On 13 Nov 2008, at 14:21, Torgny Tholerus wrote:
> 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
>> 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
>> paper on discrete mathematics.
> I will look closer at the Kauffman paper on Non-commutative Calculus
> Discrete Physics. It seems interesting, but not quite what I am
> 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*.
Like with comp + occam. Look I think I will concentrate on the MGA
thread for a period.
Meanwhile I will ask one of my student, who has a craving for discrete
math, to take a look on your finite calculus, and he will contact you
in case he find it interesting. Sorry but I have not so much time
> 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,
> 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
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