Prove that if two polynomial functions have equal values over a closed interval, they are equal.

Question

I want to prove that two polynomial functions that are equal over a specific interval, $(a, b) \in \mathbb R$ (closed interval with more than one point if that condition is necessary) are equal over $\mathbb R$. I had an attempt that intuitively makes sense but I want to make the argument more formal.

These were my attempts:

I have tried expanding the general equation of a polynomial as a product of its roots but got stuck.

I then simplified the question. I examined the case of two linear functions equivalent at a point. I figured if it is equivalent at another point (not necessarily over an interval), it would be trivial to prove that the two functions are the same.

I moved to extend it to quadratic functions. I figured all we need is three points to pinpoint the expression of the polynomial. Given that the functions are equal over an interval, we have more than enough.

I figured that, intuitively, this should continue for any two polynomials of arbitrary degree. Therefore, if two polynomials of $n$-degree are equal over $n + 1$ points on the plane, then the two polynomials are the same. However, something feels very wishy-washy with this proof. How can I make this proof more formal and are there any alternate proofs?

Answer 1

If the closed interval comprises a single point, then the result is false. Otherwise the closed interval contains infinitely many points and this reduces to the fact that a polynomial of degree $n$ can have at most $n$ distinct roots. This is a standard result that is usually proved using polynomial division. I phrase the proof below using the real numbers for definiteness, but it works over any field or any integral domain.

If $f(x)$ is a polynomial over $\Bbb{R}$ of degree $n$ and $f(x_1) = 0$ for some real number $x_0$, then if we divide $f(x)$ by $(x - x_1)$ we get a polynomial $g_1(x)$ of degree $n-1$ and a real number $r$ such that:

$$ f(x) = (x - x_1) g_1(x) + r $$

where, as $f(x_1) = 0$, we must have $r = 0$. I.e., the polynomial $(x - x_1)$ of degree $1$ divides $f(x)$.

If $f(x)$ has another root $x_2 \neq x_1$, then $x_2$ must also be a root of $g(x)$, so $(x - x_2)$ divides $g(x)$. Repeating this process, if $f(x)$ has $n$ distinct roots, $x_1, \ldots, x_n$, we will get:

$$ f(x) = (x - x_1)(x - x_2) \cdots (x - x_n)g_n(x) $$

where $g_n(x)$ has degree zero, i.e., it is a constant. So if $f(x)$ has another root $x_{n+1}$ (differing from each of $x_1, \ldots, x_n$), we have $(x_{n+1} - x_1)(x_{n+1} - x_2) \cdots (x_{n+1} - x_n) \neq 0$ and the only possibility is that $g_n(x) = 0$ for all $x$, implying that $f(x) = 0$ for all $x$.

Answer 2

Your idea is correct: If two polynomials of degree $n$ agrees on $(n+1)$ points, then they are the same. Suppose that we have two polynomials $$p(x)=a_nx^n + \cdots + a_1 x+ a_0$$ and $$q(x)=b_nx^n + \cdots + b_1 x+ b_0.$$ Also assume that we have $(n+1)$ points $x_0$, $\cdots$, $x_n$ such that $p(x_0)=q(x_0)$, $\cdots$, $p(x_n)=q(x_n)$. It gives you a system of linear equations of variable $(b_k-a_k)$ given by

$$ \begin{cases} (a_n-b_n) x_0^n + \cdots + (a_1-b_1) x_0 + (a_0-b_0)=0\\ (a_n-b_n) x_1^n + \cdots + (a_1-b_1) x_1 + (a_0-b_0)=0\\ \qquad\qquad\qquad\qquad\qquad\vdots\\ (a_n-b_n) x_n^n + \cdots + (a_1-b_1) x_n + (a_0-b_0)=0. \end{cases}$$ Could you convince yourself that it has the unique solution, say, $$(a_0-b_0,\cdots,a_n-b_n)=(0,\cdots, 0)?$$

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Or we may work on this problem from a completely different direction: We know that polynomials are analytic and it has a Taylor expansion

$$p(x) = p(x_0) + p'(x_0) (x-x_0) + \cdots \frac{p^{(n)}(x_0)}{n!}(x-x_0)^n$$ if $p$ has degree $n$. If $p$ and $q$ are equal over a neighborhood of $x_0$, then $p^{(i)}(x_0)=q^{(i)}(x_0)$ for all $i$, so they are equal over the whole real line.

Answer 3

Since this is tagged real analysis but not algebra, and OP attempted to build this up from small cases, a proof by induction involving derivatives seems relevant.

A polynomial has degree $n\in \big\{-\infty, 0, 1, 2, 3, \dots\big\}$ where $-\infty$ only occurs for the zero polynomial, $0$ for constant non-zero polynomials, degree 1 is an affine map, and so on. The below assumes the closed interval $[a,b]$ is not a singleton (which isn't really needed for the Base Case but is needed to make the derivative argument). For convenience I work on the open interval $(a,b)$.

