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How could I obtain the largest and smallest roots of this equation.

$$Y^3 - (1 - C_{1}) Y^2 + (A - C_{1} + C_{2}) Y - C_{2} = 0 $$

The three roots are real, positive, between $0$ and $1$.

I need to be sure they are the smallest and the largest in order to continue with really tedious work.

Any help wll be welcome. Thanks.

cool
  • 1,089
Claude Leibovici
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1 Answers1

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You can use the formula of Cardano for the roots of an equation.But before using you should make a substitution $y=x+\frac{1-C_{1}}{3}$.You will get an equation $$x^3+px+q=0,$$ which can easy be dealed with in Maple,Mathematica,Matlab(choose one). There you can get the formula for the roots and minimize (maximize) them.

cool
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  • thanks but what I need is to pick the largest and the smallest root before lauching a quite enormous calculation, all of that being symbolic and not numeric. – Claude Leibovici Jul 15 '13 at 10:12
  • @Paul.Is there a trig method which provides the roots in the proper order ? – Claude Leibovici Jul 15 '13 at 10:28
  • There is a formula, but it does not provide the proper order. I meaned that you could get those explicit formulas and deal with then in depence on the values of $C_{1},C_{2}$ – cool Jul 15 '13 at 10:31
  • Go ahead, please – Claude Leibovici Jul 15 '13 at 10:40
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    $$x_{k=1,2,3}=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p} \sqrt{\frac{-3}{p}}\right)-k\frac{2\pi}{3}\right)$$ This is formula. But is does not provide, which of roots is bigger or less. As you said that there are exactly 3 roots, $p<0$ anyway, so this formula is correct – cool Jul 15 '13 at 10:50
  • if you make k=0, you get the largest root. If you make k=2, you get the smallest. Thanks again. – Claude Leibovici Jul 16 '13 at 03:09