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Suppose we have a family of functions $\{f_1,f_2,\cdots,f_n\}$, all are $ f_i:\mathbb{R}\to\mathbb{R}$. For simplicity, write $f_i(x)$ as $f_i x$. So $f_if_jx=f_i(f_j(x))$. Let $i_j$ be a number in $\{1,2,\cdots n\}$, and $j$ is the index. Now introduce the set for some $x_0\in\mathbb{R}$: $$S\equiv \{f_{i_m}f_{i_{m-1}}f_{i_{m-2}}\cdots f_{i_1}x_0 | m\in~\mathbb{N} \}$$ Are there any mathematics studying the behaviour of this set?

For example, when we have only one function $f(x)=2x$, and let $x_0=1$. Then the set $S$ is just $\{2^m|m\in\mathbb{N}\}$. When we have only two functions $f_1(x)$ and $f_2(x)$, then we can imagine a tree like structure below:

Tree

Then, $S$ is all the values appearing in these nodes.

Background

I ask this bizzare question because my friend asked me this:

If we have only functions $x^2,x^{1/2},e^x,\ln(x),\sin(x),\cos(x)$, can we compute all the numbers $\mathbb{Z}$, starting with $0$? No additional function is given, such as addition $f(x,y)=x+y$, or multiplication $f(x,y)=xy$.

I could only manage to get $0,1,2^m$, etc. There is no way to show that my method would exhaust all numbers though.

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    Do you mean multiplication or function composition? for example does $f_1f_2x_0$ mean $f_1(f_2(x_0))$ or $f_1(x_0) \cdot f_2(x_0)$2017-01-10
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    @mathreadler Composition.2017-01-10
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    The functions that can be written as compositions of a set $F$ of functions is the monoid $M$ generated by $F$. I'm fairly sure one would notate your set as $M\cdot x_0$ and call it the orbit of $x_0$ under $M$, as applying a function is a monoid action. (But I'm not sure; I've only really seen these things with groups. Maybe someone will give a more useful and authoritative answer)2017-01-10

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