Handle a function of itself I have a tricky question. I want to handle a function y(x) defined in this way:
$y(x)=f(y(x))g(x)$
Here, $f(y(x))$ and $g(x)$ are smooth functions: is there any way/method to express $y(x)$ as a (complicated) function of $x$ solely?
Maybe it is impossible if $f$ and $g$ are generic.
Edit:
If I have the numerical functions which describe $f(y)$ and $g(x)$, is there any approximated way to express $y(x)$? For example with recursive formulae?
My point is not to guess what form $y(x)$ should have, but deriving it or, at least, inferring it approximately.
 A: This is generally impossible. For example let $$g(x)={1\over 1+x^2}$$and $$f(u)=\cos u$$then $$y={\cos y\over 1+x^2}$$In the most general form, if $f(x)$ is a polynomial of degree at most $4$ or $f(x)={1\over ax^3+bx^2+cx+d}$ where $a,b,c,d$ are proper real numbers, then $y$ can be explicitly represented as a function of $x$.
A: If $f$ and $g$ are any smooth functions, there is no way of obtaining $y$ as an explicit function of $x$. $g$ here may be dropped, so the real question is whether you can solve
$$
y(x)=f(y(x))
$$
for $y$ explicitly, which is impossible in general.
Notice that $y$ may not even be a function. Take $x=0$, then if $y(0)=0$ and $f(0)=1$, we also have that $y(0)=1$, thus $y$ can't be a function.
A: You can find many functions to satify your conditions.
For example,on $x\in[0,\infty)$ define  $$y(x)=x^2,f(x)=\sqrt x, g(x)=x$$
We get $$f(y(x))g(x) =x\sqrt {y(x)} = x^2= y(x)$$
You may design other examples with little effort.    
A: Assuming $y(x)$ and $g(u)$ admit inverse functions (which is quite restrictive), we can replace $y(x) = u$:
$$
u = f(u) g\left(y^{-1}(u)\right)
$$
$$
\frac{u}{f(u)} = g\left(y^{-1}(u)\right)
$$
$$
g^{-1}\left(\frac{u}{f(u)}\right) = y^{-1}(u)
$$
Now the left hand side only depends on a free-variable $u$ and on known functions $f$ and $g$. So if you can find $g^{-1}$, then you can find a closed expression for the left hand side as a function of $u$. Call it $w(u)$:
$$
w(u) = g^{-1}\left(\frac{u}{f(u)}\right) = x
$$
Now your $y$ function is given by:
$$
y(x) = w^{-1}(x)
$$
As others pointed out, finding such $y$ is not always possible, and this particular solution is quite restrictive. It will likely only work for cherry picked examples of $f$ and $g$.
