Cubic function

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This article is about cubic equations in one variable. For cubic equations in two variables, see cubic plane curve.

Graph of a cubic function with 3 real roots (where the curve crosses the horizontal axis—where y = 0). It has 2 critical points. Here the function is ƒ(x) = (x³ + 3x² − 6x − 8) / 4.

In mathematics, a cubic function is a function of the form

$f(x)=ax^3+bx^2+cx+d,\,$

where a is nonzero; or in other words, a function defined by a polynomial of degree three. The derivative of a cubic function is a quadratic function. The integral of a cubic function is a quartic function.

Setting ƒ(x) = 0 produces a cubic equation of the form:

$ax^3+bx^2+cx+d=0.\,$

Usually, the coefficients a, b,c, d are real numbers. However, most of the theory is also valid if they belong to any field of characteristic other than 2 or 3. To solve a cubic equation is to find the roots (zeros) of a cubic function. There are various ways to solve a cubic equation. The roots of a cubic, like those of a quadratic or quartic (fourth degree) function but no higher degree function (by the Abel–Ruffini theorem), can always be found algebraically (as a formula involving simple functions like the square root and cube root functions). The roots can also be found trigonometrically. Alternatively, one can find a numerical approximation of the roots in the field of the real or complex numbers. This may be obtained by any root-finding algorithm, like Newton's method.

Solving cubic equations is a necessary part of solving the general quartic equation, since solving the latter requires solving its resolvent cubic equation.

[ History

Cubic equations were known to ancient Greek mathematician Diophantus;^[1] even earlier to ancient Babylonians who were able to solve certain cubic equations;^[2] and also to the ancient Egyptians. Doubling the cube is the simplest and oldest studied cubic equation, and one which the ancient Egyptians considered to be impossible.^[3] Hippocrates reduced this problem to that of finding two mean proportionals between one line and another of twice its length, but could not solve this with a compass and straightedge construction,^[4] a task which is now known to be impossible. Hippocrates, Menaechmus and Archimedes are believed to have come close to solving the problem of doubling the cube using intersecting conic sections,^[4] though historians such as Reviel Netz dispute whether the Greeks were thinking about cubic equations or just problems that can lead to cubic equations. Some others like T. L. Heath, who translated all Archimedes' works, disagree, putting forward evidence that Archimedes really solved cubic equations using intersections of two cones, but also discussed the conditions where the roots are 0, 1 or 2.^[5]

Two-dimensional graph of a cubic, the polynomial ƒ(x) = 2x³ − 3x² − 3x + 2.

In the 7th century, the Tang dynasty astronomer mathematician Wang Xiaotong in his mathematical treatise titled Jigu Suanjing systematically established and solved 25 cubic equations of the form $x^3+px^2+qx=N$ , 23 of them with $p,q \ne 0$ , and two of them with $q = 0$ .^[6]

In the 11th century, the Persian poet-mathematician, Omar Khayyám (1048–1131), made significant progress in the theory of cubic equations. In an early paper he wrote regarding cubic equations, he discovered that a cubic equation can have more than one solution and stated that it cannot be solved using compass and straightedge constructions. He also found a geometric solution.^[7]^[8] In his later work, the Treatise on Demonstration of Problems of Algebra, he wrote a complete classification of cubic equations with general geometric solutions found by means of intersecting conic sections.^[9]^[10]

In the 12th century, the Indian mathematician Bhaskara II attempted the solution of cubic equations without general success. However, he gave one example of a cubic equation:^[11]

$x^3+12x=6x^2+35$

In the 12th century, another Persian mathematician, Sharaf al-Dīn al-Tūsī (1135–1213), wrote the Al-Mu'adalat (Treatise on Equations), which dealt with eight types of cubic equations with positive solutions and five types of cubic equations which may not have positive solutions. He used what would later be known as the "Ruffini-Horner method" to numerically approximate the root of a cubic equation. He also developed the concepts of a derivative function and the maxima and minima of curves in order to solve cubic equations which may not have positive solutions.^[12] He understood the importance of the discriminant of the cubic equation to find algebraic solutions to certain types of cubic equations.^[13]

Leonardo de Pisa, also known as Fibonacci (1170–1250), was able to find the positive solution to the cubic equation x³ + 2x² + 10x = 20, using the Babylonian numerals. He gave the result as 1,22,7,42,33,4,40 (equivalent to 1 + 22/60 + 7/60² + 42/60³ + 33/60⁴ + 4/60⁵ + 40/60⁶),^[14] which differs from the correct value by only about three trillionths.

