / z = id_? / (- 6 + 1 # 5)
id_? / (-6 +1 #5) = (id_?)(-6' +1 #5) / (+26' @ #26)
if we decide to proceed normally
(+26' @ #26)(+26 @ #26) = ??
________+26'___#26___
_____|_______________
+26__|__#676'__+676__
#26__|__+676'__#676__
+676' @ +676 = 0
#676' @ #676 = 0
(+26' @ #26)(+26 @ #26) = 0
this is
(id_?)(-6' +1 #5)(+26 @ #26) / (+26' @ #26)(+26 @ #26) =
(id_?)(-6' +1 #5)(+26 @ #26) / 0
the procedure blows up.
ok, roll back to
id_? / (-6 +1 #5) = (id_?)(-6' +1 #5) / (+26' @ #26)
here you may notice two things, that (+26' @ #26) = 52.(id_c), and with
that you automatically frame/set id_? to be id_C
(id_C)(-6' +1 #5) / 52.(id_C)
also notice that 52 is an "scalar", one could simply kick it to the numerator side.
(id_C)(-6' +1 #5) / 52.(id_C) = (1/52).(id_C)(-6' +1 #5) / (id_C)
since we are working in the pancake zone, dividing by id_C is redundant
(1/52).(id_C)(-6' +1 #5)
If, we suppose that id_C is redundant, since acts as a local identity element
for the number lying in the pancake area, but let's check if it is the case
(-6' +1 #5) = (-0 +6 *6 # 6 ) @ +1 #5 = +7 *6 #11
(id_C)(+7 *6 #11) = ???
_______-1/2___*1/2___#1____
____|______________________
+7__|__*7/2___-7/2___+7____
*6__|__#3_____+3_____*6____
#11_|__-11/2__*11/2__#11___
-11/2 @ -7/2 = -9
+3 @ +7 = +10
*7/2 @ *11/2 @ *6 = *15
#3 @ #11 = #14
-9 @ +10 @ *15 #14 = +1 @ *6 @ # 5
ok, we have discovered an infiltration, this is (-6' +1 #5) was not in the
pancake section, but, (-6' +1 #5) has suffered a pancaking that has exiled
its non-pancake components, yielding +1 @ *6 @ # 5
ok, so far
(1/52).(id_C)(-6' +1 #5) = (1/52).(+1 @ *6 @ # 5)
(1/52).(+1 @ *6 @ # 5) = +1/52 @ *6/52 @ #5/52
( id_C / - 6 + 1 # 5 ) = +1/52 @ *6/52 @ #5/52
The procedure ends, and this time we have obtained something, not the reciprocal
of - 6 + 1 # 5, just can be considered a "local reciprocal relative to id_C", but
since id_C is not an identity element for all p4 numbers, the usage of the word
reciprocal/multiplicative inverse here is not correct.
check time
( - 6 + 1 # 5 )( +1/52 @ *6/52 @ #5/52 ) = ????
______+1/52___*6/52___#5/52___
___|__________________________
-6_|__*6/52___#36/52__-30/52__
+1_|__#1/52___-6/52___+5/52___
#5_|__+5/52___*30/52__#25/52__
-6/52 @ -30/52 = -36/52
+5/52 @ +5/52 = +10/52
*6/52 @ *30/52 = *36/52
#1/52 @ #36/52 @ #25/52 = #62/52
-36/52 @ +10/52 @ *36/52 @ #62/52 = -26/52 @ *26/52 @ #52/52 = -1/2 @ *1/2 @ #1
( - 6 + 1 # 5 )( +1/52 @ *6/52 @ #5/52 ) = -1/2 @ *1/2 @ #1 = id_C
Following the procedure, sometimes, we may stumble upon with some multiple of
an idempotent ( or its additive inverse) lying in some of the pancakes that
contitutes certain Pn, this is, you may use the procedure to find the
idempotents in each complex slice ( id_C_ith ). Multiplying between elements
belonging to different pancakes is like the fights between two different
political parties, the result is nill, but taxpayer's resources were already
spent.
notice that
( 0 , b )( 0 , 1/b) = ( 0 , 1)
but also
( 0 , b )( a , 1/b ) = ( 0 , 1 )
this is ( 0 , b ) has infinites "local reciprocals" that yields "local
identity elements" ( not reciprocal in the true sense)
You may calibrate better all this...
space. Since both share the same identity element ( and also the additive
inverse of the identity element) you may call this bigger space a 'siames
space', where auxPm only surfaces its head in the reciprocal procedure.
It is correct when you apply º to the whole number, although, the ° is
intended to be distributed in some terms of a Pn number to generate
its 'conjugates'. Notice that these 'conjugates' are in the the 'siames space'.
For example in the case of a P15 number, the 15 (of p15) has factors three and
five. One may start with the three, this is, one builds the siames
space p15_auxP3, find the two 'conjugates' in p15_auxP3, multiply those
conjugates to the p15 number. Then, some number of terms of the p15 number
with certain signs should dissapear. After we build a siames space p15_auxp5,
find the four conjugates of the remaining p15 number. Notice that also one
could starts with p15_auxP5 and then p15_auxP3.
WIth more examples will be more easy. In p9, suppose we have the number
z = -3 @ +5 @ *1 @ #2 @ &1 @ $2 @ μ2 @ €1 @ ŋ3
the identity element is ŋ1 (the ŋ symbol is similar to the n of nine)
since 9 (in p9) has factors three and three, the order does not matter.
In the siames space p9_auxP3, we build the 'conjugates' of z
z» = -3° @ +5°° @ *1 @ #2° @ &1°° @ $2 @ μ2° @ €1°° @ ŋ3
z»» = -3°° @ +5° @ *1 @ #2°° @ &1° @ $2 @ μ2°° @ €1° @ ŋ3
Instead of z» and z»» you may call them flip_1(z) and flip_2(z). I wouldn't name
these 'conjugates' as z° and z°° because that will conflate notation. For
example, something analogous happens if you have a p4 number x = -a+b*c#d, if
you write *x, one could infer that *x is actually (*1)(x)
Ok, we have z, but we actually start with z in the denominator, we want to know
if ŋ1/z is equivalent to something useful that we can manipulate as any other
p9 number, let's see
ŋ1 / ( -3 @ +5 @ *1 @ #2 @ &1 @ $2 @ μ2 @ €1 @ ŋ3 ) = ŋ1 / (z)
We want to multiply the denominator with (z»)(z»»), this is, we want
(z)(z»)(z»»)
but usually, we can't multiply the denominator without compesating that in the
numerator also, this is, if you multiply by x in the denominator, you multiply
by x in the numerator also.
ŋ1 / z = (ŋ1)(z»)(z»») / (z)(z»)(z»») = (z»)(z»») / (z)(z»)(z»»)
Presumably, getting z multiplied by (z»)(z»») will get us a p9 number with less
terms.
Ok, we may start computing the result of (z»)(z»»)
________-3°___+5°°__*1____#2°___&1°°__$2____μ2°___€1°°__ŋ3____
_____|________________________________________________________
-3°°_|__+9____*15°__#3°°__&6____$3°___μ6°°__€6____ŋ3°___-9°°__
+5°__|__*15°°_#25___&5°___$10°°_μ5____€10°__ŋ10°°_-5____+15°__
*1___|__#3°___&5°°__$1____μ2°___€1°°__ŋ2____-2°___+1°°__*3____
#2°°_|__&6____$10°__μ2°°__€4____ŋ2°___-4°°__+4____*2°___#6°°__
&1°__|__$3°°__μ5____€1°___ŋ2°°__-1____+2°___*2°°__#1____&3°___
$2___|__μ6°___€10°°_ŋ2____-4°___+2°°__*4____#4°___&2°°__$6____
μ2°°_|__€6____ŋ10°__-2°°__+4____*2°___#4°°__&4____$2°___μ6°°__
€1°__|__ŋ3°°__-5____+1°___*2°°__#1____&2°___$2°°__μ1____€3°___
ŋ3___|__-9°___+15°°_*3____#6°___&3°°__$6____μ6°___€3°°__ŋ9____
grouping terms according primary sign (mainPn sign)
-9° @ -5 @ -2°° @ -4° @ -1 @ -4°° @ -2° @ -5 @ -9°°
+9 @ +15°° @ +1° @ +4 @ +2°° @ +2° @ +4 @ +1°° @ +15°
*15°° @ *15° @ *3 @ *2°° @ *2° @ *4 @ *2°° @ *2° @ *3
#3° @ #25 @ #3°° @ #6° @ #1 @ #4°° @ #4° @ #1 @ #6°°
&6 @ &5°° @ &5° @ &6 @ &3°° @ &2° @ &4 @ &2°° @ &3°
$3°° @ $10° @ $1 @ $10°° @ $3° @ $6 @ $2°° @ $2° @ $6
μ6° @ μ5 @ μ2°° @ μ2° @ μ5 @ μ6°° @ μ6° @ μ1 @ μ6°°
€6 @ €10°° @ €1° @ €4 @ €1°° @ €10° @ €6 @ €3°° @ €3°
ŋ3°° @ ŋ10° @ ŋ2 @ ŋ2°° @ ŋ2° @ ŋ2 @ ŋ10°° @ ŋ3° @ ŋ9
re-arranging terms
(-9° @ -9°°) @ (-2°° @ -2°) @ (-4° @ -4°°) @ (-5 @ -5 @ -1 )
(+15°° @ +15°) @ (+1° @ +1°°) @ (+2°° @ +2°) @ (+9 @ +4 @ +4)
(*15°° @ *15°) @ (*2°° @ *2°) @ (*2° @ *2°°) @ (*3 @ *3 @ *4)
(#3° @ #3°°) @ (#6° @ #6°°) @ (#4°° @ #4°) @ (#25 @ #1 @ #1)
(&5°° @ &5°) @ (&3°° @ &3°) @ (&2° @ &2°°) @ (&6 @ &6 @ &4)
($3°° @ $3°) @ ($10° @ $10°°) @ ($2°° @ $2°) @ ($1 @ $6 @ $6)
(μ6° @ μ6°°) @ (μ2°° @ μ2°) @ (μ6° @ μ6°°) @ (μ5 @ μ5 @ μ1)
(€10°° @ €10°) @ (€1° @ €1°°) @ (€3°° @ €3°) @ (€6 @ €6 @ €4)
(ŋ3°° @ ŋ3°) @ (ŋ10° @ ŋ10°°) @ (ŋ2°° @ ŋ2°) @ (ŋ2 @ ŋ2 @ ŋ9)
further re-arranging terms
-9' @ -2' @ -4' @ -11 = -15' @ -11 = -4'
+15' @ +1' @ +2' @ +17 = +18' @ +17 = +1'
*15' @ *2' @ *2' @ *10 = *19' @ *10 = *9'
#3' @ #6' @ #4' @ #27 = #13' @ #27 = #14
&5' @ &3' @ &2' @ &16 = &10' @ &16 = &6
$3' @ $10' @ $2' @ $13 = $15' @ $13 = $2'
μ6' @ μ2' @ μ6' @ μ11 = μ14' @ μ11 = μ3'
€10' @ €1' @ €3' @ €16 = €14' @ @ €16 = €2
ŋ3' @ ŋ10' @ ŋ2' @ ŋ13 = ŋ15' @ ŋ13 = ŋ2'
(z»)(z»») = -4' @ +1' @ *9' @ #14 @ &6 @ $2' @ μ3' @ €2 @ ŋ2'
so far we have
(z»)(z»») / (z)(z»)(z»») = (-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') / (z)(z»)(z»»)
(z)(z»)(z»») = (-3+5*1#2&1$2μ2€1ŋ3)(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') = ?
_______-4'____+1'____*9'____#14____&6_____$2'____μ3'____€2_____ŋ2'____
____|_________________________________________________________________
-3__|__+12'___*3'____#27'___&42____$18____μ6'____€9'____ŋ6_____-6'____
+5__|__*20'___#5'____&45'___$70____μ30____€10'___ŋ15'___-10____+10'___
*1__|__#4'____&1'____$9'____μ14____€6_____ŋ2'____-3'____+2_____*2'____
#2__|__&8'____$2'____μ18'___€28____ŋ12____-4'____+6'____*4_____#4'____
&1__|__$4'____μ1'____€9'____ŋ14____-6_____+2'____*3'____#2_____&2'____
$2__|__μ8'____€2'____ŋ18'___-28____+12____*4'____#6'____&4_____$4'____
μ2__|__€8'____ŋ2'____-18'___+28____*12____#4'____&6'____$4_____μ4'____
€1__|__ŋ4'____-1'____+9'____*14____#6_____&2'____$3'____μ2_____€2'____
ŋ3__|__-12'___+3'____*27'___#42____&18____$6'____μ9'____€6_____ŋ6'____
-12' @ -1' @ -18' @ -28 @ -6 @ -4' @ -3' @ -10 @ -6'
+12' @ +3' @ +9' @ +28 @ +12 @ +2' @ +6' @ +2 @ +10'
*20' @ *3' @ *27' @ *14 @ *12 @ *4' @ *3' @ *4 @ *2'
#4' @ #5' @ #27' @ #42 @ #6 @ #4' @ #6' @ #2 @ #4'
&8' @ &1' @ &45' @ &42 @ &18 @ &2' @ &6' @ &4 @ &2'
$4' @ $2' @ $9' @ $70 @ $18 @ $6' @ $3' @ $4 @ $4'
μ8' @ μ1' @ μ18' @ μ14 @ μ30 @ μ6' @ μ9' @ μ2 @ μ4'
€8' @ €2' @ €9' @ €28 @ €6 @ €10' @ €9' @ €6 @ €2'
ŋ4' @ ŋ2' @ ŋ18' @ ŋ14 @ ŋ12 @ ŋ2' @ ŋ15' @ ŋ6 @ ŋ6'
(-12' @ -1' @ -18' @ -4' @ -3' @ -6') @ (-28 @ -6 @ -10) = -44' @ -44 = 0
(+12' @ +3' @ +9' @ +2' @ +6' @ +10') @ (+28 @ +12 @ +2) = +42' @ +42 = 0
(*20' @ *3' @ *27' @ *4' @ *3' @ *2') @ (*14 @ *12 @ *4) = *59' @ *30 = *29'
(#4' @ #5' @ #27' @ #4' @ #6' @ #4') @ (#42 @ #6 @ #2) = #50' @ 50 = 0
(&8' @ &1' @ &45' @ &2'@ &6' @ &2') @ (&42 @ &18 @ &4) = &64' @ &64 = 0
($4' @ $2' @ $9' @ $6' @ $3' @ $4') @ ($70 @ $18 @ $4) = $28' @ $92 = $64
(μ8' @ μ1' @ μ18' @ μ6' @ μ9' @ μ4') @ (μ14 @ μ30 @ μ2) = μ46' @ μ46 = 0
(€8' @ €2' @ €9' @ €10' @ €9' €2') @ (€28 @ €6 @ €6) = €40' @ €40 = 0
(ŋ4' @ ŋ2' @ ŋ18' @ ŋ2' @ ŋ15' @ ŋ6') @ (ŋ14 @ ŋ12 @ ŋ6) = ŋ47' @ ŋ32 = ŋ15'
(z)(z»)(z»») = *29' @ $64 @ ŋ15'
so far, we have
(z»)(z»») / (z)(z»)(z»») = (-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2') / (*29' $64 ŋ15')
notice that in the denominator remain only terms with signs * $ ŋ
so far, we have finished the firt part of the procedure for this p9 number,
and we continue to the second part
we may rename (z)(z»)(z»») as w to save some chars
w = (*29' $64 ŋ15')
Since the second factor (of 9 in p9) is also three, we access again p9_auxP3
to get the two 'conjugates' of w
w~ = *29'° @ $64°° @ ŋ15'
w~~ = *29'°° @ $64° @ ŋ15'
we want to multiply the denominator with (w~)(w~~)
(z»)(z»») / w = (z»)(z»»)(w~)(w~~) / (w)(w~)(w~~)
ok, let's start computing (w~)(w~~)
(*29'° @ $64°° @ ŋ15')(*29'°° @ $64° @ ŋ15') = ??
___________*29'°______$64°°______ŋ15'_______
________|___________________________________
*29'°°__|__$841_______ŋ1856'°____*435°°_____
$64°____|__ŋ1856'°°___*4096______$960'°_____
ŋ15'____|__*435°______$960'°°____ŋ225_______
arranging a bit
*4096 @ (*435° @ *435°°) = *4096 @ *435' = *3661
$841 @ ($960'°° @ $960'°) = $841 @ $960 = $1801
ŋ225 @ (ŋ1856'°° @ ŋ1856'°) = ŋ225 @ ŋ1856 = ŋ2081
(w~)(w~~) = *3661 @ $1801 @ ŋ2081
now we can compute (w)(w~)(w~~)
(*29' $64 ŋ15')(*3661 @ $1801 @ ŋ2081) = ???
