The cubics
Contents
The cubics¶
Polynomials of degree three are referred to as cubics. In one variable, cubic univariates. They are generically of the form
for \(a\neq0\). As with the quadratics, cubic polynomials are polynomials. They are coded in the module cubics.py as a subclass of Poly.
Library import¶
from class_scripts import cubics as cbc
Attributes¶
Pre-initialisation¶
The __new__ method is run before initialisation and verifies whether a given array is indeed a member of the class Cubic. Recall, it will be a cubic if it has degree three with non-zero leading term. The following two instances are not cubic, as detected by __new__.
cbc_test_1 = cbc.Cubic([1])
cbc_test_2 = cbc.Cubic([-4, 5, -2, 0])
cbc_test_3 = cbc.Cubic([-3, 3, 1, -4, 0])
print(cbc_test_1)
print(cbc_test_2)
print(cbc_test_3)
Not a cubic
Not a cubic
Not a cubic
Initialisation¶
For an honest cubic \(f(x) = d + cx + bx^2 + ax^3\), it comes with all the attributes of Poly. Furthermore, as with quadratics, it is much faster to calculate its discriminant from the coefficients passed rather than first calculating its resultant.
To illustrate, on the cubic \(f(x) = -3 + 5x + x^2 -12x^3\), when initialised it has attribues:
cbc_1 = cbc.Cubic([-3, 5, 1, -12])
print(f"The cubic polynomials is: {cbc_1}")
print(f"Its coefficient list is: {cbc_1.coeffs}")
print(f"The degree of the cubic is: {cbc_1.degree}")
print(f"Its discriminant is: {cbc_1.disc}")
The cubic polynomials is: -3x^0 + 5x^1 + 1x^2 - 12x^3
Its coefficient list is: [-3, 5, 1, -12]
The degree of the cubic is: 3
Its discriminant is: -25715
As a check, to see that its discriminant coincides with the discrinimant call discriminant() when viewed as an instance of Poly, see that:
print(cbc_1.discriminant())
-25715.0
Cubic depression¶
Preliminaries¶
An important transformation for cubics is the depression. A depressed cubic \(g(x)\) is a cubic of the form \(g(x) = q + px + x^3\) for coefficients \(p, q\). From the Wikipedia entry, any cubic can be transformed into depressed form.
For a general cubic \(f(x) = d + cx + bx^2 + ax^3\), recall that we know \(a\neq0\). As such we can divide through by the leading coefficient \(a\) without cause for heart palpatations. The transformation \(x \mapsto x - \frac{b}{3a}\) will send \(f\) to the form \(\tilde f(x) = aq + apx + ax^3\). Dividing through by \(a\) gives the depressed cubic \(g(x) = q + px + x^3\). The new coefficients \((p, q)\) can be expressed in terms of the old coefficients \((a, b, c, d)\) thusly,
Implementation¶
In the class Cubic, the method toDepressed() takes any instance of Cubic and, through the above formula for the coefficients, returns its depression which is another instance of Cubic. To illustrate, consider the cubic,
Its depression is:
undepressed = cbc.Cubic([1, 7, -3, 2])
print(undepressed.toDepressed())
2.0x^0 + 2.75x^1 + 1x^3
We obtained the above cubic by applying the formula. It is also possible to recover it through composing it with the transformation \(x \mapsto x - \frac{b}{3a}\) and dividing the resulting polynomial by \(a\). This can be implemented through the composeWith() method callable on any instance of Poly. In doing so:
leading_coeff = undepressed.coeff(undepressed.degree)
coeff_b = undepressed.coeff(undepressed.degree - 1)
composite = cbc.Poly([-coeff_b/(3*leading_coeff), 1])
factor = cbc.Poly([1/leading_coeff])
print(factor * undepressed.composeWith(composite))
2.0x^0 + 2.75x^1 + 1.0x^3
Zeroes¶
Background¶
A root of a cubic univariate \(f(x) = d + cx + bx^2 + ax^3\) is a value \(x_0\) such that
The problem of expressing roots of a cubic univariate in terms of its coefficients and radicals, in a manner analogous to the case of quadratic univariates, is a classical problem in the history of mathematics. Over the complex numbers, there will always exist three roots of \(f\), counted with multiplicity.
