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Enumerate the rationals as $b_1,b_2,\dots$ and define the (set) function: $$f(x) = (x-b_1)^2 + (x-b_1)^2(x-b_2)^2 + \dots.$$ At any particular $x$, only finitely many terms are non zero so this is perfectly well defined as a (set) function but surely, it is not equal to any polynomial! (or is it?) How do I show that there is no polynomial $p(t) \in \mathbb Q[t]$ such that $p(x) = f(x)$ for all $x \in \mathbb Q$?

If $f(x)$ were defined as $(x-b_1) + (x-b_1)(x-b_2) + \dots$, then this question is not so hard. If $p(x)$ has degree $n$, then testing on $b_1,\dots,b_n$ would show that $p(x)$ is necessarily $(x-b_1) + (x-b_1)(x-b_2) + \dots + (x-b_1)\dots(x-b_n)$ but then $x=b_{n+1}$ derives a contradiction.

I don't know how to adapt this approach. Trying to guess the polynomial seems hard even if we think $p(x)$ is degree $1$.

I posted a follow up to this question here: (Variation of an old question) Are these functions polynomials?.

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    $\begingroup$ It is even conceivable that the question of whether $f(x),\mathcal{O}$ (where $\mathcal{O}$ represents some ordering in the enumeration of the rationals,) is expressible as a polynomial, depends on the choice taken for $\mathcal{O}$. So a good question is whether there is any $\mathcal{O}$ and $p(t)\in\Bbb Q[t]$ for which $p(x) = f(x)$ for all rational $x$. $\endgroup$ Dec 11 '17 at 19:27
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    $\begingroup$ It's easy to see that then $f_1=f/(x-b_1)^2$ would have to be a polynomial too, but of lower degree. Then again $f_2=(f_1-1)/(x-b_2)^2$ would also be a polynomial of lower degree yet... and so on. $\endgroup$ Dec 11 '17 at 21:08
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    $\begingroup$ There may be a flow in my scheme: I'm no longer sure that it's obvious that $b_1$ is a zero of $f/(x-b_1)$ and thus that $f/(x-b_1)^2$ is indeed a polynomial... $\endgroup$ Dec 11 '17 at 21:37
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    $\begingroup$ @ChristianRemling, but (modulo my comment) this doesn't seem to show that the polynomial $p/(x - b_1)$ has a root. That is, we can divide $f$ by $x - b_1$ 'canonically' to obtain a function $g$ (EDIT: oops, sorry, not your $g$), and we can divide $p$ by $x - b_1$ canonically to obtain a polynomial $q$, but it isn't a priori clear to me that $g = q$ (in other words, that these two kinds of division preserve equality). $\endgroup$
    – LSpice
    Dec 11 '17 at 23:52
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    $\begingroup$ I agree that $f/(x - b_1)$ is not always well defined for a function vanishing at $b_1$. In this case, though, each term in the defining sum for $f$ is a polynomial divisible by $x - b_1$, and so there is a natural sum of polynomials that deserves to be called $f/(x - b_1)$. $\endgroup$
    – LSpice
    Dec 12 '17 at 1:15
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For each positive integer $n$ and any rational $x$, we have
$$f(x)\geq (x-b_1)^2(x-b_2)^2\dots(x-b_n)^2.$$ For large $x$, we then have $f(x)\gg x^{2n}$, which implies that if $f$ is a polynomial, it must have degree $≥2n$.

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    $\begingroup$ Just curious: Do you see a way to prove something similar if we replace $\mathbb Q$ by $\overline{ \mathbb F}_p$ ? $\endgroup$
    – Asvin
    Dec 11 '17 at 19:50
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You can adapt the same approach as follows: Say $p_n(x)$ is an $n$-th degree polynomial matching $f(x)$ at all rational points, then in particular, $$ p_n(b_1) = 0\\ p_n(b_2) = (b_2-b_1)^2 \\ \vdots\\ p_n(b_k) = \sum_{i\in\Bbb Z+, i<k}\prod_{j\in\Bbb Z+, j\leq i} (b_k-b_i)^2 $$ For a given fixed sequencing of the rationals as $b_1, b_2, \cdots$, and for any given $k$, the latter expression is just some fixed rational number.

So $p(n)$ is fixed by its values at $b_1 \ldots b_n$, and now consider $-f(b_{n+1}-p_n(b_{n+1}))$. Since all the terms past the $n+2$ term in $f(b_{n+1})$ are zero, $$f(b_{n+1}) = p_n(b_{n+1}) + \prod_{i\leq n}(b_{n+1}-b_i)^2 > p_n(b_{n+1})$$ which contradicts the statement that $f$ matches $p_n$ at all rationals.

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  • $\begingroup$ why do $n$ values determine a degree $n$ polynomial? Don't you need $n+1$ terms? $\endgroup$
    – Asvin
    Dec 11 '17 at 20:20
  • $\begingroup$ Do you really mean to consider $-f(b_{n+1} - p_n(b_{n+1}))$ and not $-f(b_{n+1}) - p_n(b_{n+1})$? Is the idea to use that $f$ is always positive? How do you show that the difference between f(b_{n+1}) and $p_n(b_{n+1})$ is that expression? $\endgroup$
    – Asvin
    Dec 11 '17 at 20:29
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"If $f(x)$ were defined as $(x−b_1) + (x−b_1)(x−b_2) + …,$ then this question is not so hard."

Does taking the derivative of the given function get you to this simpler case?

"At any particular $x$, only finitely many terms are non zero"

But at irrational values of $x$, none of the terms are zero. Now, you may respond "But I'm talking about it over $\mathbb Q.$" But unless you're going to claim that $f$ isn't continuous, if you take a sequence of irrationals approaching a rational, the function evaluated at that rational must be the limit of the function evaluated at those irrational numbers. You could also construct a sequence of rationals approaching $b_1$ such that there exists some $\epsilon > 0$ such that for all $x$ in the sequence, $f(x) > \epsilon$. If $f(x)$ is continuous, $f(b_1)$ must be $> 0$, but clearly by the definition of $f$, $f(b_1) = 0$. I suppose you'll still have to argue that $f$ must be continuous, but that should follow from it being a polynomial, even with the restriction to $\mathbb Q$.

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    $\begingroup$ It is not clear that the function is continuous, much less continuously extendible to $\mathbb R$, or differentiable. (Also, the termwise derivative doesn't match this easier function.) Even if the function extended continuously, your argument only shows that the same formula doesn't naïvely define the extension. $\endgroup$
    – LSpice
    Dec 11 '17 at 23:48
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    $\begingroup$ It's a proof by contradiction. I'm certainly not claiming that it is continuous, merely that if it were defined by a polynomial over Q, then it would be continuous. $\endgroup$ Dec 12 '17 at 0:04

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