When you are going to do exact mathematical computations for the
discrete space-time, then the continuous mathematics is not enough,
because then you will only get an approximation of the reality. So
there is a need for developing a special calculus for a discrete
mathematics.

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One difference between continuous and discrete mathematics is the rule
for how to derĂvate the product of two functions. In continuous
mathematics the rule says:
D(f*g) = f*D(g) + D(f)*g.
But in the discrete mathematics the corresponding rule says:
D(f*g) = f*D(g) + D(f)*g + D(f)*D(g).
In discrete mathematics you have difference equations of type: x(n+2) =
x(n+1) + x(1), x(0) = 0, x(1) = 1, which then will give the number
sequence 0,1,1,2,3,5,8,13,21,34,55,... etc. For a general difference
equation you have:
Sum(a(i)*x(n+i)) = 0, plus a number of starting conditions.
If you then introduce the step operator S with the effect: S(x(n)) =
x(n+1), then you can express the difference equation as:
Sum((a(i)*S^i)(x(n)) = 0.
You will then get a polynom in S. If the roots (the eigenvalues) to
this polynom are e(i), you will then get:
Sum(a(i)*S^i) = Prod(S - e(i)) = 0.
This will give you the equations S - e(i) = 0, or more complete: (S -
e(i))(x(n)) = S(x(n)) - e(i)*x(n) = x(n+1) - e(i)*x(n) = 0, which have
the solutions x(n) = x(0)*e(i)^n.
The general solution to this difference equation will then be a linear
combination of these solutions, such as:
x(n) = Sum(k(i)*e(i)^n), where k(i) are arbitrary constants.
To get the integer solutions you can then build the eigenfunctions:
x(j,n) = Sum(k(i,j)*e(i)^n) = delta(j,n), for n < the grade of the
difference equation.
With the S-operator it is then very easy to define the difference- or
derivation-operator D as:
D = S-1, so D(x(n)) = x(n+1) - x(n).
What do you think, is this a good starting point for handling the
mathematics of the discrete space-time?
--
Torgny Tholerus
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