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Math Help - Roots of equation

  1. #1
    Super Member malaygoel's Avatar
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    Roots of equation

    How many negative roots has the equation
    x^6 - bx^5 - 2ax^3 - cx + a^2=0, b\geq0, c\geq0?

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  2. #2
    Eater of Worlds
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    Since f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2} has no variation in signs, there are no negative roots.

    It's been awhile since I thought about Descartes rule. Isn't that the way it goes?.
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  3. #3
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by galactus
    Since f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2} has no variation in signs, there are no negative roots.

    It's been awhile since I thought about Descartes rule. Isn't that the way it goes?.
    It would be...
    f(-x)=-x^{6}+bx^{5}+2ax^{3}+cx+a^{2}
    which does have a variation.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by galactus
    Since f(-x)=x^{6}+bx^{5}+2ax^{3}+cx+a^{2} has no variation in signs, there are no negative roots.

    It's been awhile since I thought about Descartes rule. Isn't that the way it goes?.
    I thought for Descartes rule to be applicable the coefficients of all the
    powers up to the maximum must be non-zero.

    RonL
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  5. #5
    Eater of Worlds
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    That would be true, Cap'n. I forgot about that. In Descartes rule, it is assumed that terms with 0 coefficients are deleted and the constant term is not 0. Also, descartes rule says the number of negative roots is equal to the change in signs or less than that by an even integer.

    Let's say b and c are 0, then we'd have (-x)^{6}+2a(-x)^{3}+a^{2}=x^{6}-2ax^{3}+a^{2}

    2 change of signs. Has 2 negative roots or 0 negative roots.

    If b and c were not 0 , then from before. No change of signs. No negative zeros.

    BTW, Quick, (-x)^{6}=x^{6}. May I ask where you got the negative from?. Maybe I am missing something.
    Last edited by galactus; July 18th 2006 at 02:17 AM.
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  6. #6
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by galactus
    BTW, Quick, (-x)^{6}=x^{6}. May I ask where you got the negative from?. Maybe I am missing something.
    I assumed that since you changed all the negatives to positive than you would change the positives to negative, but I forgot that even-numbered exponents are automatically positive
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  7. #7
    Super Member malaygoel's Avatar
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    I am not familiar with Descartes rule. Please guide me.

    What is the answer?What is the use of f(-x)?

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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by malaygoel
    I am not familiar with Descartes rule. Please guide me.

    What is the answer?What is the use of f(-x)?

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    Malay
    Thw Mathworld link is here

    RonL
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  9. #9
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    Thw Mathworld link is here

    RonL
    What is meant by sign changes?How will you calculate sign changes in x^7+x^6-x^4-x^3-x^2+x-1and -x^7+x^6-x^4+x^3-x^2-x-1?
    What is the logic behind Descartes rule?

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  10. #10
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    Thw Mathworld link is here

    RonL
    What is meant by sign changes?How will you calculate sign changes in
    x^7+x^6-x^4-x^3-x^2+x-1and
    -x^7+x^6-x^4+x^3-x^2-x-1?
    What is the logic behind Descartes rule?

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  11. #11
    Grand Panjandrum
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    Quote Originally Posted by malaygoel
    What is meant by sign changes?How will you calculate sign changes in
    x^7+x^6-x^4-x^3-x^2+x-1and
    -x^7+x^6-x^4+x^3-x^2-x-1?
    What is the logic behind Descartes rule?

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    Malay
    First I believe Descartes rule does not work for these polynomials. To work I
    believe that it requires all the powers less than the maximum appear with non-
    zero coefficients. So it applies to:

    x^7+x^6+2x^5-x^4-x^3-x^2+x-1,

    where the signs are +,+,+,-,-,-,+,-, which change three times, and so this
    polynomial has at most three positive roots, and so has either 1 or 3 positive
    roots.

    Now for the negative roots we switch the signs of the odd powers to give:

    -x^7+x^6-2x^5-x^4+x^3-x^2-x-1,

    now the signs are -,+,-,-,+,-,-,-, which change sign 4 times, so there
    are at most four negative roots to the original polynomial. So the original
    polynomial has 4, 2 or 0 negative roots.

    RonL
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  12. #12
    Super Member malaygoel's Avatar
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    Quote Originally Posted by CaptainBlack
    First I believe Descartes rule does not work for these polynomials. To work I
    believe that it requires all the powers less than the maximum appear with non-
    zero coefficients. So it applies to:

    x^7+x^6+2x^5-x^4-x^3-x^2+x-1,

    where the signs are +,+,+,-,-,-,+,-, which change three times, and so this
    polynomial has at most three positive roots, and so has either 1 or 3 positive
    roots.

    Now for the negative roots we switch the signs of the odd powers to give:

    -x^7+x^6-2x^5-x^4+x^3-x^2-x-1,

    now the signs are -,+,-,-,+,-,-,-, which change sign 4 times, so there
    are at most four negative roots to the original polynomial. So the original
    polynomial has 4, 2 or 0 negative roots.

    RonL
    The expressions were given in your link

    Malay
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by Malay
    Quote Originally Posted by CaptainBlack
    First I believe Descartes rule does not work for these polynomials. To work I
    believe that it requires all the powers less than the maximum appear with non-
    zero coefficients.

    RonL
    The expressions were given in your link

    Malay
    You will note the what we call weasel words in my earlier post about
    the applicability of the rule of signs to polynomials with missing powers.

    (weasel words - wording which alows the author to subsequently disavow
    what they wrote).

    It appears that the rule of signs is applicable . I have looked at the
    problem that I thought might exist with such polynomials again and
    it turns out they are not real so we can proceed:

    <br />
x^7+x^6-x^4-x^3-x^2+x-1<br />

    Has signature +,+,-,-,+,-. The signs chenge from + to - or - to + three
    times in this signature so there are at most three positive roots.

    Now to investigate the negative roots we change the signs of all the
    odd power terms in the polynomial and then proceed as before:

    <br />
-x^7+x^6-x^4+x^3-x^2-x-1<br />

    which has signature -,+-,+,-,-. The signs change 4 times in this signature,
    so there are at most four negative roots.

    RonL
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