be a polynomial of degree $3$ with $a,b,c,d,e\in\mathbb{R}$ and $a>0$. Suppose that the cubic equation $g(x)=0$ has three distinct real roots, i.e. the discriminant $\Delta>0$.

Let $f(x)=\frac{a}{4}x^4+\frac{b}{3}x^3+\frac{c}{2}x^2+dx$. Can we express $M:=\min(f(r_1),f(r_2),f(r_3))$ in terms of the parameters $a,b,c,d$ without directly inserting $r_1$, $r_2$ and $r_3$ (given by solution formula) into $f(x)$ and compare them? In other words, we have explicit solution as shown, for instance in here. Can we give conditions on $a,b,c,d$ and thus determine which $r_i$ is the one that satisfies $M$?

Any reference, suggestion, idea, or comment is welcome. Thank you!

  • $\begingroup$ It's probably possible to track the real and imaginary parts of all of the parts of the formula (noting that the solutions, assumed real, are given by the real parts of the formula). However, it will probably be easier to identify the minimum using the trigonometric solution for three real roots. $\endgroup$ – Greg Martin Feb 1 at 19:47
  • $\begingroup$ I'm a bit confused: if you are defining $r_1$ etc to be the roots then $g(r_1)$ is zero. Do you mean $M=min(r_1,r_2,r_3)$ ? $\endgroup$ – MilesB Feb 1 at 23:34
  • $\begingroup$ Ah, sorry! I made a mistake. I want an anti-derivative of $g$. Now I include $f$. $\endgroup$ – LCH Feb 2 at 2:45
  • $\begingroup$ What you're asking, basically, is which is greater in absolute value, the integral from $r_1$ to $r_2$ or the integral from $r_2$ to $r_3$. I wouldn't be surprised if there's some clever way of comparing, but it's not immediately obvious to me. $\endgroup$ – Steven Stadnicki Feb 2 at 2:50
  • $\begingroup$ @StevenStadnicki Yes you are right. I'm looking for a clever way of comparing. Now I assume $a>0$. $\endgroup$ – LCH Feb 2 at 2:54

So, (lots of this is on the Wikipedia page linked in above comments), if we set $$x=t-\frac{b}{3a} ..[1]$$ then $g(x)$ becomes $G(t)=t^3+pt+q$ where $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3-9abc+27a^2d}{27a^3}$.

Provided $4p^3+27q^2<0$, it has three real roots given by setting $k=0, 1, 2$ in $$t_k=2\sqrt{\frac{-p}{3}}\cos\left(\frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{3}{-p}}\right)-\frac{2\pi k}{3}\right) ...[2]$$

Then the integral of G(t) becomes $$F(t)=\frac{t^4}{4}+\frac{pt^2}{2}+qt$$ which will only differ from f(x) by a constant. Assuming $a>0$, any deduction about which root gives the minimum value for F(t) will translate to the same conclusion for f(x).

So it remains to evaluate F(t) at each of the roots of G(t). As we are only concerned with the minimum value, we need only consider the lowest and highest root which are respectively given by $k=2$ and $k=0$ in [2]. (This can be established by recognising that if $\theta = \frac13\cos^{-1}\left(\frac{3q}{2p}\sqrt{\frac{3}{-p}}\right)$ then $0< \theta < \frac{\pi}{3}$).

By definition, at a root of G(t), $t^3 = -pt-q$ and so at the roots, we can show $$F(t_k)=\frac{t_k}{4}(pt_k+3q).$$ The two values of interest are $F(t_2)$ and $F(t_0)$. Once you have established which of these is lower, you can translate the relevant value of t into a value for x using [1] and evaluate f(x) there.

Given $0< \theta < \frac{\pi}{3}$, it is possible to show that $$F(t_2)>F(t_0)$$ $$\Leftarrow\Rightarrow \cos(\theta-\frac{4\pi}{3})+\cos(\theta)+\cos(3\theta)>0$$ $$\Leftarrow\Rightarrow \theta<\frac{\pi}{6}$$ $$\Leftarrow\Rightarrow q<0.$$

So, putting it all together, if $q<0$ ie if $$2b^3-9abc+27a^2d<0 $$ then the minimum value is given by $$f(t_0-\frac{b}{3a})$$ and if $$2b^3-9abc+27a^2d>0 $$ then the minimum value is given by $$f(t_2-\frac{b}{3a})$$

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  • $\begingroup$ Playing with it numerically, it seems as simple as $F(t_2)<F(t_0)$ if and only if $q>0$. $\endgroup$ – MilesB Feb 2 at 15:58
  • $\begingroup$ Could you give more information, thank you! $\endgroup$ – LCH Feb 2 at 16:26
  • $\begingroup$ @ MilesB I mean your numerical experiment $\endgroup$ – LCH Feb 2 at 17:12
  • $\begingroup$ Will look later. Also spotted an error in expression of theta and in [1], which I'll correct. $\endgroup$ – MilesB Feb 3 at 17:36
  • $\begingroup$ I don't think my numerical conclusion is quite right, so will need to consider further. $\endgroup$ – MilesB Feb 3 at 18:23

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