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 

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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