Floundering and clearing things up.

First point about the spectral chart.  In it I matched powers of 3/2 against powers of Phi.  Null are inert, they do not quantize. I just knew that 3/2 was likely and I looked for matches with powers of Phi.

Second, I threw out a comment that the whole numbers in the Phi powers are likely Nulls.  They likely are. The powers of Phi I have in the chart are the result of the first Lagrange mode, they are really combinations of two Fibonacci series. They should be viewed as recursive additions of the form, below:

Third, the reason we hit a high match at Phi^91 and a bottom at Phi^107 is simply that Phi got too accurate to match 3/2 powers.  Phi, as  gets higher in power it is making the continued fraction.  Its minimum phase solution is simply:

(Phi^n - F[n-1])/Fn = Phi.

The 3/2 halves effect, making spheres, has not yet happened in the spectral chart.

Finally, I said the smallest globe of nulls was (2^17-1), I probabl should be the fraction, 1/[b^17], the base depending on the characteristic function. 1/17 is  the mass of the electron when the proton is 108. But I am not sure the proton even did the second Lagrange number, root(8), just haven't figured all that out.

Lets make a Fibonacci polynomial group

I will hand wave it.
I take the Pell equation and let my short but order polynonial, R and Q. R,Q is a smaller polynomial in Phi irrational number. It is two Finacci sequencse powers  of the form :   mR + n. I have to make a :

u^2 - x^2 (u)^2 = 4

Which looks like this from the Wiki  hyperbolic functions. Except that Cosh(u) is compressed by the factor ln(r). And the unit circle has become the unit oval.

I like unit circles. I resolve the polynomials into a:  R^k =  mR + n, a resolved finbonacci polynomial, of a number of digits  acually, Some shift to get mR + nR  which makes a circle.  I get a Large Lagrange, in exchange for a large shift in the two polynomials.  So, we make a set of these long shifts. But we get a difit for each shift. Lagrangerions have confirmed that the large Lagrange will be less than some O(n+1) error., so I know my digits length. the error is the smallest bag of nulls that stabilize, having the least kinetic energy. Then we thro in the null bubble, the polynomial is composed of bubble 1, the bubble 2 and bubble zero, our favorite. b1 and b2 exchange with b0. In the process they leave little chunks of null bubbles b0.  The b0 out number b1 and b2 inside the unit circle. Outside, from the surface of the unit sphere, the noise is some radially invariant set of conics, rotationally invariant, perpendicular to the sphere surface, and from the outside surface, they count null bubbles by exchange. The mass of nulls is the constant in our Rk, n.  R is exchanged alternating between b1 and b2. The small m, is the shift or charge, spin thing. It tells us when the b1 out number the b2.  But big n makes longer digits systems.

 Bags of packed nulls have surface deformations, noise plus offset. Phase offsets make the b0  packed slightly oval perpendicular, flat  side to the line of motion the sphere. Packed nulls scoot around.

Two bubbles will attempt to cancel, wave moves toward a line of symmetry where Nulls increase in number, counting in our digit system.. Three quarks, havine large nulls and long digit systems will have large offsets cancelling in the center and a tubs of b0 sloshing around  The three unit spheres, do what units spheres do, but are anchored to the tub of b0. Out side the three, the quarks have much more b1 and b2, and much fewer b0.

Rotations

Rotations of the center 'angle' the point Phi^N, about which the fractions 1/Phi^N and set, that point has to rotate in space to preserve volume and orientation and maintain the special linear group.  When N goes as 1,2,3; then U changes by quants, and each kU represent a digit in the system of the form: Phi^kU + 1/Phi^kU.  Changing the Lagrange, the parameters r, takes us from one, two and three degrees of freedom, each degree representing a rotation about the center line of U. The third Lagrange gives us three roatations, hence the quarks. Each roatation should cause a spiral up the center line, I think.

We end up with this compound, spherical worm gear system which maintains gear teeth using batches of Nulls, some packed into matter.  The system goes out to about 180 meters, and the net effect is to keep the proton centers inside a slightly curved salad bowl. When protons coordinate, they tend to curve space toward the quasars a bit, communication space curvature thru a broadly distributed network of protons. Eventually, the protons in direct management of the quasar center, tilt the local gradient to adjust the spew of new baryons for local galactic debris. The universe becomes a huge salad bowl.

