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190 5. Approximation and Location of Zeros

3.6. (a) A standard way to isolate the difference of the coefficients is to

multiply the polynomial by 1 - t. Show that, if lwl < 1 and w # 1,

then l(l-w)g(w)l > bo-(bo-b~)-(b~-b~)-~~~-(b,-l-b,)-b,.

(b) Use the result of Exercise (a) along with Exercise 4(a).

(c) The conditions imply that 0 < ana” 5 an-run-l <... < ur’l~ 5

Qo-

3.8. (a) Consider two cases, according as the zeros are real or nonreal.

Observe that, if l+c 5 lbl, then both zeros are real. Examine (l+c)fb

as a function of the zeros.

(b) This solution needs some careful arguing. Again, look at two cases

according as one or three zeros are real. Write out 1 f b + c f d and

1 - c + bd - d2 in terms of the zeros. It may be useful to recognize

the values of the polynomials at t = -1 and t = 1.

4.1. If u and v are distinct real zeros, find a relation between them which

indicates that 0 < u, w < 1 cannot occur. Alternatively, look at what

the graph of the polynomial must be.

4.2. Show that the left side exceeds 1 at x = l/n!. Express (x + k) as

q1+ x/k).

4.3. Multiply by x-“.

4.4. Separate the even and odd powers of x; this is a “quickie”.

4.5. Prove more generally by induction that fn(t) = h has 2” real roots

when Ikl < 2.

4.6. Multiply the equation by (t - a)(t - b). Rolle’s Theorem will be useful.

(Exploration E.28).

4.7. Rewrite the polynomial as (t-al)(t-a3)(t-as)+(t-u2)(t-a4)(t-as>.

4.8. The hard part is to show that f(2(ra + b + c)/3) 5 0. This is equiv-

alent to showing that lOXa - 6Ca2b + 6abc is nonnegative. Since

this quantity vanishes when a = b = c, try to write it in the form

A(b - c)~ + B(c - d)” + C(u - b)2. A, B and C will of course be linear

and homogeneous in a, b, c.

4.9. Factor H(x - ei) - 1.

4.10. Factorp(t) andcomputep’(ar). Note that Iei-erl _< [ai-bl+lb-all.

4.11. Let Q < b 5 c and u 5 V, where u and v are the zeros of (t - e)
(t - c) - e2. f(u) can be written as a perfect square.

4.12. The left side can be written as a quadratic in (2” - lOlo).