Existence of n distinct (real) roots of an orthogonal polynomial

I'm trying to get my head around the proof that an orthogonal polynomial ($P_n$ say) has at least n distinct roots. My understanding of the proof http://en.wikipedia.org/wiki/Orthogonal_polynomials#Existence_of_real_roots so far is that (by contradiction):

• Assume we have $m \le n$ roots. We'll show $m=n$
• Let $\displaystyle S(x) = \prod_{j=1}^m (x-x_j)$
• Gives us that $S(x)$ is an nth degree polynomial
• $S(x)$ changes sign at each of the $x_j$

My problem is this statement:

$S(x)P_n(x)$ is therefore strictly positive, or strictly negative, everywhere except at the $x_j$.

The $x_j$? What $x_j$? The lecture notes I have also say "except at $x_i$" so I'm pretty confused.

If someone can help me out here I'd greatly appreciate it. Thank you!

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The polynomial $P_n$ is of the form $P_n=S Q$ where $S$ is the product of the roots and $Q$ is a factor without roots that therefore cannot change its sign. – Phira May 12 '11 at 23:07
If you multiply by $S$, you have the factor $Q$ that is strictly positive or strictly negative and the factor $S^2$ that is only zero for $x=x_i$. – Phira May 12 '11 at 23:08
When you say "for $x=x_i$", is that $x_i$ for each $0 \leq i \leq m$? – Tim Green May 12 '11 at 23:30
Since your starting assumption is that $P_n$ might have multiple roots, what you do is to take all those roots and ignore multiplicity, form a polynomial from them, and call it $S(x)$; something like forming $(x-1)(x-3)$ as the $S$ if $P$ is $(x-1)^5(x-3)^2$... – J. M. May 13 '11 at 0:37
Did your lecture notes include: - "Let $\displaystyle S(x) = \prod_{i=1}^m (x-x_i)$ - Gives us that $S(x)$...changes sign at each of the $x_i$"? I'm simply asking if your confusion about the $x_j$ is simply due to the particular letter chosen for the index.? – amWhy May 13 '11 at 0:52