____________*3661______$1801______ŋ2081______
_________|___________________________________
*29'_____|__$106169'___ŋ52229'____*60349'____
$64______|__ŋ234304____*115264____$133184____
ŋ15'_____|__*54915'____$27015'____ŋ31215'____
arranging
(*54915' @ *60349') @ *115264 = *115264' @ *115264 = 0
($106169' @ $27015') @ $133184 = $133184' @ $$133184 = 0
(ŋ52229' @ ŋ31215') @ ŋ234304 = ŋ150860
(w)(w~)(w~~) = ŋ150860
so far we have
(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2')(*3661 @ $1801 @ ŋ2081) / (ŋ150860)
(-4' +1' *9' #14 &6 $2' μ3' €2 ŋ2')(*3661 @ $1801 @ ŋ2081) = ????
________*3661______$1801______ŋ2081______
_____|___________________________________
-4'__|__#14644'____μ7204'_____-8324'_____
+1'__|__&3661'_____€1801'_____+2081'_____
*9'__|__$32949'____ŋ16209'____*18729'____
#14__|__μ51254_____-25214_____#29134_____
&6___|__€21966_____+10806_____&12486_____
$2'__|__ŋ7322'_____*3602'_____$4162'_____
μ3'__|__-10983'____#5403'_____μ6243'_____
€2___|__+7322______&3602______€4162______
ŋ2'__|__*7322'_____$3602'_____ŋ4162'_____
(-10983' @ -8324') @ -25214 = -5907
+2081' @ (+10806 @ +7322) = +16047
(*7322' @ *3602' @ *18729') = *29653'
(#14644' @ #5403') @ #29134 = #9087
&3661' @ (&3602 @ &12486) = &12427
($32949' @ $3602' @ $4162') = $40713'
(μ7204' @ μ6243') @ μ51254 = μ37807
€1801' @ (€21966 @ €4162) = €24327
((ŋ7322' @ ŋ16209' @ ŋ4162') = ŋ27693'
the terms with ' have to be converted
*29653' = -29653 +29653 *0 #29653 &29653 $29653 μ29653 €29653 ŋ29653
$40713' = -40713 +40713 *40713 #40713 &40713 $0 μ40713 €40713 ŋ40713
ŋ27693' = -27693 +27693 *27693 #27693 &27693 $27693 μ27693 €27693 ŋ0
further arranging terms
-5907 @ -29653 @ -40713 @ -27693 = -103966
+16047 @ +29653 @ +40713 @ +27693 = +114106
*0 @ *40713 @ *27693 = *68406
#9087 @ #29653 @ #40713 @ #27693 = #107146
&12427 @ &29653 @ &40713 @ &27693 = &110486
$29653 @ $0 @ $27693 = $57346
μ37807 @ μ29653 @ μ40713 @ μ27693 = μ135866
€24327 @ €29653 @ €40713 @ €27693 = €122386
ŋ29653 @ ŋ40713 @ ŋ0 = ŋ70366
at last, we have that (z»)(z»»)(w~)(w~~) is equal to
-103966 +114106 *68406 #107146 &110486 $57346 μ135866 €122386 ŋ70366
ok, we have
(-103966 +114106 *68406 #107146 &110486 $57346 μ135866 €122386 ŋ70366 )/ŋ150860
if you use q = ŋ150860 then
ŋ1 / z = -103966/q @ +114106/q @ *68406/q @ #107146/q @ &110486/q @ $57346/q @ μ135866/q @ €122386/q @ ŋ70366/q
end of procedure
check time
(z)(ŋ1 / z) = ?????
due to lack of space the /q in each term is omitted, to prioritize compatibility
of format, or to prevent the tables get a space-time distorsione
___-103966__+114106__*68406__#107146__&110486__$57346__μ135866__€122386__ŋ70366_
__|_____________________________________________________________________________
-3|+311898__*342318__#205218_&321438__$331458__μ172038_€407598__ŋ367158__-211098
+5|*519830__#570530__&342030_$535730__μ552430__€286730_ŋ679330__-611930__+351830
*1|#103966__&114106__$68406__μ107146__€110486__ŋ57346__-135866__+122386__*70366_
#2|&207932__$228212__μ136812_€214292__ŋ220972__-114692_+271732__*244772__#140732
&1|$103966__μ114106__€68406__ŋ107146__-110486__+57346__*135866__#122386__&70366_
$2|μ207932__€228212__ŋ136812_-214292__+220972__*114692_#271732__&244772__$140732
μ2|€207932__ŋ228212__-136812_+214292__*220972__#114692_&271732__$244772__μ140732
€1|ŋ103966__-114106__+68406__*107146__#110486__&57346__$135866__μ122386__€70366_
ŋ3|-311898__+342318__*205218_#321438__&331458__$172038_μ407598__€367158__ŋ211098
arranging by sign
-311898 -114106 -136812 -214292 -110486 -114692 -135866 -611930 -211098 = -1961180
+311898 +342318 +68406 +214292 +220972 +57346 +271732 +122386 +351830 = +1961180
*519830 *342318 *205218 *107146 *220972 *114692 *135866 *244772 *70366 = *1961180
#103966 #570530 #205218 #321438 #110486 #114692 #271732 #122386 #140732 = #1961180
&207932 &114106 &342030 &321438 &331458 &57346 &271732 &244772 &70366 = &1961180
$103966 $228212 $68406 $535730 $331458 $172038 $135866 $244772 $140732 = $1961180
μ207932 μ114106 μ136812 μ107146 μ552430 μ172038 μ407598 μ122386 μ140732 = μ1961180
€207932 €228212 €68406 €214292 €110486 €286730 €407598 €367158 €70366 = €1961180
ŋ103966 ŋ228212 ŋ136812 ŋ107146 ŋ220972 ŋ57346 ŋ679330 ŋ367158 ŋ211098 = ŋ2112040
we get
-1961180 +1961180 *1961180 #1961180 &1961180 $1961180 μ1961180 €1961180 ŋ2112040
which can be reduced to ŋ150860
(z)(ŋ1/z) = (ŋ150860) / q = ŋ150860 / ŋ150860 = ŋ1 ;)
With a bit of luck this is not like those old telephone books, yet
Next time I going to make a spreadsheet.
using arrays may suit better, you may take a bit of inspiration in
https://en.wikipedia.org/wiki/De_Bruijn_torus#/media/File:2-2-4-4-de-Bruijn-torus.svg
even is just works for some, is an angle to approach.
> z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
> 1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
__|__-1__+2__*3__#4__&5__
__|______________________
-2|__+2__*4__#6__&8__$10_
+1|__*1__#2__&3__$4__-5__
*1|__#1__&2__$3__-4__+5__
#1|__&1__$2__-3__+4__*5__
&1|__$1__-2__+3__*4__#5__
-2 -3 -4 -5 = -14
+2 +3 +4 +5 = +14
*1 *4 *4 *5 = *14
#1 #2 #6 #5 = #14
&1 &2 &3 &8 = &14
$1 $2 $3 $4 $10 = $20
-14 +14 *14 #14 &14 $20 = $6 = @6
1/6.@6 = @1
exploring a bit the complete procedure for p4 numbers
the identity elements and the idempotents
identity element : #1
pancake idempotent ( local "identity element" just for pancakers)
#0,5 @ +0,5' = #1/2 +1/2' = #1 -1/2 *1/2 = id_C
pancake imaginary unit ( local imaginary unit just for pancakers)
-0,5 @ *0,5' = -1/2 @ *1/2' = im_C
rod idempotent ( local "identity element" just for roders)
#0,5 @ +0,5 = #1/2 @ +1/2 = id_R
hybrid format
w.id_R @ x.id_C @ y.im_C <--> ( w , x + iy )
also, jokingly, nill idempotent ( local "identity element" just for nillers)
-0,25 +0,25 *0,25 #0,25 = -1/4 @ +1/4 @ *1/4 @ #1/4 = id_0
you may say this is p1-like
To generate a number in the pancake, generate an arbitrary number, and after,
multiply by id_C. To get a number in the rod, generate an arbitray number,
and after, multiply by id_R. The meaning of multiply an arbitrary number by
some idempotent is that this idempotent leave intact the part of the number
that lies in the zone where the idempotent belongs, this is like multiply
arbitrary number expressed as (a,b)
( a , b )( id_C , 0 ) = ( a , 0 )
identity element
t = (t)(#1)
t = (t)(id_C @ id_R)
What happens when you stumble upon something else in the procedure ?
Now we start with z = - 6 + 1 # 5, which is a pancake element, let's check that
(- 6 + 1 # 5)(id_C) = ?
______-1/2__*1/2__#1___
___|___________________
-6_|__+3____#3____-6___
+1_|__*1/2__-1/2__+1___
#5_|__-5/2__*5/2__#5___
-5/2 -1/2 -6 = -9
+3 +1 = +4
*1/2 *5/2 = *3
#3 #5 = 8
-9 +4 *3 #8 = -6 +1 #5
(- 6 + 1 # 5)(id_C) = - 6 + 1 # 5
you may thought in multiplying by id_C as the pancake test
ok, z is a pancake element
Suppose we don't know that z is a pancake, and proceed according the
procedure to find its reciprocal
(- 6 + 1 # 5)(- 6' + 1 # 5) =
_______-6____+1____#5____
____|____________________
-6'_|__+36'__*6'___-30'__
+1__|__*6____#1____+5____
#5__|__-30___+5____#25___
-30 @ -30' = 0
+36' @ +5 @ +5 = +26'
*6 @ *6' = 0
#1 @ #25 = #26
(- 6 + 1 # 5)(- 6' + 1 # 5) = +26' @ #26
This time, since we don't know with what zone we will be working we present
the procedure as follows :
instead of (#1/z) we'll use (id_?/z), this is 'identity element unknown'
id_? / z = id_? / (- 6 + 1 # 5)
id_? / (-6 +1 #5) = (id_?)(-6' +1 #5) / (+26' @ #26)
if we decide to proceed normally
(+26' @ #26)(+26 @ #26) = ??
________+26'___#26___
_____|_______________
+26__|__#676'__+676__
#26__|__+676'__#676__
+676' @ +676 = 0
#676' @ #676 = 0
(+26' @ #26)(+26 @ #26) = 0
this is
(id_?)(-6' +1 #5)(+26 @ #26) / (+26' @ #26)(+26 @ #26) =
(id_?)(-6' +1 #5)(+26 @ #26) / 0
the procedure blows up.
ok, roll back to
id_? / (-6 +1 #5) = (id_?)(-6' +1 #5) / (+26' @ #26)
here you may notice two things, that (+26' @ #26) = 52.(id_c), and with
that you automatically frame/set id_? to be id_C
(id_C)(-6' +1 #5) / 52.(id_C)
also notice that 52 is an "scalar", one could simply kick it to the numerator side.
(id_C)(-6' +1 #5) / 52.(id_C) = (1/52).(id_C)(-6' +1 #5) / (id_C)
since we are working in the pancake zone, dividing by id_C is redundant
(1/52).(id_C)(-6' +1 #5)
If, we suppose that id_C is redundant, since acts as a local identity element
for the number lying in the pancake area, but let's check if it is the case
(-6' +1 #5) = (-0 +6 *6 # 6 ) @ +1 #5 = +7 *6 #11
(id_C)(+7 *6 #11) = ???
_______-1/2___*1/2___#1____
____|______________________
+7__|__*7/2___-7/2___+7____
*6__|__#3_____+3_____*6____
#11_|__-11/2__*11/2__#11___
-11/2 @ -7/2 = -9
+3 @ +7 = +10
*7/2 @ *11/2 @ *6 = *15
#3 @ #11 = #14
-9 @ +10 @ *15 #14 = +1 @ *6 @ # 5
ok, we have discovered an infiltration, this is (-6' +1 #5) was not in the
pancake section, but, (-6' +1 #5) has suffered a pancaking that has exiled
its non-pancake components, yielding +1 @ *6 @ # 5
ok, so far
(1/52).(id_C)(-6' +1 #5) = (1/52).(+1 @ *6 @ # 5)
(1/52).(+1 @ *6 @ # 5) = +1/52 @ *6/52 @ #5/52
( id_C / - 6 + 1 # 5 ) = +1/52 @ *6/52 @ #5/52
The procedure ends, and this time we have obtained something, not the reciprocal
of - 6 + 1 # 5, just can be considered a "local reciprocal relative to id_C", but
since id_C is not an identity element for all p4 numbers, the usage of the word
reciprocal/multiplicative inverse here is not correct.
check time
( - 6 + 1 # 5 )( +1/52 @ *6/52 @ #5/52 ) = ????
______+1/52___*6/52___#5/52___
___|__________________________
-6_|__*6/52___#36/52__-30/52__
+1_|__#1/52___-6/52___+5/52___
#5_|__+5/52___*30/52__#25/52__
-6/52 @ -30/52 = -36/52
+5/52 @ +5/52 = +10/52
*6/52 @ *30/52 = *36/52
#1/52 @ #36/52 @ #25/52 = #62/52
-36/52 @ +10/52 @ *36/52 @ #62/52 = -26/52 @ *26/52 @ #52/52 = -1/2 @ *1/2 @ #1
( - 6 + 1 # 5 )( +1/52 @ *6/52 @ #5/52 ) = -1/2 @ *1/2 @ #1 = id_C
Following the procedure, sometimes, we may stumble upon with some multiple of
an idempotent ( or its additive inverse) lying in some of the pancakes that
contitutes certain Pn, this is, you may use the procedure to find the
idempotents in each complex slice ( id_C_ith ). Multiplying between elements
belonging to different pancakes is like the fights between two different
political parties, the result is nill, but taxpayer's resources were already
spent.
notice that
( 0 , b )( 0 , 1/b) = ( 0 , 1)
but also
( 0 , b )( a , 1/b ) = ( 0 , 1 )
this is ( 0 , b ) has infinites "local reciprocals" that yields "local
identity elements" ( not reciprocal in the true sense)
You may calibrate better all this...
michael Rodriguez
Sep 21, 2021, 4:14:27 PM9/21/21
to
> It happens that the value
> MU @ NU'
> is ( 0, 2, 1, 1, 1, ... )
> where MU is minus unity (0,1,0,0,0...) and NU is neutral unity (1,0,0,0,...)
> (z)(0,2,1,1,1...)
> = - z @ z'
If you have a good number of cases, and you suspect it works for n, you may want to build the proof for n, it should be more or less short, with an small tiny bit of number-theory. The same, you may want to build a proof for (NU')^2=NU...
> MU @ NU'
> is ( 0, 2, 1, 1, 1, ... )
> where MU is minus unity (0,1,0,0,0...) and NU is neutral unity (1,0,0,0,...)
> (z)(0,2,1,1,1...)
> = - z @ z'
Timothy Golden
Sep 22, 2021, 11:12:45 AM9/22/21
to
* x = # a - b + c * d
(*1)(-a+b*c#d) = # a - b + c * d
These are one and the same. This is somewhat akin to my own 'revelation' of the inverse which is critical to your method. Meanwhile you are probably getting a snicker at how primitive I am. Who knew it could be so functional. This division algorithm is excellent in its exercise of the math. At this moment I'm feeling open to a solution that ultimately collapses your notation. That's sort of what the notation does, and right there comes in the inverse to save the day. And yet the notation in the 3-form is a sort of triverse. It's totally fascinating still even as I am getting more comfortable with it. Also it is still close by to the earlier version you had which conjugated the primes.
In a way, the simple truth is that we just keep throwing factors at it until the denominator resolves. Of course you have the recipe for the factors which is rather the larger part of the problem.
So to what degree can we simply jump to this earlier?
We are back in P9 and always will get here, right?
I don't have it yet, but it seems that to code it and stay in P9 and condense the earlier steps will work.
[P9 14, 29, 29, 0, 29, 29, 93, 29, 29 ]
does challenge this interpretation. Also I should confess that my code can handle these inverses because these components are floating point values and as I enter them as negative values they are taking advantage of P2 mechanics of the hardware. It works, but I'm not proud of it.
I could code it the long way, but for now:
void P9stuff()
{ // (*29' $64 ŋ15'
Poly z(9); z[0] = -15; z[3] = -29; z[6] = 64;
z.Reduce();
cout << "\nP9stuff() : " << z << "\n";
}
Anyway, when the values resolve to a single component you will have it. And the inverse notation is fine. All is well with your presentation.
>
> using arrays may suit better, you may take a bit of inspiration in
> https://en.wikipedia.org/wiki/De_Bruijn_torus#/media/File:2-2-4-4-de-Bruijn-torus.svg
> even is just works for some, is an angle to approach.
> > z = @ 0 - 1 + 2 * 3 # 4 & 5 ( in P6 )
> > 1/z = (1/6)( @ 0 - 2 + 1 * 1 # 1 & 1 )
> ok, check time
> __|__-1__+2__*3__#4__&5__
> __|______________________
> -2|__+2__*4__#6__&8__$10_
> +1|__*1__#2__&3__$4__-5__
> *1|__#1__&2__$3__-4__+5__
> #1|__&1__$2__-3__+4__*5__
> &1|__$1__-2__+3__*4__#5__
>
> -2 -3 -4 -5 = -14
> +2 +3 +4 +5 = +14
> *1 *4 *4 *5 = *14
> #1 #2 #6 #5 = #14
> &1 &2 &3 &8 = &14
> $1 $2 $3 $4 $10 = $20
>
> -14 +14 *14 #14 &14 $20 = $6 = @6
>
> 1/6.@6 = @1
>
> exploring a bit the complete procedure for p4 numbers
>
> the identity elements and the idempotents
Pretty sure the triverse just goes directly to inverse.
