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This is a followup to the question here: How to show that the following function isn't a polynomial over Q?.

As before, let $b_1,b_2,\dots$ be an enumeration of $\mathbb Q$. The question might be sensitive to the enumeration (but probably not).

Question 1: Suppose I define $$f_3(x) = (x-b_1)^3 + (x-b_1)^3(x-b_2)^3 + \dots,$$

how do I show that this function is not a polynomial? I do not believe the answers to the old question extend to this case since they all seem to use the positivity of squares.

Can we show that $f(x)$ is not a polynomial?

There is a even more general version (and the one I was interested in from the beginning) that I was interested in, however I forgot to omit the trivial cases/there was some ambiguity in the second question, so I asked it again here:

Is the following function a polynomial?

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  • $\begingroup$ Clearly you can ask an infinite family of these questions, and probably most people don't want to answer each new variant individually. What is the motivation? (E.g., if this question is answered negatively, would that settle your curiosity, or is there some deeper question that these questions are meant to approximate?) $\endgroup$
    – LSpice
    Commented Dec 11, 2017 at 23:35
  • $\begingroup$ The tone of my question was misleading, I think these functions are not polynomials and would like to show that. Question 2 is what I am really interested in. Also, YaakovBaruch himself retracted his comment. It is not clear why $(x-b_1)$ divides the polynomial twice... $\endgroup$
    – Asvin
    Commented Dec 11, 2017 at 23:43
  • $\begingroup$ (Sorry, I saw @YaakovBaruch's retraction and deleted my comment about it before you responded.) $\endgroup$
    – LSpice
    Commented Dec 11, 2017 at 23:49
  • $\begingroup$ I answered both of your questions now. $\endgroup$
    – GH from MO
    Commented Dec 12, 2017 at 3:09

1 Answer 1

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$f_3(x)$ is not a polynomial function, which can be seen as follows. Call $b_n$ a champion if it exceeds all the earlier terms $b_1,\dots,b_{n-1}$. Clearly, there are infinitely many champions, and they tend to infinity. Let $m$ be fixed, and consider the champions $b_n$ with $n>m$. We get, as $b_n$ tends to infinity, $$ f_3(b_n)>\prod_{k=1}^m (b_n-b_k)^3 = (1+o(1))\ b_n^{3m}. $$ This shows that $f_3$ has degree at least $3m$. As $m$ is arbitrary, the claim follows.

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    $\begingroup$ Over $\mathbb Q$, that is enough, of course. I'm curious though if we can take care of the convergence issue on the whole $\mathbb R$ as well (choosing a reasonable enumeration, of course) and make it the identity map everywhere. $\endgroup$
    – fedja
    Commented Dec 12, 2017 at 0:40
  • $\begingroup$ @fedja: I don't know. I thought $G(x,y)$ was defined on $\mathbb{Q}\times\mathbb{Q}$. $\endgroup$
    – GH from MO
    Commented Dec 12, 2017 at 1:21
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    $\begingroup$ The unique solution to this recurrence is $a_0=b_1$, $a_1=1$, and $a_n=0$ for $n\geq 2$. So the expression for $g$ is actually a finite sum. $\endgroup$
    – Julian Rosen
    Commented Dec 12, 2017 at 2:04
  • $\begingroup$ @JulianRosen: Thanks a lot! I added this in the "Added" section. $\endgroup$
    – GH from MO
    Commented Dec 12, 2017 at 2:30
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    $\begingroup$ Erm... If we allow zeroes, I can just put all $a_k=0$ and get the identically $0$ function, which is certainly a polynomial. This, of course, is formally OK, but it is pretty clear that it is not what was meant... $\endgroup$
    – fedja
    Commented Dec 12, 2017 at 4:05

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