Cubic equation

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Image:Polynomialdeg3.png In mathematics, a cubic equation is a polynomial equation in which the highest occurring power of the unknown is the third power. An example is the equation

2x³ − 4x² + 3x − 4 = 0

and the general form may be written as:

α₃x³ + α₂x² + α₁x + α₀ = 0.

Usually, the coefficients α₀, ..., α₃ are real numbers. However, most of the theory is also valid if they belong to a field of characteristic other than two or three. We will always assume that α₃ is non-zero (otherwise it is a quadratic equation).

Solving a cubic equation amounts to finding the roots of a cubic function.

(This article discusses cubic equations in one variable. For a discussion of cubic equations in two variables, see elliptic curve.)

Contents 1 History 2 The nature of the roots 3 Cardano's method 4 Lagrange resolvents 5 Factorization 6 Chebyshev radicals 7 See also 8 External links 9 References

History

Cubic equations were first discovered by Jaina mathematicians in ancient India sometime between 400 BC and 200 CE.

The Persian mathematician Omar Khayyám (1048–1123) constructed solutions of cubic equations by intersecting a conic section with a circle. He showed how this geometric solution could be used to get a numerical answer by consulting trigonometric tables.

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 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 at that time. Dal Ferro kept his achievement secret until just before his death, when he told his student Antonio Fiore about it.

In 1530, Niccolo Tartaglia (1500-1557) received two cubic equation problems from Zuanne da Coi and announced that he could solve them. He was soon challenged by Fiore and this led to a famous contest between the two. Each contestants had to deposit certain amount of money and 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 which he had worked out a general method to solve all of them. Fiore received question in the form x³ + mx² = n, which was proved to be to be too difficult for him to solve. Eventually Tartaglia won the contest.

However, Tartaglia was later persuaded by Gerolamo Cardano (1501-1576) to tell him the secret for solving cubic equations. Tartaglia did so only on condition that Cardano would never reveal it. Few years later Cardano learned about Ferro's prior work and he broke the promise by publishing it in his book Ars Magna (1545) with credit given to Tartaglia. This led to another competition between Tartaglia and Cardano, which the latter did not show up but represented by his student Lodovico Ferrari (1522-1565). Ferrari did better than Tartaglia in the competition, and Tartaglia lost both his prestige and income.

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.

The nature of the roots

Every cubic equation 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,

<math> \Delta = 4\alpha_1^3\alpha_3 - \alpha_1^2\alpha_2^2 + 4\alpha_0\alpha_2^3 - 18\alpha_0\alpha_1\alpha_2\alpha_3 + 27\alpha_0^2\alpha_3^2. </math>

The following cases need to be considered.

• If Δ < 0, then the equation has three distinct real roots.
• If Δ > 0, then the equation has one real root and a pair of complex conjugate roots.
• If Δ = 0, then (at least) two roots coincide. To decide how many distinct roots there are, we define

<math> \Delta_2 = 2\alpha_2^3 - 9\alpha_1\alpha_2\alpha_3 + 27\alpha_0\alpha_3^2, </math>

and consider two further cases.

• If Δ₂ = 0, then all three roots coincide and we have a triple real root.
• Otherwise, the equation has a double real root and a single real root.

The number Δ₂ is the resultant of the cubic and its second derivative.

See also: multiplicity of a root of a polynomial

Cardano's method

The solutions can be found with the following method due to Scipione dal Ferro and Tartaglia, published by Gerolamo Cardano in 1545.

We first divide the given equation by α₃ to arrive at an equation of the form

<math>x^3 + ax^2 + bx +c = 0. \qquad (1) </math>

The substitution x = t - a/3 eliminates the quadratic term; in fact, we get the equation

<math> t^3 + pt + q = 0, \quad\mbox{where } p = b - \frac{a^2}3 \quad\mbox{and}\quad q = c + \frac{2a^3-9ab}{27}. \qquad (2) </math>

This is called the depressed cubic.

