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cubic polynomials for time passengers...

.. Welcome in the early 16th century ..

The ancient mathematicians (such as Euclid: around 330 BC - 275 BC) encountered the problem of solving a cubic polynomial a very long time ago, while working on various geometric challenges (such as doubling the volume of a cube or trisecting an angle). These problems cannot be solved using instruments like compasses or rulers. There is evidence that attempts to find solutions for this particular problem date back to ancient Babylonia (around 2000 BC). Arabian mathematicians influenced the famous Italian mathematician Leonardo of Pisa, also known as Fibonacci (1170 - 1250), who discovered a solution for a specific cubic equation (where c1 = 2, see below). He likely used an Arabic method of successive approximations.

The general case(except from the pardonable coefficient 1 with x3, in today's language we talk about a normed polynomial):
P(x) = x3 + c1x2 + c2x + c3 = 0

Girolamo Cardano (1501 - 1576);

Girolamo Cardano

The solution of a cubic polynomial is linked with the Renaissance mathematician Girolamo Cardano: 24.9.1501(Pavia,Italy) - 21.9.1576(Rome,Italy). But now lets look at our polynomial P(x) above (assuming only REAL coefficients). By the substitution z = x - c1/3 we can transform P(x) into the reduced form Q(z) = z3 + pz + q = 0. We assume that p is unequal to 0. If z is a solution of Q(z), we can represent z as: z = u + v. When we do that, our equation Q(z) turns into: (u + v)3 + p(u + v) + q =0, eg. u3 + v3 + (3uv + p)(u + v) + q = 0. As u + v has the dimension 1, we can choose 1 parameter in order to make the middle term vanish by 3uv = - p. (I) 3uv = - p and (II) u3 + v3 = - q. This means u3 and v3 are roots of the quadratic equation s2 + qs - (p/3)3 = 0. The solutions of these equations are:

u3 = (-q/2) + [(q/2)2 + (p/3)3]1/2 and v3 = (-q/2) - [(q/2)2 + (p/3)3]1/2

This way we obtain 3 solutions each for u and v, leaving us with 9 solutions in total for our z = u + v at first glance. But we remember that 3uv = - p, eg once we choose one solution for u, v is determined, which reduces the total number of solutions for u + v from 9 to 3.

So we arrived at what is called "Cardano formula": u + v = z1 = ((-q/2) + D1/2)1/3 + ((-q/2) - D1/2)1/3 and where D = q2/4 + p3/27 , D is called the discriminant of the Cardano formula.

In todays terms, we can say that these 3 must have the form z1, z1r and z1r2 where r = e2πi/3. Starting with solution u1 and the relation 3u1v1 = - p, our complete solution will look like this:
z1 = u1 + v1 , z2 = r u1 + r2v1 , z3 = r2u1 + rv1 (+), of course, we would need to verify that these are indeed the solutions of Q(z), but let's take a coffee break and skip this analytical exercise for now. This triple (+) we can name as the "extented Cardanos formulas" (at least if there is any free speech left)

It should be noted that solving a cubic polynomial was bordering on a miracle, as mathematical symbolism did not exist in Cardano's time, and the concept of complex numbers was unknown. Therefore, the formula we referenced above was not actually in Cardano's handwriting. Cardano relied on geometric shapes and proofs to assist himself. Although Cardano's contribution to cubic equations was very significant, he was not the first to discover a partial solution. Nicolas Tartaglia (around 1500 or 1499 - 1557) appears to have found a general solution even before Cardano; however, Cardano published Tartaglia's idea first in his book Ars Magna (Nuremberg, 1545), and did not adhere to their original agreement. Tartaglia never forgave Cardano for what he perceived as treason. Cardano argued that since Scipione del Ferro (Italian mathematician, 1465 - 1526) had found a solution before Tartaglia, he (Cardano) felt no obligation to keep Tartaglia's secret any longer and regarded the solution as a 'factum' - a publicly available achievement. It is important to mention that Cardano's Ars Magna presented a more generalized view of the cubic equation compared to Tartaglia's results. Cardano even mentioned in Ars Magna that the solution to the cubic equation he published was, in fact, taken from Scipione del Ferro.

