Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Or in other words, polynomial relation of the function rather than the argument. I've worked out that in general $ f(x+1)={f(x)}^n $ implies $$f(x) = C^{n^x} $$ for some C, but I would like to know if there's a general form for more complicated polynomials, and I'd especially like to know how many arbitrary constants are involved.

share|cite|improve this question

There are (up to conjugacy by linear functions) two sequences of polynomials $P_n$ such that $P_n$ has degree $n$ and $P_n(P_m(x)) = P_{nm}(x)$ (i.e. $P_n$ form a semigroup under conjugation). These are the monomials $P_n(x) = x^n$ and the Chebyshev polynomials $P_n = \cos(n \arccos(x))$. Each of these give rise to closed-form general solutions of your recurrence:

$f(x+1) = P_k(f(x))$ with $f(0)=c$ has solution $f(x) = P_{k^x}(c)$.

On the other hand, any polynomial $f$ of degree $> 1$ has fixed points and periodic points, and these result in particular closed-form solutions of your recurrence.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.