# Tetration Forum

Full Version: Functional power
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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.
Or, how about functional zeration or functional negative rank hyper-operations?
(07/11/2022, 01:50 AM)Catullus Wrote: [ -> ]$\mathbb{What \: about \: functional \: addition?}$

$\mathbb{How \: would \: that \: work?}$

$\mathbb{Or,\: How \: about \: functional \: zeration,\: or \: negative \: rank \: hyper-operations?}$

$\mathbb{Please\:remember\:to\:stay\:hydrated.}$

$\mathbb{Sincerely:Catullus}$

Catullus, are you trolling us? Do you have an issue with writing readable posts? Why didn't you try something readable in code and actually displaying tex, like the following?
Code:
```[tex]\mathbb{What \: about \: functional \: addition?}[/tex] [tex]\mathbb{How \: would \: that \: work?}[/tex] [tex]\mathbb{Or,\: How \: about \: functional \: zeration,\: or \: negative \: rank \: hyper-operations?}[/tex] [tex]\mathbb{Please\:remember\:to\:stay\:hydrated.}[/tex] [tex]\mathbb{Sincerely:Catullus}[/tex]```

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Okay, I changed it.