3
$\begingroup$

I know the formulas for solving cubic equation, but when I try to use them in both Cardano's method and https://en.wikipedia.org/wiki/Cubic_function#Algebraic_solution, I usually encounter something like Cube root of numbers such as $2+11i$ or nested radicals, where square root is inside of cube root. According to hypergeometric's answer to my previous question, I would need to solve another cubic equation to trisects the argument angle to find the cube root and there is similiar problem with the denesting the cube root. So is there some algorithm to analytically solve cubic equation without guessing and rounding (I want to solve it symbolically)? How do the computer algebra systems like Wolfram Alpha do it?

$\endgroup$
  • 7
    $\begingroup$ What is unsatisfactory about a result such as $\sqrt[3]{2+11i}$? $\endgroup$ – JMoravitz Jul 2 '18 at 18:15
  • 2
    $\begingroup$ The general way to solve them is the Cardano formula (which was discovered by Tartaglia). If a simpler method would exist, likely it had been already found. There are significant simplifications for many specific cases. Before the discovery of the Cardano-formula, the solution was considered generally unlikely to exist, and many math contests had the tasks of solving 3rd-grade equations with ad hoc, improvised methods. These may be even today useful, because too many nested radicals may be unfeasible. $\endgroup$ – peterh Jul 2 '18 at 18:19
  • 2
    $\begingroup$ See the Wikipedia page Casus irreducibilis and this google search. For a lot of literature references and historical information, see my 26 September 2005 sci.math post. $\endgroup$ – Dave L. Renfro Jul 2 '18 at 18:44
  • 2
    $\begingroup$ Some of these Stack Exchange questions/answers might be of interest: How to check if a quadratic surd is a perfect cube? AND Simplification of expressions containing radicals AND How does one evaluate $\sqrt[3]{x + iy} + \sqrt[3]{x - iy}$? AND others under "Linked" (see right side) at this places. $\endgroup$ – Dave L. Renfro Jul 3 '18 at 15:39
  • 1
    $\begingroup$ Bear in mind that $r:=\sqrt{a^2+b^2},\,\theta:=\operatorname{atan2}(y,\,x)$ implies $\sqrt[3]{x+yi}=\sqrt[3]{r}\exp \frac{i\theta}{3}\omega^n$ with $\omega:=\exp\frac{2\pi i}{3},\,n\in\{0,\,1,\,2\}$. The point of Cardano's method isn't to avoid cube roots; it's to reduce the problem of solving arbitrary cubics to the problem of solving $z^3-w=0$. $\endgroup$ – J.G. Jul 8 '18 at 8:09
1
$\begingroup$

There are a number of similar questions, for example

Cubic roots and Cardano formula

Cube root of numbers such as $2+11i$

The core of all questions seems to be an uneasy feeling about cubic roots of complex numbers because "they are not really computable".

What is the difference between a purely real expression like $\sqrt[3]{2}$ and a complex expression like $\sqrt[3]{2+11i}$?

$\sqrt[3]{2}$ seems to be very familiar whereas one might be a little skeptical about $\sqrt[3]{2+11i}$ - how to compute the latter?

But is it really "easier"to compute $\sqrt[3]{2}$ than $\sqrt[3]{2+11i}$? I believe the answer is "no". If you want to compute $\sqrt[3]{2}$ you need an algorithm producing a sequence of rational approximations $x_n$ converging to $\sqrt[3]{2}$ plus a concrete estimate of $\lvert \sqrt[3]{2} - x_n \rvert$ so that you know when to stop the computation. There are very simple algorithms to compute cubic roots of real numbers, but there are also algorithms to compute cubic roots of complex numbers. These are somewhat more complicated, but there is no philosophical difference concerning the iterative approach.

Let me close with some remarks concerning the Cardano formula. If you have a cubic equation $x^3 + a_2x^2 +a_1x + a_0 =0$, you can transform it to $y^3 +ay = b$ (by substituting $x = y - a_2/3$). The solutions are given as follows. Set

$$R = \frac{b^2}{4} + \frac{a^3}{27} ,$$

$$w = \sqrt[3]{\frac{b}{2} + \sqrt{R}}.$$

Although at first glance $w$ seems to be uniquely determined, it involves two choices: The square root has two values, the cubic root three values in $\mathbb{C}$. Let us adopt the following conventions: For $x \ge 0$ we let $\sqrt{x}$ denote the nonnegative square root of $x$, for $x <0$ we define $\sqrt{x} = i\sqrt{-x}$. For any $x \in \mathbb{R}$ we let $\sqrt[3]{x}$ denote the real cubic root of $x$ . For a non-real $z \in \mathbb{C}$ we do not get explicit about the choice of $\sqrt[3]{z}$ among the the three complex cubic roots of $z$.

