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Oct 29, 2018 at 19:41 vote accept Student
Oct 29, 2018 at 18:25 answer added François Brunault timeline score: 7
Oct 29, 2018 at 18:18 comment added Robert Israel According to your definition, if $f$ is not identically $0$ you must have $n$ even, because $f(x) = f(-(-x)) = (-x)^n f(-1/x) = (-x)^n (1/x)^n f(x)$.
Oct 29, 2018 at 17:18 comment added Philipp Lampe Is $n$ the degree of $f$? In the literature, authors typically use the functional equation $-f(x)=x^nf(x^{-1})$ to define antipalindromic polynomials.
Oct 29, 2018 at 17:00 history asked Student CC BY-SA 4.0