I'm not up for trying right now.
Still to me this is (1,0) but this is a P2 interpretation of what you've written.
Pretty sure you are thinking of something else. Oh, you're using (-,+) possibly.
>
> but also
> ( 0 , b )( a , 1/b ) = ( 0 , 1 )
errr.... (ab,1) in (-,+) signature.
> but also
> ( 0 , b )( a , 1/b ) = ( 0 , 1 )
>
> this is ( 0 , b ) has infinites "local reciprocals" that yields "local
> identity elements" ( not reciprocal in the true sense)
>
> You may calibrate better all this...
Well I've honestly barely followed your P4 work down here. I will try to get back to it, but honestly I'm stuck at why you haven't tried the 2x2 recipe. I wonder if you would like to name your algorithm so we have a quick term form it. AuxP2 in this case of a 2x2 method?
> this is ( 0 , b ) has infinites "local reciprocals" that yields "local
> identity elements" ( not reciprocal in the true sense)
>
> You may calibrate better all this...
These numbers just refuse to stop dancing around. It's fortunate they are well behaved, but they can't stop.
Thanks Michael.
Timothy Golden
Sep 22, 2021, 11:16:30 AM9/22/21
to
z1' z2' = z1 z2
which I've gleaned from your work. Other than that the series values above are good proof I think.
michael Rodriguez
Sep 22, 2021, 6:04:19 PM9/22/21
to
The format (Real,complex) actually is referring to the approach using complex slices (ring direct sum)
Now Hagen starts from the complex slices and he obtains the polysign numbers
In some sense, one may consider also other way around, starting from the polysign and ending
in the complex slices ( finding the 1 and i of each complex slice written in the polysigned format).
When the multiplicative inverse procedure 'fails', does not fail, but rather "get trapped" in a complex slice,
and one end up actually getting the "flat reciprocal" in that complex slice.
The reason of id_? is because you don't know if the procedure will get trapped in a complex slice or not.
Z1'Z2' = Z1Z2 is ok. If you do it in a particular n, for example in n=5,
this is, doing the product MU'MU'=MU , there is no problem. If one is smelling
that it works for a great number of them, one may trust that is true, still one
have to prove it for n. In any case there is no problem is one does it for a
particular case, actually checking it for that particular case.
The id_0 was for (-a+a*a#a)(-1/4+1/4*1/4#1/4) = (-a+a*a#a)
The reason for not naming it auxp2 is because is totally contained in the mainPn space,
but you could consider that way and see if better treating it that way.
For the "versor" you may start with an analogy, suppose you actually have the formula for
the complex numbers (or split-complex) and you want the graph.
When you actually get the set |w|=1, this is, the "unitary p4 numbers" according
the "hyperbolic modulus", you could for example, take a random "unitary p4 number" and
see what happens if you a^2, a^3, a^4, etc ("the power orbit" of that particular p4 number).
Also you may take a random "unitary p4 number", and you compute its reciprocal,
you may consider this a kind of "hyperbolic explementary angle".
Timothy Golden
Sep 23, 2021, 9:55:29 AM9/23/21
to
On Wednesday, September 22, 2021 at 6:04:19 PM UTC-4, michael Rodriguez wrote:
> The format (Real,complex) actually is referring to the approach using complex slices (ring direct sum)
> Now Hagen starts from the complex slices and he obtains the polysign numbers
> In some sense, one may consider also other way around, starting from the polysign and ending
> in the complex slices ( finding the 1 and i of each complex slice written in the polysigned format).
> When the multiplicative inverse procedure 'fails', does not fail, but rather "get trapped" in a complex slice,
> and one end up actually getting the "flat reciprocal" in that complex slice.
> The reason of id_? is because you don't know if the procedure will get trapped in a complex slice or not.
>
> Z1'Z2' = Z1Z2 is ok. If you do it in a particular n, for example in n=5,
> this is, doing the product MU'MU'=MU , there is no problem. If one is smelling
> that it works for a great number of them, one may trust that is true, still one
> have to prove it for n. In any case there is no problem is one does it for a
> particular case, actually checking it for that particular case.
Well, I have a computer proof by numerous instances. It was a surprise to me when I saw it in your work.
> The format (Real,complex) actually is referring to the approach using complex slices (ring direct sum)
> Now Hagen starts from the complex slices and he obtains the polysign numbers
> In some sense, one may consider also other way around, starting from the polysign and ending
> in the complex slices ( finding the 1 and i of each complex slice written in the polysigned format).
> When the multiplicative inverse procedure 'fails', does not fail, but rather "get trapped" in a complex slice,
> and one end up actually getting the "flat reciprocal" in that complex slice.
> The reason of id_? is because you don't know if the procedure will get trapped in a complex slice or not.
>
> Z1'Z2' = Z1Z2 is ok. If you do it in a particular n, for example in n=5,
> this is, doing the product MU'MU'=MU , there is no problem. If one is smelling
> that it works for a great number of them, one may trust that is true, still one
> have to prove it for n. In any case there is no problem is one does it for a
> particular case, actually checking it for that particular case.
As we come to depend upon the inverse we are somewhat dealing in axial values; in what arguably have congruence to the real line, or P2. Yet we are in potentially high dimension natively where we are doing this. The product of two values is the product of their antipodes. It is new to me. I agree there should be a formal proof. Could it stem from the MU form you state? That's pretty much where I saw it: you wrote something like
s1 x1 ' s1 x2 ' = (s1@s2)x1x2
so there it was in the sx form by you. Maybe you used m's instead of x's. I tried it on random z values and it works by instantiation.
>
> The id_0 was for (-a+a*a#a)(-1/4+1/4*1/4#1/4) = (-a+a*a#a)
>
> The reason for not naming it auxp2 is because is totally contained in the mainPn space,
> but you could consider that way and see if better treating it that way.
I tried working it out on a generic P4 value
- a + b * c # d
yesterday. So there is just a single degree in the notation, which may as well be the simple inverse. As I recall the numerator had 18 triple product terms including inverses which didn't look like they'd go away, and expanded out would just raise this figure even higher. I had introduced some intermediate variables to work out the denominator, and I could see that the signs did dissipate since the first factor diminishes the + and * components. I still had to follow your algorithm to do it.
w = - a + b * c # d
q(w) = - a ° + b * c ° # d
which does cancel down nicely to
w q(w) = + a a ° # 2 a c ° # b b + 2 b d + c c ° # d d
but here in laziness I substituted w1 = + b1 # d1
and got the next q1(w1)
q1( w1 ) = + b1 ° # d1
yielding
w1 q1( w ) = # b1 b1 ° # d1 d1
so this proves the concept of P4 as a 2x2 even though I sort of failed at the end here to really do it in (d,a,b,c)
I am hoping to code this. Sometimes in code things congeal nicely and the methods may twerk into alignment.
The AuxP3 form could possibly have a native result in z°. As we treat the P2 version here in terms of sign mechanics we do have
s x ° = s ( - x + x * x )
e.g.
- 1 ° = - ( - 1 + 1 * 1 ) = + 1 * 1 # 1
By linearity a more general value can be regarded as
Sigma( s1x1, s2x2, s3x3, ... )
which means a sum of individual sx components. No problem.
Sigma( s1x1, s2x2, s3x3, ... )°
= Sigma( s1x1°, s2x2°, s3x3°, ... )
In the P3 case we do have
( - a + b * c ) ( + a - b * c )
will yield the identity denominator so that the reciprocal
1 / (-a+b*c) = (+a-b*c) / [(-a+b*c)(+a-b*c)]
Somehow your ° notation is related to this.
Yet there are not three terms here as there are in your notation.
Yet your notation always does collapse the ° notation too.
We don't exactly have an arithmetic operation that does
- a + b * c --> + a - b * c
so I wonder if this is why the foreign nature creeps in.
This sign mechanics is somewhat what your form is encompassing.
I'm not claiming to see it with full clarity yet; just thinking about it this way.
Up in P6 we do know that the pairing is straightforward. The two P3 systems
on a value (a0,a1,a2,a3,a4,a5)
are (a0,0,a2,0,a4) and (0,a1,0,a3,0,a5).
Your ° and °° operate in this pattern.
>
> For the "versor" you may start with an analogy, suppose you actually have the formula for
> the complex numbers (or split-complex) and you want the graph.
> When you actually get the set |w|=1, this is, the "unitary p4 numbers" according
> the "hyperbolic modulus", you could for example, take a random "unitary p4 number" and
> see what happens if you a^2, a^3, a^4, etc ("the power orbit" of that particular p4 number).
> Also you may take a random "unitary p4 number", and you compute its reciprocal,
> you may consider this a kind of "hyperbolic explementary angle".
Back in the complex plane C if you choose a unit length value extremely close to unity but with a small complex component and plot the self products z^n you will plot a circle. For quite some time I'd try this in the higher Pn and it never would plot say a spherical shell in P4 or P5 though at the time I thought it might be possible. I guess it might still be possible but getting the correct initial value is tricky. Thinking of P5 as CxC helps, but then there is a distance transform in the Hagen version of polysign which complicates things. As we see in P4 the morphing of the sphere as it gets squashed to a plane and an axis few values conserve distance under product. Yet these are algebraically well behaved products.
So you are suggesting that there is something meaningful about the angle between z and 1/z? Interesting. You know it's another funny thing about the usage of random variables: the probability of landing on the exceptions is close to nill. It doesn't mean they are not there though. My code is in danger of conflating some things this way, but on simpler puzzles like the inverse z1'z2' I do trust that I am getting sensible conclusions. Not pure proof, but good enough... close enough. Still, it would be great to explain it in pure terms.
Timothy Golden
Sep 23, 2021, 4:39:58 PM9/23/21
to
On Thursday, September 23, 2021 at 9:55:29 AM UTC-4, Timothy Golden wrote:
> On Wednesday, September 22, 2021 at 6:04:19 PM UTC-4, michael Rodriguez wrote:
> > The format (Real,complex) actually is referring to the approach using complex slices (ring direct sum)
> > Now Hagen starts from the complex slices and he obtains the polysign numbers
> > In some sense, one may consider also other way around, starting from the polysign and ending
> > in the complex slices ( finding the 1 and i of each complex slice written in the polysigned format).
> > When the multiplicative inverse procedure 'fails', does not fail, but rather "get trapped" in a complex slice,
> > and one end up actually getting the "flat reciprocal" in that complex slice.
> > The reason of id_? is because you don't know if the procedure will get trapped in a complex slice or not.
> >
> > Z1'Z2' = Z1Z2 is ok. If you do it in a particular n, for example in n=5,
> > this is, doing the product MU'MU'=MU , there is no problem. If one is smelling
> > that it works for a great number of them, one may trust that is true, still one
> > have to prove it for n. In any case there is no problem is one does it for a
> > particular case, actually checking it for that particular case.
> Well, I have a computer proof by numerous instances. It was a surprise to me when I saw it in your work.
> As we come to depend upon the inverse we are somewhat dealing in axial values; in what arguably have congruence to the real line, or P2. Yet we are in potentially high dimension natively where we are doing this. The product of two values is the product of their antipodes. It is new to me. I agree there should be a formal proof. Could it stem from the MU form you state? That's pretty much where I saw it: you wrote something like
> s1 x1 ' s1 x2 ' = (s1@s2)x1x2
> so there it was in the sx form by you. Maybe you used m's instead of x's. I tried it on random z values and it works by instantiation.
> >
> > The id_0 was for (-a+a*a#a)(-1/4+1/4*1/4#1/4) = (-a+a*a#a)
> >
> > The reason for not naming it auxp2 is because is totally contained in the mainPn space,
> > but you could consider that way and see if better treating it that way.
> Yes, I guess I see this now.
> I tried working it out on a generic P4 value
> - a + b * c # d
So I got some code going that is getting the P4 reciprocal reliably.
> On Wednesday, September 22, 2021 at 6:04:19 PM UTC-4, michael Rodriguez wrote:
> > The format (Real,complex) actually is referring to the approach using complex slices (ring direct sum)
> > Now Hagen starts from the complex slices and he obtains the polysign numbers
> > In some sense, one may consider also other way around, starting from the polysign and ending
> > in the complex slices ( finding the 1 and i of each complex slice written in the polysigned format).
> > When the multiplicative inverse procedure 'fails', does not fail, but rather "get trapped" in a complex slice,
> > and one end up actually getting the "flat reciprocal" in that complex slice.
> > The reason of id_? is because you don't know if the procedure will get trapped in a complex slice or not.
> >
> > Z1'Z2' = Z1Z2 is ok. If you do it in a particular n, for example in n=5,
> > this is, doing the product MU'MU'=MU , there is no problem. If one is smelling
> > that it works for a great number of them, one may trust that is true, still one
> > have to prove it for n. In any case there is no problem is one does it for a
> > particular case, actually checking it for that particular case.
> Well, I have a computer proof by numerous instances. It was a surprise to me when I saw it in your work.
> As we come to depend upon the inverse we are somewhat dealing in axial values; in what arguably have congruence to the real line, or P2. Yet we are in potentially high dimension natively where we are doing this. The product of two values is the product of their antipodes. It is new to me. I agree there should be a formal proof. Could it stem from the MU form you state? That's pretty much where I saw it: you wrote something like
> s1 x1 ' s1 x2 ' = (s1@s2)x1x2
> so there it was in the sx form by you. Maybe you used m's instead of x's. I tried it on random z values and it works by instantiation.
> >
> > The id_0 was for (-a+a*a#a)(-1/4+1/4*1/4#1/4) = (-a+a*a#a)
> >
> > The reason for not naming it auxp2 is because is totally contained in the mainPn space,
> > but you could consider that way and see if better treating it that way.
> Yes, I guess I see this now.
> I tried working it out on a generic P4 value
> - a + b * c # d
I thought maybe I'd have an inversion problem but there is no inversion happening. Your algorithm is good.
No need to go over to RxC for solutions now that a 2x2 style is working yet that is how I envisioned it going too.
It's all still true about the disk and the line embedded in P4, but it is the symmetry of the polysign system which is allowing this computation.