Base Case:
we have two polynomial functions, $p,q$ and
$\text{degree}\big(p(x)-q(x)\big)\leq 0\implies p(x)=q(x)$
is immediate since $\big(p(x)-q(x)\big)$ is constant everywhere and equal to zero on the interval.

Inductive Case:
$\text{degree}\big(p(x)-q(x)\big)\leq n$
apply the derivative operator to get
$\big(p'(x)-q'(x)\big)$ which has degree $\leq n-1$ and is identically zero on $(a,b) \implies p'(x)=q'(x)$ by induction hypothesis, i.e. $\text{degree}\big(p'(x)-q'(x)\big)= -\infty\implies \text{degree}\big(p(x)-q(x)\big)\leq 0\implies p(x)=q(x)$ by the Base Case

Answer 4

Let $p\left(x\right)=a_{P}x^{P}+a_{P-1}x^{P-1}+\dots+a_{0}$ and $q\left(x\right)=b_{Q}x^{Q}+b_{Q-1}x^{Q-1}+\dots+b_{0}$, where $P,Q\in\mathbb{Z}_{0}^{+}$ and $a_{P},b_{Q}\neq0$.

Assume without loss of generality $P\ge Q$.

Suppose $p\left(x\right)-q\left(x\right)\overset{0}{=}0$ for all $x\in\left[a,b\right]$ where $a<b$.

Then $p^{\prime}\left(x\right)-q^{\prime}\left(x\right)\overset{1}{=}0$ for all $x\in\left(a,b\right)$.

So, $p^{\prime\prime}\left(x\right)-q^{\prime\prime}\left(x\right)\overset{2}{=}0$ for all $x\in\left(a,b\right)$.

$\vdots$

So, $p^{\left(P-1\right)}\left(x\right)-q^{\left(P-1\right)}\left(x\right)\overset{P-1}{=}0$ for all $x\in\left(a,b\right)$.

So, $p^{\left(P\right)}\left(x\right)-q^{\left(P\right)}\left(x\right)\overset{P}{=}0$ for all $x\in\left(a,b\right)$.

Case 1. Suppose $P>Q$. Then $p^{\left(P\right)}\left(x\right)=P!a_{P}\neq0$ and $q^{\left(P\right)}\left(x\right)=0$---contradicting $\overset{P}{=}$.

Case 2. Suppose $P=Q=0$. Then $p\left(x\right)=a_{0}$ and $q\left(x\right)=b_{0}$. By $\overset{0}{=}$, $0=p\left(x\right)-q\left(x\right)=a_{0}-b_{0}$. So, $a_{0}=b_{0}$. So, $p\left(x\right)$ and $q\left(x\right)$ are the same polynomial.

Case 3. Suppose $P=Q\ge1$.

First, $p^{\left(P\right)}\left(x\right)=P!a_{P}$ and $q^{\left(P\right)}\left(x\right)=P!b_{Q}$. So, by $\overset{P}{=}$, we have $a_{P}=b_{Q}$.

Next, $p^{\left(P-1\right)}\left(x\right)=P!a_{P}x+\left(P-1\right)!a_{P-1}$ and $q^{\left(P-1\right)}\left(x\right)=P!b_{Q}x+\left(P-1\right)!b_{Q-1}$. So, by $\overset{P-1}{=}$ and $a_{P}=b_{Q}$, we have $a_{P-1}=b_{Q-1}$.

$\vdots$

... So, $a_{0}=b_{0}$.

So, $p\left(x\right)$ and $q\left(x\right)$ are the same polynomial.

Answer 5

Saying that two polynomials $f(x)$ and $g(x)$ is the same as saying that $P(x)=f(x)-g(x)$ is the zero polynomial.

Thus your statement becomes

if $P(x)$ is a polynomial function that vanishes over a nontrivial closed interval, then $P(x)$ is the constant zero function.

The “nontrivial” clause is necessary, because $f(x)=x(x+1)$ and $g(x)=x^2$ coincide over the closed interval $[0,0]$, but aren't the same. But this depends on how you define a closed interval to begin with.

Theorem. Let $P(x)$ be a polynomial of degree $n$. Then $P$ has at most $n$ distinct roots.

(Note: the zero polynomial hasn't a degree or, if one prefers, has degree $-\infty$. In the statement above $n$ is meant to be a nonnegative integer.)

Proof. By induction on $n$. If $n=0$, then the polynomial has indeed zero roots, because it's a nonzero constant.

Suppose that the statement holds for polynomials of degree $n$. If $P$ has degree $n+1$ and no roots, then we're done. Suppose it has a root $r$. Then $P(x)=(x-r)Q(x)$ and by the induction hypothesis $Q(x)$ has at most $n$ distinct roots; moreover a root of $P(x)$ is either $r$ or a root of $Q(x)$. If $r$ is also a root of $Q(x)$ we see that $P(x)$ has at most $n$ roots. Otherwise $P(x)$ has at most $n+1$ roots. ◻

Your statement follows from the fact that a nontrivial closed interval has infinitely many points, so more than any possible degree.