Make web site - 570 ANSWERSTOEXERCISES 4.2.1 LD2 TEMP(EXP) r12 t e,.
570 ANSWERSTOEXERCISES 4.2.1 LD2 TEMP(EXP) r12 t e,. DEC2 Q J2NP ++3 SLA 0.2 Remove integer part of U. ENT2 0 JANN l.F ENN2 0.2 Fraction is negative: find SRAX 0,2 its complement. ENT2 0 JAZ ++2 INCA 1 ADD WMl Add word size minus one. 1H INC2 Q Prepare to normalize the answer. JMP NORM Normalize, round, and exit. 8H EQU l(l:l) WMl CON 8B-1,8B-l(1: 4) Word size minus one 1 16. If Jc] 2 Id], then set r + d @ c, s + c @ (r @I d); x +-(a @ (b @ r)) @ s, y t (bG(a@r))0s. 0th erwise set r +-c@d, s t $@(T@c); z t ((a@r)$b)@s, y + ((b@r)@a)@s. Then x+iy is the desired approximation to (a+bi)/(c+di). [CACM 5 (1963) 435. Other algorithms for complex arithmetic and function evaluation are given by P. Wynn, BIT 2 (1962), 232-255; see also Paul Friedland, CACM 10 (1967), 665.1 17. See Robert Morris, IEEE Transactions C-20 (1971) 1578-1579. Error analysis is more difficult with such systems, so interval arithmetic is correspondingly more desirable. 18. For positive numbers: shift fraction left until fi = 1, then round, then if the fraction is zero (rounding overflow) shift it right again. For negative numbers: shift fraction left until fi = 0, then round, then if the fraction is zero (rounding underflow) shift it right again. 19. (43f(l if e, < e,)-(1 if fraction overflow)-(10 if result zero)+(4 if magnitude is rounded up)+(l if first rounding digit is b/2)+(5 if rounding digits are b/20.. .0)+(7 if rounding overflow) + 7N + A(-1 + (11 if N > O)))u, where N is the number of left shifts during normalization and A = 1 if rX receives nonzero digits (otherwise A = 0). The maximum time of 73~ occurs for example when v. = +50 01000000, v == -4649999999, b = 100. [The average time, considering the data in Section 4.2.4, will be about 4512~1 SECTION 4.2.2 1.u~v=u$-w=-v~u=-(v$–2L)=-(w~u). 2. u @ x 2 u @ 0 = u, by (8), (2), (6); h ence by (8) again, (U @ x) $ v 2 u $ v. Similarly, (8) and (6) together with (2) imply that (U @ x) $ (V $ y) 2 (U @ x) @ v. 3. u = 8.0000001, v = 1.2500008, w = 8.0000008; (u @ v) @ w = 80.000064, u @ (w @ m) = 80.000057.