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.

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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! -- Torgny Tholerus --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---