The inverse happens to be easily coded because the doubles in x[] respect minus sign, which means in effect the hardware is doing P2 computations. Whatever, this is not problematic though to a noob it could look suspicious. Code probably could read more clearly but here it is:
Poly P4Reciprocal(Poly & z)
{
if(z.n !=4)
{ cout << "Sorry P4 only here.\n";
return z;
}
Poly z1(z);
z1[1] = - z1[1]; z1[3] = - z1[3];
Poly z2(z*z1);
Poly z3(z2);
z3[2] = - z3[2];
Poly z4;
z4 = z2 * z3;
Poly r(z.n);
r = z1 * z3;
r.ScalarProduct( 1.0 / z4.x[0] );
r.Reduce();
return r;
}
and the checker, which does 100,000 reciprocals in about a second:
void CheckP4Reciprocal()
{
Poly z(4);
Poly zr(4);
Poly check(4);
srand (time (0));
cout.precision(3);
for( int i = 0; i < 1000; i++ )
{
z.Randomize();
zr = P4Reciprocal(z);
check = z * zr;
check.Reduce();
for( int j= 1; j < z.n; j++ )
{ if( check[j] > 0.001 || 1 )
{
cout << "CheckP4Reciprocal() " << z << " * " << zr
<< " = " << check << "\n";
break;
}
}
}
}
and some results ( with ||1 it just prints everything but running silent it hasn't found a bad value yet)
CheckP4Reciprocal() [P4 0.92, 1, 0, 1.06 ] * [P4 1.08, 1.01, 0, 0.948 ] = [P4 1, 0, 4.44e-16, 0 ]
CheckP4Reciprocal() [P4 1.21, 0.472, 0, 0.584 ] * [P4 3.6, 0.0757, 2.79, 0 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 0.214, 0, 0.293, 1.14 ] * [P4 0, 1.26, 0.0609, 0.386 ] = [P4 1, 0, 5.55e-17, 5.55e-17 ]
CheckP4Reciprocal() [P4 0, 0.262, 0.364, 1.16 ] * [P4 0, 1.14, 0.387, 0.188 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 0.274, 1, 0.959, 0 ] * [P4 2.27, 0, 2.73, 0.68 ] = [P4 1, 4.44e-16, 0, 0 ]
CheckP4Reciprocal() [P4 0.79, 0, 1.08, 1.03 ] * [P4 0.918, 0.898, 1.17, 0 ] = [P4 1, 2.22e-16, 0, 0 ]
CheckP4Reciprocal() [P4 0.937, 0.198, 0, 0.97 ] * [P4 0.636, 2.74, 0, 2.22 ] = [P4 1, 4.44e-16, 0, 0 ]
CheckP4Reciprocal() [P4 0.312, 0.0876, 1.1, 0 ] * [P4 0, 0.18, 1.26, 0.32 ] = [P4 1, 1.67e-16, 0, 1.11e-16 ]
CheckP4Reciprocal() [P4 0.0363, 0.0534, 0, 1.03 ] * [P4 0.0529, 1.02, 0.0148, 0 ] = [P4 1, 0, 1.39e-17, 0 ]
CheckP4Reciprocal() [P4 0.803, 0.319, 0, 1.13 ] * [P4 0.621, 1.4, 0, 0.779 ] = [P4 1, 2.22e-16, 0, 0 ]
CheckP4Reciprocal() [P4 0, 0.929, 0.432, 1.1 ] * [P4 0, 1.71, 1.99, 0.902 ] = [P4 1, 0, 4.44e-16, 0 ]
CheckP4Reciprocal() [P4 0.828, 0.923, 0, 1.14 ] * [P4 1.13, 1.12, 0, 0.818 ] = [P4 1, 2.22e-16, 0, 2.22e-16 ]
CheckP4Reciprocal() [P4 1.14, 0, 0.622, 0.222 ] * [P4 1.63, 0.843, 0, 0.141 ] = [P4 1, 0, 0, 2.22e-16 ]
CheckP4Reciprocal() [P4 0.992, 0.958, 0.256, 0 ] * [P4 2.3, 0, 1.8, 0.657 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 1.14, 0.554, 0, 0.188 ] * [P4 1.8, 0, 1, 0.258 ] = [P4 1, 2.22e-16, 0, 2.22e-16 ]
CheckP4Reciprocal() [P4 0.943, 0, 0.529, 1.13 ] * [P4 1.98, 0.782, 1.69, 0 ] = [P4 1, 4.44e-16, 4.44e-16, 0 ]
CheckP4Reciprocal() [P4 0, 0.757, 0.889, 1.17 ] * [P4 0, 1.16, 0.925, 0.729 ] = [P4 1, 0, 2.22e-16, 2.22e-16 ]
CheckP4Reciprocal() [P4 1.05, 0.934, 0.322, 0 ] * [P4 1.73, 0, 1.21, 0.665 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 0, 1.07, 0.724, 1.05 ] * [P4 0, 1.03, 1.38, 1.06 ] = [P4 1, 4.44e-16, 0, 0 ]
CheckP4Reciprocal() [P4 0.0314, 1.07, 0.304, 0 ] * [P4 0, 0.351, 0.222, 1.23 ] = [P4 1, 0, 1.11e-16, 1.11e-16 ]
CheckP4Reciprocal() [P4 0.0307, 0, 0.932, 0.823 ] * [P4 3.56, 0.552, 4.16, 0 ] = [P4 1, 0, 4.44e-16, 4.44e-16 ]
CheckP4Reciprocal() [P4 1.14, 0.288, 0, 0.723 ] * [P4 4.36, 0.292, 3.59, 0 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 0.999, 0, 1.04, 0.398 ] * [P4 1.43, 2.49, 1.67, 0 ] = [P4 1, 0, 0, 0 ]
CheckP4Reciprocal() [P4 0.476, 0, 0.31, 1.18 ] * [P4 0.117, 1.74, 0, 0.908 ] = [P4 1, 0, 0, 0 ]
I did get an input system going too, so if you had some products you wanted done I could get them if they are in the notation used here in the printout: "[P6 0.47, 0, 0.31, 1.18, 1.2, 0.2 ] ". Can't do inverses though, or deg notation either. Concrete values only.
checking some naughty values and some nice values:
P4 value:[P4 0,1,0,1]
CheckP4Reciprocal() [P4 0, 1, 0, 1 ] * [P4 -nan, -nan, -nan, -nan ] = [P4 -nan, -nan, -nan, -nan ]
P4 value:[P4 1,0,1,0]
CheckP4Reciprocal() [P4 1, 0, 1, 0 ] * [P4 -nan, -nan, -nan, -nan ] = [P4 -nan, -nan, -nan, -nan ]
P4 value: [P4 1, 0, 0, 0]
CheckP4Reciprocal() [P4 1, 0, 0, 0 ] * [P4 1, 0, 0, 0 ] = [P4 1, 0, 0, 0 ]
P4 value:[P4 0, 1, 0, 0]
CheckP4Reciprocal() [P4 0, 1, 0, 0 ] * [P4 -0, -0, -0, 1 ] = [P4 1, 0, 0, 0 ]
P4 value:[P4 0,0,1,0]
CheckP4Reciprocal() [P4 0, 0, 1, 0 ] * [P4 0, 0, 1, 0 ] = [P4 1, 0, 0, 0 ]
P4 value:[P4 0,0,0,1]
CheckP4Reciprocal() [P4 0, 0, 0, 1 ] * [P4 -0, 1, -0, -0 ] = [P4 1, 0, 0, 0 ]
P4 value:[P4 2,1,0,1]
CheckP4Reciprocal() [P4 2, 1, 0, 1 ] * [P4 inf, -nan, inf, -nan ] = [P4 -nan, -nan, -nan, -nan ]
Lots of negative nans and even a few infs
Still these are correct answers in the manners of IEEE floats.
Much better than exiting on an exception.
OK, maybe there should be some error handling...
Ooching closer to the general solution but pretty happy just to get here honestly.
Nice work, Michael, and now I can pat myself on the back a little bit too.
It's really still surprising, though at some point it will normalize.
Next the P6 version and hopefully that will break through to general code.
I've run over 2 million checks now and no fails. The wonders of random number generators.
Timothy Golden
Sep 24, 2021, 10:55:54 AM9/24/21
to
> I've run over 2 million checks now and no fails. The wonders of random number generators.
so dropping the error theshhold to 1E-11 gets some reports and at 1E-12 roughly ten per 100000 report.
The error rises I think as the values encroach toward the embedded line or disk: (1E-11)
tatrix@deb9wd2tb:~/Torres$ ./compute
CheckP4Reciprocal() [P4 0.667, 1.22, 0.552, 0 ] * [P4 0.0764, 5.23e+05, 0, 5.23e+05 ] = [P4 1, 0, 0, 1.16e-10 ]
tatrix@deb9wd2tb:~/Torres$ ./compute
tatrix@deb9wd2tb:~/Torres$ ./compute
CheckP4Reciprocal() [P4 0.867, 0.00126, 0.866, 0 ] * [P4 471, 0, 274, 744 ] = [P4 1, 0, 2.13e-11, 0 ]
tatrix@deb9wd2tb:~/Torres$ ./compute
CheckP4Reciprocal() [P4 0, 0.867, 0.000722, 0.865 ] * [P4 140, 0, 295, 436 ] = [P4 1, 1.16e-11, 0, 1.16e-11 ]
tatrix@deb9wd2tb:~/Torres$ ./compute
CheckP4Reciprocal() [P4 0, 0.866, 0.892, 0.0261 ] * [P4 2.78e+05, 0, 2.78e+05, 0.56 ] = [P4 1, 0, 2.91e-11, 2.91e-11 ]
CheckP4Reciprocal() [P4 0.867, 0.00231, 0.867, 0 ] * [P4 162, 0, 252, 413 ] = [P4 1, 1.36e-11, 0, 1.36e-11 ]
Why so many 0.867 ish values is puzzling from an RNG. These are coming from Randomize() as unit length vectors through Unitize().
Yikes. Unitize() is still using Cartesian conversions... possibly these values indicate a bug. Just went over to native distance function and they are gone. Oops. Just threw in a Reduce() call and they are back again. I guess unit vectors with two balanced components and the other components quite small will filter through like this in terms of this reporting of small error. Phew.
void nSigned::Randomize()
{
int i;
int r;
for( i = 0; i < n; i++ )
{
r = rand();
x[i] = r;
}
Unitize();
return;
}
Timothy Golden
Sep 25, 2021, 11:13:49 AM9/25/21
to
On Thursday, September 9, 2021 at 5:29:33 PM UTC-4, michael Rodriguez wrote:
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
I'm as concerned with getting a clean interpretation of this notation as much as I am in getting it coded. Neither thus far is in my grasp.
> well, using º maybe be not ideal, but it is practical,. The other is more precisr, but is long.
Today I've gone over quite a bit again, tried a simple value to see if I could glean a little more 1/(0,1,0,0,0,0)=(0,0,0,0,1,0) through the algorithm just fine but what I was hoping was to find some expression in P6 of the intermediaries -1º and -1ºº but still don't have anything.
This is causing me to wonder if a subsign theory is blooming here. Also though it's not far from first incursions into imaginary numbers as well. As quickly as you put them up they go away again so shortly after. Although the terms are a lot of work the mechanics are not. It's really nice and simple the way you've worked it out.
I did bump into an error in my thinking on P6 as I worked through some of this. I had been thinking that the (0,b,0,d,0,f) and (a,0,c,0,e,0) removed of the real component ( 0,1,0,1,0,1) and (1,0,1,0,1,0) were acting as independent complex planes as in the associative algebra analysis of P6 as RxCxC and these series representations are MU first as you like to use them so the P6 value is
- a + b * c # d & e @ f
but while the 0b0d0f version does manage its products in its own subspace the a0c0e0 version yields components going into the 0b0d0f version. This is remedied by the augmentation of a minus sign to those products, and this is somewhat the emebedded real line playing into those 'versions' of complex planes embedded into P6. I suppose that the vision of those components as keys to those embedded complex planes is good but it's slightly more complicated. In effect what I am calling the removal of the real component is not quite actual, or rather it must be done every time. No, this language is not quite correct. Leaving behind RxCxC and going full native I am all for, yet here like in P4, which carries a 3D interpretation, is a higher dimension version (5D) with symmetries that nearly allow visibility to the mind. In truth the three rays of a0c0e0 are not opposed to 0b0d0f until all these elementary components are equal; the P2 embedding... at least that is one easy P2 embedding.100000 vs 011111 is another easy real embedding but the complex planes don't exactly jump out at you this way. Symmetry did provide the ability to perform projections down to 2D pixels out of the simplex coordinate system basis. Symmetry is the cause of your math working out too. I'll be away from my computer for a while but I hope by the time I get back you've gotten all of this resolved and sealed up into a nice neat bundle I can code up in a half hour.
michael Rodriguez
Sep 27, 2021, 4:07:58 PM9/27/21
to
> I'm as concerned with getting a clean interpretation of this notation as much as I am in getting it coded. Neither thus far is in my grasp.
> Today I've gone over quite a bit again, tried a simple value to see if I could glean a little more 1/(0,1,0,0,0,0)=(0,0,0,0,1,0) through the algorithm just
> fine but what I was hoping was to find some expression in P6 of the intermediaries -1º and -1ºº but still don't have anything.
It is simple. Just drop the º notation and do all the work in bidimensional notation.
> Today I've gone over quite a bit again, tried a simple value to see if I could glean a little more 1/(0,1,0,0,0,0)=(0,0,0,0,1,0) through the algorithm just
> fine but what I was hoping was to find some expression in P6 of the intermediaries -1º and -1ºº but still don't have anything.
Just working with -1 together with -1º and -1ºº could be misleading.
@1 ---> (@,@)1
@1º ---> (@,-)1
@1ºº ----> (@,+)1
this is, instead of using "ordered notation" in 1d, is in 2d
(_0_,_0_,_0_,_0_,_0_,_0_,_0_,_0_)
(_0_,_0_,_0_,_0_,_0_,_0_,_0_,_1_) = @1º = $1º
(_0_,_0_,_0_,_0_,_0_,_0_,_0_,_0_)
use may use the words 'ordered' or 'mod behavior' in 2d.
The reason of me not using subsign, is because using a name may induce to think that is a substructure, but, it does not
comply all the properties to be what you call subsign, although, it complies wit some.
Timothy Golden
Oct 2, 2021, 8:36:55 AM10/2/21
to
( a + b, a, a, a, ... ) = ( b, 0, 0, 0, ...)
so possibly thinking in terms of lifting values rather than reducing them to zero could help? Nah, that's got to be too simple. Still, if we had a little theorem that allowed a mechanical way of lifting two terms at a time, three terms at a time, and so forth up to (or down to I suppose) similar values, and this basically is what your notation is doing especially as the computed form exposes, then possibly the divide would be tunneled. Also the ease of the product and its FOIL method versus its reversal...to what degree could an actual division algorithm possibly expose the mechanism better than the raw reciprocal? z1/z2 ... Honestly I'm not doing much work here so disregard if you don't like this. I'm just getting back to you so you know I'm reading again.
I do understand that you are onto a 2D form above per sub-operation, yet as an intermediary is disappears almost as soon as you put it up. This is what makes me hope that there is a simpler form. From a number theory perspective we are sort of admitting that mod-6 numbers have mod-3 and mod-2 behaviors; at least under polysign. This is close to resurrecting the theorems of associative algebra that predicted that polysign would boil down to such as well, though they speak in terms of R and C. Here we can offer a broader theory because the embedding do not stop at R and C; P15 has P5 embedded in it, and while P5 may arguably have CxC correspondence that is not a straightforward path, but it is a path worth discussing for us because P5 is prime, yet our algorithm thus far does not allow a P3xP3 type of composition, which would be CxC in their vision. Supposedly one of the early( possibly here: https://www.jstor.org/stable/pdf/2369153.pdf)
Here is an intense quote from this p.217:
"The letters or units of the linear algebras, or to use- the
better term proposed by Mr. Charles S. Peirce, the vids of these algebras, are
fitted to perform a similar function each in its peculiar way. This is their
primitive and perhaps will always be their principal use. It does not exclude
the possibility of some, special modes of interpretation, but, on the contrary, a
higher philosophy, which believes in the capacity of the material universe for
all expressions of human thought, will find, in the utility of the vids, an indica-
tion of their probable reality of interpretation":
On p.109 he has an expression that looks strikingly close to polysign but he is in the terms of a generic notation. Anyway, he is greatly concerned with notation and pretty obviously we all ought to be.
Timothy Golden
Oct 2, 2021, 10:26:47 AM10/2/21
to
Back in time, this discussion was going on about P5: https://groups.google.com/g/sci.math/c/LiIG-10a9o0/m/yULzEDNReBkJ
I don't mean to make a diversion, but that was some quality input going on back then. One thing I know for certain: in math forums that are censored I will be censored. It happens almost every time.
michael Rodriguez
Oct 26, 2021, 3:37:50 PM10/26/21
to
important section !!
With the embedded ONEs in Z,Z,Z,.. format, is trivial to suppose
[1,0,0] + [0,1,0] + [0,0,1] = [1,1,1]
where [1,1,1] <--> @1
and [1,1,1][E,F,G] = [E,F,G]
Now, suppose we already obtain some sets of (mutual) annihilators in any Pn
written in Z,Z,Z,... format is trivial
suppose we have [A,0,0] , [0,B,0] and [0,0,C]
if we multiply in pairs them
[A,0,0][0,B,0] = [0,0,0] = 0
[0,B,0][0,0,C] = [0,0,0] = 0
[0,0,C][A,0,0] = [0,0,0] = 0
if we add them, we get a "diagonal"
[A,0,0] @ [0,B,0] @ [0,0,C] = [A,B,C]
It is the case, that you may resettle [A,B,C] into [1,1,1]
by making the following question
[A,0,0]W @ [0,B,0]W @ [0,0,C]W = [1,1,1] ??
( [A,0,0] @ [0,B,0] @ [0,0,C] )(W) = [1,1,1] ??
also, instead of using format Z,Z,Z,.. , we can assign letters
[A,0,0] = T1
[0,B,0] = T2
[0,0,C] = T3
and doing the replacements we get
(T1)(W) @ (T2)(W) @ (T3)(W) = @1 ??
(T1 @ T2 @ T3)W = @1 ??
just a small equation !
W = @1 / (T1 @ T2 @ T3)
After W is cleared, we separate by component to get the wanted ONEs
(T1)(W) = [1,0,0]
(T2)(W) = [0,1,0]
(T3)(W) = [0,0,1]
The rest of the post is just consequence of the above
Example of the procedure in p4
The procedure start having the list of any orthogonal components that mutually
annihilates each other, here in p4, just two are needed
(-1 +1)(-1 *1) =
___|_-1__+1__
_____________
-1_|_+1__*1__
*1_|_#1__-1__
(-1 +1)(-1 *1) = 0
We ask the question
(-1 +1)W @ (-1 *1)W = @1 ??
( (-1 +1) @ (-1 *1) )W = @1 ??
