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Math Help - polinomial solution question..

  1. #1
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    polinomial solution question..

    <br />
x^{13}+x^7-x-2006=0\\<br />
    prove that there is a solution on this interval
    <br />
\left ( \sqrt[6]{\frac{\sqrt{101}-7}{26}},0\right )<br />

    i tried like this
    f(x)=x^13 +x^7 -x -2006
    f'(x)=13x^2+7x^6 -1
    f''(x)=156x^11+42x^5
    t=x^6
    the extreme points are
    minimum point is x1= \sqrt[6]{\frac{\sqrt{101}-7}{26}}=0.699
    maximum point is x2= -\sqrt[6]{\frac{\sqrt{101}-7}{26}}=-0.669
    f(x1) and f(x2) are negative
    then they just input 10 and its positive
    so we have one solution between x=0.699 and x=10 (the mid something theorem)
    but its far away from the asked range
    ??
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  2. #2
    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by transgalactic View Post
    <br />
x^{13}+x^7-x-2006=0\\<br />
    prove that there is a solution on this interval
    <br />
\left ( \sqrt[6]{\frac{\sqrt{101}-7}{26}},0\right )<br />

    i tried like this
    f(x)=x^13 +x^7 -x -2006
    f'(x)=13x^2+7x^6 -1
    f''(x)=156x^11+42x^5
    t=x^6
    the extreme points are
    minimum point is x1= \sqrt[6]{\frac{\sqrt{101}-7}{26}}=0.699
    maximum point is x2= -\sqrt[6]{\frac{\sqrt{101}-7}{26}}=-0.669
    f(x1) and f(x2) are negative
    then they just input 10 and its positive
    so we have one solution between x=0.699 and x=10 (the mid something theorem)
    but its far away from the asked range
    ??
    Check f(\sqrt[6]{(\sqrt{101}-7)/26})f(0), and if this value is negative, because of the continuity and the intermediatevalue theorem you must have a root in the interval (\sqrt[6]{(\sqrt{101}-7)/26},0).
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  3. #3
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    f(0) is negative the is no solution between then
    but beteen there root and f(10)
    ??
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  4. #4
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by transgalactic View Post
    f(0) is negative the is no solution between then
    but beteen there root and f(10)
    ??
    There must be sth wrong, cuz the power of the root is even, hence it is positively defined, but you wrote it to the left side of 0.
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  5. #5
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    what calculation you think is wrong
    ??
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  6. #6
    Senior Member bkarpuz's Avatar
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    Exclamation

    Using Mathematica 6.0, I got the following roots:
    x_{1}=-1.7411-0.433334 i
    x_{2}= -1.7411+0.433334 i
    x_{3}=-1.3471-1.18815 i
    x_{4}=-1.3471+1.18815 i
    x_{5}=-0.632548-1.67713 i
    x_{6}=-0.632548+1.67713 i
    x_{7}=0.21387- 1.78518 i
    x_{8}=0.21387+1.78518 i
    x_{9}=1.01904- 1.47315 i
    x_{10}=1.01904+ 1.47315 i
    x_{11}=1.59242- 0.83695 i
    x_{12}=1.59242+ 0.83695 i
    x_{13}=1.79084
    You see there is only one real root in the interval (1,2).
    Check that f(1)f(2)=-12655560<0?

    Note. Since the order of the polynomial is odd, we must have at least one real root because of f(-\infty)f(\infty)<0.
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