6
$\begingroup$

There is a slight part of the following proof in my textbook which I don't quite get.

THEOREM

If $x_0, x_1,...,x_n$ are distinct real numbers, then for arbitrary values $y_0, y_1,...,y_n$, there is a unique polynomial $p_n$ of degree at most $n$ such that

$p_n(x_i) = y_i \quad (0 \leq i \leq n)$

PROOF

Let us prove the uniqueness or unicity first. Suppose there were two such polynomials, $p_n$ and $q_n$. Then the polynomial $p_n - q_n$ would have the property $(p_n - q_n)(x_i) = 0$ for $0 \leq i \leq n$. Since the degree of $p_n - q_n$ can be at most $n$, this polynomial can have at most $n$ zeros if it is not the $0$ polynomial. Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros; it must therefore be $0$. Hence, $p_n = q_n$.

OK, so the only thing I don't quite here is the part:

"Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros"

How does it follow that this must have exactly $n+1$ zeros. I have pondered this for some time, but just can't see why this is obvious. I'm probably overlooking something simple here, but if anyone can help me out, I would greatly appreciate it!

  • 0
    Yes! See below. I understand it now. Thanks!2012-11-19

2 Answers 2

2

You don't know that it has exactly $n+1$ zeros, just that it has at least that many because you have found them: $x_0, x_1, \ldots, x_n$. But an $n^{\text{th}}$ degree polynomial can only have $n$ zeros, unless it is identically zero...

  • 0
    Ah, thanks. I get it now :). Really appreciate your answer!2012-11-19
2

The phrase "Since the $x_i$ are distinct, $p_n - q_n$ has $n+1$ zeros" is intended to mean that there are at least $n+1$ zeros. It would have been a little better to say "at least," but being absolutely precise can be a bit tedious.

  • 0
    Thanks a lot! Yes, I get it now :)2012-11-19