Map size in minimum redundancy encoding

It goes as nLog2(n) in Shannon because his channels are all single, bitstream pipes, and signal has two degrees of freedom. The map has to be transmitted.  It is just an anomaly of where Shannon worked. But if he had considered three pipes, then his map size goes down, to nLog3(n), and signal has three degrees of freedom.

There is nothing sacred about SNR going as the square of frequency. SNR can easily go as the third power, no  problem. We just get a different basis function. In a three pipe system, each pipe has root2(noise) , but when combined they have root3/2(noise) (and 3 times signal). In physics it makes a lot of difference, noise is kinetic energy, the more degrees of freedom kinetic energy has, the more stable is packed matter.

Cubes, squares and Fibonacci shifts

Why is there a 1/2 in the Golden Ratio?  Mainly because the recurrence relation is a shift once and add, and if you look at the derivation, you will see the 1/2 is needed to avoid a coefficient in the relationship: R^2 -R-1 = 0.

So, the canonical digits (Phi)^N look like :

2^(-N) * (1+root(5))^N, and can be put further into:

2^(-N) * 2^(N*log2(1+root5)) or 2^[N(log2(1+root(5))-1)], if you like twos binary.

So, the 2 results from sampling the shift, once for each shift. This Nyquist sampling is sampling to be orthogonal to  phase error introduced by the shift, which has phase noise falling at twice the quant rate.

Anyway, I bring this up because there is a cubic being solved, in the proton to generate the quarks.  It is mainly related to curvature of the unit circle, it is too curved. So phase which cannot meet the curvature fly off, going Gamma, as we say.  Call it quantization noise.  The unit circles that survive meet some minimum curvature requirement, mainkly satisfyin a relationship in R^3 that keeps curvature optimum. So the system has this perfect number with solutions that are interleaved base three and base two. Somewhere, in getting this 2/3 charge we ended up with a shift three times in the Fibonacci series.

Physicists call a unit circle which is too curved a unit circle which curves faster than light travels. I call it a curve that curves faster than light samples.

This ruler, it gets expanded and accommodates the base three curvature management. Its maximum entropy when the available bits finite

Noise power and the optimum digits, with corrections

Warning again: My math skills.  A subtraction became a division down there somewhere, and added in the identity where it belongs.  I will try to straighten this out, later. But is is just  the Schrodinger all over again.

When Noise drops of at the quant of signal, my SNR is S/(N*r)  I  get this equation:

(1/N) * M - 1/N = S

M is my measurement, a scalar, the the linear equations makes my measurement orthoganal to noise. I subtract off the 1/N because my measurements start with 1, and count up in a quantized system.  When I know my maximum qwuants, my maximum integer, then M is a digit system in base 1, is is simply adding up the signal one by one.

When the same systme has Noise dropping as the square then SNR goes as S/N^2, and the equation is:

[N^1/2]*M -I*(N^1/2) = S

M,N are vectors of dimension m and we take the dot product. Since noise falls as the square, the vector N^1/2 look like:

N*(1, 2^-2,2^-3,...), and so my digits go as the base 2.  So in the equation above, subtracting the noise is equivalent to sampling at twice the baud rate. Also, any scale on signal returns to the exponents as a measurement baud. So this will get the Shannon equivalent.  In the general case, all the vectors are really diagonal matrices, except where N falls as some polynomial in quant number.


Noise falls when dealing with whole numbers, noise dropping as the digits go from high to low.   Noise is a polynomial in the some k degrees where k is the order of the differential in the system.

Proton

This are slightly simplified for the proton.  Our base is always Phi and our exponents are matrix polynomials of integer quants representing a shift.  Further, the proton in interested in making the unit circle, and we can work in log Phi. So if M(phi) is a block diagonal polynomial matrix, and SNR is a solution to three differentials, one for each order of the Lagrange, then we have:

[M(phi) - I ] / [SNR(phi)] = [0],  where the block diagonals in M are the three orders of Fibonacci shifts, which I think are periodic. The S are really the Zeta zeros for the prime 2^17-1, times the mass quant for each particle and the system is trying to zero them out.   However, the length of the digits must be long enough to count out one unit circle for each particle which appears simultaneously. We have no reason to think the particle will be spherical, only that their boundaries are near unit one.
Any non sperical shape is part of kinetic energy, the motion required to zero out the inter particle functions.

