Hyper operator space - Printable Version

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Hyper operator space - Printable Version

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Hyper operator space - JmsNxn - 08/12/2013

Well I've been muddling this idea around for a while. I have been trying to create a hyper operator space and I recently realized the form of this. I'll start as follows:

If $[n]$ is a hyper operator then, $[n] [m]$ is a hyper operator created by forming left composition. I.e:

$[n] = x [n] y$ for all $x,y \in \mathbb{N}$ then

$[n][m] = x[n] (x[m] y)$

Associate to every function that is a finite product a number as follows:

$[e_1] [e_2] ... [e_n] = (\ p_1^{e_1} \cdot p_2^{e_2} \cdot ... \cdot p_n^{e_n})$

Where $e_n \in \mathbb{N}$ and p_n is the nth prime.

Now hyper operator space is the following:

$\frac{1}{(n)} \in \mathbb{H}$
$f,g \in \mathbb{H} \,\, \alpha,\beta \in \mathbb{C}$
$\alpha f + \alpha g \in \mathbb{H}$

Now define the inner product as follows:

$(f,g) = \sum_{x=1}^\infty \sum_{y=1}^{\infty} f(x,y) \bar{g(x,y)}$

Where quite clearly (f,f) converges for all elements since the terms decay to zero across x and y faster or just as fast as $\frac{1}{(x+y)^2}$

We say all the functions $\{ 1/(1), 1/(2), 1/(3),...,\}$ are dense in $\mathbb{H}$

Orthonormalize them to get $\Delta_n$ such that:

$(\Delta_i, \Delta_j) = \delta_{ij}$

$f = \sum_{i=0}^\infty (f, \Delta_i) \Delta_i$

Now we have the advantage of being in a Hilbert space and having an orthonormal basis.

The first operator we have is the transfer operator:

$T f = f(x,y+1)$

Since $[n-1][n] = T [n]$ this operator is well defined for any element of $\mathbb{H}$ where $T [a_1][a_2]...[a_n] = [a_1][a_2]...[(a_n) - 1][a_n]$

Suppose:
$[s]$ exists such that $[s-1][s] = T [s]$ for all values that [s] returns natural numbers at, this is our solution to hyper operators.

I think the key is to invesetigate the inner product.