Fermions and Bosons

Just thinking about the adapted hyperbolic system:

1/2 *(b+1/b)^2 - 1/2 * (b-1/b)^2 = 1

What is this except a match set of bosons and fermions.  Call it the Shannon condition, why not. The bandlimited number of bosons fermions minus the bandlimited number of fermions bosons will be one unit of exchange energy.  This comfims my view that the Cosh is the cold position, and cosh(0) is the winning positoon, the bubble that does not exchange.  But the sinh are the odd functions and would have though they would be the fermions, I guess something is not quite right in my thinking, but never mind, go one.

 I am cheating here, certainly, because I assume the bubbles have already done the job of figuring out how to pack a sphere, and I am just dumbly decoding what they did. The numbers of things are built right into the exponents.

Now we can get the statistics because we have energy levels quantized, by assumption. So, b is really:

kT is what you make it, the transaction rate of hiting a tennis racket or the transaction rate of hot molecules, who cares, the system is adapted. That exponent is closely related to how close the energy levels get to tanh = 1.0 before the capacity is exceeded.



The statistics will always look something like:

Out of scale, as usual, but these are just the second derivative plots for Coth (red fermion) and Tanh (blue boson) , they meet the flow conditions for the two groups, and give the probability for each of the integer exponents.

I mean, this is all about encoding the groups to meet Shannon.  In one direction you are mapping to SNR and including the optimum sampler, in the other the fermion have their own sampler, you your are mapping NSR and excluding the sampler.

So lets play the game of Wythoff.

F is the number of fermions, B the number of bosons. For some energy states, the number of Fermions to Bosones is F/B.  Now it takes three Wythoff moves for a Boson to maintain a cold position, of F/B^3. The same holds for the Fermions, B/F^3. At low energy levels, the fermions start with at least one. And the probability of finding a fermion is F/B^3, is large as B, sinh, is small.  At high energys, the number of fermions is close to the number of bosons.   

At high energy states, he denominator dominates.
 So do I understand this? No, still confused. But it seems clear the game of Wythoff is happening and its takes three moves to exit the sphere. The connection with fermion and boson statistics most likely has to do with the connected system, so we end up with a rational approximation, of small order, for the exponential..