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.