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Answers to Exercises; Chapter (^5 323)
ai - c < ai + c < 2ai (for n 1 2), we have that u:+~ - c < (a: - c)/2, so
that u:+~ - c < 2 -(“-‘)(a~ - c). From this estimate, the result follows.
1.8. Let the polynomial be p(t). When t 5 -1, p(t) > 3t4 - t2 > 0 and
when -1 < t < 0, p(t) > -t2 - 3t = -t(t + 3) > 0. Hence all of the real
zeros of p(t) are positive. Two zeros are 0.295 and 1.336. As can be seen
by the methods of Section 2, there are only two real zeros.
1.9. (b) Each successive u, seems to be farther from P t.ha.n its predecessor.
(d, e) Beginning with 3, the sequence of successive approximations is:
(3, 3.10723, 3.121968, 3.123983, 3.124258, 3.124297, 3.124302,.. .}. Begin-
ning with 4, the sequence is: (4, 3.239609, 3.139972, 3.126440, 3.124595,
3.124342, 3.124307, 3.124303). Th e zero, to five decimal places, is 3.12430.
2.6. (c) The proof is by induction on the degree of p(t). To give a feel for
the argument, examine the case that degp(t) = 1. Since p(t) = at is trivial
to analyze, let p(t) = at + b, b # 0. Then (t - r)p(t) = at2 + (b - ra)t - rb.
If ab > 0, then a and -rb differ in sign; if ab < 0, then a and -rb have one
sign and b - ra the opposite sign. In any case, the result holds.
Suppose the result holds for polynomials of degree up to n - 1 1 0. Let
p(t) = &It” + q(t), where q(t) = an-#-l +... + alt + ac. Then
(t - r)p(t) = a,t”+l - ra,P + (t - r)q(t)
= ant n+l + (a,-~ - ran)t” +....
Suppose p(t) has k sign changes.
Case (i): a, has the same sign as the next nonzero coefficient. Then q(t)
has k sign changes. The leading coefficients of (t - r)p(t) and (t - r)q(t)
have the same sign and all other corresponding coefficients except that of
tn have the same value. Hence (t - r)p(t) has at least as many sign changes
as (t - r)q(t), and the result follows.
Case (ii): a, differs in sign from the next nonzero coeficient. Then q(t)
has k - 1 sign changes. Whether or not a,-1 vanishes, the coefficients of
tnS1 and t” in (t - r)p(t) h ave opposite signs, so that this polynomial has
one more sign change than (t - r)q(t), i.e. at least k + 1 sign changes.