Suppose that we can find numbers u and v such that

<math> u^3-v^3 = q \quad\mbox{and}\quad uv = \frac{p}{3}. \qquad (3) </math>

A solution to our equation is then given by

<math>t = v - u, \,</math>

as can be checked by directly substituting this value for t in (2), as a consequence of the third order binomial identity

<math> (v-u)^3+3uv(v-u)+(u^3-v^3)=0 \ . </math>

The system (3) can be solved by solving the second equation for v, which gives

<math> v = \frac{p}{3u}. </math>

Substituting this in the first equation in (3) yields

<math> u^3 - \frac{p^3}{27u^3} = q. </math>

This can be seen as a quadratic equation for u³. If we solve this equation, we find that

<math> u=\sqrt[3]{{q\over 2}\pm \sqrt{{q^{2}\over 4}+{p^{3}\over 27}}}. \qquad (4) </math>

Since t = v − u and t = x + a/3, we find

<math>x=\frac{p}{3u}-u-{a\over 3}.</math>

Note that there are six possibilities in computing u with (4), since there are two solutions to the square root (<math>\pm</math>), and three complex solutions to the cubic root (the principal root and the principal root multiplied by <math>-1/2 \pm i\sqrt{3}/2</math>). However, which sign of the square root is chosen does not affect the final resulting x, although care must be taken in two special cases to avoid divisions by zero. First, if p = 0, then one should choose the sign of the square root that gives a nonzero value for u, i.e. <math>u = \sqrt[3]{q}</math>. Second, if p = q = 0, then we have the triple real root x = −a/3.

Lagrange resolvents

The symmetric group S₃ of order three has the cyclic group of order three as a normal subgroup, which suggests making use of the discrete Fourier transform of the roots, an idea due to Lagrange. Suppose that r₀, r₁ and r₂ are the roots of equation (1), and define <math>\zeta = (-1+i\sqrt{3})/2</math>, so that ζ is a primitive third root of unity. We now set

<math>s_0 = r_0 + r_1 + r_2,\,</math>

<math>s_1 = r_0 + \zeta r_1 + \zeta^2 r_2,\,</math>

<math>s_2 = r_0 + \zeta^2 r_1 + \zeta r_2.\,</math>

The roots may then be recovered from the three s_i by inverting the above linear transformation, giving

<math>r_0 = (s_0 + s_1 + s_2)/3,\,</math>

<math>r_1 = (s_0 + \zeta^2 s_1 + \zeta s_2)/3,\,</math>

<math>r_2 = (s_0 + \zeta s_1 + \zeta^2 s_2)/3.\,</math>

We already know the value s₀ = −a, so we only need to seek values for the other two. However, if we take the cubes, a cyclic permutation leaves the cubes invariant, and a transposition of two roots exchanges s₁³ and s₂³, hence the polynomial

<math>(z-s_1^3)(z-s_2^3) \qquad (5) </math>

is invariant under permutations of the roots, and so has coefficients expressible in terms of (1). Using calculations involving symmetric functions or alternatively field extensions, we can calculate (5) to be

<math>{z}^{2}+ \left( -9\,ba+2\,{a}^{3}+27\,c \right) z+ \left( {a}^{2}-3\,b\right)^{3}.</math>

The roots of this quadratic equation are

<math>\frac92\,ab-{a}^{3}- \frac{27}{2}\,c \pm \frac32\,\sqrt{3\Delta}, </math>

where Δ is the discriminant defined above. Taking cube roots give us s₁ and s₂, from which we can recover the roots r_i of (1).

Factorization

If r is any root of (1), then we may factor using r to obtain

<math>(x-r)(x^2+(a+r)x+b+ar+r^2) = x^3+ax^2+bx+c.</math>

Hence if we know one root we can find the other two by solving a quadratic equation, giving

<math> \frac12 \left(-a-r \pm \sqrt{-3r^2-2ar+a^2-4b}\right) </math>

for the other two roots. If we are finding the roots of a polynomial with real coefficients and one real root, we can find the real root purely in terms of the real (rather than complex) cube root function, or alternatively stated we can find the root by extracting cube roots only of positive quantities. The complex conjugate roots can then be found as above.

Chebyshev radicals

The cube root function is in some respects not a well-behaved function, or one convenient for the purposes of finding the roots of a cubic equation. While cube roots are well-known and traditional, it is possible to use other algebraic functions to determine the roots, and avoid some of the problems of cube roots. The cube root function has a branch singularity at zero, as a result of which the real cube root function does not extend nicely to a complex cube root function. Moreover, when using cube roots to find the roots of a polynomial with three real roots we must take the roots of complex numbers, which introduces complex numbers into a situation which does not, in fact, require them.

