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I'm trying to prove that the cubic equation

$_3 ^3 + _2 ^2 + _1 + _0=0$

has three real roots. The coefficients are

$_3=−1−−−$

$_2=−2(++)$

$_1=(+)+(−3)$

$_0=2$

where each of , and are greater than zero. Applying the Rule of Descartes indicates that the polynomial has one real positive root and zero or two real negative roots. I've played around with forming the discriminant for the cubic, and a 3-D contour plot in Mathematica suggests that the discriminant is positive for ,,>0 (indicating three real roots). However, I've hit a wall trying to come up with a formal proof that applies for any (positive) choice of ,,.

One thing of note is that the polynomial coefficients are unchanged on interchange of and , and so the discriminant is a symmetric function of and .

Rich T
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  • What is your explicit computation of the discriminant? It seems like this is an assigned problem, and that you are using the method intended by the problem composer. Normally one would expect the math in the composition's intended solution to be reasonably nice. Also, it is conceivable that (for example) the composer intended that you take the alternative approach of exploring the first and second derivatives of the function. – user2661923 Dec 03 '23 at 18:15
  • No, this isn't an assigned problem. It's part of a proof I'm assembling to determine the fastest-growing mode in a hydro instability.

    In any case: my discriminant is the usual cubic discriminant

    $\Delta = 18 a_{3} a_{2} a_{1} a_{0} - 4 a_{2}^3 a_{0} + a_{2}^2 a_{1}^2 - 4 a_{3} a_{1}^3 - 27 a_{3}^2 a_{0}^2$

    The problem is to confirm that $\Delta$ is non-negative for non-negative $\tau, \sigma, \chi$. I've made some progress by demonstrating that $\Delta$ is non-negative for $\chi=0$, and now I'm working on showing that $d\Delta/d\chi$ (which is independent of $\chi$) is also non-neg.

     – Rich T Dec 03 '23 at 22:41
  • In that case, I suggest that you strongly consider the alternative approach of exploring the first and second derivative, which involve 2nd degree and 1st degree equations, respectively. It is easy to use a computer program to chart (for example) $~f(x), f'(x), ~$ and $~f''(x) ~: ~x \in {-20,-19,\cdots, 0,1,2,\cdots, 20 ~},~$ as a function of the coefficients of the cubic equation. Further, the analysis is fairly easy, since $~f'(x) = 0,~$ is a 2nd degree equation. So, the combination of computer assistance + manual analysis should defeat the problem. ...see next comment – user2661923 Dec 03 '23 at 23:27
  • Depending on the boundaries of the values that are used to construct $~a_3, ~a_2, ~a_1, ~$ and $~a_0~$ you may wish to extend the computer assistance to (for example) $~{ ~-200, ~-190, \cdots, ~0, ~10, ~20, \cdots, ~200 ~}.$ – user2661923 Dec 03 '23 at 23:30

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