Polynomials — commuting and cubics

Let us start delving deeper in Algebra. But, I will be providing only an outline to you in the present article. I encourage you to fill in the  details. This is a well-known way to develop mathematical aptitude/thinking. (This method of  learning works even in Physics and esoteric/hardcore programming).

Definition. Commuting polynomials. Two polynomials are said to commute under composition if and only if (p \circ q)(t)=(q \circ p)(t)

(i.e., p(q(t)=q(p(t)))). We define the composition powers of a polynomial as follows



and in general, p^{(k)}(t)=p(p^{(k-1)}(t) for k=2,3 \ldots

Show that any two composition powers of the same polynomial commute with each other.

One might ask whether two commuting polynomials must be composition powers of the same polynomial. The  answer is no. Show that any pair of  polynomials in the  following  two sets commute

I. {t^{n}: n=1,2 \ldots}

II. {T^{n}(t):n=1,2 \ldots}

Let a and b be any constants with a not equal to zero. Show that, if p and q are two polynomials which commute under composition, then the polynomials 

(t/a-b/a) \circ p \circ (at+b) and (t/a-b/a) \circ q \circ (at+b) also  commute under the composition. Use this fact to find from sets I and II other  families which commute under composition.

Can you find pairs of polynomials not  comprised in the foregoing discussion which commute under composition? Find families of polynomials which commute under composition and within which  there is exactly one polynomial of each positive degree.

The Cubic Equation. Cardan’s Method. An elegant way to solve the general cubic is due to Cardan. The strategy is to replace an equation in one variable by one in two variables. This provides an extra degree of  freedom by which we can impose a convenient second constraint, allowing us to  reduce the problem to that of solving a quadratic.

(a) Suppose  the given equation is t^{3}+pt+q=0. Set t=u+v andn obtain the equation u^{3}+v^{3}+(3uv+p)(u+v)+q=0.

Impose the second condition 3uv+p=0 (why do we do  this?) and argue that we can obtain solutions for  the cubic by solving the system


uv = -p/3

(b) Show that u^{3} and v^{3} are roots of  the quadratic equation


(c) Let D=27q^{2}+4p^{3}. Suppose that p and q are both real and that D>0. Show that the quadratic in (b) has real solutions, and that if

u_{0} and v_{0} are the real cubic roots of  these solutions, then the system in (a) is satisfied by

(u,v)=(u_{0},v_{0}), (u_{0}\omega, v_{0}\omega^{2}), (u_{0}\omega^{2}, v_{0}\omega)

where \omega is the imaginary cube root (0.5)(-1+\sqrt{-3}) of  unity. Deduce that the cubic polynomial t^{3}+pt+q has one real and two nonreal zeros.

(d) Suppose that p and q are both real and that D=0. Let u_{0} be the real cube root of the solution of  the quadratic in (b). Show  that, in this case, the cubic has all its zeros real, and in fact can be written in the form

sy^{2} where y=(t+u_{0}) and s=t-2u_{0}

(e) Suppose that p and q are both real and that D<0. Show that the solutions  of  the quadratic equation in (b) are nonreal complex conjugates, and that it is possible to choose cube roots u and v of  these solutions which are complex conjugates and satisfy the system in (a). If

u=r(cos \theta + isin \theta) and v=r(cos \theta - isin \theta), show that the three roots of  the cubic equation are the reals

2r cos \theta

2r cos (\theta + (2/3)\pi)

2r cos(\theta + (4/3)\pi) .

(f) Prove that every cubic equation with real coefficients has at least one real root.

Use Cardan’s Method to solve the cubic equation.

(a) x^{3}-6x+9=0

(b) x^{3}-7x+6=0.

Part (b) above will require  the use of a pocket calculator and some trigonometry. You will also need De Moivre’s Theorem and give a solution to an accuracy of 3 decimal places.

More later…


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