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On Calculating Entropy

June 10th, 2014 3 comments

When talking about Pasquali patches that converge and stationary surfaces that they converge to, we discussed that such would stabilize because it wouldn't loop in any shape or form nor would it diverge.  Such Claim we called Entropy.  We didn't provide a mechanism to calculate such. We do that now.

The way we propose we calculate the entropy is by measuring the deviation from the stationary surface.  Thus for the case where we have a Pasquali patch p_t(x,y), t \in \mathbb{Z}^+ we could measure entropy by:

 S(t) = \int_{[0,1]^2} \vert p_t(x,y) - p_\infty(x) \vert \, dA

that is, by taking the absolute deviation from the stationary surface (notice we use the area element dA).  Since in this case the time element is configured at integer times, rather than on continuous time, letting  t \in \mathbb{R} implies using the Pasqualian.  Thus we would have 

 S(t) = \int_{[0,1]^2} \vert p(x,y,t) - p_\infty(x) \vert \, dA

It seems fairly clear that 0 < S(t) < \infty , but here we must agree that 0 means that the system is stable and has the most entropy.

We may however, rather than the absolute difference from the stationary surface, choose to emphasize larger changes and (heavily) discount smaller ones.  Thus we could suggest

 S^{\square}(t) = \int_{[0,1]^2} \left( p(x,y,t) - p_\infty(x) \right)^2 \, dA

for t \in \mathbb{R}.

 

More on Claim 5. The Grand Classification Theorem of Limiting Surfaces

June 4th, 2014 No comments

In my previous post I wrote about Claim 5. The Grand Classification Theorem of Limiting Surfaces (A General and Absolutely Complete Classification of Pasquali patches which are functions of x alone).  In words, it says that we can classify polynomials that converge in area between 0 and 1 by tying it to any convergent sum.  Moreover, the polynomials that converge in area to one create equivalence classes in the convergent sums.

For example, take

 A_i = \left[ \begin{array}{cccc} 1 & 1 & 0 & \ldots \\ \end{array} \right]

(for a particular chosen i).  Here the sum of the convergent sum is 2, and the polynomial we are talking about is

A_i \cdot D = 1 + 2x

On the other hand take  

 A_i =\left[ \begin{array}{cccc} 2 & 2 & 0 & \ldots \\ \end{array} \right]

and generated polynomial 

A_i \cdot D = 2 + 4x

These two polynomials are equivalent when we make sure that the convergent area under the integral in the interval from 0 to 1 is equal to 1 exactly, which implies dividing each entry of A_i by the convergent sum and generating

A_i^* = \left[ \begin{array}{cccc} \frac{1}{2} & \frac{1}{2}& 0 & \ldots \\ \end{array} \right]

 Thus, we obtain the polynomial  

A_i^* \cdot D = \frac{1}{2} + x

which we can declare equivalent to the other two.

We quickly check that

 \int_0^1 \left( \frac{1}{2} + x \right) \, dx = 1

is indeed the case.

In essence, we have a criterion for making converging sums equivalent (by normalizing) which in turn generates polynomials that are equivalent.

I will expound on this a bit more later but that's the gist of it.

Categories: Mathematics