Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

Sequences of polynomials?

Let f(n) be a sequence of polynomials defined simply as f(n)(x)=x^n.

The f(m)(f(n)) (x) = f(mn)(x) for all m,n, and hence f(m)(f(n)) = f(n)(f(m)). So far, so obvious.

So what about OTHER sequences of polynomials with those properties? I've found one other sequence, also with integer coefficients, of polynomials in one variable with the properties:

A. f(n) is a polynomial of degree n.

B. f(n)(f(m)) = f (mn)

C. f(n)(f(m)) = f(m)(f(n))

So my questions are:

1. Prove, as I have, that there exists at least one other sequence of polynomials with properties A, B, and C beyond the trivial case given above.

2. Are there any other sequences of polynomials with those properties?

3. Are there any with properties A and C but not B?

4. Do the answers change if we allow coefficients to be, say, from the field of real (or, if you prefer, complex) numbers?

Update:

Great start!.

The example I was thinking of, by the way, is that cos(nx) turns out to always be an nth-degree polynomial in cos x.

Update 2:

OK. I was lost in your notation for a moment, but now I get it. You're conjugating f by another transformation. Duh! Good work!

2 Answers

Relevance
  • 1 decade ago
    Favorite Answer

    Okay, I finished the proof. See [1] for the pdf. It is a bit hard to reproduce the proof here in this space.

    To summarize: up to a conjugation (but we have to allow conjugation over rational numbers... but fret not, this is enough for the existence and uniqueness over integers) we can show that any such sequence is

    f_n = x^n

    or

    f_n = 2 cos nθ

    with

    x = 2 cos θ

    This second case corresponds exactly to the case f_2 = x^2 - 2. (I didn't quite understand your additional comment before. Some of my friends helped me parse what you were saying.)

    I was able to prove uniqueness using the polynomial formulation: basically once you fix f_2, any f_n will be unique. Therefore conditions A and C does imply B.

  • 1 decade ago

    The proof of the classification of commuting sequences of polynomials is sketched on page 436 of "Mathematical Omnibus: Thirty Lectures on Classic Mathematics" by D. B. Fuks and Serge Tabachnikov. You can find it by searching Google Books with the phrases "omnibus" and "sequence of commuting polynomials."

Still have questions? Get your answers by asking now.