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I thought a lot at this problem: a polynomial $p(x) = a_{n}x^n + a_{n - 1}x^{n - 1} + ... + a_1x + a_0$, where $\forall i : a_{i} > 0$ (so that means that $p(x)$ includes all non-negative integer powers of $x$ lesser than $n + 1$) is given. But you don't know any of $a_{i}$ and you don't know $n$ too. You may choose any $x_0$ and ask a question: "what is the value of this polynomial at point $x=x_0$?". And the question is what is the least count of questions you have to ask to guaranteed get to know all coefficients?

I find out that 2 questions is enough: firstly we asking a value of $p(1)$ and getting $s = \sum_{i = 0}^{n} a_i$. Then we ask $p(s + 1) = q$ and simply getting all coefficients converting $q$ into $s + 1$ base system. Coefficients will be represented at digits in this notation. (This works because $s + 1$ is greater than any coefficient for sure).

But I heard that this solution is not optimal and 1 question is actually enough. As I understood, the solution with one question is not useful and can't be used in real life because it is correct but formal. I could not think of this solution (and I don't know it as well) but I know that it somehow uses irrational numbers. I thought that maybe there is such a theorem which claims that at irrational number all $p(x)$ that satisfies our limitations have got different values or something like this, but I did not find it.

So I ask you to help me with that and either explicate the solution or tell the title of such theorem (if it exists). And a big request for you to attach proof or links to useful related materials. Thanks!

Kirill Afentev
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  • Do you know any ring theory? Generally $,R[x]\cong R[\alpha]\iff \alpha $ is transcendental over $R,,$ i.e. $\alpha$ is not a root of a nonzero polynomial with coef's in $R,,$ e.g. $,\Bbb Z[x]\cong \Bbb Z[\pi].,$ Informally: the indeterminate in a polynomial ring can be represented by any transcendental. This is reified structurally by the universal property of polynomial rings. – Bill Dubuque May 09 '20 at 17:34
  • That's what I've written in my answer – ShBh May 09 '20 at 17:40
  • Here is said universal mapping property of polynomial rings. – Bill Dubuque May 09 '20 at 17:41
  • @Gone here you can find a nice article on an application of the transcendence of $\pi$ – ShBh May 09 '20 at 18:12

1 Answers1

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Rather looking at only irrational numbers, look at transcendental numbers which are not roots of a polynomial with rational coefficients. Let $P(X)$ and $Q(X)$ be two polynomials of degree $n$ having coefficients in $\mathbb{N}$. Let $P(\pi)=Q(\pi)$. If $P$ and $Q$ are different polynomials then $P-Q$ is a non-constant polynomial in $\mathbb{Q}[X]$ with $\pi$ as a root. A contradiction to the fact that $\pi$ is transcendental. So different polynomials in $\mathbb{Q}[X]$ have different values at transcendental numbers.

But the question is how can we get the coefficients from the value of $P(\pi)$.

ShBh
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  • We need to figure out the algorithm if any, to find the coefficients – ShBh May 09 '20 at 17:41
  • I'm pretty sure we'll need an oracle that, given an integer-coefficient polynomial $Q$, returns whether $Q(\pi)$ is equal to the sampled value $P(\pi)$. Otherwise, we will never be sure we got the right polynomial, after any finite amount of time. Even with an oracle, I doubt any method is significantly better than going over all (countably many) candidate polynomials one by one and testing them for equality. – Meni Rosenfeld May 09 '20 at 18:48
  • Yes, I think you're right with checking all polynomials. I'm thankful for your explanation! – Kirill Afentev May 09 '20 at 18:56
  • @MeniRosenfeld Here we have the restriction that the coefficients are positive (non-negative would suffice) integers, so there are only finitely many candidates. Of course if $P(\pi)$ is large, there will still be too many candidates for it to be practical. – Daniel Fischer May 09 '20 at 22:28
  • @DanielFischer: Oh, right. I was in a weird state where I remembered the premise that the coefficients are positive, but didn't think clearly of the consequences. Not only are there finitely many possibilities, you also don't need an oracle - you can simply calculate one digit at a time for each candidate, and eventually all will drop out except for the correct one. And perhaps there are better ways that make use of this fact... – Meni Rosenfeld May 10 '20 at 12:10