In the early 16th century, the Italian mathematician Scipione del Ferro (1465–1526) found a method for solving a class of cubic equations, namely those of the form x³ + mx = n. In fact, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known to him at that time. Del Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it.

Niccolò Fontana Tartaglia

In 1530, Niccolò Tartaglia (1500–1557) received two problems in cubic equations from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore, which led to a famous contest between the two. Each contestant had to put up a certain amount of money and to propose a number of problems for his rival to solve. Whoever solved more problems within 30 days would get all the money. Tartaglia received questions in the form x³ + mx = n, for which he had worked out a general method. Fiore received questions in the form x³ + mx² = n, which proved to be too difficult for him to solve, and Tartaglia won the contest.

Later, Tartaglia was persuaded by Gerolamo Cardano (1501–1576) to reveal his secret for solving cubic equations. In 1539, Tartaglia did so only on the condition that Cardano would never reveal it and that if he did reveal a book about cubics, that he would give Tartaglia time to publish. Some years later, Cardano learned about Ferro's prior work and published Ferro's method in his book Ars Magna in 1545, meaning Cardano gave Tartaglia 6 years to publish his results (with credit given to Tartaglia for an independent solution). Cardano's promise with Tartaglia stated that he not publish Tartaglia's work, and Cardano felt he was publishing del Ferro's, so as to get around the promise. Nevertheless, this led to a challenge to Cardano by Tartaglia, which Cardano denied. The challenge was eventually accepted by Cardano's student Lodovico Ferrari (1522–1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and income.^[15]

Cardano noticed that Tartaglia's method sometimes required him to extract the square root of a negative number. He even included a calculation with these complex numbers in Ars Magna, but he did not really understand it. Rafael Bombelli studied this issue in detail and is therefore often considered as the discoverer of complex numbers.

François Viète (1540–1603) independently derived the trigonometric solution for the cubic with three real roots, and René Descartes (1596–1650) extended the work of Viète.^[16]

[ Derivative

Graph showing the relationship between the roots, turning points, stationary points, inflection point and concavity of a cubic polynomial x³ - 3x² - 144x + 432 and its first and second derivatives.

Through the quadratic formula the roots of the derivative f ′(x) = 3ax² + 2bx + c are given by

$x=\frac{-b \pm \sqrt {b^2-3ac}}{3a}$

and provide the critical points where the slope of the cubic function is zero. If b² − 3ac > 0, then the cubic function has a local maximum and a local minimum. If b² − 3ac = 0, then the cubic's inflection point is the only critical point. If b² − 3ac < 0, then there are no critical points. In the cases where b² − 3ac ≤ 0, the cubic function is strictly monotonic.

[ Roots of a cubic function

The general cubic equation has the form

$ax^3+bx^2+cx+d=0 \qquad(1)$

with $a\neq 0\,.$

This section describes how the roots of such an equation may be computed. The coefficients a, b, c, d are generally assumed to be real numbers, but most of the results apply when they belong to any field of characteristic not 2 or 3.

[ The nature of the roots

Every cubic equation (1) with real coefficients has at least one solution x among the real numbers; this is a consequence of the intermediate value theorem. We can distinguish several possible cases using the discriminant,

$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2. \,$

The following cases need to be considered: ^[17]

If Δ > 0, then the equation has three distinct real roots.
If Δ = 0, then the equation has a multiple root and all its roots are real.
If Δ < 0, then the equation has one real root and two nonreal complex conjugate roots.

