03/11/2017, 10:22 AM

Let f and g be total functions (so e. g. C -> C) and N and M be complexes.

Then (f o g)(x) and f o a = f(a) are so-called functional multiplications. But the interesting thing is the following: functional power:

\( f^{oN} = f o f o ... o f (N-times) \)

When N is an integer, it is trivial, just look:

\( f^{o0} = x \)

\( f^{o1} = f \)

\( f^{o2} = f o f \)

\( f^{o3} = f o f o f \)

...

\( f^{o-1} = f^{-1} \)

We have rules for it, like these ones:

\( (f^{oN}) o (f^{oM}) = f^{o N+ M} \)

\( (f^{oN})^{oM} = f^{o N M} \)

\( f o (f^{oN}) = (f^{oN}) o f = f^{o N+1} \)

But for instance:

\( (f^{oN}) o (g)^{oN} != (f o g)^{oN} \)

(Also functional tetration exists.)

My theory is that if we can get an explicit formula for \( f^{oN} \) with x and N, then N is extendable to any total function.

For example:

\( (2x)^{oN} = 2^N x

N := log_2 (x)

(2x)^{o log_2 (x)} = x^2 \)

And in the same way, theoritacelly you could do the same with all the functions.

But how?

My concept is that by Carleman matrices.

Then (f o g)(x) and f o a = f(a) are so-called functional multiplications. But the interesting thing is the following: functional power:

\( f^{oN} = f o f o ... o f (N-times) \)

When N is an integer, it is trivial, just look:

\( f^{o0} = x \)

\( f^{o1} = f \)

\( f^{o2} = f o f \)

\( f^{o3} = f o f o f \)

...

\( f^{o-1} = f^{-1} \)

We have rules for it, like these ones:

\( (f^{oN}) o (f^{oM}) = f^{o N+ M} \)

\( (f^{oN})^{oM} = f^{o N M} \)

\( f o (f^{oN}) = (f^{oN}) o f = f^{o N+1} \)

But for instance:

\( (f^{oN}) o (g)^{oN} != (f o g)^{oN} \)

(Also functional tetration exists.)

My theory is that if we can get an explicit formula for \( f^{oN} \) with x and N, then N is extendable to any total function.

For example:

\( (2x)^{oN} = 2^N x

N := log_2 (x)

(2x)^{o log_2 (x)} = x^2 \)

And in the same way, theoritacelly you could do the same with all the functions.

But how?

My concept is that by Carleman matrices.

Xorter Unizo