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Thread: Descartes Rule and Fundamental Theorem

  1. #1
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    Descartes Rule and Fundamental Theorem

    Hello Hello!

    I've been doing some homework and I really don't understand a few things.

    Can anyone explain to me how the Descartes' Rule and Fundamental theorem can be used to determine the number of complex roots (in a polynomial)?
    Examples would be absolutely stellar!

    Also, does anyone know what setting a "good window" (on a graphing calculator) is?
    and what features of a polynomial should be given?

    Thank you! :]
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Deleted
    Last edited by Mathstud28; Dec 22nd 2008 at 08:37 PM.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Descartes rule of signs: Let $\displaystyle p(x)$ be defined as before. The amount of sign changes in $\displaystyle p(x)$ is the number of real, positive roots of $\displaystyle p(x)$. The number of sign changes in $\displaystyle p(-x)$ is the number of real, negative roots of $\displaystyle p(x)$.

    So let's combine these two rules. As before let the polynomial $\displaystyle p(x)$ be of degree $\displaystyle n$. Let there be $\displaystyle m$ sign changes in $\displaystyle p(x)$ and $\displaystyle e$ sign changes in $\displaystyle p(-x)$. By the FTA we must have $\displaystyle n$ roots exactly, so the amount of roots left is $\displaystyle n-m-e=i$. This number $\displaystyle i$ is the number of imaginary roots of $\displaystyle p(x)$
    a little comment here..

    the $\displaystyle m$ stated there is the maximum number ONLY of positive real roots and not necessarily the number of positive real roots. (and $\displaystyle e$ respectively..)

    why maximum and not the exact? if it is $\displaystyle \geq 2$, the polynomial may possibly have complex roots.. (complex roots always come in pairs.)

    Quote Originally Posted by Mathstud28 View Post
    ... so the amount of roots left is $\displaystyle n-m-e=i$. This number $\displaystyle i$ is the number of imaginary roots of $\displaystyle p(x)$
    this statement is not necessarily true..

    counter-example: $\displaystyle p(x)=x^4-2x^3+2x^2-2x+1$ here, we have $\displaystyle m=4$ and $\displaystyle e=0$.. our $\displaystyle n=4$.. thus $\displaystyle i=0$?

    however, $\displaystyle p(x)=(x^2+1)(x-1)^2$ for which there are 2 positive real roots, no negative real roots and 2 imaginary number roots.. thus, the equation is not necessarily true..
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    a little comment here..
    Thank you...I have deleted my post. I definitely made an error as you pointed out it should be there are AT LEAST $\displaystyle i$ imaginary roots.
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