(-2 +1 *1)W = @1
W = @1 / (-2 +1 *1)
we use the procedure for the reciprocal of -2 +1 *1
W = @1 / (-2 +1 *1) = 1/4(-1 +2 *3)
we separate by components to get the ONEs
(-1 +1)( 1/4(-1 +2 *3) ) = (1/2)(-1 *1 #2) <--- [1,0]
(-1 *1)( 1/4(-1 +2 *3) ) = (1/2)(+1 #1) <-----[0,1]
Suspected ONEs for P5 and P7
[Z,Z,..,Z] for odd n
p5 "embedded" ONEs :
( 1/sqrt(5) )(@F -1 #1) <---> [1,0]
( 1/sqrt(5) )(@F +1 *1) <---> [0,1]
F = 1,61803398... https://oeis.org/A001622
p7 "embedded" ONEs :
(1/(@ 3Q @ 2P' @ 1) )(@P -1 $1)(@P +1 &1) <---> [1,0,0]
(1/(@ 3Q @ 2P' @ 1) )(@P +1 &1)(@P *1 #1) <---> [0,1,0]
(1/(@ 3Q @ 2P' @ 1) )(@P *1 #1)(@P -1 $1) <---> [0,0,1]
P = 1,8019377... https://oeis.org/A160389
Q = 2,2469796... https://oeis.org/A231187
(@ 3Q @ 2P' @ 1) can be considered an scalar, like ( 1/sqrt(5) )
in both cases, p5 and p7, is presented in (1/K)(T1 @ T2 @ T3 @ ...) format
because of concision
The factor K is what allows resize [A,0,0] into [1,0,0]
notice that ( (1/K)(T1 @ T2 @ T3 @ ...) )^2 = (1/K)(T1 @ T2 @ T3 @ ...)
this is, if we square any "embedded" ONE, and we get the same
In the case of the Pn with n even, there is a chance that what you call
identity axis contains an embedded ONE, like the p6 case
+1/3 @ #1/3 @ $1/3
> Back in time, this discussion was going on about P5: https://groups.google.com/g/sci.math/c/LiIG-
10a9o0/m/yULzEDNReBkJ
(@F -1 #1)(@F +1 *1) = 0
( 1/sqrt(5) )(@F -1 #1) @ ( 1/sqrt(5) )(@F +1 *1) = @1
notice
(@F -1 #1) @ (@F +1 *1) = @sqrt(5)
( ( 1/sqrt(5) )(@F -1 #1) )^2 = ( 1/sqrt(5) )(@F -1 #1)
( ( 1/sqrt(5) )(@F +1 *1) )^2 = ( 1/sqrt(5) )(@F +1 *1)
Regarding p7 embedded Ones
short math, See chapter 4 and chapter 5 of :
http://www.sacred-geometry.es/?q=en/content/golden-trisection
instead of the lowercase sigma, a Q will be used
P =aprox= 1.80194 and
Q =aprox= 2.24698
P^2 = 1 @ Q
QP = P @ Q
Q^2 = 1 @ PQ = 1 @ P @ Q
P^3 = (1 @ Q)(P) = P @ PQ = P @ P @ Q = 2P @ Q
P^4 = (2P @ Q)(P) = 2P^2 @ PQ = 2(1 @ Q) @ (P @ Q) = 2 @ 2Q @ P @ Q = P @ 3Q @ 2
(@P -1 $1)(@P +1 &1) =
___|_@P____-1__$1__
@P_|_@P^2__-P__$P__
+1_|_+P____*1__-1__
&1_|_&P____$1__#1__
(@P +1 &1)(@P *1 #1) =
___|_@P____+1__&1__
@P_|_@P^2__+P__&P__
*1_|_*P____&1__-1__
#1_|_#P____$1__+1__
(@P *1 #1)(@P -1 $1) =
___|_@P____*1__#1__
@P_|_@P^2__*P__#P__
-1_|_-P____#1__&1__
$1_|_$P____+1__*1__
(@P -1 $1)(@P +1 &1)(@P *1 #1) = 0
(@P -1 $1)(@P +1 &1) @ (@P +1 &1)(@P *1 #1) @ (@P *1 #1)(@P -1 $1) = @3Q @2P' @1
ask wolfram x^3 - x^2 - 2x + 1 = 0 and click 'exact form' with roots
notice that, although (@P +1 &1) does not annihilates with just one other partner,
the representation in C slices well could be something in the lines of [A,A,0]
Some check relative to the p7 ONEs
( (1/K)T )^2 = (1/K)T
( 1/(K^2) )(T^2) = (1/K)T
T^2 = (K)(T)
First part, calculate T^2
(@P^2 -P +P &P $P -1 *1 #1 $1 )^2 =
____|_@p^2__-P____+P____&P____$P____-1____*1____#1____$1____
____________________________________________________________
@P^2|_@P^4__-P^3__+P^3__&P^3__$P^3__-P^2__*P^2__#P^2__$P^2__
-P__|_-P^3__+P^2__*P^2__$P^2__@P^2__+P____#P____&P____@P____
+P__|_+P^3__*P^2__#P^2__@P^2__-P^2__*P____&P____$P____-P____
&P__|_&P^3__$P^2__@P^2__*P^2__#P^2__$P____-P____+P____#P____
$P__|_$P^3__@P^2__-P^2__#P^2__&p^2__@P____+P____*P____&P____
-1__|_-P^2__+P____*P____$P____@P____+1____#1____&1____@1____
*1__|_*P^2__#P____&P____-P____+P____#1____$1____@1____+1____
#1__|_#P^2__&P____$P____+P____*P____&1____@1____-1____*1____
$1__|_$P^2__@P____-P____#P____&P____@1____+1____*1____&1____
grouping the terms with mag 1 (16 terms)
+1 #1 &1 @1 #1 $1 @1 +1 &1 @1 -1 *1 @1 +1 *1 &1
cancelling terms (7 terms cancelled)
#1 @1 +1 &1 @1 @1 +1 *1 &1
ordering and collapsing terms
@1 @1 @1 +1 +1 *1 #1 &1 &1
@3 +2 *1 #1 &2
grouping the terms with mag P (16x2 terms), notice that they are duplicated
2( +P #P &P @P *P &P $P -P $P -P +P #P @P +P *P &P )
cancelling terms (14x2 terms cancelled)
2( +P &P ) = +2p +&2P
grouping terms with mag P^2 ( 16+(2x4) terms)
+P^2 *P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 &p^2 -P^2
*P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
cancelling terms (7 terms cancelled)
*P^2 @P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 -P^2 *P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
ordering
@P^2 @P^2 @P^2 -P^2 -P^2 -P^2 *P^2 *P^2 *P^2 *P^2 #P^2 #P^2 #P^2 #P^2 $P^2 $P^2 $P^2
grouping terms with mag P^3 (2x4 terms)
2(-P^3 +P^3 &P^3 $P^3) = -2P^3 +2P^3 &2P^3 $2P^3
grouping terms with mag P^4
@P^4
So far, we have :
@P^4
-2P^3 +2P^3 &2P^3 $2P^3
*P^2 @P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 -P^2 *P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
+2p &2P
@3 +2 *1 #1 &2
We proceed to expand
Expanding p^4 in @P^4
(@P @3Q @2)
Expanding P^3 in -2P^3 +2P^3 &2P^3 $2P^3
-2(2P @ Q) +2(2P @ Q) &2(2P @ Q) $2(2P @ Q)
-4P -2Q +4P +2Q &4P &2Q $4P $2Q
Expanding P^2 in *P^2 @P^2 $P^2 @P^2 ...
@(1 @ Q) @(1 @ Q) @(1 @ Q) -(1 @ Q) -(1 @ Q) -(1 @ Q) *(1 @ Q) *(1 @ Q) *(1 @ Q) *(1 @ Q)
#(1 @ Q) #(1 @ Q) #(1 @ Q) #(1 @ Q) $(1 @ Q) $(1 @ Q) $(1 @ Q)
@3 @3Q -3 -3Q *4 *4Q #4 #4Q $3 $3Q
So far, we have :
@P @3Q @2
-4P -2Q +4P +2Q &4P &2Q $4P $2Q
@3 @3Q -3 -3Q *4 *4Q #4 #4Q $3 $3Q
+2p &2P
@3 +2 *1 #1 &2
we divide in three sets (Ps,Qs and 1s)
@P -4P +4P +2p &4P &2P $4P
@3Q @3Q -2Q -3Q +2Q *4Q #4Q &2Q $3Q $2Q
@2 @3 @3 -3 +2 *1 *4 #4 #1 &2 $3
collapsing terms
@P -4P +6p &6P $4P
@6Q -5Q +2Q *4Q #4Q &2Q $5Q
@8 -3 +2 *5 #5 &2 $3
cancelling terms if needed
@P -4P +6p &6P $4P
@4Q -3Q *2Q #2Q $3Q
@6 -1 *3 #3 $1
At last, (@P^2 -P +P &P $P -1 *1 #1 $1 )^2 =
@4Q -3Q *2Q #2Q $3Q @P -4P +6p &6P $4P @6 -1 *3 #3 $1
(@4Q -3Q *2Q #2Q $3Q) @ (@P -4P +6p &6P $4P) @ (@6 -1 *3 #3 $1)
Second part, calculate (K)(T)
(3Q @ 2P' @ 1)(@1 @Q -P +P &P $P -1 *1 #1 $1) =
_____|_@1____@Q_____-P_____+P_____&P_____$P_____-1____*1____#1____$1____
________________________________________________________________________
@3Q__|_@3Q___@3Q^2__-3PQ___+3PQ___&3PQ___$3PQ___-3Q___*3Q___#3Q___$3Q___
@2P'_|_@2P'__@2PQ'__-2P^2'_+2P^2'_&2P^2'_$2P^2'_-2P'__*2P'__#2P'__$2P'__
@1___|_@1____@Q_____-P_____+P_____&P_____$P_____-1____*1____#1____$1____
we have
@1 -1 *1 #1 $1
@2P' -2P' *2P' #2P' $2P' -P +P &P $P
@Q @3Q -3Q *3Q #3Q $3Q
@2PQ' -3PQ +3PQ &3PQ $3PQ
-2P^2' +2P^2' &2P^2' $2P^2'
@3Q^2
collapsing and cancelling terms
@1 -1 *1 #1 $1
@2P' -P' *2P' #2P' $P' +P &P
@4Q -3Q *3Q #3Q $3Q
@2PQ' -3PQ +3PQ &3PQ $3PQ
-2P^2' +2P^2' &2P^2' $2P^2'
@3Q^2
expanding PQ in @2PQ' -3PQ +3PQ &3PQ $3PQ
@2(P @ Q)' -3(P @ Q) +3(P @ Q) &3(P @ Q) $3(P @ Q)
@2P' @2Q' -3P -3Q +3P +3Q &3P &3Q $3P $3Q
expanding P^2 in -2P^2' +2P^2' &2P^2' $2P^2'
-2(1 @ Q)' +2(1 @ Q)' &2(1 @ Q)' $(1 @ Q)'
-2' -2Q' +2' +2Q' &2' &2Q' $2' $2Q'
expanding Q^2 in @3Q^2
@3(1 @ P @ Q)
@3 @3P @3Q
So far, we have :
@1 -1 *1 #1 $1
@2P' -P' *2P' #2P' $P' +P &P
@4Q -3Q *3Q #3Q $3Q
@2P' @2Q' -3P -3Q +3P +3Q &3P &3Q $3P $3Q
-2' -2Q' +2' +2Q' &2' &2Q' $2' $2Q'
@3 @3P @3Q
ordering terms in Ps,Qs and 1S
@1 @3 -1 -2' +2' *1 #1 &2' $1 $2'
@3P @2P' @2P' -3P -P' +P +3P *2P' #2P' &P &3P $3P $P'
@4Q @3Q @2Q' -3Q -3Q -2Q' +3Q +2Q' *3Q #3Q &3Q &2Q' $3Q $3Q $2Q'
collapsing and cancelling terms
@4 -1' +2' *1 #1 &2' $1'
@P' -2P +4P *2P' #2P' &4P $2P
@5Q -4Q +Q *3Q #3Q &Q $4Q
separting depending if have or not the ' symbol
(@4 *1 #1) @ (-1' +2' &2' $1')
(-2P +4P &4P $2P) @ (@P' *2P' #2P')
@5Q -4Q +Q *3Q #3Q &Q $4Q
expanding -1' +2' &2' $1'
-1' = @1 -0 +1 *1 #1 &1 $1
+2' = @2 -2 +0 *2 #2 &2 $2
&2' = @2 -2 +2 *2 #2 &0 $2
$1' = @1 -1 +1 *1 #1 &1 $0
ordering
@1 @2 @2 @1 -2 -2 -1 +1 +2 +1 *1 *2 *2 *1 #1 #2 #2 #1 &1 &2 &1 $1 $2 $2
collapsing terms
@6 -5 +4 *6 #6 &4 $5
expanding @P' *2P' #2P'
@P' = @0 -P +P *P #P &P $P
*2P' = @2P -2P +2P *0 #2P &2P $2P
#2P' = @2P -2P +2P *2P #0 &2P $2P
ordering
@2P @2P -P -2P -2P +P +2P +2P *P *2P #P #2P &P &2P &2P $P $2P $2P
collapsing terms
@4P -5P +5P *3P #3P &5P $5P
So far we have :
(@4 *1 #1) @ (@6 -5 +4 *6 #6 &4 $5)
(-2P +4P &4P $2P) @ (@4P -5P +5P *3P #3P &5P $5P)
@5Q -4Q +Q *3Q #3Q &Q $4Q
collapsing terms
@10 -5 +4 *7 #7 &4 $5
@4P -7P +9P *3P #3P &9P $7P
@5Q -4Q +Q *3Q #3Q &Q $4Q
cancelling terms
@6 -1 *3 #3 $1
@P -4P +6P &6P $4P
@4Q -3Q +0 *2Q #2Q &0 $3Q
At last
(3Q @ 2P' @ 1)(@1 @Q -P +P &P $P -1 *1 #1 $1) =
(@4Q -3Q *2Q #2Q $3Q) @ (@P -4P +6P &6P $4P) @ (@6 -1 *3 #3 $1)
In both cases, the result is the same
Other pertinent topic
https://en.wikipedia.org/wiki/Catastrophic_cancellation
With the embedded ONEs in Z,Z,Z,.. format, is trivial to suppose
[1,0,0] + [0,1,0] + [0,0,1] = [1,1,1]
where [1,1,1] <--> @1
and [1,1,1][E,F,G] = [E,F,G]
Now, suppose we already obtain some sets of (mutual) annihilators in any Pn
written in Z,Z,Z,... format is trivial
suppose we have [A,0,0] , [0,B,0] and [0,0,C]
if we multiply in pairs them
[A,0,0][0,B,0] = [0,0,0] = 0
[0,B,0][0,0,C] = [0,0,0] = 0
[0,0,C][A,0,0] = [0,0,0] = 0
if we add them, we get a "diagonal"
[A,0,0] @ [0,B,0] @ [0,0,C] = [A,B,C]
It is the case, that you may resettle [A,B,C] into [1,1,1]
by making the following question
[A,0,0]W @ [0,B,0]W @ [0,0,C]W = [1,1,1] ??
( [A,0,0] @ [0,B,0] @ [0,0,C] )(W) = [1,1,1] ??
also, instead of using format Z,Z,Z,.. , we can assign letters
[A,0,0] = T1
[0,B,0] = T2
[0,0,C] = T3
and doing the replacements we get
(T1)(W) @ (T2)(W) @ (T3)(W) = @1 ??
(T1 @ T2 @ T3)W = @1 ??
just a small equation !
W = @1 / (T1 @ T2 @ T3)
After W is cleared, we separate by component to get the wanted ONEs
(T1)(W) = [1,0,0]
(T2)(W) = [0,1,0]
(T3)(W) = [0,0,1]
The rest of the post is just consequence of the above
Example of the procedure in p4
The procedure start having the list of any orthogonal components that mutually
annihilates each other, here in p4, just two are needed
(-1 +1)(-1 *1) =
___|_-1__+1__
_____________
-1_|_+1__*1__
*1_|_#1__-1__
(-1 +1)(-1 *1) = 0
We ask the question
(-1 +1)W @ (-1 *1)W = @1 ??
( (-1 +1) @ (-1 *1) )W = @1 ??
