# How to show that the following function isn't a polynomial over Q?

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?.

• 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 $x$. Dec 11 '17 at 19:27
• 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. Dec 11 '17 at 21:08
• 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... Dec 11 '17 at 21:37
• @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). Dec 11 '17 at 23:52
• 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 $f/(x - b_1)$. Dec 12 '17 at 1:15

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 $≥2n$.

• Just curious: Do you see a way to prove something similar if we replace $\mathbb Q$ by $\overline{ \mathbb F}_p$ ? Dec 11 '17 at 19:50

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.

• why do $n$ values determine a degree $n$ polynomial? Don't you need $n+1$ terms? Dec 11 '17 at 20:20
• 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? Dec 11 '17 at 20:29

"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 $\mathbb Q$.

• 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. Dec 11 '17 at 23:48
• 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. Dec 12 '17 at 0:04