Web hosting india - 572 ANSWERSTOEXERCISES 42.2 15. u < 2, implies

572 ANSWERSTOEXERCISES 42.2 15. u < 2, implies that (u @ u) @ 2 5 (u @ u) @ 2 5 (v @ v) @ 2, so the condition holds for all u and v iff it holds whenever ZL = v. For base b = 2, the condition is therefore always satisfied (barring overflow); but for b > 2 there are numbers v # w such that v $ v = w $ w, hence the condition fails. [On the other hand, the formula u CD ((v 8 4 0 2) does g ive a midpoint in the correct range. Proof: It suffices to show that 2~ + (v 8 u) @ 2 5 v, i.e., (v 8 u) @ 2 2 v -u; and it is easy to verify that round(+round(z)) 5 x for all x 2 0.1 16. (a) Exponent changes occur at C,, = 11.111111, Es, = 101.11111, Es,, = 1001.1102, c,,,, = 10001 .020, c9000, = 100000.91, ~soosls = 1000000.0; therefore = 1109099.1. c 1000000 (b) C l n. It suffices to find the coefficient of xl, since the coefficient of xk will be just the same except with all subscripts increased by k -1. Let (fk, gk) denote the coefficient of 21 in (Sk -ck, ck) respectively. Then fr = (~+rll)(~-~Y1-~l6l-~l~l–61a~-~lbl~l), 91 =(1+61)(1+rll)(yl+al+ylal), and fk = (1 -7kffk -bkck -ykbkok)fk-1 + (?k -77k + Yk6k + ykqk + ykbkqk + ykqkck + 6kvkak + ykbkqkgk)gk-1, gk = gk(l + yk)(l + bk)fk-1 -(1 + 6k)(-/k +

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