Solving Cubic Equations

Ask anyone who took freshman math, and they will tell you of their introduction to algebra - which is one of the most useful branches of mathematics, since it paves the way for properly doing derivatives, and integrations in calculus. And almost everyone taking it would have been exposed to quadratic equations of the form:

ax^2 + bx + c = 0

which are then solved, either by factoring and solving for x, or - very often - by using the quadratic formula:

x = [-b +/- {b^2 - 4ac}^1/2]/ 2a

But what about solving basic cubic equations? Here we solve equations of the form:


x^3 – px^2 + qx – r = 0



Example:

x^3 – x^2 - x + 2 = 0

What I will show here, or rather demonstrate, is an algorithm for simple solution provided the numerical coefficient of the cubic term is 1.

We begin by writing: p = 1, q = -1, r = -2

And let:

DELTA = SQRT [(- 4p^3 r – 27r^2 + 18 pqr – 4 q^3 + p^2 q^2)]

With:

Epsilon1 = [p^3 –9/2 (pq – 3r) – 3/2 x SQRT(-3)x DELTA]^1/3

Epsilon2 = (p^2 – 3q)/ Epsilon1


The ROOTS are then – in turn:

alpha1 = 1/3(p + rho2 x Epsilon1 + rho x Epsilon2)

WHERE:

rho = -1/2 + SQRT(-3)/2 x SQRT(-3)x DELTA

rho2 = -1/2 – SQRT(-3)/2


Last TWO roots:

alpha2 = 1/3( p + rho2 x Epsilon1 + rho x Epsilon2)

alpha3 = 1/3( p + rho x Epsilon1 + rho x rho2 x Epsilon2)


For the original eqn. in question, one obtains:

DELTA = 7.681i

Epsilon1 = 0.578 + 1.001i

Epsilon2 = 1.73 – 2.997i

and

alpha1 = 1.103 - 0.665i

alpha2 = 1.103 + 0.665i

alpha3 = -1.206

IF Roots are correct THEN:

alpha1 + alpha2 + alpha3 = 1

AND

(alpha1 x alpha2) + (alpha 2 x alpha3) + (alpha3 x alpha1) = q

AND

alpha1 x alpha2 x alpha3 = r

AND:

(alpha 1 – alpha2) x (alpha2 – alpha3)x (alpha3 – alpha1) = DELTA


We will just check the first: alpha1 + alpha2 + alpha3 = 1

(and let interested readers check the others):

We have then: [1.103 - 0.665i] + [1.103 + 0.665i] + (-1.206) = 2.206 - 1.206 = 1

All of the above will be found to check out for the given equation!
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Solution to previous fractal problem:

to find generator of Z1(x1,y1)

from the diagram given set the boundary in terms of:

@(na1a2 + ma3a1 + C1(Y)) = @1(na1a2) + @1(ma3a1) + C_o(Y)

where as before, @ denotes the differential boundary (Greek alpha) sign

Then:

= n(a2 - a1) + m(a1 - a3) + C_o(Y)

= (m - n)a1 + C_o(Y)

(Since a2, a3 belong to C_o(Y))

and, provided we demand m=n for a cycle, then the generator of Z1(x,y) is:

a1a2 + a3a1 + C1(Y)