Graphs and entropy

The idea that a signal of the form x^(1/b), can always be represented in a base b system of finite size digits should be easy to prove. b is integer, make it simple. make logb(x)  finite in base b. So there exists a finite, balanced decoding graph with b branches at each node. From there it is simple to show a base b counter can represent all paths through the tree, up to the precision of my finite log. Shannon did no different, except he started with noise.

Computing the quantization error is simple. If the number of base b digits is n, then add up the inverses of the finite set 1/b^n, to get the finite log base b, then compare that to the Taylor series of the real log.  The last term is the error term.

Also, there will exist an encoding that is minimum redundancy. I know there is a general information theory about this in Wiki. If b is not integer, then make it so.

Why do I care? Because we are going to deal with cubic roots. So instead of 0,1,2 as my coefficients, I can use r1,r2,r3; I think, which would be degrees of freedom, say for the l numbers in atomic orbitals. I can rotate though the roots as I count the digits. The L numbers in the orbitals seem to be base three. I think the 'one' spots on the unit oval will rotate about by cubic roots.
What about digits for cyclic  graphs? Dunno, I just though about it. Digits on the interior of a sphere? They would not be ordered, necessarily. But more likely three independent wave actions since quarks are three.

The quants on the orbitals: Principal, angular, and magnetic; all seem to be the Lagrange degrees of freedom, in base 1,2 and 3; So they are interleaved in any digits system.  The principal quant does not have bu one value per digit, it is the first Lagrange. But it does a wave shift with energy, shortening and leaving the charge to be counted by l in 2 degrees.  Then they all shorten for the magnetic, it is almost a shift of the decimal point. More degrees of freedom added out from the electron as energy increases. But those quantum numbers are not completely orthogonal, I can tell. They also do not include the quarks.

Also, I notice, when the principal quant is at its lowest energy level, my picture of the hyperbola says the radial charge is a straight line, and the electron mass is flat.  No, small Lagrange, and small q in the denominator of the minimum error. So error cone from the electron is wide, the packed nulls nearly spherical. But for principal quant 2, there are two spheres, where did the extra degree of freedom come from? Spin? No. The first few bits must be powers of the first Lagrange. Still a bit confused, I am. But it looks like they assign Markov numbers straight from the list, starting with the most least significant bit. I guess the spin took all of the first Lagrange.