This claim that Scipione del Ferro discovered a solution to the reduced form Q(z) earlier than Tartaglia is not entirely tenable; some argue that del Ferro only managed to solve a specific cubic equation. Del Ferro kept his solution secret and passed his mathematical notes shortly before his death to his son-in-law and mathematician, Hannibal della Nave (he married del Ferro's daughter, Filippa). Hannibal later took over del Ferro's lecturing duties at the University of Bologna. Unfortunately, del Ferro's notes were lost. However, two years before Cardano published Ars Magna, he (Cardano) traveled to Bologna along with Lodovico Ferrari (see my notes about the quartic equation) to visit Hannibal della Nave and view the late Scipione del Ferro's notes. If he really found there a rudimentary or complete form of what is now known as the 'Cardano formula' for the cubic equation, we can only speculate.

Some evidence suggests that Tartaglia and del Ferro arrived at their solutions independently.

Niccolo Tartaglia (1499 - 1557);

Niccolo Tartaglia



Scipione del Ferro (1465 - 1526);

Scipione del Ferro



Now lets look at one example, an example our Renaissance mathematicians could have solved for us:

x3 - 3x2 + 7x - 21 = Q(x) = 0 (but let's transform it first to the reduced form)

The transform of x = r + 1 leads us to : r3 + 4r - 16 = P(r) = 0, thus p = 4 and q = -16, so u + v = z1 = 2 (as 1st leaf of Riemann's surface), eg x1 = 3, we easily see, that the other solutions are complex. However, if D = (q/2)2 + (p/3)3 is negative, and we do not know the classical concept of complex numbers (or trigonometry), we would have to definitely surrender here.

In a letter dated August 4th, 1539, Cardano indicated to Tartaglia that the condition (q/2)2 + (p/3)3 < 0 could be valid, but at the time, he did not know how to handle it. Tartaglia's written reply to Cardano was elusive and did not offer any useful suggestions regarding the problem. It is quite likely that Cardano was one of the first mathematicians to encounter, either directly or indirectly, some sort of complex numbers when attempting to solve the cubic equation. Since this situation posed an insurmountable obstacle, Cardano referred to it as casus irreducibilis. There is no algebraic way to transform (a + bi)1/3 into x + ξi, without resorting to trigonometric methods; otherwise, we would end up with a more complicated equation than the original cubic polynomial. This unfortunate reality occurs precisely where we might have least expected it - namely, in cases where there are 3 real solutions! The "casus irreducibilis" can only occurr for a negative p and it must hold: q2/4 < |p3/27|. In order to continue with this negative D, there are at least two (though not very different) alternatives available, neither of which was known during Cardano's lifetime.

[a] we continue with the negative D and use complex numbers ..
[b] using trigonometry, we use the equivalence of cos 3α = 4cos3 α - 3cos α to Q(x) and expand the equation by a factor ω as |cos α | ≤ 1. This way we obtain: z = ω cos α , ω = 2 (-p/3)1/2 and cos (α/3) = - 4q / ω 3.

But the period of cosine is 2π so if we set: 3α = φ + 2πn we will get: z1 = ω cos φ/3 , z2 = ω cos (φ + π)/3 , z3 = ω cos (φ + 2π)/3

Below our code in an educational style

/brings the cubic polynomial to the reduced form; the deal: red 1 -3 7 -21 ... 1 0 4 -16 (see Q(x), Q(z) above)
red:{c:x%(*)x;a2:c 2;a11:a1*a1:c 1;p:a2-a11%3;q:(c 3)+(a1*(2*a11)-9*a2)%27f;:(1 0f),p,q}
/3rd root
root:{(-1f+2*x>0f)*(abs x) xexp 1%3}
/the discriminant D
D:{c:0.25*c*c:last x;c+(d*d*d:(*)x)%27f}
/UV returns the pair (u,v)
UV:{c:-0.5*last x;e:sqrt D x;(root c+e),root c-e}
/Cardano formula for z1
sum UV (p,q)
/extended Cardano formula for z1, z2 and z3
eps1:0.5*-1f,b:sqrt 3f
eps2:0.5*-1f,neg b
uv:UV (p,q)
/z2
sum uv*(eps1;eps2)
/z3
sum uv*(eps2;eps1)
/the final step would be to subtract a1/3 from our z1, z2 and z3


