EFTA01155621.pdf

Moments of {x„} The purpose of this subsection is to see if something can be said about the ratio (Ek kx,)/(1k X0 when {xjn, is a non-zero solution to the system 0 = -q) n,1 Oarx - (TS, + di)x, 0 = q San., xn.1 - (qta,, + d„) x„. (1.28) with a suitable constant. To this end, introduce by way of notation; = I ,,A)a„x„. The equations in (1.28) can be used to derive two expressions for ;, these being • Xn = oisani+ do (110.iska (goatir+k do )Citaj Xj for n 2. • xy, — mani+ do (niska (qoacitak+kdk) )(1 q)S. (1.29) Note that 4) must be such that _q v oak - 1. q k=1 (Oak +dk ) (1.30) This last condition can be restated as saying that ins2(qOan+dn)x• = cic- (1.31) and therefore q; - (with, +di); + En.g, d„ x„ = q; (1.32) This tells us that E„,d„x„ =(qtal +dax, = (1 -q); , (1.33) where the left hand inequality comes via the n = 1 version of (1.29). What is written in (1.33) is of at least two identities involving `moments' of @ca. To elaborate, introduce a variable t and use (1.29) to see the equality between the following two formal series: ((clOan+d,,)x.) = En, t° (94)a„x,,) (1.35) Let Q(t) denote the series En, ta(qta„x„) and let p(t) denote En, Then (1.35) says that EFTA01155621  f l Q(t) + p(t) = (2(0+ (q. a +d 1)x1 . (1.36) This in turn can be rewritten using (1.33) as p(t) = (t- 1) Q(t) + -q); (1.37) Taking t = I on both sides recovers (1.33): E,„ dnxn = (1- q);. Differentiating once and setting t = 1 finds E.Irldnx„= C2(1)+0 -(05. (1.38) To go further, use (1.31) to see that Q(1)= - F.aIdn + q; + (qt al +di)x, =qS, (1.39) Granted this last equality, then (1.38) asserts that En:Indnx„ = ; . (1.40) This with (1.33) says that nd„x„ 1 Ea' d„x„ (1.q) (1.41) In the case d„ = d for all n, this asserts what is conjectured by Martin. Identities for `moments' of the form Eno.?d„ x„ for p ≥ 2 require knowing something of the (p-1)'st derivative of Q at t = 1. I don't know any good way to obtain these. EFTA01155622