(-2 +1 *1)W = @1
W = @1 / (-2 +1 *1)
we use the procedure for the reciprocal of -2 +1 *1
W = @1 / (-2 +1 *1) = 1/4(-1 +2 *3)
we separate by components to get the ONEs
(-1 +1)( 1/4(-1 +2 *3) ) = (1/2)(-1 *1 #2) <--- [1,0]
(-1 *1)( 1/4(-1 +2 *3) ) = (1/2)(+1 #1) <-----[0,1]
Suspected ONEs for P5 and P7
[Z,Z,..,Z] for odd n
p5 "embedded" ONEs :
( 1/sqrt(5) )(@F -1 #1) <---> [1,0]
( 1/sqrt(5) )(@F +1 *1) <---> [0,1]
F = 1,61803398... https://oeis.org/A001622
p7 "embedded" ONEs :
(1/(@ 3Q @ 2P' @ 1) )(@P -1 $1)(@P +1 &1) <---> [1,0,0]
(1/(@ 3Q @ 2P' @ 1) )(@P +1 &1)(@P *1 #1) <---> [0,1,0]
(1/(@ 3Q @ 2P' @ 1) )(@P *1 #1)(@P -1 $1) <---> [0,0,1]
P = 1,8019377... https://oeis.org/A160389
Q = 2,2469796... https://oeis.org/A231187
(@ 3Q @ 2P' @ 1) can be considered an scalar, like ( 1/sqrt(5) )
in both cases, p5 and p7, is presented in (1/K)(T1 @ T2 @ T3 @ ...) format
because of concision
The factor K is what allows resize [A,0,0] into [1,0,0]
notice that ( (1/K)(T1 @ T2 @ T3 @ ...) )^2 = (1/K)(T1 @ T2 @ T3 @ ...)
this is, if we square any "embedded" ONE, and we get the same
In the case of the Pn with n even, there is a chance that what you call
identity axis contains an embedded ONE, like the p6 case
+1/3 @ #1/3 @ $1/3
> Back in time, this discussion was going on about P5: https://groups.google.com/g/sci.math/c/LiIG-
10a9o0/m/yULzEDNReBkJ
( 1/sqrt(5) )(@F -1 #1) @ ( 1/sqrt(5) )(@F +1 *1) = @1
notice
(@F -1 #1) @ (@F +1 *1) = @sqrt(5)
( ( 1/sqrt(5) )(@F -1 #1) )^2 = ( 1/sqrt(5) )(@F -1 #1)
( ( 1/sqrt(5) )(@F +1 *1) )^2 = ( 1/sqrt(5) )(@F +1 *1)
Regarding p7 embedded Ones
short math, See chapter 4 and chapter 5 of :
http://www.sacred-geometry.es/?q=en/content/golden-trisection
instead of the lowercase sigma, a Q will be used
P =aprox= 1.80194 and
Q =aprox= 2.24698
P^2 = 1 @ Q
QP = P @ Q
Q^2 = 1 @ PQ = 1 @ P @ Q
P^3 = (1 @ Q)(P) = P @ PQ = P @ P @ Q = 2P @ Q
P^4 = (2P @ Q)(P) = 2P^2 @ PQ = 2(1 @ Q) @ (P @ Q) = 2 @ 2Q @ P @ Q = P @ 3Q @ 2
(@P -1 $1)(@P +1 &1) =
___|_@P____-1__$1__
@P_|_@P^2__-P__$P__
+1_|_+P____*1__-1__
&1_|_&P____$1__#1__
(@P +1 &1)(@P *1 #1) =
___|_@P____+1__&1__
@P_|_@P^2__+P__&P__
*1_|_*P____&1__-1__
#1_|_#P____$1__+1__
(@P *1 #1)(@P -1 $1) =
___|_@P____*1__#1__
@P_|_@P^2__*P__#P__
-1_|_-P____#1__&1__
$1_|_$P____+1__*1__
(@P -1 $1)(@P +1 &1)(@P *1 #1) = 0
(@P -1 $1)(@P +1 &1) @ (@P +1 &1)(@P *1 #1) @ (@P *1 #1)(@P -1 $1) = @3Q @2P' @1
ask wolfram x^3 - x^2 - 2x + 1 = 0 and click 'exact form' with roots
notice that, although (@P +1 &1) does not annihilates with just one other partner,
the representation in C slices well could be something in the lines of [A,A,0]
Some check relative to the p7 ONEs
( (1/K)T )^2 = (1/K)T
( 1/(K^2) )(T^2) = (1/K)T
T^2 = (K)(T)
First part, calculate T^2
(@P^2 -P +P &P $P -1 *1 #1 $1 )^2 =
____|_@p^2__-P____+P____&P____$P____-1____*1____#1____$1____
____________________________________________________________
@P^2|_@P^4__-P^3__+P^3__&P^3__$P^3__-P^2__*P^2__#P^2__$P^2__
-P__|_-P^3__+P^2__*P^2__$P^2__@P^2__+P____#P____&P____@P____
+P__|_+P^3__*P^2__#P^2__@P^2__-P^2__*P____&P____$P____-P____
&P__|_&P^3__$P^2__@P^2__*P^2__#P^2__$P____-P____+P____#P____
$P__|_$P^3__@P^2__-P^2__#P^2__&p^2__@P____+P____*P____&P____
-1__|_-P^2__+P____*P____$P____@P____+1____#1____&1____@1____
*1__|_*P^2__#P____&P____-P____+P____#1____$1____@1____+1____
#1__|_#P^2__&P____$P____+P____*P____&1____@1____-1____*1____
$1__|_$P^2__@P____-P____#P____&P____@1____+1____*1____&1____
grouping the terms with mag 1 (16 terms)
+1 #1 &1 @1 #1 $1 @1 +1 &1 @1 -1 *1 @1 +1 *1 &1
cancelling terms (7 terms cancelled)
#1 @1 +1 &1 @1 @1 +1 *1 &1
ordering and collapsing terms
@1 @1 @1 +1 +1 *1 #1 &1 &1
@3 +2 *1 #1 &2
grouping the terms with mag P (16x2 terms), notice that they are duplicated
2( +P #P &P @P *P &P $P -P $P -P +P #P @P +P *P &P )
cancelling terms (14x2 terms cancelled)
2( +P &P ) = +2p +&2P
grouping terms with mag P^2 ( 16+(2x4) terms)
+P^2 *P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 &p^2 -P^2
*P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
cancelling terms (7 terms cancelled)
*P^2 @P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 -P^2 *P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
ordering
@P^2 @P^2 @P^2 -P^2 -P^2 -P^2 *P^2 *P^2 *P^2 *P^2 #P^2 #P^2 #P^2 #P^2 $P^2 $P^2 $P^2
grouping terms with mag P^3 (2x4 terms)
2(-P^3 +P^3 &P^3 $P^3) = -2P^3 +2P^3 &2P^3 $2P^3
grouping terms with mag P^4
@P^4
So far, we have :
@P^4
-2P^3 +2P^3 &2P^3 $2P^3
*P^2 @P^2 $P^2 @P^2 *P^2 #P^2 @P^2 -P^2 #P^2 -P^2 *P^2 #P^2 $P^2 -P^2 *P^2 #P^2 $P^2
+2p &2P
@3 +2 *1 #1 &2
We proceed to expand
Expanding p^4 in @P^4
(@P @3Q @2)
Expanding P^3 in -2P^3 +2P^3 &2P^3 $2P^3
-2(2P @ Q) +2(2P @ Q) &2(2P @ Q) $2(2P @ Q)
-4P -2Q +4P +2Q &4P &2Q $4P $2Q
Expanding P^2 in *P^2 @P^2 $P^2 @P^2 ...
@(1 @ Q) @(1 @ Q) @(1 @ Q) -(1 @ Q) -(1 @ Q) -(1 @ Q) *(1 @ Q) *(1 @ Q) *(1 @ Q) *(1 @ Q)
#(1 @ Q) #(1 @ Q) #(1 @ Q) #(1 @ Q) $(1 @ Q) $(1 @ Q) $(1 @ Q)
@3 @3Q -3 -3Q *4 *4Q #4 #4Q $3 $3Q
So far, we have :
@P @3Q @2
-4P -2Q +4P +2Q &4P &2Q $4P $2Q
@3 @3Q -3 -3Q *4 *4Q #4 #4Q $3 $3Q
+2p &2P
@3 +2 *1 #1 &2
we divide in three sets (Ps,Qs and 1s)
@P -4P +4P +2p &4P &2P $4P
@3Q @3Q -2Q -3Q +2Q *4Q #4Q &2Q $3Q $2Q
@2 @3 @3 -3 +2 *1 *4 #4 #1 &2 $3
collapsing terms
@P -4P +6p &6P $4P
@6Q -5Q +2Q *4Q #4Q &2Q $5Q
@8 -3 +2 *5 #5 &2 $3
cancelling terms if needed
@P -4P +6p &6P $4P
@4Q -3Q *2Q #2Q $3Q
@6 -1 *3 #3 $1
At last, (@P^2 -P +P &P $P -1 *1 #1 $1 )^2 =
@4Q -3Q *2Q #2Q $3Q @P -4P +6p &6P $4P @6 -1 *3 #3 $1
(@4Q -3Q *2Q #2Q $3Q) @ (@P -4P +6p &6P $4P) @ (@6 -1 *3 #3 $1)
Second part, calculate (K)(T)
(3Q @ 2P' @ 1)(@1 @Q -P +P &P $P -1 *1 #1 $1) =
_____|_@1____@Q_____-P_____+P_____&P_____$P_____-1____*1____#1____$1____
________________________________________________________________________
@3Q__|_@3Q___@3Q^2__-3PQ___+3PQ___&3PQ___$3PQ___-3Q___*3Q___#3Q___$3Q___
@2P'_|_@2P'__@2PQ'__-2P^2'_+2P^2'_&2P^2'_$2P^2'_-2P'__*2P'__#2P'__$2P'__
@1___|_@1____@Q_____-P_____+P_____&P_____$P_____-1____*1____#1____$1____
we have
@1 -1 *1 #1 $1
@2P' -2P' *2P' #2P' $2P' -P +P &P $P
@Q @3Q -3Q *3Q #3Q $3Q
@2PQ' -3PQ +3PQ &3PQ $3PQ
-2P^2' +2P^2' &2P^2' $2P^2'
@3Q^2
collapsing and cancelling terms
@1 -1 *1 #1 $1
@2P' -P' *2P' #2P' $P' +P &P
@4Q -3Q *3Q #3Q $3Q
@2PQ' -3PQ +3PQ &3PQ $3PQ
-2P^2' +2P^2' &2P^2' $2P^2'
@3Q^2
expanding PQ in @2PQ' -3PQ +3PQ &3PQ $3PQ
@2(P @ Q)' -3(P @ Q) +3(P @ Q) &3(P @ Q) $3(P @ Q)
@2P' @2Q' -3P -3Q +3P +3Q &3P &3Q $3P $3Q
expanding P^2 in -2P^2' +2P^2' &2P^2' $2P^2'
-2(1 @ Q)' +2(1 @ Q)' &2(1 @ Q)' $(1 @ Q)'
-2' -2Q' +2' +2Q' &2' &2Q' $2' $2Q'
expanding Q^2 in @3Q^2
@3(1 @ P @ Q)
@3 @3P @3Q
So far, we have :
@1 -1 *1 #1 $1
@2P' -P' *2P' #2P' $P' +P &P
@4Q -3Q *3Q #3Q $3Q
@2P' @2Q' -3P -3Q +3P +3Q &3P &3Q $3P $3Q
-2' -2Q' +2' +2Q' &2' &2Q' $2' $2Q'
@3 @3P @3Q
ordering terms in Ps,Qs and 1S
@1 @3 -1 -2' +2' *1 #1 &2' $1 $2'
@3P @2P' @2P' -3P -P' +P +3P *2P' #2P' &P &3P $3P $P'
@4Q @3Q @2Q' -3Q -3Q -2Q' +3Q +2Q' *3Q #3Q &3Q &2Q' $3Q $3Q $2Q'
collapsing and cancelling terms
@4 -1' +2' *1 #1 &2' $1'
@P' -2P +4P *2P' #2P' &4P $2P
@5Q -4Q +Q *3Q #3Q &Q $4Q
separting depending if have or not the ' symbol
(@4 *1 #1) @ (-1' +2' &2' $1')
(-2P +4P &4P $2P) @ (@P' *2P' #2P')
@5Q -4Q +Q *3Q #3Q &Q $4Q
expanding -1' +2' &2' $1'
-1' = @1 -0 +1 *1 #1 &1 $1
+2' = @2 -2 +0 *2 #2 &2 $2
&2' = @2 -2 +2 *2 #2 &0 $2
$1' = @1 -1 +1 *1 #1 &1 $0
ordering
@1 @2 @2 @1 -2 -2 -1 +1 +2 +1 *1 *2 *2 *1 #1 #2 #2 #1 &1 &2 &1 $1 $2 $2
collapsing terms
@6 -5 +4 *6 #6 &4 $5
expanding @P' *2P' #2P'
@P' = @0 -P +P *P #P &P $P
*2P' = @2P -2P +2P *0 #2P &2P $2P
#2P' = @2P -2P +2P *2P #0 &2P $2P
ordering
@2P @2P -P -2P -2P +P +2P +2P *P *2P #P #2P &P &2P &2P $P $2P $2P
collapsing terms
@4P -5P +5P *3P #3P &5P $5P
So far we have :
(@4 *1 #1) @ (@6 -5 +4 *6 #6 &4 $5)
(-2P +4P &4P $2P) @ (@4P -5P +5P *3P #3P &5P $5P)
@5Q -4Q +Q *3Q #3Q &Q $4Q
collapsing terms
@10 -5 +4 *7 #7 &4 $5
@4P -7P +9P *3P #3P &9P $7P
@5Q -4Q +Q *3Q #3Q &Q $4Q
cancelling terms
@6 -1 *3 #3 $1
@P -4P +6P &6P $4P
@4Q -3Q +0 *2Q #2Q &0 $3Q
At last
(3Q @ 2P' @ 1)(@1 @Q -P +P &P $P -1 *1 #1 $1) =
(@4Q -3Q *2Q #2Q $3Q) @ (@P -4P +6P &6P $4P) @ (@6 -1 *3 #3 $1)
In both cases, the result is the same
Other pertinent topic
https://en.wikipedia.org/wiki/Catastrophic_cancellation
michael Rodriguez
Oct 26, 2021, 3:42:56 PM10/26/21
to
> Back in time, this discussion was going on about P5: https://groups.google.com/g/sci.math/c/LiIG-10a9o0/m/yULzEDNReBkJ
More or less close, the guy almost hit them in the thread you mentioned....
Timothy Golden
Oct 28, 2021, 7:53:37 PM10/28/21
to
On Tuesday, October 26, 2021 at 3:37:50 PM UTC-4, michael Rodriguez wrote:
> important section !!
>
> With the embedded ONEs in Z,Z,Z,.. format, is trivial to suppose
> [1,0,0] + [0,1,0] + [0,0,1] = [1,1,1]
>
> where [1,1,1] <--> @1
I'm not sure how to interpret your notation here.
> important section !!
>
> With the embedded ONEs in Z,Z,Z,.. format, is trivial to suppose
> [1,0,0] + [0,1,0] + [0,0,1] = [1,1,1]
>
> where [1,1,1] <--> @1
To me (1,1,1) is zero.
This would be an ordered form where
(a,b,c)= @ a - b + c
in NU first notation (neutral unity)
Clearly your brackets are something else but I don't quite understand what.
I guess I can ask why [1,1,1] maps to @1 ?
Ahh... reading down a bit you are studying the identity vectors as ONEs I suppose.
In P5 there are only two.... why three elements in the braces?
Ah, I see, you are working P7 here.
Wouldn't doubt if you are up to something good, but it's awfully cryptic.
I am lost below and not sure how much time to spend on it.
Maybe a post for us beginners?
If z is a ONE then:
zz = z
right?
I'm finding it more meaningful to specify the EMUs than the ONEs.
The EMU cubed is the ONE.
The EMU to the fourth is the EMU again.
The EMU squared will quite possible be interpreted as the EMU itself.
Pretty sure the EMU wins.
one vector does it all. Just like MU does it all.
MU over NU in terms of fundamental status...
Oh, under this naming your ONE is ENU.
Assuming of course that I am interpreting correctly.
z:[P5 0.270481, 0.494194, 0.931841, 0, 1.0701 ]
zz:[P5 0.270481, 1.0701, 0, 0.931841, 0.494194 ]
zzz:[P5 1.02333, 0, 0.632456, 0.632456, 2.81997e-14 ]
zzzz:[P5 0.270481, 0.494194, 0.931841, 0, 1.0701 ]
Error: 7.00297e-13
print 1.02333 / 0.632456
1.61802560178
The symmetry of the EMU and the EMU^2 is interesting.
Does it have to be this way?
z:[P5 0.270481, 0.931841, 1.0701, 0.494194, 0 ]
zz:[P5 0.270481, 0, 0.494194, 1.0701, 0.931841 ]
zzz:[P5 1.02333, 0.632456, 0, 1.37668e-14, 0.632456 ]
zzzz:[P5 0.270481, 0.931841, 1.0701, 0.494194, 0 ]
Error: 5.75459e-13
And here is the other one looking almost just the same as the first one.
So the EMU square thing is almost happening again. Yet these are the unique ones.
graffiti:
https://drive.google.com/file/d/1UF1pum8eLOR09jTIpt3GdMZlgxYiH5zR/view?usp=sharing
Timothy Golden
Oct 29, 2021, 11:36:46 AM10/29/21
to
I'm going to try to reduce the cryptic nature of Michael's and my interactions here.
We are studying polysign numbers as laid out at http://bandtechnology.com
I work from my computer mostly, whereas Michael is well cross-trained in many branches of mathematics.
He has recently worked out division in polysign. As easy as the product is performing division is a bear.
I still have not assembled Michael's work into a general algorithm. I don't know how to yet.
He resolved prime Pn quite a while ago and those coded up nicely.
The more general solution involves subPf (where f is a prime factor of n) while here we investigate another sort of subP3 behavior that is going on.
Always lurking in the background is the possibility of leaning on AA realizations to perform division.
These topics interconnect though the criticisms that are possible on the standing system from the native polysign form are substantial.
From associative algebra and possibly abstract algebra it is proven that high dimensional systems will map to copies of C and possibly R.
