TITLE: How would I go about finding a closed form solution for $g(x,n) = f(f(f(...(x))))$, $n$ times?
QUESTION [9 upvotes]: A friend of mine mentioned that given the functions $g(x, n)$ and $f(x)$ such that:
$$g(x,n)=\overbrace{f(f(f(...f(x))}^\text{n}$$
For example:
$$g(x,1)= f(x)\\g(x,2) = f(f(x))$$
and that there is a closed form solution. I don't know exactly where I would begin, but any help or pointers would be very much appreciated. I am also not sure if I denoted the function $g(x,n)$ using the correct notation. 
Thank you in advance.

REPLY [3 votes]: $$g(x,n)=\overbrace{f(f(f(...f(x))}^\text{n}$$
You can express $g(x,n)$ in Abel equation. 
http://en.wikipedia.org/wiki/Abel_equation
Let's define $h(f(x))=h(x)+1$ 
$h(f(f(x)))=h(f(x))+1=h(x)+1+1=h(x)+2$
$h(f(f(f(x))))=h(f(x))+2=h(x)+1+2=h(x)+3$
If you continue in that way
$h(\overbrace{f(f(f(...f(x))}^\text{n})=h(g(x,n))=h(x)+n$
If you can find $h(x)$  , you can easily  find $g(x,n)$
$h^{-1}(h(g(x,n))=h^{-1}(h(x)+n)$
$g(x,n)=\overbrace{f(f(f(...f(x))}^\text{n}=h^{-1}(h(x)+n)$
Let's try to find $h(x)$

$$\cfrac{\partial}{\partial n }(h(g(x,n)))=\cfrac{\partial}{\partial n }(h(x)+n)$$
$$h'(g(x,n))\cfrac{\partial(g(x,n))}{\partial n }=1  \tag1$$

$$\cfrac{\partial}{\partial x }(h(g(x,n)))=\cfrac{\partial}{\partial x }(h(x)+n)$$
$$h'(g(x,n))\cfrac{\partial(g(x,n))}{\partial x }=h'(x) \tag2$$

If we divide two equations (1) and (2) then $h'(g(x,n))$ dissapears
$$\cfrac{\partial(g(x,n))}{\partial n }=\frac{1}{h'(x)}\cfrac{\partial(g(x,n))}{\partial x }$$
$$\cfrac{\partial^2(g(x,n))}{\partial n^2 }=\frac{1}{h'(x)}\cfrac{\partial}{\partial x }(\cfrac{\partial(g(x,n))}{\partial n })=\frac{1}{h'(x)}\cfrac{\partial}{\partial x }(\frac{1}{h'(x)}\cfrac{\partial(g(x,n))}{\partial x })$$
$$\cfrac{\partial^3(g(x,n))}{\partial n^3 }=\frac{1}{h'(x)}\cfrac{\partial}{\partial x }(\cfrac{\partial(g(x,n))}{\partial n })=\frac{1}{h'(x)}\cfrac{\partial}{\partial x }(\frac{1}{h'(x)}\cfrac{\partial}{\partial x }(\frac{1}{h'(x)}\cfrac{\partial(g(x,n))}{\partial x }))$$
Taylor series method.
$g(x,n)=f(g(x,n-1))$
$g(x,1)=f(x)=f(g(x,0))$
$g(x,0)=x$
$\cfrac{\partial(g(x,n))}{\partial n }|_{n=0}=\cfrac{1}{h'(x)}\cfrac{\partial(x+n.u(x)+n^2.p(x,n)))}{\partial x }|_{n=0}=\cfrac{1}{h'(x)}(1+nu'(x)+n^2\cfrac{\partial(p(x,n)))}{\partial x }|_{n=0}=\cfrac{1}{h'(x)}$
$\cfrac{\partial^2(g(x,n))}{\partial n^2 }|_{n=0}=\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)})'$
$\cfrac{\partial^3(g(x,n))}{\partial n^3 }|_{n=0}=\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)})')'$
$$g(x,n)= g(x,0)+n\cfrac{\partial(g(x,n))}{\partial n }|_{n=0}+\frac{n^2}{2!}\cfrac{\partial^2(g(x,n))}{\partial n^2 }|_{n=0}+...$$
$$g(x,n)= x+n\cfrac{1}{h'(x)}+\frac{n^2}{2!}\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)})'+\frac{n^3}{3!}\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)}(\cfrac{1}{h'(x)})')'...$$
If $\cfrac{1}{h'(x)}=w(x)$ then
$$g(x,n)= x+nw(x)+\frac{n^2}{2!}w(x)w'(x)+\frac{n^3}{3!}w(x)(w(x)w'(x))'...$$
$g(x,1)=f(x)=x+w(x)+\frac{1}{2!}w(x)w'(x)+\frac{1}{3!}w(x)(w(x)w'(x))'+...$
$g(x,-1)=f^{-1}(x)=x-w(x)+\frac{1}{2!}w(x)w'(x)-\frac{1}{3!}w(x)(w(x)w'(x))'+...$

To show how it works  in easy example:
If $h(x)=\ln(x)$
$$w(x)=\frac{1}{h'(x)}=x$$
$h(f(x))=h(x)+1$ 
$$\ln(f(x))=ln(x)+1$$
$$f(x)=e^{ln(x)+1}=ex$$
$$g(x,n)=e^n x$$
$$g(x,n)= x+nw(x)+\frac{n^2}{2!}w(x)w'(x)+\frac{n^3}{3!}w(x)(w(x)w'(x))'...$$
$$g(x,n)= x+nx+\frac{n^2}{2!}x (x)'+\frac{n^3}{3!}x (x)'+...=x(1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+...)=e^n x$$

$$g(x,n)= x+nw(x)+\frac{n^2}{2!}w(x)w'(x)+\frac{n^3}{3!}w(x)(w(x)w'(x))'...$$
$$g(x,-n)= x-nw(x)+\frac{n^2}{2!}w(x)w'(x)-\frac{n^3}{3!}w(x)(w(x)w'(x))'+...$$
$$\frac{g(x,n)-g(x,-n)}{2}= nw(x)+\frac{n^3}{3!}w(x)(w(x)w'(x))'+...$$
$$\lim_{n=0}\frac{g(x,n)-g(x,-n)}{2n}= w(x)=\frac{1}{h'(x)}$$
$$h(x)=\lim_{n\to 0}\int\frac{2n}{g(x,n)-g(x,-n)} dx$$
or
$$h(x)=\lim_{n\to 0}\int\frac{n}{g(x,n)-x} dx$$
It is still hard problem to find $h(x)$.
There can be approximation for $h(x)$
$$g(x,1)=f(x)= x+w(x)+\frac{1}{2!}w(x)w'(x)+\frac{1}{3!}w(x)(w(x)w'(x))'...$$
$$g(x,-1)=f^{-1}(x)= x-w(x)+\frac{1}{2!}w(x)w'(x)-\frac{1}{3!}w(x)(w(x)w'(x))'+...$$
After 3rd terms , If we have small values and so we can write
$$f(x)+f^{-1}(x)-2x \approx w(x)w'(x)$$
$$2\int f(x)+f^{-1}(x)-2x dx \approx w^2(x)+c$$
I still do not know how to write $h(x)$ exactly as series of $f(x)$.