[ General formula for roots

For the general cubic equation

$a x^3 + b x^2 + c x + d = 0$

the general formula for the roots, in terms of the coefficients, is as follows:[18]^[19]

$x_i = - \frac{1}{3 a}(b\ +\ u_i C\ +\ \frac{\Delta_0}{u_iC})\ , \qquad i \in \{1,2, 3\}$

where

$u_1 = 1\ ,\qquad u_2 = {-1 + i\sqrt{3} \over 2}\ ,\qquad u_3 = {-1 - i\sqrt{3} \over 2}$

are the three cubic roots of unity, and where

$C = \sqrt[3]{\frac{\Delta_1 + \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}$

with

$\Delta_0 = b^2-3 a c$

$\Delta_1 = 2 b^3-9 a b c+27 a^2 d$

and

$\Delta_1^2 - 4 \Delta_0^3 = -27\,a^2\,\Delta$ , where $\Delta$ is the discriminant discussed above.

In these formula, $\sqrt {~~}$ and $\sqrt[3]{~~}$ denote any choice for the square or cubic roots. Changing of choice for the square root amounts of exchanging $x_2$ and $x_3$ . Changing of choice for the cubic root amounts to circularly permute the roots. Thus the freeness of choosing a determination of the square or cubic roots corresponds exactly to the freeness for numbering the roots of the equation.

Four centuries ago, Gerolamo Cardano proposed a similar formula (see below), which appears yet in many textbooks:

$x_i = - \frac{1}{3a}(b\ +\ u_i C\ +\ \bar u_i \bar C)$

where

$\bar C = \sqrt[3]{\frac{\Delta_1 - \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}}$

and $\bar u_i$ is the complex conjugate of $u_i$ (note that $C\bar C=\Delta_0$ ).

However, this formula is applicable without further explanation only when a, b, c, d are real numbers and the operand of the square root $\Delta_1^2 - 4 \Delta_0^3 \ge 0$ is non-negative. When this operand is real and non-negative, the square root refers to the principal (positive) square root and the cube roots in the formula are to be interpreted as the real ones. Otherwise, there is no real square root and one can arbitrarily choose one of the imaginary square roots (the same one everywhere in the solution). For extracting the complex cube roots of the resulting complex expression, we have also to choose among three cube roots in each part of each solution, giving nine possible combinations of one of three cube roots for the first part of the expression and one of three for the second. The correct combination is such that the two cube roots chosen for the two terms in a given solution expression are complex conjugates of each other (whereby the two imaginary terms in each solution cancel out).

The next sections describe how these formulas may be obtained.

[ Special cases

If $\Delta \neq 0$ and $\Delta_0 = 0$ , the sign of $\sqrt{\Delta_1^2 - 4 \Delta_0^3}=\sqrt{\Delta_1^2}$ has to be chosen to have $C \neq 0$ , that is one should define $\sqrt{\Delta_1^2} =\Delta_1,$ whichever is the sign of $\Delta_1.$

If $\Delta = 0$ and $\ \Delta_0 = 0$ , the three roots are equal:

$x_1=x_2=x_3=-\frac{b}{3a}.$

If $\Delta=0$ and $\Delta_0 \neq 0$ , the above expression for the roots is correct but misleading, hiding the fact that no radical is needed to represent the roots. In fact, in this case, there is a double root,

$x_1=x_2=\frac{9ad-bc}{2\Delta_0},$

and a simple root

$x_3=\frac{4abc-9a^2d-b^3}{a\Delta_0}.$

[ Reduction to a depressed cubic

Dividing Equation (1) by $a$ and substituting $x$ by $t-\frac{b}{3a}$ (the Tschirnhaus transformation) we get the equation

$t^3+pt+q=0 \qquad(2)$

where

$\begin{align} p=&\frac{3ac-b^2}{3a^2}\\ q=&\frac{2b^3-9abc+27a^2d}{27a^3}. \end{align}$

The left hand side of equation (2) is a monic trinomial called a depressed cubic.

Any formula for the roots of a depressed cubic may be transformed into a formula for the roots of Equation (1) by substituting the above values for $p$ and $q$ and using the relation $x=t-\frac{b}{3a}$ .

Therefore, only Equation (2) is considered in the following.

