Finite systems are always the case. So, any condition where some finite derivatives of F are within some bound of F itself. Each derivative is a degree o freedom in finite systems, in the limit, the natural log has infinite degrees of freedom.
Take my N-body, for example. For N bodies, the local gravity vecor is a polynomial of Nth degree. The first N derivatives are bound and real with respect to the zeroeth derivative. So each body will have a root equadistant from the dominate root, Gravity. The polynomial is: (1-Gx^N)*(Vx+Px)*(Vx+Gx) = 0
Using my 'everything is a unit' method. Implicit in Gx is a p/q, a finite approximation of e. So ln(r) is the root of an Nth degree polynomial, and is approximated by the sum of (f^ -n), keep one (and sometimes 2) of the inverses of a set F being the finite sum of exponents in p/q. Tanh bounds the error of p/q.
So any prototype system of finite order is to define an Nth recursive approximation of the natural log. That equation of motion with the N+1 derivative must be zero. The wave otion is finite, and local. The system process is local exchanges of units 1/F and F for some F(n) n = quant number. The spectrum is then bound by f^k and f, where k is small. The number of root determined by coupling between the different differential from k=0,1,2,..N.
The component used in estimating ln(r), in the N-Body come from mostly orthogonal polynomials: (Gx -1)^n * (px-1)*(vx-1) = approximation of e. There is a limited set of local spectrum needed to solve.
Time is simple. local time = local time^ -1 + 1, it is phi when you want the most accurate time. It means the approximation p/q is a power of Phi. Less acurate time is Tl = Tl^ -3 + 1 is within bonds. So time is a count of quants. Quants are markers on the yard stick, and the yard stick is appropriately curved. So in my N-body, the solution, in units of time, is going to be: T(logF(q) + P(3*logF(q).
But you can see time can use a different bas, as can G, as can the third degree os motion. They are independent.
The Efimov state.
That is a ste where the approximation of LogF(p/q) has higher order terms within the approximating error. Those terms are eliminated in Tanh functions. So the three body Efimov solution is a square root, there are two root among the three bodies and all equantions can be referenced to that one and use the p/q approximation. This means you can use a cube root as one partition in the finite log network, and stay within the unit^n finite band bandstops. So, log(p/q) has two recursive moments but is a cubic. That means a free degree of freedom for motion in the solution. The Efimov bodies get a region of simultaneous action. Hence the quarks and gluons, they have a range of motion within some band that is white noise, well encoded.
So, you take these quarks, within the proton. Well the band stops are whit, the largest shift in exponent. It is broken into three groups by the quark cubit root, whihc is brokent in a polynomial root. The color white is likely a +3 to -3. That root reduces in error faster than some square root. So, within the cubic band stops, a quark can operate about some square root, fewer transactions needed to maintain stability.
The error, then, the band width allocated for the quarks will be as follows:
e3 = 1-(p/q)^3, e2 = 1-(p/q)^2, e1 = (p/q)^1 The error decrease from right to left, higher order powers have less approximation error. That means the quants are separable by common multiples. Quants, the exponents, for threes digits go as 3*k, and for two digits as 2*i. Any linear combination of twos and threes digits is accommodated as long as all linear combinations of error are less than p/q - e^ln(p/q).
In the quark, the quants digits count as a threes digit and twodigit: [3^n * 2^m] in shifts of n + m. n is 3*k and m is 2*i. The i and j represent independent degrees of freedom. So the canonical form of a system is a system that produces the optimum integer set. It simply counts things with an integer output counting machine. These integer are the optimum markers in our yardstick manufacturing house.
So, yes indeed, I can write the polynomial solution to the N body in a polynomial, P(i), i 1 to N. i can be units of time, p(i) being some position in an independent vectors of dimension N. But you have to use the integer counting system natural to the system.
This works in proton/electron mass, 17*108. That means the count size is 1^17, and it counts b^(2*17) of them, and produces 108 equally distinguishable integers. The proton is making a yard stick. A universe wide plot to get a better approximation to the log. The idea is to generate as many distinguishable integers as possible. How accurate? The proton can count the largest integer to 9e-5. I think it manages 16(2^16) groups of nulls, altogether.
The system returns an integer count of all actions possible, in increasing order. This is arcTanh, actually, the quant number. Each digits can count unique actions relative to the degrees of freedom. It is a not quite invertable counter for groups and their relationships. It has an almost inverse. It approximates roots using groups of Higgle mass spread around as band stops.
So we see the proton count 3^3 * 2^2 * 1^17 = 27*4 = 108 times 17 electrons.
Spectral function
Consider only S(n) = S(n-1)^(-k) + 1 = 0; Any S(n) is a recursive function of S(n-k). However, by making my N into subsets of n*3 and n*2; I can find all the solution for the roots of r. I need only solve for the S(n) as a quadric root and S(n) as a cubic root. Then I can mix and match and get all the roots for the system.
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