I am looking for an expression $f(x)$ involving only constants, $x$, multiplication, exponentiation, and division such that for in as large an interval [1,a) as possible, the function closely approximates $g(x)=x+1$. Or maybe even a recursive sequence of increasingly good approximations $f_i(x)$.
$f(x)$ should not include any addition and logarithms. Allowed operators are: $$ x^{n}, nx, \frac {x}{n}$$ where $n$ can be $x$ or $const$. Any composition of these operators is allowed. Here is something I tried (graph):
$$ f(x) = 1.8 (x^{\frac{1}{2}}) (x^{\frac{1}{2*3}})^{(x\frac{1}{2*3})} (x^{\frac{1}{2*3*5}})^{{(x\frac{1}{2*3*5})}^{(x\frac{1}{2*3*5})}} $$
But continuing this pattern does not make $f(x)$ any closer to $g(x)$.
The way I understand the question, you're allowed to do the following:
Take $f(x)^{g(x)}$, if $f$ and $g$ are allowed functions
Take $f(x)g(x)$, if $f$ and $g$ are allowed functions
Take $\frac{f(x)}{g(x)}$, if $f$ and $g$ are allowed functions
(starting with the base allowed functions of constants and $f_0(x)=x$). With this, one can express
$$e^{f(x)}=\frac{(ex)^{f(x)}}{x^{f(x)}}.$$
Thus, we can use the technique of kingW3 and approximate
$$x+1=\exp\left(\ln(x+1)\right)\approx x\exp\left(\sum_{n=1}^{N} \frac{(-1)^{n+1}}{nx^n}\right)=x\prod_{n=1}^N \exp\left(\frac{(-1)^{n+1}}{nx^n}\right).$$
As this is simply the product of allowed functions (as $\exp(f(x))$ is an allowed function where $f(x)$ is), this is also allowed.