340 Answers to Exercises and Solutions to Problemsfollowing: if f(t) = Ca,t’ and g(t) = CZ,+,t’ satisfy a,g(t) = Eof(t), then
all of the zeros of f(t) lie on the unit circle.
4.28. If k = 0, the equation is 2” + t”-l +... + t + 1 = 0 and the result
holds. Suppose the result has been established for k < r. Let
f(z) = (n + 1)-@+‘)p + n-+-+1)$-1 +... + 2-(‘+1)2 + 1Then
g’(c) = (n + l)--‘Z? + n-‘z”-l +... + 2-‘2 + 1.
Suppose n is odd. Then, by the induction hypothesis, g’(z) has exactly
one real zero. By Rolle’s Theorem, g(z) has at most two real zeros. Since
g(O) = 0 and f(O) # 0, f(z) h as at most one real zero. But, since deg f(z)
is odd, f(z) has a real zero.
If n is even, g’(z) has no real zero, so g(e) has at most one real zero.
Since g(0) = 0 and f(0) # 0, f(z) can have no real zero.
4.29. The equation can be rewritten as f(z) = 0 where
f(z) = n(al - 2)(a2 - x)... (a, - Z)Now f(oi) = (-l)‘aipi where pi > 0. If all the ai have the same sign, then
the signs of the f(ai) alternate and each interval (ai, oi+i) (1 < i 5 n - I)
contains a root of the equation. Since 0 is an additional root, the equation
has n distinct real roots.
Suppose ok < 0 < ak+i. Then the above argument provides for a root
of the equation in each interval (ai,ai+l) (1 < i < k - 1, k + 1 5 i 5 n).
Now f(Q) and f(ak+i) have the same sign and f(0) = 0. Hence, either
f(z) has at least two distinct roots in (ak,ak+i) or else a double root at 0.
Hence the equation in this case has at least n - 1 distinct real roots, or n
roots if we count multiplicity. The case n = 2, al = -1, a2 = 1 yields an
equation with a double root at 0.
4.30. Let ]z] < l/(k + 1). Then11 + alz+a2z2+...+anznl
2 1 - Iall 1.~1 - Ia21 lz12 - ... - IanI 1-V
1 1-k[(k+1)-‘+(k+1)-2+~~~+(k+1)-“]
> 1 - k(l/k) = 0.