I was watching this Mathologer video (https://youtu.be/YuIIjLr6vUA?t=1652) and he says at 27:32

First, suppose that our initial chunk is part of a parabola, or if you like a cubic, or any polynomial. If I then tell you that my mystery function is a polynomial, there's always going to be exactly one polynomial that continues our initial chunk. In other words, a polynomial is completely determined by any part of it. [...] Again, just relax if all this seems a little bit too much.

So he didn't give a proof of the theorem in bold text – I think this is very important.

I understand that there always exists a polynomial of degree $n$ that passes through a set of $n+1$ points (i.e. there are finitely many custom points to be passed by, the chunk has to be discrete, like $(1,1),(2,2),(3,3),(4,5)$). But there also exists some polynomial of degree $m$ ($m\ne n$) that passes through the same set of points.

But how do I prove that there exists one and only one polynomial that passes through a set of infinitely many points?


3 Answers 3


If $p$ and $q$ are polynomials agreeing on infinitely many points, then $p-q$ is a polynomial that’s 0 on infinitely many points.

But if a polynomial $f$ of degree $n$ is $0$ on more than $n$ points, then it’s zero everywhere. (If it has zeroes $a_1, \ldots a_n$, then by repeated division it’s of the form $c(x-a_1)\cdots(x-a_n)$; if it’s zero at some other point as well, then we get $c=0$.)

So a polynomial that’s 0 on infinitely many points is 0 everywhere. So, going back to the beginning, if $p$ and $q$ are polynomials agreeing on infinitely many points, then $p-q$ is zero everywhere, i.e. $p=q$.


Polynomials are analytic functions. If two analytic functions agree on a set having a limit point, they must be equal by the Identity Theorem.

  • 2
    $\begingroup$ @PoderRac Identity principle/theorem $\endgroup$
    – egorovik
    Dec 14, 2019 at 13:18
  • 1
    $\begingroup$ @PoderRac Identity theorem: en.wikipedia.org/wiki/Identity_theorem?wprov=sfla1 $\endgroup$
    – user515010
    Dec 14, 2019 at 13:18
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    $\begingroup$ And more generally for polynomials (as in Alberto's answer): if two polynomials agree on an infinite set (even one with no limit point), then they are equal. Also holds for polynomials with coefficients in any field. $\endgroup$
    – GEdgar
    Dec 14, 2019 at 13:28
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    $\begingroup$ Boom! That's using a nuclear bomb to kill a fly! ;-) $\endgroup$ Dec 14, 2019 at 13:55
  • 2
    $\begingroup$ But surely it's harder to prove this for analytic functions than to merely prove it for polynomials. $\endgroup$ Dec 14, 2019 at 21:23

“There is one and only one polynomial” means two things:

1) There is at most one polynomial.

2) There is at least one polynomial.

Only the first affirmation is true.

1) There is at most one polynomial:

Proof by contradiction.

Assume $P$ and $Q$ are two different polynomials passing trough $(x_i,y_i)$, $i\in\mathbb N$. Let $n$ be the maximum of their degrees.

There is a unique polynomial of degree at most $n$ through $(x_i,y_i)$ for $i=1,...,n+1$. But we already know that $P$ and $Q$ do. So $P=Q$.

2) There may be no polynomial.

Example: let us consider the points $(n,e^n)$, $n\in\mathbb N$. Suppose $P$ is a polynomial passing through them.

Notice that $f(x)=e^x$ passes through them as well.

Hence if $$\lim_{x\to+\infty}\frac{e^x}{P(x)}$$ exists, it must be $1$, since it is $1$ when $x\in\mathbb N$. But (as it is easily proved using De L'Hospital), that limit exists and is $\pm\infty$. Contradiction. Hence, there is no such polynomial.

Conclusion: It is false that there exists a polynomial that passes through any infinite set of points, but if you know that the function is a polynomial beforehand, then it is uniquely determined.

  • 3
    $\begingroup$ This is all true and the video agrees. He starts by saying "If I then tell you that my mystery function is a polynomial...". So his conclusion is really, if we are given that there is at least one polynomial, then there is exactly one polynomial. $\endgroup$
    – kotoole
    Dec 19, 2019 at 1:47

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