Talk:Sturm's theorem: Difference between revisions

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Error in def of Sturm chain?

The definition here is inconsistent with the Sturm chain (Sturm's one) given later in the article: Second point in the definition says:

if $p(\xi )=0$ , then $\operatorname {sign} (p_{1}(\xi ))=-\operatorname {sign} (p'(\xi ));$

However, $p_{1}(x)=p'(x)$ as defined several lines after that, so $\operatorname {sign} (p_{1}(\xi ))=+\operatorname {sign} (p'(\xi ))$ (this combined with the point above would imply that $\xi$ is a double root, which we assumed to not exist).

I just fixed this. Baccala@freesoft.org (talk) 20:06, 15 May 2009 (UTC)[reply]

Question by Krishnavedala

What if the sequence evaluates to '0' (zero)? Zero is neither negative nor positive. So, What will it be considered as and why? I have seen it being considered as positive value .. but, did not find any reason. --Electron Kid (talk) 21:43, 12 March 2008 (UTC)[reply]

For counting sign changes, treat the zero as not there. I.e. +1, 0, +1 has 0 sign changes. -1, 0, -1 has 0 sign changes. +1, 0, -1 has one sign change. 81.32.98.44 (talk) 19:13, 5 October 2008 (UTC)[reply]

@@ Line 4: / Line 4: @@
 * if <math>p(\xi)=0</math>, then <math>\operatorname{sign}(p_1(\xi))= -\operatorname{sign}(p'(\xi));</math>
 However, <math>p_1(x) = p'(x)</math> as defined several lines after that, so <math>\operatorname{sign}(p_1(\xi))= +\operatorname{sign}(p'(\xi))</math> (this combined with the point above would imply that <math>\xi</math> is a double root, which we assumed to not exist).
+: I just fixed this. [[User:Baccala@freesoft.org|Baccala@freesoft.org]] ([[User talk:Baccala@freesoft.org|talk]]) 20:06, 15 May 2009 (UTC)
 ==Question by Krishnavedala==