Web server info - 4.6.4 ANSWERS TO EXERCISES 643 22. 0Z6 =
4.6.4 ANSWERS TO EXERCISES 643 22. 0Z6 = 1, ff,, = -1, 01 = 1, p1 = -2, p2 = -2, /& = -2, @z, = 1, Cy3 = -4, ~2 = 0, ~4 = 4, CQ = -2. We form z = (X -1)X + 1, w = z + X, and u(X) = ((z -x -4)w + 4)~ -2. In th IS special case we see that one of the seven additions can be saved if we compute w = x2 + 1, z = w -X. 23. (a) We may use induction on n; the result is trivial if n < 2. If f(0) = 0, then the result is true for the polynomial f(z)/ Z, SO it holds for f(z). If f(Q) = 0 for some real y # 0, then g(fiy) = h&y) = 0; since the result is true for f(z)/(z2 + y ), it hold8 also for f(z). Therefore we may assume that f(z) has no root8 whose real part is zero. NOW the net number of times the given path circles the origin is the number of roots of f(z) inside the region, which is at most 1. When R is large, the path f(ReZt) for r/2 5 t < 37~/2 will circle the origin clockwise approximately n/2 times; so the path f(it) for -R 5 t < R must go counterclockwise around the origin at least n/2 -I times. For n even, this implies that f(it) crosses the imaginary axis at least n-2 times, and the real axis at least n -3 times; for n odd, f(it) crosses the real axis at least n -2 times and the imaginary axis at least n -3 times. These are roots respectively of g(it) = 0, h(it) = 0. (b) If not, g or h would have a root of the form a + bi with o # 0 and b # 0. But this would imply the existence of at least three other such roots, namely o -bi and –a f bi, while g(z) and h(z) have at most n roots. 24. The roots of u are -7, -3 * i, -2 f i, and -1; permissible value8 of c are 2 and 4 (but not 3, since c = 3 makes the sum of the roots equal to zero). Case I, c = 2: p(x) = (x + 5)(x2 + 2x + 2)(x + 1)(X -1) = x6 +6X5 +6×4 +4X3 -5×2 -2x -10; q(x) = 6×2 +4x -2 = 6(s + 1)(x -4). L et (~2 = -1, cyi = 5; pr(x) = x4 f 6×3 + 5×2 -2x -10 = (x2 + 6x + q)(x -3) -9 ; CYO= 6, PO = 9, p1 = -y. Case 2, c = 4: A similar analysis gives oz = 9, crl = -3, are = -6, PO = 12, pi = -26. 25.h = 02, p2 = 201, /33 = a-r, /34 = a6, /35 = PEG = 0, /& = CY~, ,&, = 0, p9 = 2ffl -as. 26. (a) Xl = cyl x XO, X2 = cy2+Xl, X3 =X2 xX0, X4 = cy3+X3, X5 =X4 x X0, x6 =04+x5. (b)n~ = l+piX, ~2 = l-l-P~KIX, 1c3 = l+ 03~25, U(X)= p4~3 = ,&/32~3~4×3 +p2p3p4×2 +/33/34x+/?4. (c) If any coefficient is zero, the coefficient of x3 must also be zero in (b), while (a) yields an arbitrary polynomial 01×3+~2×2+03x+~4 of degree 5 3. 27. Otherwise there would be a nonzero polynomial f(qn, . . . , 41, qo) with integer coeffi- cients such that q,, . f(qn, , 41, qo) = 0 for all sets (qn, . . . , qo) of real numbers. This cannot happen, since it is easy to prove by induction on n that a nonzero polynomial always takes on some nonzero value. (Cf. exercise 4.6.1-16. However, this result is false for finite fields in place of the real numbers.) 28. The indeterminate quantities CY~, . . . , cyLl form an algebraic basis for the polynomial domain &[cyi, . . , cys], where & is the field of rational numbers. Since s + 1 is greater than the number of elements in a basis, the polynomials f3(01, . , a,) are algebraically dependent; this means that there is a nonzero polynomial g with rational coefficients such that g(fo(cui, . , as), . . . , fs(cyl,. , as)) is identically zero. 29. Given j8, . . . , j, E (0, 1, . . . , n}, there are nonzero polynomials with integer coefficients such that gj(qj,, . . . , qjt) = 0 for all (qn,. . . , qo) in Rj, 1 5 j 5 m. The product glg2 . ..g.isthereforezeroforall(q,,…,qo)inR1U…UR,.