We can get around these problems by using Chebyshev cube roots in place of ordinary cube roots. The polynomial <math>C_3 = x^3 - 3x</math> is the third Chebyshev polynomial normalized to obtain a monic polynomial. The Chebyshev cube root is then defined as a (suitably chosen) root (depending on t) of the polynomial equation

<math>x^3 - 3x = t \ .</math>

The polynomial <math> C_3(x) </math> satisfies the third order addition relations

<math>\, 2\,\cos(3x)= C_3(2 \cos x)</math>

and (as <math>\, 2\cos(ix)=2\cosh(x) </math>)

<math> 2\,\cosh(3x)=C_3(2\cosh x) \ .</math>

If t is represented as <math> t=2\cos y </math>, then the polynomial equation <math>x^3 - 3x = t</math> can now be transformed into

<math> t=2\,\cos y=C_3(2\cos (y/3)) \ .</math>

The function <math>C_{1\over3}(t)</math> is then defined as (a branch of) the algebraic function of the third order which transforms <math> 2\cos(x) </math> into <math> 2\cos(x/3) </math>. It is given (inverting the relation <math> t=2\cos(x)</math> to <math> x=\arccos(t/2)</math>) as

<math>C_{1\over3}(t) = 2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right) , </math>

if t lies in the real interval [−2, 2]. If t lies in the interval <math> [2,\infty] </math>, then the Chebyshev root is given as

<math>C_{1\over3}(t) = 2 \,\operatorname{cosh}\left(\operatorname{arccosh}\left({t\over2}\right)/3\right) \ . </math>

The branch is uniquely defined by the value at <math>t=0</math>, which is <math> 2\, \operatorname{cos}\left(\operatorname{arccos}(0)/3\right)=2\, \operatorname{cos}(\pi/6)=\sqrt{3}</math>, corresponding to the positive solution of <math> x^3-3x=x(x^2-3)=0 </math>.

This procedure is precisely analogous to the definition of the cube root in terms of logarithms and exponentials, with arccosh(x/2) resp. arccos(x/2) in the place of ln(x), and 2cosh(x) resp. 2cos(x) in the place of exp(x). The Chebyshev cube root can be constructed as an analytic function on the cut plane <math>\mathbb{C}\setminus [-\infty,-2]</math> and is the unique branch of the algebraic function <math>C_{1\over3}(t)</math> with this property. In the domain <math>D_1 := \{z \in \mathbb{C}\, | \, \Re{z}>2\} </math> it can be defined as

<math> C_{1\over3}(t)= 2\,\operatorname{cosh}\left(\operatorname{arccosh}\left({t\over2}\right)/3\right) </math>

where <math>\operatorname{arccosh}(z/2)=\ln{{z+\sqrt{z^2-4}}\over 2}, </math> using the branch of the logarithm which is real on the positive real line and the branch of the square root which is positive on the real axis. On the domain <math>D_2 :=\mathbb{C}\setminus{\{[-\infty,-2] \cup [2,\infty]\}}</math> it can be defined as

<math> C_{1\over3}(t)= 2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right) </math>,

where <math>\operatorname{arccos}(z/2)={\pi \over 2}+i\ln{{iz+\sqrt{4-z^2}}\over 2} \ .</math> Both D₁ and D₂ are simply-connected domains in <math>\mathbb{C}</math> on which the functions <math> \operatorname{arccos}(z) </math> and <math> \operatorname{arccosh}(z) </math> are well-defined analytic functions (because the square roots <math> \sqrt{\pm (z^2-4)} </math> exist as analytic functions on D₁ resp. D₂ and the argument functions <math> {z+\sqrt{z^2-4}}\over 2 </math> and <math> {iz+\sqrt{4-z^2}}\over 2 </math> of the logarithm do not vanish on each domain). Both (partially overlapping) definitions of the Chebyshev cube root on the domains D₁ and D₂ can be put together to define the Chebyshev cube root unambiguously as an analytic function on the larger domain <math>D= \mathbb{C}\setminus [-\infty,-2]</math>. In fact, if one approaches the critical value <math> t=2 </math> from either the left or the right on the real axis the value of each representative will tend to 2. Because <math> x=2 </math> is a simple root of the polynomial <math> x^3-3x-2 </math> the branch of the Chebyshev root (defined as the algebraic function F(t)=2+G(t) satisfying

<math>\, F(t)^3-3F(t)-2=9G(t)+6G(t)^2+G(t)^3=0 </math>

and <math> F(2)=2 </math> exists locally as an analytic function in a (sufficiently small) neighbourhood U of <math> t=2 </math> (according to the (complex-analytic ) inverse function theorem) and takes real values if <math> t=2\pm \epsilon, \,\epsilon >0 </math>. Then it must coincide (on the intersection <math> U \cap D_1 </math> and <math> U \cap D_2 </math>) with each of the two representatives (in terms of arccos z resp. arccosh z) constructed above. Therefore the Chebyshev cube root is in fact an analytic function on the whole of the domain D.