P4 are three dimensional and so would map to RxC which is in polysign P2 P3
P5 are four dimensional and so would map to CxC or P3 P3
P6: RxCxC
P7: CxCxC
...
Normally the specification of one of these complex planes in a higher dimensional space would require two vectors, but in this discussion this is not true. Just one vector suffices in Pn to expose the embedded P3 plane: the P3- component will suffice since its powers will expose the other two rays. These P3- components (or P2- ) are the EMUs below: Embedded Minus Unity (EMU). These EMUs below are not sorted and further they have potential pairs for the P3 versions since the P3+ vector is exactly symmetrical to the P3- vector. Arguably for numerical analysis those squares of the EMU should be presented here, and even the cube which is the identity vector of each planar subsystem, but this is the most compact version of the data. These are only printed out to six places but the algorithm works down to 1.0E-12 in error right now. This is the error between the fourth power of the EMU and the EMU itself as a geometric distance. Also this data is quite fresh, but most of the kinks have been worked out. This will not be satisfying to most mathematicians since there is no human analysis present. My computer did all the work. These are all unit vectors.
---------- P4 EMUs ---------
EMU0: [P4 0.224144, 1.06066, 0.836516, 0 ]
EMU1: [P4 0.224144, 0, 0.836516, 1.06066 ]
---------- P5 EMUs ---------
EMU0: [P5 0.270481, 0, 0.494194, 1.0701, 0.931841 ]
EMU1: [P5 0.270481, 1.0701, 0, 0.931841, 0.494194 ]
---------- P6 EMUs ---------
EMU0: [P6 0.263523, 0, 0.263523, 0.790569, 1.05409, 0.790569 ]
EMU1: [P6 0.263523, 0.790569, 1.05409, 0.790569, 0.263523, 0 ]
EMU2: [P6 1.58813e-13, 2.41695e-13, 0.790569, 7.82551e-14, 0, 0.790569 ]
---------- P7 EMUs ---------
EMU0: [P7 0.241909, 0.670141, 0.96223, 0.898227, 0.526326, 0.126578, 0 ]
EMU1: [P7 0.241909, 0.126578, 0.898227, 0.670141, 0, 0.526326, 0.96223 ]
EMU2: [P7 0.241909, 0.526326, 0.670141, 0.126578, 0.96223, 0, 0.898227 ]
---------- P8 EMUs ---------
EMU0: [P8 0.171193, 0, 0.6389, 0.810093, 0.171193, 3.48974e-13, 0.6389, 0.810093 ]
EMU1: [P8 0.217917, 0.903541, 0.0467241, 0.572822, 0.685624, 0, 0.856817, 0.330719 ]
EMU2: [P8 0.217917, 0, 0.0467241, 0.330719, 0.685624, 0.903541, 0.856817, 0.572822 ]
EMU3: [P8 0.171193, 0.810093, 0.6389, 5.45284e-13, 0.171193, 0.810093, 0.6389, 0 ]
---------- P9 EMUs ---------
EMU0: [P9 1.92801e-13, 3.41966e-13, 0.666667, 7.88102e-14, 0, 0.666667, 7.8812e-14, 0, 0.666667 ]
EMU1: [P9 0.195419, 0.758105, 1.5939e-13, 0.862086, 0, 0.758105, 0.195419, 0.494818, 0.494818 ]
EMU2: [P9 0.195419, 3.1513e-14, 0.494818, 0.862086, 0.494818, 0, 0.195419, 0.758105, 0.758105 ]
EMU3: [P9 2.49745e-13, 0.666667, 5.6478e-13, 1.36787e-13, 0.666667, 0, 1.36786e-13, 0.666667, 8.67362e-19 ]
---------- P10 EMUs ---------
EMU0: [P10 0.202861, 0.829986, 0.370645, 0.0274084, 0.698881, 0.627125, 0, 0.459341, 0.802577, 0.131105 ]
EMU1: [P10 0.202861, 0, 0.370645, 0.802577, 0.698881, 0.202861, 0, 0.370645, 0.802577, 0.698881 ]
EMU2: [P10 0.202861, 0.131105, 0.802577, 0.459341, 0, 0.627125, 0.698881, 0.0274084, 0.370645, 0.829986 ]
EMU3: [P10 0.202861, 0.370645, 0.698881, 0, 0.802577, 0.202861, 0.370645, 0.698881, 0, 0.802577 ]
EMU4: [P10 0.202861, 0.698881, 0.802577, 0.370645, 0, 0.202861, 0.698881, 0.802577, 0.370645, 0 ]
---------- P11 EMUs ---------
EMU0: [P11 0.201438, 0.640544, 0.803928, 0.500567, 0.0851409, 0.0433539, 0.424062, 0.782152, 0.698957, 0.271745, 0 ]
EMU1: [P11 0.201438, 0.698957, 0.0433539, 0.803928, 0, 0.782152, 0.0851409, 0.640544, 0.271745, 0.424062, 0.500567 ]
EMU2: [P11 0.201438, 0.0851409, 0.698957, 0.640544, 0.0433539, 0.271745, 0.803928, 0.424062, 0, 0.500567, 0.782152 ]
EMU3: [P11 0.201438, 0.500567, 0.424062, 0.271745, 0.640544, 0.0851409, 0.782152, 0, 0.803928, 0.0433539, 0.698957 ]
EMU4: [P11 0.796647, 0.732108, 0.55898, 0.33223, 0.123851, 4.42701e-15, 0, 0.123851, 0.33223, 0.55898, 0.732108 ]
---------- P12 EMUs ---------
EMU0: [P12 0.195434, 0.390868, 0.586302, 0.0523664, 0.781736, 0.0523664, 0.586302, 0.390868, 0.195434, 0.72937, 0, 0.72937 ]
EMU1: [P12 0.195434, 0.0523664, 0, 0.0523664, 0.195434, 0.390868, 0.586302, 0.72937, 0.781736, 0.72937, 0.586302, 0.390868 ]
EMU2: [P12 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302 ]
EMU3: [P12 0.143068, 0, 0.533936, 0.677003, 0.143068, 2.10337e-13, 0.533936, 0.677003, 0.143068, 1.73472e-18, 0.533936, 0.677003 ]
EMU4: [P12 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0 ]
EMU5: [P12 0.143068, 0.677003, 0.533936, 0, 0.143068, 0.677003, 0.533936, 0, 0.143068, 0.677003, 0.533936, 0 ]
---------- P13 EMUs ---------
EMU0: [P13 0.1872, 0.676888, 0.636672, 0.137289, 0.0571173, 0.537173, 0.733073, 0.300244, 0, 0.360449, 0.747587, 0.480467, 0.0289333 ]
EMU1: [P13 0.1872, 0.480467, 0.360449, 0.300244, 0.537173, 0.137289, 0.676888, 0.0289333, 0.747587, 0, 0.733073, 0.0571173, 0.636672 ]
EMU2: [P13 0.1872, 0.137289, 0.733073, 0.360449, 0.0289333, 0.636672, 0.537173, 0, 0.480467, 0.676888, 0.0571173, 0.300244, 0.747587 ]
EMU3: [P13 0.1872, 0.636672, 0.0571173, 0.733073, 0, 0.747587, 0.0289333, 0.676888, 0.137289, 0.537173, 0.300244, 0.360449, 0.480467 ]
EMU4: [P13 0.1872, 0.747587, 0.300244, 0.0571173, 0.676888, 0.480467, 0, 0.537173, 0.636672, 0.0289333, 0.360449, 0.733073, 0.137289 ]
EMU5: [P13 0.1872, 0, 0.137289, 0.480467, 0.733073, 0.676888, 0.360449, 0.0571173, 0.0289333, 0.300244, 0.636672, 0.747587, 0.537173 ]
---------- P14 EMUs ---------
EMU0: [P14 0.17804, 0, 0.0931587, 0.387366, 0.661077, 0.708182, 0.493211, 0.17804, 0, 0.0931587, 0.387366, 0.661077, 0.708182, 0.493211 ]
EMU1: [P14 0.17804, 0.227085, 0.708182, 0.0592186, 0.387366, 0.627137, 0, 0.542256, 0.493211, 0.0121131, 0.661077, 0.33293, 0.0931587, 0.720295 ]
EMU2: [P14 0.17804, 0.708182, 0.387366, 0, 0.493211, 0.661077, 0.0931587, 0.17804, 0.708182, 0.387366, 0, 0.493211, 0.661077, 0.0931587 ]
EMU3: [P14 0.17804, 0.0121131, 0.387366, 0.720295, 0.493211, 0.0592186, 0.0931587, 0.542256, 0.708182, 0.33293, 0, 0.227085, 0.661077, 0.627137 ]
EMU4: [P14 0.17804, 0.720295, 0.0931587, 0.33293, 0.661077, 0.0121131, 0.493211, 0.542256, 0, 0.627137, 0.387366, 0.0592186, 0.708182, 0.227085 ]
EMU5: [P14 0.17804, 0.387366, 0.493211, 0.0931587, 0.708182, 0, 0.661077, 0.17804, 0.387366, 0.493211, 0.0931587, 0.708182, 0, 0.661077 ]
EMU6: [P14 0.17804, 0.661077, 0, 0.708182, 0.0931587, 0.493211, 0.387366, 0.17804, 0.661077, 0, 0.708182, 0.0931587, 0.493211, 0.387366 ]
---------- P15 EMUs ---------
EMU0: [P15 0.168675, 0.454069, 0.667327, 0.667327, 0.454069, 0.168675, 0, 0.0596637, 0.308184, 0.581105, 0.697825, 0.581105, 0.308184, 0.0596637, 1.67089e-14 ]
EMU1: [P15 0.168675, 0.581105, 0.0596637, 0.667327, 0, 0.697825, 8.77999e-13, 0.667327, 0.0596637, 0.581105, 0.168675, 0.454069, 0.308184, 0.308184, 0.454069 ]
EMU2: [P15 0.168675, 0.0596637, 0, 4.21619e-13, 0.0596637, 0.168675, 0.308184, 0.454069, 0.581105, 0.667327, 0.697825, 0.667327, 0.581105, 0.454069, 0.308184 ]
EMU3: [P15 0.168675, 0.308184, 0.454069, 0.581105, 0.667327, 0.697825, 0.667327, 0.581105, 0.454069, 0.308184, 0.168675, 0.0596637, 4.55131e-14, 0, 0.0596637 ]
EMU4: [P15 0.168675, 0.308184, 0.581105, 4.70013e-13, 0.667327, 0.168675, 0.308184, 0.581105, 4.70013e-13, 0.667327, 0.168675, 0.308184, 0.581105, 0, 0.667327 ]
EMU5: [P15 0.168675, 1.38778e-17, 0.308184, 0.667327, 0.581105, 0.168675, 1.38778e-17, 0.308184, 0.667327, 0.581105, 0.168675, 0, 0.308184, 0.667327, 0.581105 ]
EMU6: [P15 0.168675, 0.0596637, 0.581105, 0.581105, 0.0596637, 0.168675, 0.667327, 0.454069, 0, 0.308184, 0.697825, 0.308184, 9.26442e-14, 0.454069, 0.667327 ]
---------- P16 EMUs ---------
EMU0: [P16 0.159499, 0.242061, 0.627126, 4.0766e-17, 0.501825, 0.419263, 0.0341986, 0.661324, 0.159499, 0.242061, 0.627126, 0, 0.501825, 0.419263, 0.0341986, 0.661324 ]
EMU1: [P16 0.1253, 8.67362e-19, 0.467627, 0.592927, 0.1253, 0, 0.467627, 0.592927, 0.1253, 8.67362e-19, 0.467627, 0.592927, 0.1253, 1.73472e-18, 0.467627, 0.592927 ]
EMU2: [P16 0.159499, 0.661324, 0.0341986, 0.419263, 0.501825, 7.93879e-13, 0.627126, 0.242061, 0.159499, 0.661324, 0.0341986, 0.419263, 0.501825, 0, 0.627126, 0.242061 ]
EMU3: [P16 0.168235, 0, 0.250797, 0.610984, 0.635861, 0.294715, 0.00873584, 0.131003, 0.510561, 0.678796, 0.427999, 0.067812, 0.0429344, 0.384081, 0.67006, 0.547793 ]
EMU4: [P16 0.168235, 0.294715, 0.427999, 0.547793, 0.635861, 0.678796, 0.67006, 0.610984, 0.510561, 0.384081, 0.250797, 0.131003, 0.0429344, 0, 0.00873584, 0.067812 ]
EMU5: [P16 0.168235, 0.067812, 0.00873584, 0, 0.0429344, 0.131003, 0.250797, 0.384081, 0.510561, 0.610984, 0.67006, 0.678796, 0.635861, 0.547793, 0.427999, 0.294715 ]
EMU6: [P16 0.1253, 0.592927, 0.467627, 0, 0.1253, 0.592927, 0.467627, 1.73472e-18, 0.1253, 0.592927, 0.467627, 1.73472e-18, 0.1253, 0.592927, 0.467627, 0 ]
EMU7: [P16 0.168235, 0.384081, 0.427999, 0.131003, 0.635861, 0, 0.67006, 0.067812, 0.510561, 0.294715, 0.250797, 0.547793, 0.0429344, 0.678796, 0.00873584, 0.610984 ]
We are studying polysign numbers as laid out at http://bandtechnology.com
I work from my computer mostly, whereas Michael is well cross-trained in many branches of mathematics.
He has recently worked out division in polysign. As easy as the product is performing division is a bear.
I still have not assembled Michael's work into a general algorithm. I don't know how to yet.
He resolved prime Pn quite a while ago and those coded up nicely.
The more general solution involves subPf (where f is a prime factor of n) while here we investigate another sort of subP3 behavior that is going on.
Always lurking in the background is the possibility of leaning on AA realizations to perform division.
These topics interconnect though the criticisms that are possible on the standing system from the native polysign form are substantial.
From associative algebra and possibly abstract algebra it is proven that high dimensional systems will map to copies of C and possibly R.
P4 are three dimensional and so would map to RxC which is in polysign P2 P3
P5 are four dimensional and so would map to CxC or P3 P3
P6: RxCxC
P7: CxCxC
...
Normally the specification of one of these complex planes in a higher dimensional space would require two vectors, but in this discussion this is not true. Just one vector suffices in Pn to expose the embedded P3 plane: the P3- component will suffice since its powers will expose the other two rays. These P3- components (or P2- ) are the EMUs below: Embedded Minus Unity (EMU). These EMUs below are not sorted and further they have potential pairs for the P3 versions since the P3+ vector is exactly symmetrical to the P3- vector. Arguably for numerical analysis those squares of the EMU should be presented here, and even the cube which is the identity vector of each planar subsystem, but this is the most compact version of the data. These are only printed out to six places but the algorithm works down to 1.0E-12 in error right now. This is the error between the fourth power of the EMU and the EMU itself as a geometric distance. Also this data is quite fresh, but most of the kinks have been worked out. This will not be satisfying to most mathematicians since there is no human analysis present. My computer did all the work. These are all unit vectors.
