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