246 ARITHMETIC 4.2.4 It is not difficult to (Remote web server)
246 ARITHMETIC 4.2.4 It is not difficult to solve the recurrence formula (11) for R, : We have Z&(s) = -1, RI(s) = -1 + T/S, Rz(s) = -1 + (r/s)(l + ln(s/r)), and in general For the stated range of s, this converges uniformly to -1 + (T/s) exp(ln(s/r)) = 0. The recurrence (11) for Qm takes the form &m(s) = ;(cm + 1 + 1 &m-l(t) ,t), (20) where cm=-1 9(l QmAl(t) dt + I &-1(t) dt) - -(21) The solution to recurrence (20) is easily found by trying out the first few cases and guessing at a formula that can be proved by induction; we find that c,+Ac,-ihis+…+ . (22) > It remains for us to calculate the coefficients c,, which by (19), (21), and (22) satisfy the relations Cl = (T -10)/9; 1 9( In 10 + $c,–i(ln 1O)2 + . . . + -ci(ln 10) (23) Gn+1=-Gn +~~~+~ln~+…+~~n~)..)-~O~. This sequence appears at first to be very complicated, but actually we can analyze it without difficulty with the help of generating functions. Let C(z) = Cl2 + c2.2 + c3z3 + . . . ; then since 10 = 1 + z In 10 + (1/2!)(2 In 1O)2 + . . . , we deduce that 1 9 Gn+1= yp+1+-c??%+1 10 = h cn+i +c,lnlO+~~~+~cl(lnlO)m ( m! > +k l+…+-$ lny m -1 ( .( >>