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336 Answers to Exercises and Solutions to Problems

4.17. Let f(x) denote the left side. We make use of Rolle’s Theorem,

f’(x) = u~(x)xyx + 1)ss + Vi(X)X8(X + 1)26 + Wi(Z)Z3(X + l)‘,

where degur(x) = degvr(x) = degwi(x) = 1. The positive zeros of f’(x)
are the same as those of

g(x) = ul(x)x’2(x + 1)56 + 2)1(x)x5(x + 1)19 + w(x).

Differentiating twice yields

g”(X) = u3(x)x10(x + 1)54 + 2)3(x)x3(x + 1)17

where degus(x) = degvs(x) = 3. The positive zeros of g”(x) are the same
as those of
h(x) = 213(x)x7(x + 1)s7 + v3(x).

Differentiating four times yields

d4)(x) = 21,(x)x3(x + 1)34

where degur(x) = 7.
The number of positive zeros of hc4)(x) does not exceed 7 = deg u?(x).
Hence the number of positive zeros of h(x) and g”(x) does not exceed 11,
the number of positive zeros of g(x) and f’(x) does not exceed 13, and the
number of positive zeros of f(x) does not exceed 14.

4.18. This problem could be solved by using Exercise 3.8 and the delin-
eation of the cases for real and nonreal zeros made through Exercise 6.1.9.
However, let us take a more elementary approach.
Let f(t) = t3-t2+a. For f(t) not to have a real zero outside of the closed
interval [-l,l], it is necessary that a = f(1) 10 and -2+a = f(-1) 5 0,
i.e. 0 5 a 2 2. Since the zeros of f’(t) are 0 and 213, f(t) has a maximum
at t = 0 and a minimum at t = 213. All zeros are real iff a = f (0) > 0 and
-4127 + a = f (213) 5 0 iff 0 5 a < 4127. For all a in this range, all three
zeros lie in [-l,l].
(To get an intuitive idea of what is happening, we imagine how the zeros
of f(t) vary with a near a = 0 and a = 4127. Let a = 0; f(t) has a double
zero at t = 0 and a simple zero at t = 1. As a decreases through zero, two
real zeros of f(t) converge to 0 and then split into a complex conjugate
pair moving away from 0 while one real zero increases along the real axis
through 1. Thus we can see that for a < 0, the real zero will lie outside the
unit disc. Now let a = 4127. Then f(t) has a double zero at t = 213 and a
simple zero at t = -l/3. As a increases through 4127, two real zeros off(t)
converge to 213 and then split into a complex conjugate pair moving away
from 213 while one real zero moves to the left through -l/3. It will be
possible for a to increase further above 4127 before the zeros leave the disc.