1.4. Equations of Low Degree^17Exploration
E.6. Commuting Polynomials. Two polynomials are said to commute
under composition if and only if (poq)(t) = (qop)(t) (i.e. p(q(t)) = q(p(t)).
We define the composition powers of a polynomial as follows
Pi2V) = p(p(t))pL31(t) = p(p(p(t)))and, in general, pIk](t) = p(plk-‘](t)) for Ic = 2,3,....
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. {P : 12 = 1,2,.. .}
II. {T,(t) : n = 1,2,.. .}.
Let a and b be any constants with a # 0. Show that, if p and q are
two polynomials which commute under composition, then the polynomials
(t/u - b/a) o p o (at + b) and (t/u - b/a) o q o (at + b) also commute under
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 dis-
cussion which commute under composition? Find families of polynomials
which commute under composition and within which there is exactly one
polynomial of each positive degree.
1.4 Equations of Low Degree
With access to complex numbers, we are able to determine the solutions
to any quadratic equation whose coefficients are real, or even complex. It
is a notable result, realized by mathematicians in the sixteenth century,
that one does not have to extend the number system any further in order
to solve real cubic or quartic equations. One phenomenon which led to the
adoption of complex numbers was the use of nonreal roots of a quadratic
equation in determining the real roots of a real cubic equation.
Exercises
- (a) Let p(t) b e a cubic polynomial. Show that r is a root of the
polynomial equation p(t) = 0 if and only if p(t) = (t - r)q(t) for
some quadratic polynomial q(t).