Order of Recursion?

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.

1. 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)}=$$1\over {x} F(x, -1, 5) = x\uparrow\uparrow (-1) F(x, -1, 6) = x\uparrow\uparrow\uparrow (-1) etc... Then it is interesting to note that: F(F(x, -1 , q), -1, q) = x for all q > 3 I can prove this for multiplication, exponentiation, tetration, and pentation (as individual cases). How does one go about extending this to my defined set of all extended algebraic functions? Can this be used to solve the general functional equation: f(f(x)) = x for all possible functions? • At least the answer to the latter is "no". Let A,B be any disjoint subsets of \mathbb R of equal cardinality and \phi\colon A\to B a bijection. Then$$f(x)=\begin{cases}\phi(x)&\text{if }x\in A,\\\phi^{-1}(x)&\text{if }x\in B,\\x&\text{otherwise}\end{cases}$$is a solution to the functional equation$f(f(x))=x$. – Hagen von Eitzen May 9 '13 at 21:39 • That makes sense looking at it... I guess I'm curious about the (forgive me) nice looking solutions such as 1/x, -x, etc... Since they can be defined in terms of functions that themselves can be defined in terms of sequentiation – frogeyedpeas May 9 '13 at 21:44 • @frogeyedpeas can you show me haw do you prove it for tetration? if$x\uparrow\uparrow n={^n}x$, how can you prove that${^{(-1)}}({^{(-1)}}x)=x$? we have${^{(-1)}}x=0$and${^{(-1)}}0\neq x$– MphLee Jun 10 '13 at 14:26 • x^^(-1) = log_x(1)... Which we often simplify to 0 but is not necessarily equal to that – frogeyedpeas Jun 10 '13 at 14:29 • @frogeyedpeas why not? for logarithms$\log_x(a/b)=\log_x(a)-\log_x(b)$holds: then if$a=b$and$a\neq 0$we have$\log_x(a/a)=\log_x(1)=\log_x(a)-\log_x(a)=0\$, then how can you prove your statement in your question? – MphLee Jun 10 '13 at 20:41