Utility of depression¶
Recall that any cubic can be transformed into a much simpler cubic, name its depression. The problem of finding roots of the cubic can therefore be reduced to finding roots of its depression, which is significantly easier. Since we know the transformation \(x \mapsto x - \frac{b}{3a}\) sends the original cubic \(f\) to (a multiple of) its depressed form, we can therefore obtain the root of \(f\) by simply translating back any root of its depressed form. More explicitly, if \(x_{depr}\) is a root of the depressed form of \(f\), then the translate \(x_{depr} + \frac{b}{3a}\) will be a root of \(f\).
For any (real) depressed cubic \(g(x) = q + px + x^3\) with negative discriminant Cardano famously obtained a formula for one of the roots in terms of the coefficients \(p, q\) and their radicals (specifically, square- and cube-roots).
Note
Recall that it suffices to find only one root. The others can be obtained on multiplying by an appropriate root of unity. Once we have three roots we know, by the fundamental theorem of algebra, that we have found all the roots. With Cardano’s formula and a cube root of unity it is possible to derive all the roots of a cubic.
See this Wikipedia entry for the general formula for roots of a cubic.
Note
The formula from the above entry is valid in the same generality as in the quadratic case, i.e., for complex cubics, not just real cubics.
Implementation¶
The method zeroes() is callable on instances of the class Cubic. When called on an instance it returns a list of its zeroes. To illustrate, for the following cubics:
\(-10 + x + 5x^3\);
\(x -9x^2 + x^3\);
\(-34 + 6x^2 + 2x^3\);
\(1 + x + x^2 + x^3\)
Their zeroes are:
test_cubic_1 = cbc.Cubic([-10, 1, 0, 5])
test_cubic_2 = cbc.Cubic([0, 1, -9, 1])
test_cubic_3 = cbc.Cubic([-34, 0, 6, 2])
test_cubic_4 = cbc.Cubic([1, 1, 1, 1])
test_cubics = [test_cubic_1, test_cubic_2, test_cubic_3, test_cubic_4]
for c in test_cubics:
print(f"Zeroes of {c} are:")
c_zros = c.zeroes()
print(*c_zros, sep='\n')
print('\n')
Zeroes of -10x^0 + 1x^1 + 5x^3 are:
1.207 + 0.0i
-0.6035 + 1.137i
-0.6035 - 1.137i
Zeroes of 1x^1 - 9x^2 + 1x^3 are:
0.0 + 0.0i
0.1125 + -0.0i
8.8875 + 0.0i
Zeroes of -34x^0 + 6x^2 + 2x^3 are:
-2.4343 + 1.7809i
-2.4343 - 1.7809i
1.8686 + 0.0i
Zeroes of 1x^0 + 1x^1 + 1x^2 + 1x^3 are:
-1.0 + 0.0i
0.0 + 1.0i
-0.0 - 1.0i
The value of the cubic at each zero must of course be zero.
And so, as a sanity check on the method zeroes() we need to check, for each cubic in test_cubics above, that the cubic at that value is indeed zero. As the method zeroes() returns complex numbers, i.e., instances of cplx, to evaluate a polynomial thereon requires calling cplxEval() from Poly. We find:
for c in test_cubics:
c_zros = c.zeroes()
for zro in c_zros:
print(c.cplxEval(zro))
print('\n')
-0.0 + 0.0i
-0.0 + 0.0i
-0.0 + 0.0i
0.0 + 0.0i
0.0 + 0.0i
0.0 + 0.0i
-0.0 + 0.0i
-0.0 + 0.0i
-0.0 + 0.0i
0.0 + 0.0i
0.0 + 0.0i
0.0 + 0.0i
And so for each cubic in test_cubics, each of their zeroes indeed returns zero upon evaluation, as expected.