1. ## polinomial solution question..

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

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 $\displaystyle x1= \sqrt[6]{\frac{\sqrt{101}-7}{26}}=0.699$
maximum point is $\displaystyle x2= -\sqrt[6]{\frac{\sqrt{101}-7}{26}}=-0.669$
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
??

2. Originally Posted by transgalactic
$\displaystyle x^{13}+x^7-x-2006=0\\$
prove that there is a solution on this interval
$\displaystyle \left ( \sqrt[6]{\frac{\sqrt{101}-7}{26}},0\right )$

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 $\displaystyle x1= \sqrt[6]{\frac{\sqrt{101}-7}{26}}=0.699$
maximum point is $\displaystyle x2= -\sqrt[6]{\frac{\sqrt{101}-7}{26}}=-0.669$
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 $\displaystyle f(\sqrt[6]{(\sqrt{101}-7)/26})f(0)$, and if this value is negative, because of the continuity and the intermediatevalue theorem you must have a root in the interval $\displaystyle (\sqrt[6]{(\sqrt{101}-7)/26},0)$.

3. f(0) is negative the is no solution between then
but beteen there root and f(10)
??

4. Originally Posted by transgalactic
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 $\displaystyle 0$.

5. what calculation you think is wrong
??

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

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