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Math Help - newtons method(finding solutions)

  1. #1
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    newtons method(finding solutions)

    we first used newtons method to find the approximate value of 200^(1/7) and got it equal to the awnser in the caclulator which is = 2.131663117

    i need some help in the right direction on how to solve the question after solving the above equation using newtons method!

    Question:
    the fundamental theorem of algebra guarantees that the equation x^7-200 = 0 must have 7 solutions. You have just found one of these. Are there any more real solutions? Use your tools of calculus to sustain your awnser.
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  2. #2
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    Do you know Descartes rule of sign?
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  3. #3
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    k

    ok so in other words using decartes rule states that
    x^(7)-200=0

    has only 1 possible positive real zero? the other 6 are imaginary/complex values?
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  4. #4
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    Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates
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  5. #5
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    y

    thx
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  6. #6
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    I don't think you need DesCarte's rule of signs. Just observe that the derivative of x^7- 200, 7x^6 is always non-negative so it is a "one to one" function on the real numbers.
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  7. #7
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    Quote Originally Posted by dwsmith View Post
    Yes because we obtain 1 (+), 0 (-), so the last solutions are 3 pairs of complex congujates
    Descartes rule of signs tells you there is exactly one positive solution. Substitute u=-x and you get -u^7-1 and then Descartes rule of signs tells you there are no positive solutions for u and so no negative solutions for x to the original equation. Hence there is exactly 1 real solution to x^7-1=0.

    But the question requires that you use HallsofIvy's method not the rule of signs.

    CB
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