Solution to $e^{e^x}=x$ and other applications of iterated functions? While trying to solve $e^{e^x}=x$, I ran into the simple solution $x=-W(-1)$.  I found it by using the equation $$e^x=x$$Then powering both sides with a base $e$.$$e^{e^x}=e^x$$Now note that the left side of the original equation equals the right side of my new equation.  Therefore:$$e^{e^x}=e^x=x$$The solution at the beginning is very easy to solve for with the Lambert W function, unlike the actual equation I was trying to solve: $e^{e^x}=x$.
Which made me realise that if $f(x)=x$, then $f[f(x)]=x$ or more generally, $f_n(x)=x$ being equivalent to asking $f(x)=x$ for any integer $n\ne0$ (I will use subscripts to describe the amount of times a given function is iterated).
I confirmed this solution by solving $L(x)=x$ where $L(x)=mx+b$, the linear equation.  No matter how many times I do $L(L(L(L(\cdots L(x)\cdots))))=x$, the solution is always $x=-\frac b{m-1}$.
Which made me wonder if I could create a solution to the general quartic polynomial $P(x)=ax^4+bx^3+cx^2+dx+e$.
Use $F(x)=px^2+qx+r$.
$F[F(x)]=$some really big quartic.
Also note that we are trying to solve $P(x)=F[F(x)]=F(x)=x$.
If you are trying to find the roots, just add $x$ to both sides.
Now let's try to see what $F[F(x)]$ equals. (Do note that I will probably make some mistakes.)
$$F(x)=px^2+qx+r$$$$F[F(x)]=p(px^2+qx+r)^2+q(px^2+qx+r)+r$$$$=p^3x^4+(2p^2q)x^3+(p(q^2+qp+2pr))x^2+(q^2+2qpr)x^1+(pr^2+qr+r)x^0$$
And we are trying to make it equal to $P(x)$.
$$ax^4+bx^3+cx^2+dx+e=p^3x^4+(2p^2q)x^3+(p(q^2+qp+2pr))x^2+(q^2+2qpr)x^1+(pr^2+qr+r)$$
Try to equate parts?  I am unsure if that'll work, but anyways...$$a=p^3$$$$b=2p^2q$$$$c=p(q^2+qp+2pr)$$$$d=q^2+2qpr$$$$e=pr^2+qr+r$$
If anyone wants to do that, be my guest because it looks messy.
A note however, is that when we are done with this, we should get two answers.  This is because of the quadratic equation.  However, a quartic polynomial should have 4 solutions, meaning we missed 2.
However, we can make up for this.  Suppose we found $x_1=y$ and $x_2=z$.  Then $P(x)-x=0$, after which we can divide by our solutions:$$\frac{P(x)-x}{(x-y)(x-z)}=0$$
Divide and simplify to get a quadratic that is easily solved for.
Lastly, if this works, perhaps we can do this for an 8-th degree polynomial, or any $2^n$th degree polynomial for that matter.  Just use $F[F(F(x))]=x$ for a sextic polynomial and more as needed.
I also note that while $F(x)=x$ produces solutions for $F[F(x)]=x$, it does not work conversely as shown above.
So my questions are as follows:


*

*Could you have solved $e^{e^x}=x$ without iterated functions or approximations?

*Could you solve $xe^{e^x}=e$ by similar methods?

*Does the method for solving quartic polynomials work?  I have yet to make heads or tails of the ones on the Wikipedia or Wolfram.

*Is there anything I should note when using my method?  Because I have noticed the failure to find all solutions in a polynomial and doubt I have found all solutions in other functions with this method.

*How else can I use iterated functions to solve for things?

*Can I do this for an infinite amount of iterated functions?  For example:$$e^{e^{e^{e^{e^{\cdots^x}}}}}=x$$
 A: *

*No, but other perhaps can=)

*I do not think so, or can you find an iterated function for this expression?

*It does but only for certain polynomials. For a general quartic polynomial you have four degrees of freedom ($Q(x) = x^4+ax^3+bx^2+cx+d$) but for an iterated quadratic you only still have $2$ degrees of freedom ($P(x) = x^2+Ax+B$), so it is quite unlikely that you can represent a given quartic polynomial by $P(P(x))$ where $P$ is a quadratic polynomial, but if you can, nobody stops you from using that technique.

*Well the important thing to notice is just that $f(x) = x$ implies $f(f(x)) = x$ but not necessarily the other way round.

*Only numerical solutions (or for certain proofs) you can use a fixed point theorem, e.g. the Banach fixed point theorem.

*No generally, not, as $e^{e^{e^{...x}}}$ is not really well defined.
A: Concerning 3
Your method for solving quartic polynomials will only work in some very special cases. Let us try to pick coefficients $p,q,r$ that generate the quartic $ax^4+bx^3+cx^2+dx+e$. Then as you said:
$$a=p^3$$
Giving us three possible values of $p$, and then:
$$b=2p^2q$$
gives us only one possible value of $q$ for each value of $p$, and then finally 
$$d=q^2+2pqr$$
gives only one possible value of $r$ for each value of $p$. Hence you can create only create three different polynomials for any given values of $a,b,d$, but $c,e$ could obviously be chosen in infinitely many ways. So for general quartics, it will only rarely work. That being said, if one happens to find a quartic that can be solved in this way, then it would make for a very elegant solution indeed. 
