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(^322) Answers to Exercises and Solutions to Problems
and those of p(t) + 1 and q(t) + 1 be ~1,... , v,. Clearly, the Ui are distinct
from the vj. Since each ui and each vj is a zero of (p - q)(t), it follows that
n = m=(degp(t), deg q(t)) 2 dedp - q>(t) 2 r + s.
Consider the derivative p’(t), which is also the derivative of p(t)+ 1. Since
the multiplicity of each zero of p(t) and of p(t) + 1 exceeds its multiplicity
as a zero of p’(t) by 1, p’(t) has at least n - T zeros, counting multiplicity,
from among the ui and n - s zeros, counting multiplicity, from among the
vj. Hence
(n- r)+(n-s),<degp’(t)=n-l$n<r+s-l<r+s.
But this contradicts the earlier inequality, and so (p - q)(t) must be the
zero polynomial.
9.10. Suppose the polynomial is t3 + at2 + bt + c. Let u = 1 - 2ii3 + 22/3.
Then u2 = -3 + 3.22/3 and u3 = -9 + 9.2113. Since u is a zero of the
polynomial,
(-9 - 3a + b + c) + (9 - b)21’3 + (31s + b)22’3 = 0.
The coefficients in parentheses vanish when (a, b, c) = (-3,9, -9) and we
obtain the polynomial t3 - 3t2 + 9t - 9.
Answers to Exercises
Chapter 5
1.2. (a) -0.77, 1.93; (b) -0.43; (c) 0.15, 3.47.
1.3. (c) For example, the final polynomial of Exercise 2 has p(3) = -257
and p(4) = 1133. Taking a = 3, b = 4 yields the next approximation 3.293.
With a = 3.293, b = 4, we are led to 3.377.
1.5. (c) 0.148.
(e) (4
1.7. (b) a;+, - c = (a; - c)s/(4aE). It follows that for n 2 2, ui 2 c,
and, since a, - a,+1 = (u: - c)/(2a,), {a,} eventually decreases. Since