588 ANSWERS TO EXERCISES 4.3.3 Another possibility would (Web hosting domains)
588 ANSWERS TO EXERCISES 4.3.3 Another possibility would perhaps be to evaluate W, and W, at 12, 22, 42, . . . , (2 ) ; although the numbers involved are larger, the calculations are faster, since all multiplications are replaced by shifting and all divisions are by binary numbers of the form 2J(2k -1). (Simple procedures are available for dividing by such numbers.) 5. Start the q and r sequences out with qo and q1 large enough so that the inequality in exercise 3 is valid. Then we will find in the formulas like those preceding Theorem C that we have ~1 -+ 0 and r]2 = (1-/-1/(2~))2~+~-~ (Qk/Qk+i). The factor &k/&k+1 -+ 1 as k + 00, so we can ignore it if we want to show that 772 < 1-c for all large k. Now d= = d26.?k + 2[1/Z&l+ 2 > d(2Qk f 2a + 1) + 1 2 J2&;E + 1 + l/(3&). Hence 77s 5 (1 + 1/(2~))2-~/(~~~), and lgns < 0 for large enough k. Note: Algorithm C can also be modified to define a sequence qo, 41, . . of a similar type that is based on n, so that n zz qk + qkfi after step Cl. This modification leads to the estimate (19). 6. Any common divisor of 6q + di and 6q + d2 must also divide their difference d2 -dl. The (t) d ff1 erences are 2, 3, 4, 6, 8, 1, 2, 4, 6, 1, 3, 5, 2, 4, 2, so we must only show that at most one of the given numbers is divisible by each of the primes 2, 3, 5. Clearly only 6q + 2 is even, and only 6q + 3 is a multiple of 3; and there is at most one IIdtipk Of 5, Since qk $ 3 (modulo 5). 7. Let Pk-1 < N 5 pk. We have tk 2 6tk--1 +ck3k for some constant c; so tk/6k 5 tk-1/6k-1 + ck/2k 5 to + c .&1(j/23) = M. Thus tk 5 M . 6k = O(P; ~~). 8. Let 2k be the smallest power of 2 that exceeds 2K. Set at + W-t2/2ut and bt + w(~~- -~) ~, where Ut = 0 for t 2 K. We want to calculate the convolutions CT -a b,-, for r = 2K -2 -s, when 0 5 s < K. The convolutions -c OjJ