Some final words about Cardano: Although he lived during the Renaissance in Italy, the country was already in decline, haunted by the Italian Wars (1536-1538 and 1542-1544). Italy had become a battleground between France (under Francis I.) and Spain (under Charles V.). The period also saw rising prices and devastating plagues (1511-1514 and 1520). Cities like Pavia (1527) and Rome (1526) were plundered, and the Duchy of Milan became only a shadow of what the Renaissance had left there. The first 40 years of Cardano's life were a struggle for survival. He was rather ill. Cardano had a daughter named Chiara, as well as two sons - Aldo Urbano and Giovanni Battista. Aldo was a gambler who was stealing money from his father, and Giovanni was beheaded for killing his unfaithful wife. Chiara married, but she remained infertile. On October 13th, 1571, Cardano was thrown into prison as a victim of the Inquisition (charge was heresy). However, he had two influencial friends - Cardinals Giovanni Morone and Carlo Borromeo - who helped him secure a relatively mild verdict.

sample 1: x3 + 7x2 - 4 = 0 (s:1 7 0 -4) returns r as 0.719824,-6.916382,-0.8034424)
sample 2: x3 - 9x2 + 36x - 28 = 0 (s:1 -9 36 -28) : returns r as (1 0;4 -3.464102;4 3.464102) .. means x1=1 , x2,3 = 4 ± 3.464102i
sample 3: x3 + 8x2 + 5x - 50 = 0 (s:1 8 5 -50) returns r as (2 0;-5 -3.575534e-008;-5 3.575534e-008) .. means x2=2 , x2,3 = -5
sample 4: 144x3 - 216x2 + 101x - 15 = 0 (s:144 -216 101 -15) returns r as (0.75,0.3333333,0.4166667) i.e. 3/4 , 1/3 , 5/12
sample 5: z3 - 6iz + 4(1-i) = 0 .. we try z = a + ib and separate the real and complex part first: a3 - 3ab2 + 6b + 4 = 0 resp. -b3 + 3a2b - 6a - 4 = 0 ; this can be reduced to (a-b)(a+b)(a2+b2) + 4(a-b) = 0 - this means a = b and we end up with -a3 + 3a + 2 and so s = -1 0 3 2: returns r as (2 0;-1 -0;-1 0); In other words: z1 = 2 + 2i and z2,3 = -1 - i

sample 6:

A floating hollow sphere (outer radius R, inner radius r, density ρm) in a fluid with density ρf .. (Archimedes principle), will lead us to a cubic equation.

H3 - 3RH2 + 4[R3(1 - ω) + ωr3] = 0 where ω = ρm/ρf and H (the unknown we are interested in) is the distance from the fluid surface to the top of the sphere. So if ω = 1 and r = 0 it follows that H = 0. (Bear in mind 0 ≤ H ≤ 2R)

sample 7:

One special equation that can be reduced to a (bi-)cubic equation is x7 - (x + a)7 = b ; a, b real numbers.
A concrete example: x7 - (x+4)7 = -1256 will lead us to: 28(x+2)6+560(x+2)4+1344(x+2)2-1000=0
x1 ≅ -1.22985676 and x2 ≅ -2.77014324 (other solutions are complex)

sample 8:

Some higher grade polynomials can be reduced with similar methods that led us to Cardano's formula
x5 - 10x3 + 20x - 12 = 0 will lead us to: s2 - 12s + 32 = 0, so x1 = 23/5 + 22/5

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