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Consider a polynomial $f(x)$ with real coefficients having the property $f(g(x)) = g(f(x))$ for every polynomial $g(x)$ with real coefficients. Determine and prove the nature of $f(x)$.

My guess is that $f(x)$ is its own inverse. Could anyone shed some more fire on this for me?

Also, what if $f(x)$ is not its own inverse?

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up vote 4 down vote accepted

Let's take a constant polynomial for $g$, say $g(x) = g_0$. The given property reads then $$ g_0 = g\bigl(f(x)\bigr) = f\bigl(g(x)\bigr) = f(g_0) $$ So, if any $f$ has your property then $f(x) = x$. As this $f = \mathrm{id}$ obviously fulfills $f \circ g = g \circ f$ for any $g$, we see that that $f = \mathrm{id}$ is the only solution.

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does $\mathrm{id}$ mean identity? Like in Abstract Algebra meaning of identity? – yiyi Nov 17 '12 at 15:31
The $\mathrm{id}\colon \mathbb R\to \mathbb R$, that is the polynomial $f(x) = x$. – martini Nov 17 '12 at 15:32
What is the name of $\mathrm{id}$, so I can look it up, or tell me the name of a book to learn more. – yiyi Nov 17 '12 at 15:48
In this context it is simply the polynomial given by $f(x)=x$. – Hagen von Eitzen Nov 17 '12 at 16:03

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