We can see that if $f(x)=g(x)=x$ then $f(g(x))=g(f(x))$.
I would like to see other examples of functions $f(x)$ and $g(x)$ such that $f(g(x))=g(f(x))$.
P.S. By definition we also must have $D_{f\circ g}= D_{g\circ f}$
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lisyarus
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Konstantinos Gaitanas
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@Workaholic the first and the third are clearly wrong read the P.S. – Konstantinos Gaitanas Feb 20 '16 at 18:29
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1something is wrong, all functions such that f=g satisfy the equation. – runaround Feb 20 '16 at 18:30
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3@runaround Why does that mean something is wrong? It just means you've found a family of solutions. – πr8 Feb 20 '16 at 18:31
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5Relevant: 1, 2, 3 – πr8 Feb 20 '16 at 18:37
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@OmarNagib this is wrong – Konstantinos Gaitanas Feb 20 '16 at 18:58
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@KonstantinosGaitanas what do you mean wrong? – Omar Nagib Feb 20 '16 at 18:58
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Check the domain of the composition – Konstantinos Gaitanas Feb 20 '16 at 18:58
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1If you've got a one-to-one(Injective) function $f(x)$, then you can always define its inverse $g(x)=f^{-1}(x)$ such that $f(g(x))=g(f(x))$. for example, consider $f=x^3$ and $g=\sqrt [3] {x}$. – Omar Nagib Feb 20 '16 at 19:05
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1@KonstantinosGaitanas both $f(g)$ and $g(f)$ maps from the reals to the reals. I cannot discern the problem. – Omar Nagib Feb 20 '16 at 19:06
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From reals to reals?I am not sure check it again. – Konstantinos Gaitanas Feb 20 '16 at 19:12
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Note that two of the examples given so far fall into a general framework. Let $f: D \to D$ be a function, and consider composing $f$ with itself several times, possibly a negative number of times if $f$ is invertible. The generator function has been $f(x) = x + 1$ in both cases. Does the tag [tag:special-functions] imply that you really want special functions (like Bessel or Lambert W) to show up, or are you just considering functions that commute "special"? – pjs36 Feb 20 '16 at 19:15
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Functions need not be inverses of each other to commute. A simple example. Let $f(x)=x+1$ and $g(x)=x+2$. Here, of course $$f(g(x))=g(f(x))=x+3.$$
Tim Raczkowski
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Here is a family of examples: let $h$ be any function from a set to itself, and $n,m$ any positive integers. Then $f(x)=h^m(x),g(x)=h^n(x)$ (here superscripts denote functional power, so for example $h^3(x)=h(h(h(x)))$). Then clearly $f(g(x))=h^{m+n}(x)=g(f(x))$.
Wojowu
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If you generalise to $f^{-m} (x)=(f^{-1})^m (x)$, I think this encompasses all of the answers given so far. So the next natural question to ask is, "are these all the solutions?" – bilaterus Feb 20 '16 at 19:53
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@bilaterus Certainly not. If you take $f(x)=x^\alpha,g(x)=x^\beta$ for some linearly independent $\alpha,\beta$, then I'd risk betting that $f,g$ aren't expressible as powers of fixed power $h$. – Wojowu Feb 20 '16 at 19:55
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Take $f(x)=x+1$ and $g(x)=x-1$
$$f(g(x))=g(x)+1\\ =(x-1)+1\\ =x\\ \text{ and }\\ g(f(x))=f(x)-1\\ (x+1)-1\\ =x$$
3SAT
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@KonstantinosGaitanas Let $f$ and $g$ be functions from $\mathbf{R}{\geqslant0}$ to $\mathbf{R}{\geqslant0}$, where $f:x\mapsto x^2$ and $g:x\mapsto\sqrt{x}$, then $f(g(x))=g(f(x))$, and your desired condition $\operatorname{dom}(f\circ g)=\operatorname{dom}(g\circ f)$ still holds. – Workaholic Feb 20 '16 at 18:43
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I think if you take two inverse functions, defined over e.g. $\Bbb R$ (or over the same subset of $\Bbb R$),
then you trivially get the desired property, because $x = f(g(x)) = g(f(x))$.
Here is an example where the two functions are not inverse.
$f(x) = x^n$
$g(x) = x^m$
where $n$ and $m$ are e.g. integers
Also:
$f(x) = a \cdot x$
$g(x) = b \cdot x$
where $a$ and $b$ are e.g. reals
peter.petrov
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I'm not sure there's much to say.
There's $g= f^{-1}$ so $f (g (x)) = x $
Then there's $f = g $ so $f (g (x))e f (f (x)) $
Then there's $f (g (x)) = g (f (x)) = h (x)$ so $g (x) = f^{-1}(g(f (x))) = f^{-1}(h (x)) $ and/or $f (x)=g^{-1}(h (x))$.
As long as we can find invertible functions we can find these. Is there anything more that needs to be said?
fleablood
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