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Our exposition for this entire exercise is based on Greene.docx

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92.Our exposition for this entire exercise is based on Greene [3.35].(a) Define a matrixM= [M(s,t)] by settingM(s,t) =ζ(s,t)f(s,t).Clearly if we order the rows and columns ofMby some linearextension ofL, thenMis triangular and detM= Qs f(s,s). On theother hand (writingζfor the matrix of theζ-function ofLwithrespect to the basisL, that is,ζis the incidence matrix of therelationL),Mtζ= "Xf(u,s)ζ(u,s)ζ(u,t)#us,tL= Xf(u,s)# = [F(st),s].usts,tLThus det[F(st,s)] = detMtζ= detM.This formula is a result of B. Lindstr¨om,Proc. Amer. Math. Soc.20(1969), 207– 208, and (in the case whereF(s,v) dependsonly ons) H. S. Wilf,Bull. Amer. Math. Soc.74(1968), 960–964."
(b) TakeLto be the set [n] ordered by divisibility, and letf(s,v) =φ(s)(soF(s,v) =s). For a proof from scratch, see G. P´olya and G. Szeg¨o,Problems and Theorems in Analysis II, Springer-Verlag,Berlin/Heidelberg/New York, 1976 (Part VIII, Ch. 1, no. 33).(c) Whenf(s,v) =µ(ˆ0,s) we have (suppressingv).Hence the matrixR= [F(st)] is just the incidence matrix ofthe relationst= ˆ0. By (a), detR= 06 . Hence some term inthe expansion of detRmust be nonzero, and this term yields
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