Talk:Descartes' rule of signs
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pay attention to the last example: it's rong... --Arirossa 23:19, 7 February 2007 (UTC)
- fixed, thanks for spotting. DavidMcKenzie 12:37, 8 February 2007 (UTC)
- Take this polynomial has two roots with the value x=-1, and one with value x=1. So counting roots we have 2 negative roots and one positive root. --Salix alba (talk) 00:46, 8 February 2008 (UTC)
The example is all well and good, but it's not a proof.
I happen to know that the proof is rather long, so perhaps it would be apt to provide an external link to it, if anyone can find it.
Thanks, Glooper 20:24, 3 November 2007 (UTC)
- There seem to be a few neat expositions of various proofs floating around, I added the one from Cut the Knot since they cater for most audiences. Chenxlee (talk) 14:25, 20 February 2008 (UTC)
Moved back to "Descartes' rule of signs"
I have reversed the recent move to "Descartes's rule of signs". The usual possessive form of Descartes is Descartes' - this is the standard followed on other sites such as MathWorld and the Stanford Encyclopedia of Philosophy, and in the titles of books such as Descartes' Error and Descartes' Metaphysical Physics. Gandalf61 (talk) 16:53, 5 February 2008 (UTC)
"This polynomial has two sign changes, meaning the original polynomial has 2 or 0 negative roots and this second polynomial has 2 or 0 positive roots."
- Wikipedia is an encyclopedic reference, not a textbook. To see a proof you can follow the link under External links or consult an algebra textbook. Gandalf61 (talk) 09:34, 22 July 2011 (UTC)
- True. I have updated that section of the article to add the condition that polynomials must not not have a root at 0. Gandalf61 (talk) 07:45, 3 October 2013 (UTC)
How about being patent nonsense? It relies solely on "(X + i)(X - i) = X^2 + b" to work in all forms. It only detects 'purely imaginary' (not complex) numbers in that manner. (X + i)(X + i) instantly defeats it, I believe... with 1 positive sign change and 1 negative sign change. http://mathworld.wolfram.com/DescartesSignRule.html uses a general polynomial [1 1 -1 -1 -1 1 -1], which has roots of
-1.66489 + 0.00000i -0.62797 + 0.90355i -0.62797 - 0.90355i 1.16202 + 0.00000i 0.37941 + 0.53195i 0.37941 - 0.53195i
- You can't apply Dsecartes' Rule to
- because it only applies to polynomials with real coefficients. As for your other example,
- this has 3 sign changes, so f(x) has a maximum of 3 positive real roots. And we have
- which also has 3 sign changes so f(x) has a maximum of 3 negative real roots. The "complex roots" version of Descartes' Rule then says that f(x) has a minimum of 6-3-3 = 0 complex roots, which is entirely consistent with it having 6 complex roots. Gandalf61 (talk) 07:38, 3 October 2013 (UTC)
- It only detects "purely imaginary pole pairs", now you COULD say If random(5) <= 5, a minimum of 0 "purely imaginary" roots exist, but that is a silly assertion... (But just as valid as this claim)
- (x + i)(x - i)(x - 1) is another fun thing, which it fails to detect the imaginary pole pair.
- This IS patent nonsense, it works under a select number of cases and does not work consistently.
- And what is more... you don't even list the conditions under which it works! 22.214.171.124 (talk) 20:19, 3 October 2013 (UTC)
- I don't know why you think that the complex roots version of the rule only applies to purely imaginary roots. This is not correct. Perhaps you were misled by the simple example in that section. I have replaced that example with a more general one where the complex roots are not purely imaginary. The complex roots version of the rule is actually a very simple extension of the real roots version - if you know there are a maximum of p+q real roots then there must be a minimum of n - (p+q) complex roots. If you accept that the real roots version of the rule is correct, then the complex roots version is quite obviously true also. Gandalf61 (talk) 10:28, 4 October 2013 (UTC)
Original source + unchanging signs
Here is a source for Descartes Geometrie: http://www.gutenberg.org/ebooks/26400 The rule is on page 42. Maybe this is interesting: Descartes also makes the statement that you can determine the number of negative roots (called "false" by Descartes) by counting unchanging signs between consecutive coefficients. Descartes does not say what to do with coefficients that are zero, but you have to insert an arbitrary sign for any zero to make the rules work. Another point: Descartes is talking only about polynomials with all real roots. — Preceding unsigned comment added by 126.96.36.199 (talk) 17:56, 11 May 2014 (UTC)