Answers to Exercises; Chapter 5 3252.15. (a) Let q(a) = q(b) = k. Th en a and b are zeros of the polynomial
q(t) - k whose derivative is p(t).
(b) Suppose there are k distinct real zeros, r1, r2,... , rk of q(t). Since
the multiplicity of each ri as a zero of p(t) is one less than its multiplicity
as a zero of q(t), the total multiplicity of all the ri as zeros of p(t) is
m-b. However, Rolle’s theorem provides an additional zero of p(t) between
adjacent pairs of the pi for an additional k - 1 zeros.
(c) We can take 2q(t) = t4-2t3-t2+2t = (t+l)t(t-l)(t-2). By Rolle’s
theorem, p(t) has a zero in each of the intervals (-l,O), (0, l), (1,2).
(d) Let f(t) = act+ait2/2+...+antn+1 /(n+l). Since f(0) = f(1) = 0,
by Rolle’s theorem, f’(t) = a0 + alt +... + a,t” has at least one zero in
the interval (0,l).
2.16. It is straightforward to see the result by examining the graph off(t).
Here is an analytic argument. If not all the zeros of f(t) are real, we may
take k = 0. Suppose all the zeros ui of f(t) are real. Then the zeros vi of
f’(t) are all real, and we can label them so that
Ul L Vl 2 u2 1 v2 >_...^1 u,,-~^1 v,,-~ >_ u, (where n = deg f(t)).With no loss of generality, we can assume that the leading coefficient of
f(t) is 1, so that f(t) = (t - ul)(t - ~2).. + (t - IL,,). For t > ~2, we have
that
(t - ux)(t - u2) = - $1 + u2)] 2 - &l - u2)2,so that
1
f(t) >_ --(ul - u2)2(u2 - u3)(u2 - u4) * (u2 - %I).
4
Let k be any positive real exceeding the absolute value of the right side of
this inequality. Then f(t) + k d oes not vanish for t 2 ~2. If f(t) + k has
m real zeros, then all are less than u2. The derivative f’(t) of f(t) + k has
at least m - 1 real zeros less than uz. But then m - 1 < n - 2, so that
m 5 n - 1 and we have the required result.
2.17. (c) Referring to the Taylor Expansion of p(t) about 0, we see that
the number of sign changes of the coefficients is equal to the number of sign
changes in the sequence (p(O), p’(O),.. .). On the other hand, for sufficiently
large positive v, P(~)(V) has the same sign as the leading coefficient of p(t)
for each k. Take this value of v with u = 0 in the Fourier-Budan criterion;
A is the number of sign changes in the coefficients and B = 0.
(d) For t = -2, the sequence of derivative values has signs +, -, +,
-, +; fort = 0, the signs are -, +, +, 0, +. The Fourier-Budan Theorem
provides for 1 or 3 zeros in the interval [-2,0]. However, since t4 + t2 -4t - 3
has one sign change, the Descartes’ Rule of Signs puts an upper bound of
1 on the number of zeros.
[t4 + t2 + 4t - 3 = (9 + t - l)(P - t + 3).]