# 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?

• Dec 14, 2019 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 Dec 14, 2019 at 13:18
• @PoderRac Identity theorem: en.wikipedia.org/wiki/Identity_theorem?wprov=sfla1
– user515010
Dec 14, 2019 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. Dec 14, 2019 at 13:28
• Boom! That's using a nuclear bomb to kill a fly! ;-) Dec 14, 2019 at 13:55
• But surely it's harder to prove this for analytic functions than to merely prove it for polynomials. 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:

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. Dec 19, 2019 at 1:47