A polynomial is completely determined by any part of it

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?

• – egorovik Dec 14 '19 at 13:15

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.

• @PoderRac Identity principle/theorem – egorovik Dec 14 '19 at 13:18
• @PoderRac Identity theorem: en.wikipedia.org/wiki/Identity_theorem?wprov=sfla1 – Caffeine Dec 14 '19 at 13:18
• 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. – GEdgar Dec 14 '19 at 13:28
• Boom! That's using a nuclear bomb to kill a fly! ;-) – Alberto Saracco Dec 14 '19 at 13:55
• But surely it's harder to prove this for analytic functions than to merely prove it for polynomials. – Tanner Swett Dec 14 '19 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:

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.

• 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. – kotoole Dec 19 '19 at 1:47