I was trying to see if I could apply the method below to solve a certain subclass of the class of cubic equations with only one real root, and I came up with a block whose cause I couldn't find.

OK, so let $$y=ax^3+bx^2+cx+d,$$ where $a,b,c,d$ are real and $a\ne 0.$ Then $y'=3ax^2+2bx+c.$ Now the plan is to consider those cubics as defined above for which $y'$ has the same sign for all $x.$ This would ensure that an inverse function $$x=f(y)$$ exists, and the plan is to get this function by integration. Let's continue.

Now, for $y'>0,$ without loss of generality, we must have $(2b)^2-4(3a)c\lt 0,$ or $$3ac-b^2>0.$$ Thus, with $k^2=3ac-b^2,$ we have $$y'=3a\left[\left(x+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right],$$ by attempting to complete squares. It then follows, since $y'\ne 0$ for all $x,$ that $$x=\int{\frac{\mathrm d y}{3a\left[\left(y+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right]}}$$ exists for all $y.$ This integral can be easily evaluated to give $$\frac1k \arctan{\left(\frac{3ay}{k}+\frac{b}{k}\right)}+C.$$ To determine the constant $C,$ note that with the original equation, we have $y=d$ when $x=0,$ so that we have $$C=-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)},$$ so that $$x=\frac1k \arctan{\left(\frac{3ay}{k}+\frac{b}{k}\right)}-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)}.$$

It is now an easy matter to find the real root $x$ of the original equation, which is the value of $x$ when $y=0.$ This gives the number $$x=\frac1k \arctan{\left(\frac{b}{k}\right)}-\frac1k \arctan{\left(\frac{3ad}{k}+\frac{b}{k}\right)}.$$

Now consider applying this procedure to the example $$2x^3+3x^2+2x+3=0.$$ Clearly $3^2-3(2)(2)<0,$ so it is of the required type. Also, it is easy to see that it has only one real root, namely $x=-3/2.$ However, applying the method above, we obtain $k=\sqrt 3,$ so that the root as given by the arctangent should be $$x=\frac {1}{\sqrt 3} \arctan{\left(\frac{3}{\sqrt 3}\right)}-\frac{1}{\sqrt 3} \arctan{\left(\frac{21}{\sqrt 3}\right)}.$$ If this is so, then we must have $$\frac{1}{\sqrt 3} \arctan{\left(\frac{3}{\sqrt 3}\right)}-\frac{1}{\sqrt 3} \arctan{\left(\frac{21}{\sqrt 3}\right)}=-\frac 3 2,$$ which gives $$\arctan{(7\sqrt 3)}-\arctan{\sqrt 3}=\frac{3\sqrt 3}{2},$$ which is clearly false.

I have checked again and again, but have failed to see where I went wrong. Please help me spot the false step. Many thanks!

PS. This method works seamlessly well with equations of first order, as can be easily checked; so I kept wondering where the analogy breaks down.

  • $\begingroup$ Interesting approach. Could it be applied to quadratic equations? Just wondering how was "c" calculated in the expression: y'=3a[...] near the top? Thanks. $\endgroup$
    – NoChance
    Jul 7 '19 at 23:19
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    $\begingroup$ Sorry if I'm missing something obvious: how did you obtain that $$x=\int{\frac{\mathrm d y}{3a\left[\left(y+\frac{b}{3a}\right)^2+\frac{k^2}{9a^2}\right]}}?$$ $\endgroup$
    – YiFan
    Jul 7 '19 at 23:33
  • $\begingroup$ I guess this would be from the definition of the inverse function x=f(y). $\endgroup$
    – NoChance
    Jul 7 '19 at 23:39
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    $\begingroup$ Error. It should be $k^2 =3ac-b^2$. $\endgroup$ Jul 8 '19 at 0:19
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    $\begingroup$ @stevengregory Oh, my. Thank you. That's what I'd meant to type -- I actually used the correct form in my work, as you can confirm. That is, this error is not the source of the discrepancy described. I will presently correct it. $\endgroup$
    – Allawonder
    Jul 8 '19 at 0:22

The observation of @YiFan is right.

You actually have $$ \eqalign{ & y' = 3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,2} }}} \right)\quad \Rightarrow \cr & \Rightarrow \quad dx = {{dy} \over {3a\left( {\left( {x + {b \over {3a}}} \right)^{\,2} + {{k^{\,2} } \over {\left( {3a} \right)^{\,2} }}} \right)}} \cr} $$ where you shall read the $x$ at the denominator as $x(y)$.

  • $\begingroup$ My thinking was this: I integrated $1/y'$ from some fixed point $p$ to a variable $y.$ That is, I treated the $x$ in the denominator as a dummy variable. That is equivalent to seeking a primitive and using FTC as displayed above, isn't it? But I appreciate your input. Will look into it. $\endgroup$
    – Allawonder
    Jul 8 '19 at 0:15
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    $\begingroup$ @Allawonder: You can't just wishfully turn $x$ into $y$. This is meaningless. $\endgroup$ Jul 8 '19 at 0:27
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    $\begingroup$ @Allawonder: " .. I treated the $x$ in the denominator as a dummy variable .." : you can't, it depends on $y$ ! $\endgroup$
    – G Cab
    Jul 8 '19 at 0:36
  • $\begingroup$ Yes, I have convinced myself that this was indeed where my problem was. Turns out there's no easy way here too, beyond the linear equations, which is nothing doing. Alright, thank you. $\endgroup$
    – Allawonder
    Jul 8 '19 at 1:32
  • $\begingroup$ @Allawonder: sorry I had to pin down that, because otherwise your "plan of attack" was interesting. $\endgroup$
    – G Cab
    Jul 8 '19 at 9:55

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