330 Answers to Exercises and Solutions to Problemsthat the system has a solution u + vi with u > 0. Hence the given equation
is not stable.Solutions to Problems
Chapter 54.1. Let u, v be any two distinct real zeros, if such exist of t3 - 3t + k. Then
u3 - v3 = 3(u - v), whence u2 + uv + v2 = 3. If both u and v are positive
and less than 1, then the left side is less than 3, yielding a contradiction.
Hence at most one real zero lies in [0, 11.
An alternative solution can be based on the graph of the polynomial,
using the fact that there is a unique maximum when t = -1 and a unique
minimum when t = 1.
4.2. By Descartes’ Rule of Signs, there is at most one positive root. The
left side can be rewritten as
n!z(l+ x)(1+ c/2)(1 + 2/3)-a-(1 + z/n),which for z = l/n! exceeds 1. Since the left side is less than 1 for 2 = 0, it
must equal 1 for some z between 0 and l/n!.
4.3. Clearly, t = 1 satisfies the equation and x = 0 does not. Let n 2 2.
The equation can be rewritten
n = xen + xsn+l +... + 2-l.If 1x1 2 1, then ]xVn +... + z-r] < n with equality iff both 1x1 < 1 and
all powers of x-l have the same argument iff x = 1. Therefore, for all z
satisfying the equation, either z = 1 or 1x1 < 1.
A number of other solutions, some using Exercise 3.6, can be found in
Amer. Math. Monthly 66 (1959), 143-144.
4.4. With the quartic written as (z2 - 2)2 - 2(5x2 + 7), it is evident that
it is positive when x is negative.
4.5. We prove by induction that, if Ikl < 2, then f,,(t) = k has 2n distinct
real roots. The result is clearly true when n = 0. Suppose it holds for n - 1.
Then fn(t) = k _ f,+l(t) = m or fn-l(t) = -m. By the
induction hypothesis, each of the alternative equations has 2”-l distinct
real roots, and any root of one is not a root of the other. The result follows.
4.6. The equation can be rewritten
tn+’ + at” +... + a”+’ = tn+’ + bt” +... + bn+l.After multiplication by (t - a)(t - b), this becomes f(t) = 0 where
f(t) = (tnt2 - a”t2)(t - b) - (tnt2 - b”+2)(t - a)
= (a - b)t“+’ - (ant2 - b”t2)t + &(a”+’ - b”+‘).