Product property for reversing coefficients for polynomials Given a polynomial $$P(x)=a_0+a_1x+\cdots+a_nx^n,$$ define $$f(P(x))=a_n+a_{n-1}x+\ldots+a_0x^n.$$ How can we prove that $$f(P(x))f(Q(x))=f(P(x)Q(x))?$$
Expanding out the expression should be possible, but seems like a big pain. 
 A: Hint $\,\ f(P(x))\, =\, x^n P(x^{-1})\ $ is multiplicative, being the product of two multiplicative functions, namely $\,P(x)\mapsto P(x^{-1}),\,$ and $\,P(x)\mapsto x^{\,\deg P}\ $ (an "exponential" of the additive degree map)
Remark $ $ This often proves handy for irreducibility tests, e.g. here by Eisenstein.
A: Here is an easy way.
Suppose $x \neq 0$, and $P(x) = p_0+...+p_n x^{n_p}$. Then $x^{n_p} P({1 \over x}) = p_0x^{n_p}+p_1 x^{n_p-1} +...+p_{n_p} = f(P)(x)$.
Then $f(P\cdot Q)(x) = x^{n_p+n_q}(P \cdot Q)({1 \over x}) = x^{n_p}(P )({1 \over x}) x^{n_q}(Q)({1 \over x}) = (f(P) \cdot f(Q))(x)$.
A: Since you only assume $P$ to be a polynomial, it is easiest to work with the product. Let
$$
P(x) = \sum {a_{i}} x^{i} \\
Q(x) = \sum b_{i} x^{i}
$$
and assume $a_{n} b_{n} \neq 0$. It is seen that $f^{2} = Id$, hence we can prove $f(f(P(x)) f(Q(x)))$ instead. We see
$$
 f(P(x)) f(Q(x)) = \left( \sum a_{n-i} x^{i} \right) \left( \sum b_{n-i} x^{i} \right)
 = \sum_{k} \sum_{i+j = k} a_{n-i}b_{n-j} x^{k}.
$$
Then it seen that
$$
f(f(P(x)) f(Q(x))) = f(\sum_{k} \sum_{i+j = k} a_{n-i}b_{n-j} x^{k}) = \sum_{k} \sum_{i+j = 2n - k} a_{n-i}b_{n-j} x^{k} = \sum_{k} \sum_{i'+j' = k} a_{i'}b_{j'} x^{k}
$$
where in the last equality the substitution $i' = n-i$ and $j'=n-j$ is used. It is seen that this is equal to $P(x) Q(x)$.
