does a section of a polynomial have a unique continuation? Given a continuous section of some polynomial on some finite interval, does this section uniquely describe some polynomial or could several different polynomials all be described by this one section?
 A: Yes. Take the polynomial $p(x) : [a,b] \to \mathbb{R}$ where $a < b$:
$$p(x) = c_nx^n + x_{n-1}x^{n-1} + \cdots + c_1x + c_0$$
Then take any $n+1$ unique points inside the interval $[a,b]$: $x_0,x_1,\dots,x_n$, and calculate $p(x_0),p(x_1),\dots,p(x_n)$.
With these, you can uniquely write up $p(x)$ using the Lagrange-interpolation polynomial:
$$p^*(x) = \sum_{j=0}^n p(x_j) \prod_{0 \le m \le n \\ \ \ \ m \ne j} \frac{x-x_m}{x_j-x_m}$$
Then $p^*(x)$ obtained this way is a degree $n$ polynomial that goes through the points $(x_0,p(x_0)),(x_1,p(x_1)),\dots,(x_n,p(x_n))$.

The $p^*(x)$ obtained is unique, and we can easily prove this. Assume there exists another polynomial $q(x)$ that has degree $n$ and also goes through the points $(x_0,p(x_0)),(x_1,p(x_1)),\dots,(x_n,p(x_n))$, but $q(x) \ne p^*(x)$.
Then the polynomial $p^*(x)-q(x)$ is degree at most $n-1$. (The leading coefficient is cancelled out.) However, $p^*(x)-q(x)$ has $n+1$ unique roots, namely $x_0,x_1,\dots,x_n$, even though it is degree at most $n-1$, which is a contradiction. (And $p^*-q \ne 0$, since $p^* \ne q$ by assumption.)
Therefore, $q(x)$ isn't different from $p^*(x)$, and such, $p^*(x) = p(x)$ for all $x \in \mathbb{R}$.
A: A polynomial of grade n  is unique if you know all the n+1 derivatives at one point, so if you know the equation of your polynomial it is unique. If you just have a good drawing, you would not know enough.
A: First, suppose that $P:[a,b] \rightarrow \mathbb{R}$ is a polynomial defined on an interval, and $f,g:\mathbb{R} \rightarrow \mathbb{R}$ are both polynomials that extend $P$ to all of $\mathbb{R}$.
Consider the function $h = f-g$. Since $h$ is the difference of two polynomials, then $h$ is a polynomial as well! Further, this polynomial $h$ is uniformly zero on the interval $[a,b]$. Thus $h$ has infinitely many zeros. It is well known that a degree $n \geq 1$ polynomial has at most $n$ zeros, so $h$ must be degree zero. Thus $h$ is the constant polynomial $0$, making $f=g$.

We have shown that the extension is unique, but we can say more. We can even learn what is is if we don’t know $P$!
Let $P(x)$ be some polynomial Suppose we have a “black box” that is able to evaluate the polynomial on a specific range of inputs. That is, for some interval $[a,b]$, we we give the black box some value $x \in [a,b]$, it will spit out the value $P(x)$. But the black box is limited to the interval $[a,b]$. If you try to give it some input outside of $[a,b]$, it won’t tell you anything.
The question I’m answer is, can we “learn” the polynomial $P(x)$ if we have access to this black box?
As it turns out, yes we can! The argument hinges on the following fact: just as two points uniquely specify a line, $n+1$ points uniquely specifies a degree $n$ polynomial that goes through all those points. This polynomial can be explicitly constructed as Daniel P Explains in his answer. There’s an extra step though that Daniel doesn’t explain — how do we actually use this to learn $P(x)$. After all, we don’t know the degree of $P$!
We may learn the polynomial as follows. Pick a first point $x_0 \in [a,b]$. Use the black box to get the value $y_0 := P(x_0)$.
Now set $f$ to be the unique degree zero polynomial that goes goes through the point $(x_0, y_0)$. Now pick a new point $x_1 \in [a,b]$ distinct from $x_0$. Evaluate $f(x_1)$ and $y_1 := P(x_1)$ by the black box. If they agree, then $P = f$. If they disagree, set $f$ to be the unique degree 1-polynomial that passes through both of the points $(x_0, y_0)$ and  $(x_1, y_1)$.
Keep repeating this process! Since every polynomial has finite degree, this algorithm will eventually terminate with $f=P$. Hence we learn the value of $P$.

Trula’s answer is also interesting. As they point out, if the black box allows us to query not just the $P$ itself, but also it’s derivatives, we can make due with a single point!
A: If you know that a function is a polynomial of degree n then the values in N+1 points determine the polynomial completely. If you know that a function is a polynomial then the values on an infinite set of points determine the polynomial completely. The same for the values on any interval. (But if values on an infinite set are given, there might be no polynomial at all).
