polynomials « Mathematics Hothouse

Problem:

If 1, $\omega, \omega^{2}, \omega^{3}, \ldots, \omega^{n}$ are the nth roots of unity, then find the value of $(2-\omega)(2-\omega^{2})(2-\omega^{3})\ldots (2-\omega^{n-1})$ .

Solution:

Learning to think:

Compare it with what we know from our higher algebra — suppose we have to multiply out:

$(x+a)(x+b)(x+c)(x+d)$ . We know it is equal to the following:

$x^{4}+(a+b+c+d)x^{3}+(ab+ac+ad+bc+bd+cd)x^{2}+(abc+acd+bcd+abd)x+abcd$

If we examine the way in which the partial products are formed, we see that

(1) the term $x^{4}$ is formed by taking the letter x out of each of the factors.

(2) the terms involving $x^{3}$ are formed by taking the letter x out of any three factors, in every possible way, and one of the letters a, b, c, d out of the remaining factor

(3) the terms involving $x^{2}$ are formed by taking the letter x out of any two factors, in every possible way, and two of the letters a, b, c, d out of the remaining factors

(4) the terms involving x are formed by taking the letter x out of any one factor, and three out of the letters a, b, c, d out of the remaining factors.

(5) the term independent of x is the product of all the letters a, b, c, d.

Further hint:

relate the above to sum of binomial coefficients.

and, you are almost done.

More later,

Nalin Pithwa

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

$p^{(2)}(t)=p(p(t))$

$p^{(3)}(t)=p(p(p(t)))$

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

$u^{3}+v^{3}=-q$

$uv = -p/3$

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

$x^{2}+qx-p^{3}/27=0$

(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…

-Nalin

Mathematics Hothouse

Tag Archives: polynomials

Complex ain’t so complex ! Learning to think!

Polynomials — commuting and cubics

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