[ Cardano's method

The solutions can be found with the following method due to Scipione del Ferro and Tartaglia, published by Gerolamo Cardano in 1545.^[20]

This method applies to the depressed cubic

$t^3 + pt + q = 0\,. \qquad (2)$

We introduce two variables u and v linked by the condition

$u+v=t\,$

and substitute this in the depressed cubic (2), giving

$u^3+v^3+(3uv+p)(u+v)+q=0 \qquad (3)\,$ .

At this point Cardano imposed a second condition for the variables u and v:

$3uv+p=0\,$ .

As the first parenthesis vanishes in (3), we get $u^3+v^3=-q$ and $u^3v^3=-p^3/27$ . Thus $u^3$ and $v^3$ are the two roots of the equation

$z^2 + qz - {p^3\over 27} = 0\,.$

At this point, Cardano, who did not know complex numbers, supposed that the roots of this equation were real, that is that $\frac{q^2}{4}+\frac{p^3}{27} >0\,.$

Solving this equation and using the fact that $u$ and $v$ may be exchanged, we find

$u^{3}=-{q\over 2} + \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}$ and $v^{3}=-{q\over 2} - \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}$ .

As these expressions are real, their cube roots are well defined and, like Cardano, we get

$t_1=u+v=\sqrt[3]{-{q\over 2}+ \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} +\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}$

The two complex roots are obtained by considering the complex cubic roots; the fact $uv$ is real implies that they are obtained by multiplying one of the above cubic roots by $\,\tfrac{-1}{2} + i\tfrac{\sqrt{3}}{2}\,$ and the other by $\,\tfrac{-1}{2} - i\tfrac{\sqrt{3}}{2}\,$ .

If $\frac{q^2}{4}+\frac{p^3}{27}\,$ is not necessarily positive, we have to choose a cube root of $u^3$ . As there is no direct way to choose the corresponding cube root of $v^3$ , one has to use the relation $v=-\frac{p}{3u}$ , which gives

$u=\sqrt[3]{-{q\over 2}- \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}} \qquad (4)$

and

$t=u-\frac{p}{3u}\,.$

Note that the sign of the square root does not affect the resulting $t$ , because changing it amounts to exchanging $u$ and $v$ . We have chosen the minus sign to have $u\ne 0$ when $p = 0$ and $q\ne 0$ , in order to avoid a division by zero. With this choice, the above expression for $t$ always works, except when $p = q=0$ , where the second term becomes 0/0. In this case there is a triple root $t=0$ .

Note also that in several cases the solutions are expressed with fewer square or cube roots

If $p=q=0$ then we have the triple real root

$t=0.\,$

If $p=0$ and $q\ne 0$ then

$u=-\sqrt[3]{q} \text{ and } v = 0$

and the three roots are the three cube roots of $-q$ .

If $p\ne 0$ and $q=0$ then

$u=\sqrt{{p\over 3}} \qquad \text{and} \qquad v=-\sqrt{{p\over 3}},$

in which case the three roots are

$t=u+v=0 , \qquad t=\omega_1u-{p\over 3\omega_1u}=\sqrt{-p} , \qquad t={u\over \omega_1}-{\omega_1p\over 3u}=-\sqrt{-p} ,$

where

$\omega_1=e^{i\frac{2\pi}{3}}=-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i.$

Finally if $4p^3+27q^2=0 \text{ and } p\ne 0$ , there is a double root and a simple root which may be expressed rationally in term of $p \text{ and } q$ , but this expression may not be immediately deduced from the general expression of the roots:

$t_1=t_2= -\frac{3q}{2p}\quad \text{and} \quad t_3=\frac{3q}{p}\,.$

To pass from these roots of $t$ in Equation (2) to the general formulas for roots of $x$ in Equation (1), subtract $\frac{b}{3a}$ and replace $p$ and $q$ by their expressions in terms of $a,b,c,d$ .

