Are there two real-valued functions defined on the same subset of $\mathbb{R}$ that commute with each other but are not inverses of each other? (After several responses, I have to make an edit to my post. Neither function should be the identity function nor the zero function. The two functions should not be the same function.)
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1$f:x\mapsto x$, $g:x\mapsto 2x$ ? – Balloon Feb 02 '17 at 17:54
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$f(x)=0$ for all $x\in\mathbb{R}$ and $g:\mathbb{R}\to\mathbb{R}$ any function with $g(0)=0$. – ryanblack Feb 02 '17 at 17:55
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2$f(x)=x^3$, and $g(x)=x^4$. – Crostul Feb 02 '17 at 17:58
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@dxiv I wanted examples of functions defined on $\mathbb{R}$. Remember, one of the tags is pre-Calculus. – user74973 Feb 02 '17 at 18:40
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@user74973 The accepted answer to that question is about real functions - polynomials and rational functions, with the Chebyshev polynomials being a classic example. The question is indeed interesting (and I actually +1'd it) but it's hard to give a better answer without shamelessly copy/pasting the other answer. – dxiv Feb 02 '17 at 18:48
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$f$ commutes with itself, just take $f$ so that $f\circ f$ is not the identity.
for another example, take $f$ and $f\circ f$, and let $f$ be any function such that $f\circ f$ and $f\circ f \circ f$ are not the identity.
Asinomás
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For example $f(x)=x^m\,$, $\,g(x)=x^n\,$.
dxiv
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@user74973 That works, but in general you may need to restrict the domain to $\mathbb{R}^+,$ once you take non-integer exponents, for example $m=3, n=1/2,$. – dxiv Feb 02 '17 at 18:04
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$f(x)=x$ and $g(x)=1/x$ for $x \in (0, \infty)$
Fred
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I forgot to exclude the case that one of the functions is the identity function. – user74973 Feb 02 '17 at 17:57
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You can use $f, g \colon \mathbb{R} \to \mathbb{R}$ with $f(x) = \pi \, x$ and $g(x) = \mathrm{e} \, x$.
gerw
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