I am not sure how related this is to what you ask in your original post, but
as for a model (and candidate TOE) of physics which is discrete, there is a
theory known as Hiem Theory (
http://en.wikipedia.org/wiki/Heim_Theory )
which posits there are six discrete dimensions. Interestingly, the theory
is able to predict the masses of many subatomic particles entirely from some
force constants, something which even the standard model is unable to
explain.
Jason
On Thu, Nov 20, 2008 at 12:12 PM, Torgny Tholerus <torgny.domain.name.hidden> wrote:
>
> Torgny Tholerus skrev:
> >
> > What I want to know is what result you will get if you start from the
> > axiom that *everything in universe is finite*.
> >
>
> One important function in Quantum Theory is the harmonic oscillator. So
> I want to know: What is the corresponding function in discrete mathematics?
>
> In continuous mathematics you have the harmonic oscillator defined by
> the differential equation D^2(f) + k^2*f = 0, which will have one of its
> solutions as:
>
> f(t) = exp(i*k*t) = cos(k*t) + i*sin(k*t), where i is sqrt(-1).
>
> In discrete mathematics you have the corresponding oscillator defined by
> the difference equation D^2(f) + k^2*f = 0, which will have one of its
> solutions as:
>
> f(t) = (1 + i*k)^t = dcos(k*t) + i*dsin(k*t), where dcos() och dsin()
> are the corresponding discrete functions of the continuous functions
> cos() and sin().
>
> So what is dcos() and dsin()?
>
> If you do Taylor expansion of the continuos function you get:
>
> exp(i*k*t) = Sum((i*k*t)^n/n!) = Sum((-1)^m*k^(2*m)*t^(2*m)/(2*m)!) +
> i*Sum((-1)^m*k^(2*m+1)*t^(2*m+1)/(2*m+1)!)
>
> And if you do binominal expansion of the discrete function you get:
>
> (1 + i*k)^t = Sum(t!/((t-n)!*n!)*(i*k)^n) =
> Sum((-1)^m*k^(2*m)*(t!/(t-2*m)!)/(2*m)!) +
> i*Sum((-1)^m*k^(2*m+1)*(t!/(t-2*m-1)!)/(2*m+1)!)
>
> When you compare these two expession, you see a remarkable resemblance!
> If you replace t^n in the upper expression with t!/(t-n)! you will then
> get exactly the lower expression!
>
> This suggest the general rule:
>
> If the Taylor expansion of a continuous function f(x) is:
>
> f(x) = Sum(a(n)*x^n) = Sum(a(n)*Prod(n;x)),
>
> then the corresponding discrete funtion f(x) is:
>
> f(x) = Sum(a(n)*x!/(x-n)!) = Sum(a(n)*Prod(n;x-m)),
>
> where Prod(n;x-m) = x*(x-1)*(x-2)* ... *(x-n+2)*(x-n+1) is a finite
> product.
>
> I have no strict proof of this general rule. But this rule is such a
> beautifil result, that it simply *must* be true!
>
> --
> Torgny
>
>
> >
>
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Received on Fri Nov 21 2008 - 02:46:19 PST