Solutions to Problems; Chapter 5 333We solve by successive approximation, by evaluating the right side at x =- With the aid of the binomial theorem, this yields
(1) x = 105[1 + (316.73)10-‘“]1’2 = lo5 + (158.36)10-5
= 100000.0016(2) 2 = 105[1 - (315.73)10-10]“2 = lo5 - (157.86)10-5
= 99999.9984.These solutions check out.
4.13. Suppose that p(x) = ax2 - bx + c has two distinct zeros in (0, l),
and that a, b, c are integers with a positive. Then b > 0, c = p(0) > 0,
a - b + c = p(1) > 0, b2 - 4ac > 0 and the product c/u of the zeros is less
than 1. Thus, it is necessary that
(1) 0 < c < a,
(2) b < a + c,
(3) 4ac < b2.
Since a and c are positive integers, (a - l)(c - 1) 2 0, whence b2 > 4ac >
4(a + c - 1) 2 4b j b 2 5 j 2a > a + c 2 6 =P a 2 4. Now a = 4 forces
c = 2 or c = 3, either of which make (2) and (3) incompatible. Hence a 2 5.
Consider the possibilities for a = 5. If c > 2, then (2) and (3) are incom-
patible. However, c = 1 forces b = 5 and yields the polynomial 5z2 - 5x+ 1,
which has the stated property.
4.14. Since each q,,(x) has nonnegative coefficients, any real zero cannot
be positive. It is readily checked that ql(x) and q2(x) each have a negative
zero. Suppose, for some n 1 2, it has been established that qi(O) > 0 and
that each qi(x) has at least one negative zero and that the greatest such
zero xi satisfies xi-1 < xi (2 < i 5 n). Then, for 1 5 i 5 j 5 n, qi(x) > 0
for Xj < 2 5 0.
From the recursion relations, it follows that q,+l(O) > 0 and qn+l(xn) <
0, so that qn+l(x) has at least one zero in the interval (x,, 0). Thus, there
is a largest real zero x,+1 and it satisfies x, < x,+1 < 0.
In fact, it can be shown that xz,,,+2 > -l/(m + 1) for m 2 1. For, from
the recursion relations, we find that
q2m+2(-l/(m + 1)) = -q2m-l(-l/(m + 1)).If qzm+2(- l/(m + 1)) < 0, th en xzm+2 > -l/(m + 1). On the other hand,
if qzm+z(-l/(m + 1)) > 0, then qm-I(-l/(m + 1)) < 0, so that x2,+2 >
~2~-1 > -l/(m + 1).
Therefore, if r is any positive number, no matter how small, we can
choose a positive integer m so that -r < -l/(m + 1) < 0. Then, for
n 2 2m + 2, -r < x, < 0, so that the x, get closer and closer to 0.