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Let $P(x) \in \mathbb{Q}[x]$ be a polynomial satisfying $P(x)= P(r-x)$ for some $r$. Then $P(x)$ is a polynomial in $x(r-x)$. Similarly if $- P(x)= P(r-x)$ then $P(x)$ is equal to the product of $(x-r/2)$ and a polynomial in $x(r-x)$. (Such $P(x)$ is called reflective and anti-reflective respectively). We may refer to D. Knuth, Johann Faulhaber and Sums of Powers for these facts.

I am collapsed in the following (probably) simple question. Let $$P(x) = x^n \pm (1-x)^n$$ so that $P(x)$ satisfies $\pm P(x)=P(1-x)$. So
$$P(x) = Q(x(1-x)) \text{ or } P(x) = (x-1/2) Q(x(1-x))$$ for some polynomial $Q$ depending on the sign. What are the coefficients of $Q(x)$?

I computed $Q(x)$ for small values of $n$ directly, but I couldn't see a way to generalize it. Factorizing $P(x)$ does not seem to work either as I get terms like $x^k(1-x)^l$ for $k \neq l$.

Is there a formula involving some binomial terms for the coefficients of $Q(x)$? A recursive formula will also be fine for me.

Thanks in advance.

user114285
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  • maybe you can find something by computing them in order. The 0-th term ($+$ case) is 1, then take $P(x) -1$, divide by $x(1-x)$ and get again the 0-th term, that is $1-n$. Continue to see if you can get a pattern – Exodd Aug 26 '20 at 08:07
  • or you could find the roots of $P$ and transform them in roots of $Q$ – Exodd Aug 26 '20 at 08:09
  • @Exodd. Thanks for the comments. I considered the first way earlier, but couldn't get a formula. The terms do not fit to standard binomial terms that I know. I will also try the second one (it seems fancy indeed). The roots of $P(x)$ are expressed in terms of roots of $1$ (Except 1/2 appearing in the $-$ case). Then I may compute the factorization of $Q(x)$ and then maybe obtain the coefficients (the passage to coefficients may be messy). – user114285 Aug 26 '20 at 08:17

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