An alternative construction of the Chebyshev cube root in terms of hypergeometric functions is sketched in the next subsection.

The Chebyshev cube root as a hypergeometric function

The expression

<math> 2 \,\operatorname{cos}\left(\operatorname{arccos}\left({t\over2}\right)/3\right)=2\,\operatorname{cos}\left({\pi\over 6}-\operatorname{arcsin}\left({t\over2}\right)/3\right) </math>

can be transformed (using the difference-to-product trigonometric identity for the cosine) into the representation

<math> \sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin }\left(\operatorname{arcsin}\left({t\over2}\right)/3\right) \ .</math>

For general complex parameter <math> \lambda \ne 0</math> the functions <math>2\,\operatorname{cos}\,(\lambda\, \operatorname{arcsin}(x/2))</math> and <math>2\,\operatorname{sin}\,(\lambda \,\operatorname{arcsin}(x/2)) </math> are two linearly independent solutions of the second-order linear differential equation

<math>\, (4-x^2)y-xy'+\lambda^2 y=0 </math>

which can be obtained by differentiating the functional relations <math>\, f(2\sin x)=2\sin(\lambda x) </math> resp. <math>\, f(2\sin x)=2\cos(\lambda x) </math> twice with respect to x. The differential equation

<math>\, (4-x^2)y-xy'+\lambda^2 y=0 </math>

is equivalent (under the affine substitution <math> x \mapsto (2-4x) </math>) to the hypergeometric differential equation

<math> x(1-x) \,y+{{1-2x}\over 2}\,y'+\lambda^2 y=0 </math>

with parameters <math> c={1\over 2},\, a=\lambda,\, b=-\lambda </math>. According to the general theory of the hypergeometric equation it has (unless c is zero or a negative integer) a uniquely defined solution g which is analytic in x=0 and satisfies <math> \,g(0)=1 </math>. It is given by the hypergeometric series (see hypergeometric function)

<math> \,F(a,b,c;z):=\,_2F_1 (a,b;c;z) = \sum_{n=0}^\infty

\frac{(a)_n(b)_n}{(c)_n} \, \frac {z^n} {n!} \ .</math> Transforming back to the original differential equation one finds a solution <math>g(x)=F(\lambda,-\lambda,{1\over2} ;{{2-x}\over 4})</math> of the differential equation

<math>\, (4-x^2)y-xy'+\lambda^2 y=0 </math>

which is analytic at <math> x=2 </math> (unique up to scalar multiple). The representation

<math>C_Template:1\over 3(t)= \sqrt{3} \,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right) </math>

obtained above shows that the Chebyshev cube root is a solution of the differential equation

<math>\, (4-x^2)y-xy'+\lambda^2 y=0 </math>

for <math> \lambda={1\over 3} </math> which is analytic at <math>x=2 </math>. It must be proportional to the argument-shifted hypergeometric series and thus

<math>C_Template:1\over 3(t)=2F({1\over 3},-{1\over 3},{1\over2} ;{{2-t}\over 4}) = \sum_{n=0}^\infty \frac{2}{1-3n} {3n \choose n}\left(\frac{2-t}{27}\right)^n\ ,</math>

where the last series converges if <math> |t-2|<4 </math>. All three roots <math> r_1, r_2, r_3 </math> of the equation <math> x^3-3x-t=0</math> are linear combinations of the two functions <math>f_1(t)=2\operatorname{sin}\,\left({1\over 3} \operatorname{arcsin}{t\over2}\right) </math> and <math>f_2(t)=2\operatorname{cos}\,\left({1\over 3} \operatorname{arcsin}{t\over2}\right) \ .</math> By construction

<math> r_1=C_Template:1\over 3(t)=\sqrt{3}

\,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\,\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)\, , </math> the other two roots are

<math> r_2=-C_Template:1\over 3(-t)=-\sqrt{3}

\,\operatorname{cos}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right)+ \operatorname{sin}\,\left(\operatorname{arcsin}\left({t\over2}\right)/3\right) </math> and