The theory of yard sticks

Here they are, the notches on a yardstick:

    F1 = { 0/1, 1/1 }
    F2 = { 0/1, 1/2, 1/1 }
    F3 = { 0/1, 1/3, 1/2, 2/3, 1/1 }
    F4 = { 0/1, 1/4, 1/3, 1/2, 2/3, 3/4, 1/1 }
    F5 = { 0/1, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 1/1 }
    F6 = { 0/1, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6, 1/1 }



 We can make the unit circle with them.

The semi circles within the circle are the bounding value in a Lagrange theory. The size of the semi circle is really the 'frequency' in the corresponding hyperbolic functions.

But, here is the thing. If I have infinite dust, then, just like Fourier, I can make copies of the lowest bandwidth and superimpose them.  If my dust is finite, then, just like Shannon, I have to use my dust sparingly, regardless of bandwidth.  So, finite dust, maximum entropy, infinite dust, don't worry.

Lagrange, Shannon: Same sampling theory different basis functions?

I noticed that Shannon information theory was simply a restatement of Shannon sampling theory for trig functions, and that Lagrange theory is the same for Hyperbolic functions.

Given a Gaussian channel, the spectrum can be organized in integer units of trig functions and a  ring established. Same with Lagrange, just change the basis set? The trigs define the bounded multipliers, the Hyperbolic the bounded quotients.

Look, they both want deviation to fall as the square, that is the 4 on the right which is 2^2. One assumes a hyperbolic noise channel, the other a trig noise channel. So they both have canonical twos binary representations. The one wants sampling to twice the unit circle, the other to 1/2 the unit circle.

And they both really define the finite value of the natural base given the noise spectrum. Both of them all about finite number lines.  If the natural base represents noise dropping at the square, the sample rate is 2, when dropping as the cube, the sample rate is 3.

It is the natural log, or natural exponent because that is what defines the full set of math operations, especially addition of powers, in the  number line.  In real cases, the problem is the finite log in a finite number line. Without powers defined there is no digit system. So Shannon and Lagrange are about defining the length of a digit system that matches the integer errors in counting. The number of integers one gets is the minimally redundant set.

The sphere
It is a third degree polynomial in r, whih generates three orthogonal bases which make up a 16 digit worm counter, as near as I can tell.  The value one in that worm counter is maximally packed, it is the quarks and electron.  The three independent bau rates must be composted of compound shift of the Phi sequence. A
shift (shift(shift (Fn,Fm),shift (Fn,Fm),shift(shift (Fn,Fm),shift (Fn,Fm))
The compound braid concept, or spiral to the third power, and it walks the Markov tree maintaining the unit zero where it can, at the sample rate of light. The whole proton must be filled with the compound braid system. The Unit zero is the sum of the 16 bit fraction, plus one. In other words, it has to be a complete, compound worm gear system.

This picture was in my head

Hardy-Littlewood thanks for the memories. I always appreciate it when something in my head shows up as a true idea worked on before, it makes me feel less nutty. But this is where I was, trying to figure out how a bunch of bubbles could compute the value of one. I wonder, let a := 1/(8^1/2) and and let b := 1/(5^1/2), then created two sequences of a^n and b^n, for n from 1 to 16, say. Would they sort in a regular pattern? Each sequence, in its own digit system looks like: .11111  a good approximation to 1. Hmm... Then take, for all three orders of irrationality, the full list, up to 16 bits, of each.  We are quite a bit beyond my brain here.

No one has ever actually calculated the number one

Just a note, the multiplicative identity has never actually been computed, only defined. So, I am stuck, how does the proton keep at least one and  a half lines of near constant symmetry about which it can calculate the value one?

What is the difference between the unit sphere and the electron?

Speaking strictly within the atom.  If Higgs 'gave' mass, and the Higgs 'gield' is everywhere, and the electron is simply managine a number line based on this field, then what makes the electron have matter as opposed to a unit sphere?

In other words, given the electron manages the fractional and whole counts radially across the sphere, what is the difference between the value 1.0 and the value 3/4?  The counting system is shifted, and certainly the multiplicative identiy is important. But there is nothing wrong with making 3/4 the multiplicative identity for a while, and letting the previous multiplicative identity be 3/4. In other words, the rules can be shifted with a turn of the screw. the unit sphere is what we can the atom, but it is nothing more than a slight repacking of the Null within the sphere.