Define an extended algebraic function $f(a)$ as a function on $a$ that utilizes any combination of recursive extensions and inverses of sequentiation.
Example:
$a + 1$ , sequentiation.
$a + a$, addition (repeated sequence).
$a\times a$, multiplication (repeated addition).
$a^a$, exponentiation (repeated multiplication).
$a\uparrow\uparrow a$, tetration (repeated exponentiation)
etc...
The first $4$ functions are familiar to most while the later ones may be strange, for example:
$2^3 = 8$, $2\uparrow\uparrow 3 = 2^{2^2} = 2^4 = 16$,
$2\uparrow\uparrow\uparrow 3 = 2\uparrow\uparrow 2\uparrow\uparrow 2 = 2\uparrow\uparrow 16 = 2^{2^{2^{...^2}}} (16 \text{ times})$ = a big*** number.
- If you define the function $F(x, -1, c)$ where $c$ denotes the order of the operation in this recursive family (example):
$F(x, -1, 3) = -1 \times x$
$F(x, -1, 4) = x^{(-1)}=$$ 1\over {x}$
$F(x, -1, 5) = x\uparrow\uparrow (-1)$
$F(x, -1, 6) = x\uparrow\uparrow\uparrow (-1)$
etc...
Then it is interesting to note that:
$F(F(x, -1 , q), -1, q) = x$ for all $q > 3$
I can prove this for multiplication, exponentiation, tetration, and pentation (as individual cases). How does one go about extending this to my defined set of all extended algebraic functions?
Can this be used to solve the general functional equation:
$f(f(x)) = x$ for all possible functions?