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I have a quintic equation $$ x^5-a_4 x^4+a_3 x^3-a_2 x^2+a_1 x - a_0=0 $$ with $a_n>0$ real coefficients, and I know that all 5 roots are real and positive (it is a characteristic polynomial).

I'd like to find the lowest root $x_1$ of this polynomial, or at least a lower bound $m$ such that $x_1>m>0$ (not approximate). I am aware that the roots of a quintic equation cannot in general be written in terms of radicals, therefore a solution to this problem can involve non-algebraic or non-trivial functions. But I have no idea how to tackle this general case.

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    $\begingroup$ Perhaps Sturm's theorem can help you? $\endgroup$ – Sergio Parreiras Jul 29 '15 at 23:41
  • $\begingroup$ @SergioParreiras is it possible to apply the Fourier's theorem instead? $\endgroup$ – sintetico Jul 30 '15 at 0:00
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You can start by knowing that $x<1+\max(\operatorname{abs}(a_0),\operatorname{abs}(a_1),\operatorname{abs}(a_2),\operatorname{abs}(a_3),\operatorname{abs}(a_4))$, so letting $x=1/y$, we have $1/x<1+(\max(a_4,a_3,a_2,a_1,1))/a_0.$ Depending the specific values of your coefficients,a linear substitution for x may improve this. You get a lower bound for the roots of p(x) as the upper bound for roots of q(y)=$.y^5.p(1/y)/a_0$.Let M be the "max" referred to above. Then $\operatorname{abs}(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4)$ cannot exceed $M(1+z+z^2+z^3+z^4)$, (where $z=\operatorname{abs}(x)$),which is less than $\operatorname{abs}(x)^5=z^5$ when $z\not < 1+M$, so the polynomial can't be zero for such x.

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  • $\begingroup$ I am not sure to understand. It seems reasonable to me, but why I can assume that the roots are $<1+\max{(a_0,a_1,a_2,a_3,a_4)}$? Also I don't understand the role of $y$. Can you be more explanatory? $\endgroup$ – sintetico Jul 30 '15 at 10:25
  • $\begingroup$ Also, I think you provide an upper bound, but I need a lower bound for the lowest root $\endgroup$ – sintetico Jul 30 '15 at 11:38
  • $\begingroup$ $\LaTeX$ hint: if you want the proper font for a function which is not built in, you can put \operatorname before it, so \operatorname{abs} gives $\operatorname{abs}$. max is built in, so you can use \max $\endgroup$ – Ross Millikan Aug 3 '15 at 3:54
  • $\begingroup$ Let x=1/y and get an upper bound for |y| $\endgroup$ – DanielWainfleet Aug 26 '15 at 21:40

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