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I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.

The general equation can in all cases be reduced to the depressed form,

$$x^3+px+q=0.$$

By a suitable scaling of the variable, this can be further reduced to

$$x^3-\frac34x-\frac r4=0$$ where $r>0$, or

$$4x^3-3x=r.$$

Then depending on the magnitude of $r$, we write

$$4\cos^3t-3\cos t=\cos3t=r,$$ $$x=\cos\frac{\arccos(r)+2k\pi}3$$

or $$4\cosh^3t-3\cosh t=\cosh3t=r,$$ $$x=\cosh\frac{\text{arcosh}(r)}3.$$

This correctly handles the cases of $1$ and $3$ real roots.

Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?

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When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.

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  • $\begingroup$ That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas. $\endgroup$ – Yves Daoust Sep 19 at 22:30
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Stupid me, I missed the relation

$$\sinh 3t=4\sinh^3t+3\sinh t$$

which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as

$$\sinh\frac{\text{arsinh}(r)}3^*.$$

Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve

$$4y^3\pm 3y=x.$$

The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".

enter image description here

Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.

This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.

Update:

These functions are directly related to the "Chebyshev cube roots".


$$^*\sinh\dfrac{\text{arsinh x}}3=\dfrac{e^{\log(x+\sqrt{x^2+1})/3}-e^{-\log(x+\sqrt{x^2+1})/3}}2=\frac12\left(\sqrt[3]{x+\sqrt{x^2+1}}-\dfrac1{\sqrt[3]{x+\sqrt{x^2+1}}}\right)$$

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