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5.2. Tests for Real Zeros 177

Secondly, observe that if we divide the terms of Sturm’s sequence by
the same polynomial, the number of sign changes will not be affected. If
q(t) is the greatest common divisor of p(t) and p’(t), verify that the Sturm
sequence for p(t)/q(t) is th e same as the Sturm sequence of p(t) divided by
q(t). Verify also that the zeros of p(t)/q(t) are all simple and are exactly the
same as the zeros of p(t). Thus, one can deal with the polynomial p(t)/q(t)
which may have degree less than that of p(t).
How does one prove Sturm’s Theorem? Argue that it suffices to prove the
result when all the zeros of p(t) are simple. In this case, show that Sturm’s
sequence ends in a constant and that it is not possible for two consecutive
polynomials in Sturm’s sequence to have a common zero. For convenience,
we will assume that no polynomial in Sturm’s sequence vanishes at the
endpoints u and v of the interval under consideration. We look at several
cases:
(i) Suppose that no polynomial pi(t) has a zero between u and v. Argue
that each polynomial in Sturm’s sequence maintains the same sign on the
interval (u, v), and so the number of sign changes in Sturm’s sequence
evaluated at u and at v are the same.
(ii) Suppose that r lies between u and v and that r is a zero of some
pi(t) (1 5 i 5 n - 1). Note that r must be a simple zero. Show that pi-i(r)
and pi+l(r) are nonzero and have opposite signs. Choose the number s
sufficiently small that pi-1 and pi do not change sign and pi has only the
root r in the closed interval [r - s,r + s]. Show that the number of sign
changes in {pi-l(t),pi(t),pi+l(t)} is the same at r - s and r + s.
(iii) Suppose that r lies between u and v and that p(r) = 0. Show that a
small positive real s can be chosen so that

(a) u<r-s<r<r+s<v

(b) p(t) has only the simple zero r in the closed interval [r - s, r + s]

(c) p’(t) has no zero in the closed interval [r - s, r. + s]

(d) p(r - s) has sign opposite to each of p(r + s), p’(r - s) and p’(r + s).

Deduce that Sturm’s sequence has one more sign change when evaluated
at r - s than when evaluated at r + s.
Complete the proof of Sturm’s Theorem by subdividing the closed inter-
val [u, v] into subintervals at the endpoints of which none of the pi vanish
and within which there is at most one number r which is a zero of any of
the pi.

E.49. Oscillating Populations. In the beginning of this chapter, the
function f(x) = ax- 6x2 was introduced. We can think of x as representing
a population density for some species in a certain area, so that its value
will lie between 0 and 1. Consider the special case a = b. Thus, let

f(x) = ax(1 -x) (0 5 x 5 1).