Bruno Marchal skrev:
> I have to think. I think that to retrieve a Leibniz rule in discrete
> mathematics, you have to introduce an operator and some non
> commutativity rule. This can be already found in the book by Knuth on
> numerical mathematics. This has been exploited by Kauffman and one of
> its collaborator, and they have published a book which I have ordered
> already two times ... without success. It is a very interesting matter.
> Dirac quantum relativistic wave equation can almost be retrieved form
> discrete analysis on complex or quaternion. It is worth investigating
> more. Look at Kauffman page (accessible from my url), and download his
> paper on discrete mathematics.
I will look closer at the Kauffman paper on Non-commutative Calculus and
Discrete Physics. It seems interesting, but not quite what I am looking
for. Kauffman only gets the ordinary Leibniz rule, not the extended
rule I have found.
What I want to know is what result you will get if you start from the
axiom that *everything in universe is finite*.
For this you will need a function calculus. A function is then a
mapping from a (finite) set of values to this set of values. Because
this value set is finite, you can then map the values on the numbers
0,1,2,3, ... , N-1.
So a function calculus can be made starting from a set of values
consisting of the numbers 0,1,2,3, ... , N-1, where N is a very large
number, but not too large. N should be a number of the order of a
googol, ie 10^100. Because the size of our universe is 10^60 Planck
units, and our universe has existed for 10^60 Planck times. As the
arithmetic, we can count modulo N, ie (N-1) + 1 = 0. This makes it
possible for the calculus to describe our reality.
A function can then be represented as an ordered set of N numbers, namely:
f = [f(0), f(1), f(2), f(3), ... , f(N-1)].
This means that S(f) becomes:
S(f) = [f(1), f(2), f(3), ... , f(N-1), f(0)].
The sum or the product of two functions is obtained by adding or
multiplying each element, namely:
f*g = [f(0)*g(0), f(1)*g(1), f(2)*g(2), ... , f(N-1)*g(N-1)].
and to apply a function f on a function g then becomes:
f(g) = [f(g(0)), f(g(1)), f(g(2)), ... , f(g(N-1))].
Exercise: Show that the extended Leibniz rule in the discrete
mathematics: D(f*g) = f*D(g) + D(f)*g + D(f)*D(g), is correct!
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