<math> r_3=-r_1-r_2= -2 \,\operatorname{sin}\left(\operatorname{arcsin}\left({t\over2}\right)/3\right) \ .</math>

One derives the further relations

<math> r_1={\sqrt{3}\over 2}\sqrt{4-r_3^2}-{r_3\over 2} , \qquad r_2=-{\sqrt{3}\over 2} \sqrt{4-r_3^2}- {r_3\over 2} </math>

which can be verified independently by calculating the other two roots ( here <math> r_1, r_2 </math> ) given one root (here <math> r_3 </math> ) by means of the relation

<math> t=x^3-3x=y^3-3y \Longrightarrow (y-x)(y^2+xy+x^2-3)=0, </math>

solving the quadratic equation <math>\, y^2+xy+(x^2-3)=0 </math> for y, given x.

Solving a general cubic equation using Chebyshev cube roots

If we have a cubic equation which is already in depressed form, we may write it as <math>\,x^3 - 3px - q = 0</math>. Substituting <math>x = \sqrt{p} z</math> we obtain <math>z^3 - 3z - p^{-\frac{3}{2}}q = 0</math> or equivalently

<math>z^3 - 3z = p^{-\frac{3}{2}}q \ .</math>

From this we obtain solutions to our original equation in terms of the Chebyshev cube root as

<math>r_1 = \sqrt{p}\,C_{1\over3}(p^{-\frac{3}{2}}q),\,</math>

<math>r_2 = -\sqrt{p}\,C_{1\over3}(-p^{-\frac{3}{2}}q),\,</math>

<math>r_3 = -r_1 - r_2 \ .</math>

If now we start from a general equation

<math>x^3 + ax^2 + bx +c = 0 \qquad (1) </math>

and reduce it to the depressed form under the substitution x = t − a/3, we have <math>\, p = (a^2-3b)/9</math> and <math>\, q = -(2a^3-9ab+27c)/27</math>, leading to

<math>t_{a;b;c} = p^{-\frac{3}{2}}q = -\frac{2a^3-9ab+27c}{(a^2-3b)^{3/2}}.</math>

This gives us the solutions to (1) as

<math>r_1 = \sqrt{p}\,C_{1\over3}(t_{a;b;c})-{a\over 3} ,\,</math>

<math>r_2 = -\sqrt{p}\,C_{1\over3}(-t_{a;b;c})-{a\over 3},\,</math>

<math>r_3 = -r_1 - r_2 - a \ .</math>

The case of a real equation

Suppose the coefficients of (1) are real. If s is the quantity q/r from the section on real roots, then s = t²; hence 0 < s < 4 is equivalent to −2 < t < 2, and in this case we have a polynomial with three distinct real roots, expressed in terms of a real function of a real variable, quite unlike the situation when using cube roots. If s > 4 then either t > 2 and <math>C_{1\over3}(t)</math> is the sole real root, or t < −2 and <math>-C_{1\over3}(-t)</math> is the sole real root. If s < 0 then the reduction to Chebyshev polynomial form has given a t which is a pure imaginary number; in this case <math>iC_{1\over3}(-it)-iC_{1\over3}(it)</math> is the sole real root. We are now evaluating a real root by means of a function of a purely imaginary argument; however we can avoid this by using the function

<math>S_{1\over3}(t) = iC_{1\over3}(-it)-iC_{1\over3}(it) = 2 \operatorname{sinh}\left(\operatorname{arcsinh}\left({t\over2}\right)/3\right),\,</math>

which is a real function of a real variable with no singularities along the real axis. If a polynomial can be reduced to the form <math>x^3 + 3x -t</math> with real t, this is a convenient way to solve for its roots.

See also

• Linear equation
• Quadratic equation
• Quartic equation
• Quintic equation

External links

• Quadratic, cubic and quartic equations on MacTutor archive.
• Template:Planetmath reference
• Quadratic, cubic, quartic and other algebraic equations at EqWorld: The World of Mathematical Equations (PDF).

• Cardano solution calculator as java applet at some local site. Only takes natural coefficients.

References

• W. S. Anglin; & J. Lambek (1995). Mathematics in the Renaissance. In The heritage of Thales, Ch. 24. Springers.
• R.W.D. Nickalls (1993). A new approach to solving the cubic: Cardan's solution revealed, The Mathematical Gazette, 77:354–359.da:Tredjegradsligning

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