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324 Answers to Exercises and Solutions to Problems

2.7. Let r1, r2 ,... , rk be all the positive zeros of p(t), each taken as often

as its multiplicity. Then p(t) = (t - rl)(t - r2)... (t - rk)q(t) where q(t) has

no positive zeros and, say, m sign changes. Applying Exercise 2.6, we find

that p(t) has k + m sign changes. The second part follows from Exercise 5.

2.10. The two adjacent vanishing coefficients correspond to consecutive

powers oft, one even and one odd. Let the signs +, -, 0 of the remaining

n - 1 coefficients be written out in order. Denoting the polynomial by p(t),

we see that there is a sign change between two consecutive coefficients in

a given position which are nonzero for exactly one of p(t) and p(-t). If

two nonzero coefficients are separated by a zero coefficient (apart from ak

and Uk+r), then there are at most two sign changes involved with p(t) and

p(-t) together. As a consequence, there are at most n - 2 changes of sign

for p(t) and p(-t) together, and so at most n - 2 real zeros. Since p(0) # 0,

the result follows.

2.11. Multiplying the given polynomial by t - 1 yields

ant” + (a,-1 - a,)t” + e-e+ (1 - a3)t3 - 1.

By Exercise 10, this has at most n - 1 real zeros.

A similar argument shows that, if any three consecutive coefficients of a

real polynomial are equal and nonzero, the zeros of the polynomial are not

all real. See problem El283 in Amer. Math. Monthly 64 (1957), 592; 65

(1958), 286.

2.12. (a)

f(r) = a,P + (an-lrnsl +a .. + a,+lrP+l) + a,?

+ a,-19-l + -..+ air + ao.

For 0 5 i 5 p, if ai < 0, then Mr’ 2 0 > air’, while if ai > 0, Mr’ 1 air’.
For p < i < n, ai > 0. Hence

f(r) > a,P + M(rP + 9-l +... + r + 1).

(b) The first inequality is obvious. Since r > r - 1,

ad n-P-l(r - 1) > an(r - l)“-P-‘(r - 1) > M.

(cl From (b),

a,# > MTP+l(r - l)-’ > M(TJ’+’ - l)/(r - 1).

Now apply (a).

2.14. (a) 1 + fi < 2.201; 1 + 3/2 = 2.5.
(b) 1 + 3 = 4; 1 + 1 = 2.