Functional equation $P(X)=P(1-X)$ for polynomials

Question

I have encountered the following problem :

Find all polynomials $P$ such as $P(X)=P(1-X)$ on $\mathbb{C}$ and then $\mathbb{R}$.

I have found that on $\mathbb{C}$ such polynomials have an even degree. Because for each $a$ root, $1-a$ must be a root too. I struggle to find whether we can do something with polynomials of degree $2$ on $\mathbb{R}$.

I was also wondering if we could find a general solution for $P(X)=P(aX+b)$ with $a$ and $b$ complexe for $P$ defined either on $\mathbb{R}$ or $\mathbb{C}$.

Answer 1

First, if $P(X) = P(1-X)$ holds on $\Bbb R$ then the same relation holds on $\Bbb C$, due to the identity principle for holomorphic functions.

With $Q(x) := P(x + \frac 12)$, the condition $P(X) = P(1-X)$ is equivalent to $Q(x) = Q(-x)$, which is satisfied exactly by all polynomials having only terms with even powers of $x$.

Therefore the general solution is $$ P(X) = Q(X - \frac 12) = a_0 + a_1 (X - \frac 12)^2 + \ldots + a_n (X - \frac 12)^{2n} $$ with coefficients $a_0, a_1, \ldots, a_n \in \Bbb R$ or $\Bbb C$, depending on whether you want $P$ to be real-valued on $\Bbb R$ or not.

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For the general case $P(X) = P(aX + b)$ with $a \ne 1$, note that the fixed point of $X \to aX +b$ is $X = \frac{b}{1-a}$, therefore define $Q(x) := P(x + \frac{b}{1-a})$ to get $Q(x) = Q(ax)$.

Now there are two cases: If $a$ is a "root of unity", i.e. $a^k=1$ for some positive integer $k$, then the solutions are exactly the polynomials having only terms with powers which are a multiple of $k$: $$ Q(x) = a_0 + a_1 x^k + \ldots + a_n x^{kn} \, . $$ If $a$ is not a root of unity then $$ Q(1) = Q(a) = Q(a^2) = Q(a^3) = \dots $$ with all arguments being different, and the only solution are constant polynomials $Q(x) = a_0$.

For $a \in \Bbb R$, $a \ne 1$, the first case occurs only for $a= -1$.

Answer 2

It is not necessary to shift $X$ by $1/2$ to symmetrize the polynomial.

Any polynomial function of $X(1-X)$ is a solution, and all solutions can be written in that form by starting from the highest degree (which must be even) term and working downward.

The advantage is that this works "over $\mathbb{Z}$" without any powers of $\frac{1}{2}$. For any ring $R$ of coefficients, if the polynomial is $\sum a_n X^n$ with $a_n \in R$ then it is equal to $\sum b_n (X - X^2)^n$ for some coefficients $b_n \in R$ that are integer linear combinations of the $a$'s.

The generalization has the same answer. There are nonconstant solutions of $P(x)=P(f(x))$ with $f(x)=Ax+B$ if any only if $f$ has (minimum) order $n$ with respect to composition, $f^{\circ n} (x) = x$ for some $n \geq 1$, and then any solution can be written as $Q(x f(x) f(f(x)) \dots f^{\circ (n-1)}(x))$, where the coefficients of $Q$ are $\mathbb{Z}$-linear combinations of the coefficients of $P(x)$.

The linear functions of order $n$ of course are $f(x)=\omega (X-u) + u$ for $w$ an $n$th root of $1$ and $u$ arbitrary, also over any ring.