ABSTRACT
Proof. (11.1) The coefficients of Am(x) and Bm(x) in {13.3) and (13.4) are positive from (11.9). (17.2) These inequalities follow from {13.6), {17.1) and {13.5), {17.1), respectively. (11.3) Using Table 10.1, we have that 6o(x) = 1 ~ x = th(x). Also, since by {10.8), 6n(x) = dn-l(x) + (n-2)(n-1)6n-2(z) for n 2: 2 and x 2: 1, then (17.3) holds using this reduction repeatedly. Form 2: 3 and x 2: 1, then 6n(x) > x6n-t(x) 2: 6n-l(x). Thus, cSm(x) < 6n(x) for x 2: 1 and
and
(17.10)
(17.11) ~n+2(x) = ((h(x) + 8n2)~n(x) - 4n(n-1)(2n -1)2 ~n-2(x), n ~ 1.