[ Vieta's substitution

Starting from the depressed cubic

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

we make the following substitution, known as Vieta's substitution:

$x = w - \frac{p}{3w}$

This results in the equation

$w^3 + q - \frac{p^3}{27w^3} = 0.$

Multiplying by w³, it becomes a sextic equation in w, which is in fact a quadratic equation in w³:

$w^6 + qw^3 - \frac{p^3}{27} = 0$

The quadratic formula allows to solve it in w³. If w₁, w₂ and w₃ are the three cubic roots of one of the solutions in w³, then the roots of the original depressed cubic are

$x_1 = w_1 - \frac{p}{3w_1}, \quad x_2 = w_2 - \frac{p}{3w_2}\quad\text{and} \quad x_3 = w_3 - \frac{p}{3w_3}.$

[ Lagrange's method

In his paper Réflexions sur la résolution algébrique des équations ("Thoughts on the algebraic solving of equations"), Joseph Louis Lagrange introduced a new method to solve equations of low degree.

This method works well for cubic and quartic equations, but Lagrange did not succeed in applying it to a quintic equation, because it requires solving a resolvent polynomial of degree at least six.^[21]^[22]^[23] This is explained by the Abel–Ruffini theorem, which proves that such polynomials cannot be solved by radicals. Nevertheless the modern methods for solving solvable quintic equations are mainly based on Lagrange's method.^[23]

In the case of cubic equations, Lagrange's method gives the same solution as Cardano's, where the latter may seem almost magical to the modern reader. But Cardano explains in his book Ars Magna how he arrived at the idea of considering the unknown of the cubic equation as a sum of two other quantities, by drawing attention to a geometrical problem that involves two cubes of different size. Lagrange's method may also be applied directly to the general cubic equation (1) without using the reduction to the depressed cubic equation (2). Nevertheless the computation is much easier with this reduced equation.

Suppose that x₀, x₁ and x₂ are the roots of equation (1) or (2), and define $\zeta = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i$ , so that ζ is a primitive third root of unity which satisfies the relation $\zeta^2+\zeta+1=0$ . We now set

$s_0 = x_0 + x_1 + x_2,\,$

$s_1 = x_0 + \zeta x_1 + \zeta^2 x_2,\,$

$s_2 = x_0 + \zeta^2 x_1 + \zeta x_2.\,$

This is the discrete Fourier transform of the roots: observe that while the coefficients of the polynomial are symmetric in the roots, in this formula an order has been chosen on the roots, so these are not symmetric in the roots. The roots may then be recovered from the three s_i by inverting the above linear transformation via the inverse discrete Fourier transform, giving

$x_0 = \tfrac13(s_0 + s_1 + s_2),\,$

$x_1 = \tfrac13(s_0 + \zeta^2 s_1 + \zeta s_2),\,$

$x_2 = \tfrac13(s_0 + \zeta s_1 + \zeta^2 s_2).\,$

The polynomial $s_0$ is an elementary symmetric polynomial and is thus equal to $-b/a$ in case of Equation (1) and to zero in case of Equation (2), so we only need to seek values for the other two.

The polynomials $s_1$ and $s_2$ are not symmetric functions of the roots: $s_0$ is invariant, while the two non-trivial cyclic permutations of the roots send $s_1$ to $\zeta s_1$ and $s_2$ to $\zeta^2 s_2$ , or $s_1$ to $\zeta^2 s_1$ and $s_2$ to $\zeta s_2$ (depending on which permutation), while transposing $x_1$ and $x_2$ switches $s_1$ and $s_2$ ; other transpositions switch these roots and multiply them by a power of $\zeta.$

Thus, $s_1^3$ , $s_2^3$ and $s_1 s_2$ are left invariant by the cyclic permutations of the roots, which multiply them by $\zeta^3=1$ . Also $s_1 s_2$ and $s_1^3+s_2^3$ are left invariant by the transposition of $x_1$ and $x_2$ which exchanges $s_1$ and $s_2$ . As the permutation group $S_3$ of the roots is generated by these permutations, it follows that $s_1^3+s_2^3$ and $s_1 s_2$ are symmetric functions of the roots and may thus be written as polynomials in the elementary symmetric polynomials and thus as rational functions of the coefficients of the equation. Let $s_1^3+s_2^3=A$ and $s_1 s_2=B$ in these expressions, which will be explicitly computed below.