Handle a function of itself

Question

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.

Answer 1

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 $x$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $g$.