Now define

$$w' = (-a/3)/w \text{ if } w \ne 0,$$ $$w' = \sqrt[3]{b} \text{ if } w = 0 .$$

Note that $w = 0$ if and only $a = 0$ and $b \le 0$. This case (in which $y^3 = b$) is trivial and could also be omitted. Anyway, in both cases we have $3ww' + a = 0$.

It is easily verified that $w'$ is a cubic root of $\frac{b}{2} - \sqrt{R}$. If $w$ is real, then also $w'$ is real. The solutions of our equation are then

$$y_0 = w + w' , y_1 = \zeta_1 w + \zeta_2 w', y_2 = \zeta_2 w + \zeta_1 w'$$

where $\zeta_1 = -\frac{1}{2} + \frac{\sqrt{3}}{2}i, \zeta_2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i = \zeta_1^2 = \overline{\zeta_1}$ are the two complex third unit roots (verification by inserting into the equation). With $\zeta_0 = 1$ we may also write

$$y_k = \zeta_k w + \zeta_k^2 w' = \zeta_k w + \overline{\zeta_k} w' .$$

If $R > 0$, then our above choices produce a real solution $y_0$ and two non-real solutions $y_1, y_2$ which are complex conjugate. All solutions can be expressed by using only square and cubic roots of real numbers.

If $R = 0$ we obtain the three real solutions $y_0 = 2\sqrt[3]{\frac{b}{2}}$ and $y_1 = y_2 = -\sqrt[3]{\frac{b}{2}}$.

The case $R < 0$ (casus irreducibilis) is most interesting. Note that it can only occur when $a < 0$. We have three distinct real solutions, but $w, w'$ are non-real. They are cubic roots of the complex conjugate numbers $\frac{b}{2} \pm i \sqrt{-R}$, but note that these cubic roots cannot be chosen independently because they are subject to the condition $3ww' + a = 0$. If we make any choice for $w$ as a cubic root of $\frac{b}{2} + i \sqrt{-R}$, we see that $\overline{w}$ is a cubic root of $\overline{\frac{b}{2} + i \sqrt{-R}} = \frac{b}{2} - i \sqrt{-R}$. We have $\lvert w \rvert^3 = \lvert \frac{b}{2} + i \sqrt{-R} \rvert = \sqrt{-a^3/27}$, i.e. $\lvert w \rvert = \sqrt{-a/3}$. This implies $3w\overline{w} + a = 3\lvert w \rvert^2 + a = 0$ which means $w' = \overline{w}$. Therefore

$$y_k = \zeta_k w + \overline{\zeta_k} w' = \zeta_k w + \overline{\zeta_k} \overline{w} = \zeta_k w + \overline{\zeta_k w} = 2 Re(\zeta_k w) .$$

Note that $\zeta_k w$, $k = 0,1,2$, are the three complex cubic roots of $\frac{b}{2} + i \sqrt{-R}$.

It is known that it is impossible to express any of these three real solutions in terms of roots of real numbers. See

https://en.wikipedia.org/wiki/Casus_irreducibilis

Those who are historically interested and read German may also look at

Hölder, Otto. "Über den Casus irreducibilis bei der Gleichung dritten Grades." Mathematische Annalen 38.2 (1891): 307-312.

$\endgroup$
  • 1
    $\begingroup$ +1 For the casus irreducibilis wikipedia might be more accessible than the 1891 reference in German. en.wikipedia.org/wiki/Casus_irreducibilis $\endgroup$ – Ethan Bolker Jul 7 '18 at 13:05
  • $\begingroup$ @EthanBolker I agree! $\endgroup$ – Paul Frost Jul 7 '18 at 13:11
  • $\begingroup$ @EthanBolker I changed my answer, thank you again for your comment. $\endgroup$ – Paul Frost Jul 8 '18 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.