Affordable web design - 4.6.4 ANSWERS TO EXERCISES 645 that calculates a
4.6.4 ANSWERS TO EXERCISES 645 that calculates a fourth-degree polynomial has the form x1= &+x0 x2 = 02 +x0 x3 =x1x x2 x4 = ara+X3 x5 =04+x3 x6 =x4 x x5 x7 = a5+A6 Actually this chain has one addition too many, but any correct scheme can be put into this form if we restrict some of the o s to be functions of the others. Now X7 has the form (x2 +Az+B)(z2+Az+C)+D = x4+2Az3+(E+A2)x2+EAx+F, where A=cul+crz,B=crlcrz$ff~g,C=crlcrz+ar,D=(rs,E=B$C,F=BC+D; and since this involves only three independent parameters it cannot represent a general manic fourth-degree polynomial. 36. As in the solution to exercise 35, we may assume that the chain computes a general manic polynomial of degree six, using only three chain multiplications and six parameter additions. The computation must take one of two general forms x1= &+x0 x1= %+x0 x2 = orz+xo x2 = @2+X0 x3 =x1 x x2 x3 =x1 x x2 x4 =aa+Xo x4 = as+X3 x5 =ffr+XB x5 =cy4+x3 x6 =x4 x x5 x6=x4 x x5 x7 =aS+x6 x7 = a5 +X3 b3 =a6+A6 &I = a6+i6 x9 =x7 x x6 19 =x7 x x&3 AlO =&7+x9 110 =07+x9 where, as in exercise 35, an extra addition has been inserted to cover a more general case. Neither of these schemes can calculate a general sixth-degree manic polynomial, since the first case is a polynomial of the form and the second case (cf. exercise 35) is a polynomial of the form (x4 + 2Az3 + (E + A2)x2 + EAx + F)(x2 + Az + G) + H; both of these involve only five independent parameters. 37. Let u[Ol(x) = unxn + ~~-~x~–l + . . . + uc, v[ ](z) = zn + 21n-15n-1 f.. . + ~0. For 1 2 j 5 n, divide U[J- l(z) by the manic polynomial v(~-~](x), obtaining u[ - l(z) = (y3vIwl (x) + pjv[jl(x). Assume that a manic polynomial v[~](x) of degree n - j exists satisfying this relation; this will be true for almost all rational functions. Let u[jl(x) = ~[j+ l(x) -XW~~~(X). These definitions imply that deg(uln]) < 1, so we may let on+1 = u( ](z).