---------- P4 EMUs ---------
EMU0: [P4 0.224144, 1.06066, 0.836516, 0 ]
EMU1: [P4 0.224144, 0, 0.836516, 1.06066 ]
---------- P5 EMUs ---------
EMU0: [P5 0.270481, 0, 0.494194, 1.0701, 0.931841 ]
EMU1: [P5 0.270481, 1.0701, 0, 0.931841, 0.494194 ]
---------- P6 EMUs ---------
EMU0: [P6 0.263523, 0, 0.263523, 0.790569, 1.05409, 0.790569 ]
EMU1: [P6 0.263523, 0.790569, 1.05409, 0.790569, 0.263523, 0 ]
EMU2: [P6 1.58813e-13, 2.41695e-13, 0.790569, 7.82551e-14, 0, 0.790569 ]
---------- P7 EMUs ---------
EMU0: [P7 0.241909, 0.670141, 0.96223, 0.898227, 0.526326, 0.126578, 0 ]
EMU1: [P7 0.241909, 0.126578, 0.898227, 0.670141, 0, 0.526326, 0.96223 ]
EMU2: [P7 0.241909, 0.526326, 0.670141, 0.126578, 0.96223, 0, 0.898227 ]
---------- P8 EMUs ---------
EMU0: [P8 0.171193, 0, 0.6389, 0.810093, 0.171193, 3.48974e-13, 0.6389, 0.810093 ]
EMU1: [P8 0.217917, 0.903541, 0.0467241, 0.572822, 0.685624, 0, 0.856817, 0.330719 ]
EMU2: [P8 0.217917, 0, 0.0467241, 0.330719, 0.685624, 0.903541, 0.856817, 0.572822 ]
EMU3: [P8 0.171193, 0.810093, 0.6389, 5.45284e-13, 0.171193, 0.810093, 0.6389, 0 ]
---------- P9 EMUs ---------
EMU0: [P9 1.92801e-13, 3.41966e-13, 0.666667, 7.88102e-14, 0, 0.666667, 7.8812e-14, 0, 0.666667 ]
EMU1: [P9 0.195419, 0.758105, 1.5939e-13, 0.862086, 0, 0.758105, 0.195419, 0.494818, 0.494818 ]
EMU2: [P9 0.195419, 3.1513e-14, 0.494818, 0.862086, 0.494818, 0, 0.195419, 0.758105, 0.758105 ]
EMU3: [P9 2.49745e-13, 0.666667, 5.6478e-13, 1.36787e-13, 0.666667, 0, 1.36786e-13, 0.666667, 8.67362e-19 ]
---------- P10 EMUs ---------
EMU0: [P10 0.202861, 0.829986, 0.370645, 0.0274084, 0.698881, 0.627125, 0, 0.459341, 0.802577, 0.131105 ]
EMU1: [P10 0.202861, 0, 0.370645, 0.802577, 0.698881, 0.202861, 0, 0.370645, 0.802577, 0.698881 ]
EMU2: [P10 0.202861, 0.131105, 0.802577, 0.459341, 0, 0.627125, 0.698881, 0.0274084, 0.370645, 0.829986 ]
EMU3: [P10 0.202861, 0.370645, 0.698881, 0, 0.802577, 0.202861, 0.370645, 0.698881, 0, 0.802577 ]
EMU4: [P10 0.202861, 0.698881, 0.802577, 0.370645, 0, 0.202861, 0.698881, 0.802577, 0.370645, 0 ]
---------- P11 EMUs ---------
EMU0: [P11 0.201438, 0.640544, 0.803928, 0.500567, 0.0851409, 0.0433539, 0.424062, 0.782152, 0.698957, 0.271745, 0 ]
EMU1: [P11 0.201438, 0.698957, 0.0433539, 0.803928, 0, 0.782152, 0.0851409, 0.640544, 0.271745, 0.424062, 0.500567 ]
EMU2: [P11 0.201438, 0.0851409, 0.698957, 0.640544, 0.0433539, 0.271745, 0.803928, 0.424062, 0, 0.500567, 0.782152 ]
EMU3: [P11 0.201438, 0.500567, 0.424062, 0.271745, 0.640544, 0.0851409, 0.782152, 0, 0.803928, 0.0433539, 0.698957 ]
EMU4: [P11 0.796647, 0.732108, 0.55898, 0.33223, 0.123851, 4.42701e-15, 0, 0.123851, 0.33223, 0.55898, 0.732108 ]
---------- P12 EMUs ---------
EMU0: [P12 0.195434, 0.390868, 0.586302, 0.0523664, 0.781736, 0.0523664, 0.586302, 0.390868, 0.195434, 0.72937, 0, 0.72937 ]
EMU1: [P12 0.195434, 0.0523664, 0, 0.0523664, 0.195434, 0.390868, 0.586302, 0.72937, 0.781736, 0.72937, 0.586302, 0.390868 ]
EMU2: [P12 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302 ]
EMU3: [P12 0.143068, 0, 0.533936, 0.677003, 0.143068, 2.10337e-13, 0.533936, 0.677003, 0.143068, 1.73472e-18, 0.533936, 0.677003 ]
EMU4: [P12 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0, 0.195434, 0.586302, 0.781736, 0.586302, 0.195434, 0 ]
EMU5: [P12 0.143068, 0.677003, 0.533936, 0, 0.143068, 0.677003, 0.533936, 0, 0.143068, 0.677003, 0.533936, 0 ]
---------- P13 EMUs ---------
EMU0: [P13 0.1872, 0.676888, 0.636672, 0.137289, 0.0571173, 0.537173, 0.733073, 0.300244, 0, 0.360449, 0.747587, 0.480467, 0.0289333 ]
EMU1: [P13 0.1872, 0.480467, 0.360449, 0.300244, 0.537173, 0.137289, 0.676888, 0.0289333, 0.747587, 0, 0.733073, 0.0571173, 0.636672 ]
EMU2: [P13 0.1872, 0.137289, 0.733073, 0.360449, 0.0289333, 0.636672, 0.537173, 0, 0.480467, 0.676888, 0.0571173, 0.300244, 0.747587 ]
EMU3: [P13 0.1872, 0.636672, 0.0571173, 0.733073, 0, 0.747587, 0.0289333, 0.676888, 0.137289, 0.537173, 0.300244, 0.360449, 0.480467 ]
EMU4: [P13 0.1872, 0.747587, 0.300244, 0.0571173, 0.676888, 0.480467, 0, 0.537173, 0.636672, 0.0289333, 0.360449, 0.733073, 0.137289 ]
EMU5: [P13 0.1872, 0, 0.137289, 0.480467, 0.733073, 0.676888, 0.360449, 0.0571173, 0.0289333, 0.300244, 0.636672, 0.747587, 0.537173 ]
---------- P14 EMUs ---------
EMU0: [P14 0.17804, 0, 0.0931587, 0.387366, 0.661077, 0.708182, 0.493211, 0.17804, 0, 0.0931587, 0.387366, 0.661077, 0.708182, 0.493211 ]
EMU1: [P14 0.17804, 0.227085, 0.708182, 0.0592186, 0.387366, 0.627137, 0, 0.542256, 0.493211, 0.0121131, 0.661077, 0.33293, 0.0931587, 0.720295 ]
EMU2: [P14 0.17804, 0.708182, 0.387366, 0, 0.493211, 0.661077, 0.0931587, 0.17804, 0.708182, 0.387366, 0, 0.493211, 0.661077, 0.0931587 ]
EMU3: [P14 0.17804, 0.0121131, 0.387366, 0.720295, 0.493211, 0.0592186, 0.0931587, 0.542256, 0.708182, 0.33293, 0, 0.227085, 0.661077, 0.627137 ]
EMU4: [P14 0.17804, 0.720295, 0.0931587, 0.33293, 0.661077, 0.0121131, 0.493211, 0.542256, 0, 0.627137, 0.387366, 0.0592186, 0.708182, 0.227085 ]
EMU5: [P14 0.17804, 0.387366, 0.493211, 0.0931587, 0.708182, 0, 0.661077, 0.17804, 0.387366, 0.493211, 0.0931587, 0.708182, 0, 0.661077 ]
EMU6: [P14 0.17804, 0.661077, 0, 0.708182, 0.0931587, 0.493211, 0.387366, 0.17804, 0.661077, 0, 0.708182, 0.0931587, 0.493211, 0.387366 ]
---------- P15 EMUs ---------
EMU0: [P15 0.168675, 0.454069, 0.667327, 0.667327, 0.454069, 0.168675, 0, 0.0596637, 0.308184, 0.581105, 0.697825, 0.581105, 0.308184, 0.0596637, 1.67089e-14 ]
EMU1: [P15 0.168675, 0.581105, 0.0596637, 0.667327, 0, 0.697825, 8.77999e-13, 0.667327, 0.0596637, 0.581105, 0.168675, 0.454069, 0.308184, 0.308184, 0.454069 ]
EMU2: [P15 0.168675, 0.0596637, 0, 4.21619e-13, 0.0596637, 0.168675, 0.308184, 0.454069, 0.581105, 0.667327, 0.697825, 0.667327, 0.581105, 0.454069, 0.308184 ]
EMU3: [P15 0.168675, 0.308184, 0.454069, 0.581105, 0.667327, 0.697825, 0.667327, 0.581105, 0.454069, 0.308184, 0.168675, 0.0596637, 4.55131e-14, 0, 0.0596637 ]
EMU4: [P15 0.168675, 0.308184, 0.581105, 4.70013e-13, 0.667327, 0.168675, 0.308184, 0.581105, 4.70013e-13, 0.667327, 0.168675, 0.308184, 0.581105, 0, 0.667327 ]
EMU5: [P15 0.168675, 1.38778e-17, 0.308184, 0.667327, 0.581105, 0.168675, 1.38778e-17, 0.308184, 0.667327, 0.581105, 0.168675, 0, 0.308184, 0.667327, 0.581105 ]
EMU6: [P15 0.168675, 0.0596637, 0.581105, 0.581105, 0.0596637, 0.168675, 0.667327, 0.454069, 0, 0.308184, 0.697825, 0.308184, 9.26442e-14, 0.454069, 0.667327 ]
---------- P16 EMUs ---------
EMU0: [P16 0.159499, 0.242061, 0.627126, 4.0766e-17, 0.501825, 0.419263, 0.0341986, 0.661324, 0.159499, 0.242061, 0.627126, 0, 0.501825, 0.419263, 0.0341986, 0.661324 ]
EMU1: [P16 0.1253, 8.67362e-19, 0.467627, 0.592927, 0.1253, 0, 0.467627, 0.592927, 0.1253, 8.67362e-19, 0.467627, 0.592927, 0.1253, 1.73472e-18, 0.467627, 0.592927 ]
EMU2: [P16 0.159499, 0.661324, 0.0341986, 0.419263, 0.501825, 7.93879e-13, 0.627126, 0.242061, 0.159499, 0.661324, 0.0341986, 0.419263, 0.501825, 0, 0.627126, 0.242061 ]
EMU3: [P16 0.168235, 0, 0.250797, 0.610984, 0.635861, 0.294715, 0.00873584, 0.131003, 0.510561, 0.678796, 0.427999, 0.067812, 0.0429344, 0.384081, 0.67006, 0.547793 ]
EMU4: [P16 0.168235, 0.294715, 0.427999, 0.547793, 0.635861, 0.678796, 0.67006, 0.610984, 0.510561, 0.384081, 0.250797, 0.131003, 0.0429344, 0, 0.00873584, 0.067812 ]
EMU5: [P16 0.168235, 0.067812, 0.00873584, 0, 0.0429344, 0.131003, 0.250797, 0.384081, 0.510561, 0.610984, 0.67006, 0.678796, 0.635861, 0.547793, 0.427999, 0.294715 ]
EMU6: [P16 0.1253, 0.592927, 0.467627, 0, 0.1253, 0.592927, 0.467627, 1.73472e-18, 0.1253, 0.592927, 0.467627, 1.73472e-18, 0.1253, 0.592927, 0.467627, 0 ]
EMU7: [P16 0.168235, 0.384081, 0.427999, 0.131003, 0.635861, 0, 0.67006, 0.067812, 0.510561, 0.294715, 0.250797, 0.547793, 0.0429344, 0.678796, 0.00873584, 0.610984 ]
Timothy Golden
Oct 30, 2021, 10:30:51 AM10/30/21
to
On Friday, October 29, 2021 at 11:36:46 AM UTC-4, Timothy Golden wrote:
> I'm going to try to reduce the cryptic nature of Michael's and my interactions here.
> We are studying polysign numbers as laid out at http://bandtechnology.com
> I work from my computer mostly, whereas Michael is well cross-trained in many branches of mathematics.
> He has recently worked out division in polysign. As easy as the product is performing division is a bear.
> I still have not assembled Michael's work into a general algorithm. I don't know how to yet.
> He resolved prime Pn quite a while ago and those coded up nicely.
> The more general solution involves subPf (where f is a prime factor of n) while here we investigate another sort of subP3 behavior that is going on.
> Always lurking in the background is the possibility of leaning on AA realizations to perform division.
> These topics interconnect though the criticisms that are possible on the standing system from the native polysign form are substantial.
> From associative algebra and possibly abstract algebra it is proven that high dimensional systems will map to copies of C and possibly R.
> P4 are three dimensional and so would map to RxC which is in polysign P2 P3
> P5 are four dimensional and so would map to CxC or P3 P3
> P6: RxCxC
> P7: CxCxC
> ...
> Normally the specification of one of these complex planes in a higher dimensional space would require two vectors, but in this discussion this is not true. Just one vector suffices in Pn to expose the embedded P3 plane: the P3- component will suffice since its powers will expose the other two rays. These P3- components (or P2- ) are the EMUs below: Embedded Minus Unity (EMU). These EMUs below are not sorted and further they have potential pairs for the P3 versions since the P3+ vector is exactly symmetrical to the P3- vector. Arguably for numerical analysis those squares of the EMU should be presented here, and even the cube which is the identity vector of each planar subsystem, but this is the most compact version of the data. These are only printed out to six places but the algorithm works down to 1.0E-12 in error right now. This is the error between the fourth power of the EMU and the EMU itself as a geometric distance. Also this data is quite fresh, but most of the kinks have been worked out. This will not be satisfying to most mathematicians since there is no human analysis present. My computer did all the work. These are all unit vectors.
>
> ---------- P4 EMUs ---------
> EMU0: [P4 0.224144, 1.06066, 0.836516, 0 ]
> EMU1: [P4 0.224144, 0, 0.836516, 1.06066 ]
>
> ---------- P5 EMUs ---------
> EMU0: [P5 0.270481, 0, 0.494194, 1.0701, 0.931841 ]
> EMU1: [P5 0.270481, 1.0701, 0, 0.931841, 0.494194 ]
>
Pretty sure this data is not good yet. It's close, but I'm discovering some issues. The number of
> I'm going to try to reduce the cryptic nature of Michael's and my interactions here.
> We are studying polysign numbers as laid out at http://bandtechnology.com
> I work from my computer mostly, whereas Michael is well cross-trained in many branches of mathematics.
> He has recently worked out division in polysign. As easy as the product is performing division is a bear.
> I still have not assembled Michael's work into a general algorithm. I don't know how to yet.
> He resolved prime Pn quite a while ago and those coded up nicely.
> The more general solution involves subPf (where f is a prime factor of n) while here we investigate another sort of subP3 behavior that is going on.
> Always lurking in the background is the possibility of leaning on AA realizations to perform division.
> These topics interconnect though the criticisms that are possible on the standing system from the native polysign form are substantial.
> From associative algebra and possibly abstract algebra it is proven that high dimensional systems will map to copies of C and possibly R.
> P4 are three dimensional and so would map to RxC which is in polysign P2 P3
> P5 are four dimensional and so would map to CxC or P3 P3
> P6: RxCxC
> P7: CxCxC
> ...
> Normally the specification of one of these complex planes in a higher dimensional space would require two vectors, but in this discussion this is not true. Just one vector suffices in Pn to expose the embedded P3 plane: the P3- component will suffice since its powers will expose the other two rays. These P3- components (or P2- ) are the EMUs below: Embedded Minus Unity (EMU). These EMUs below are not sorted and further they have potential pairs for the P3 versions since the P3+ vector is exactly symmetrical to the P3- vector. Arguably for numerical analysis those squares of the EMU should be presented here, and even the cube which is the identity vector of each planar subsystem, but this is the most compact version of the data. These are only printed out to six places but the algorithm works down to 1.0E-12 in error right now. This is the error between the fourth power of the EMU and the EMU itself as a geometric distance. Also this data is quite fresh, but most of the kinks have been worked out. This will not be satisfying to most mathematicians since there is no human analysis present. My computer did all the work. These are all unit vectors.
>
> ---------- P4 EMUs ---------
> EMU0: [P4 0.224144, 1.06066, 0.836516, 0 ]
> EMU1: [P4 0.224144, 0, 0.836516, 1.06066 ]
>
> ---------- P5 EMUs ---------
> EMU0: [P5 0.270481, 0, 0.494194, 1.0701, 0.931841 ]
> EMU1: [P5 0.270481, 1.0701, 0, 0.931841, 0.494194 ]
>
z^4 = z
solutions is larger than I've allowed for.
Timothy Golden
Jan 7, 2022, 10:04:07 AM1/7/22
to
EP1MU[], EP2MU[], EP3MU[], EP4MU[], ...
and these are the embedded Pn Minus Unity findings in array format; preferably for instance EP3MU[2] will generate EP1MU[2] at its third power, and of course generate itself at its fourth power. This means that Michaels ONEs are relevant, for they are the first in the progression; they are the EP1MU[], which are NU (Neutral Unity). What is strange is that there could be EP4MU as outlined above, yet I don't believe that anybody has found them yet. Of course from code my job is quite easy. The EP4MU will respect:
z z z z z = z
but will they be so linear as the EP3MU? Linear? Wasn't that the EP2MU? Yes. Linear; Planar; Volumetric; ... and all that is left out is the P1 start. So is it a point or is it a ray? Is there a difference? EP1MU may be deserving of such interpretations within polysign which they cannot gain under ordinary mathematics. Certainly the native standard P1 does this. We ought to expect the embedded form does the same. Possibly this is a step in interdimensional analysis.
I am guilty of missing some of the connections and only just recently got to review some old threads here where bygone respected mathematicians did point me in a good direction:
"While many of Dedekind’s contributions to mathematics and its foundations are thus common knowledge, they are seldom discussed together. In particular, his foundational writings are often treated separately from his other writings. This entry provides a broader and more integrative survey. The main focus will be on Dedekind’s foundational writings, but they will be related to his mathematical work as a whole. Indeed, it will be argued that foundational concerns are at play throughout, so that any attempt to distinguish sharply between his “mathematical” and his “foundational” work is artificial and misleading. Another goal of the entry is to establish the continuing relevance of his contributions to the philosophy of mathematics. Their full significance has only started to be recognized, as should become evident. This is especially so with respect to methodological and epistemological aspects of Dedekind’s approach, which ground the logical and metaphysical views that emerge in his writings."
- https://plato.stanford.edu/entries/dedekind-foundations
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