# Re: Mathematical methods for the discrete space-time.

```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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