Descartes' Rule of Signs

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Descartes' Rule of Signs

MarkAU
This post was updated on .
Hi Carl,

In the Precalculus v3 presentation of Descartes' Rule of Signs, you (Carl and Jeff) put aside describing the rather "technical" proof of the Rule, referring interested readers to an external proof.  I can see why you put the proof aside beyond the bounds of your book!  Frankly, reading the external proof made my head swim, but I figured out another way (probably not meeting the full standard of mathematical rigour) to demonstrate to myself why Descartes' Rule works.  Without going into details, the initial seed to my quasi-"proof" was the end behaviour of the constant term of the polynomial, ie, would the horizontal graph line of simply p(x)=c lie above or below the x-axis.  I soon realised that I had nothing to seed my quasi-"proof" with if the polynomial lacked a (non-zero) constant term.

I then checked to see whether a polynomial strictly requires a constant to qualify as a polynomial (I didn't think so), and the solution to example problem 3.1.1(3) on page 235 of Precalculus v3 reaffirmed what I thought.

As a result of my thoughts about Descartes' Rule that followed, I'm wondering if it is worthwhile explicitly pointing out to students that, before they apply Descartes' Rule, they must first ensure that the polynomial that they are applying it to includes a constant.  For instance, if they apply the Rule directly to p(x)=2x^4 - 3x^3 + 4x^2, they will determine that p(x) has two or no positive roots, and no negative roots.  However, if they first factor p(x)=x^2(2x^2 - 3x + 4), they will still obtain the same predictions as before, but will also be reminded that p(x) has an additional two roots for x=0.  While Descartes' Rule will give some clues about how many positive and negative roots there are, it could be easy in the heat of the moment to overlook that there might also be one or more zero roots for p(x).

It's reminiscent of the issue that I raised in my query of 04-NOV-2020 about dividing by an unknown variable.  You're free to do so, but need to bear in mind that a possibly valid solution of x=0 might then not be revealed to you.
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Re: Descartes' Rule of Signs

Carl Stitz
Administrator
That's a fair point!

As always, we appreciate your careful reading of the text!

My favorite proof of Descartes uses the Factor Theorem.... what happens when you multiply a polynomial by a factor (x-a)...

A more advanced proof uses Calculus concepts: Rolle's Theorem and Fermat's Theorem (not his last...)

Carl