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2.7. FILTERING 71To summarize at this point, for sufficiently large number N of theseunderlying periodic components, we can represent a time-series q t asq t =NXj=1cos(ω j t)u j − sin(ω j t)v j , (2.104)where u j = a j cos(λ j )andv j = a j sin(λ j ), E(u 2 i )=σ2 i ,E(u iu j )=0,i 6= j, E(vi 2 )=σi 2 ,E(v i v j )=0,i6= j.Now suppose that E(u i v j ) = 0 for all i, j and let N →∞. 30 Youare carving the interval into successively more subintervals and arecramming more ω j into the interval [0, π]. Since each u j and v j isassociated with an ω j , in the limit, write u(ω) andv(ω) as functionsof ω. For future reference, notice that because cos(−a) =cos(a), wehave u(−ω) = u(ω) whereas because sin(−a) = − sin(a), you havev(−ω) =−v(ω). The limit of sums of the areas in these intervals is theintegralZ πq t = cos(ωt)du(ω) − sin(ωt)dv(ω). (2.105)0Using (2.103), (2.105) can be represented asq t =Z π0e iωt + e −iωtdu(ω) −2Z πe iωt − e −iωtdv(ω) . (2.106)2i0| {z }(a)Let dz(ω) = 1 [du(ω)+idv(ω)]. The second integral labeled (a) canbe2simpliÞed asZ π0e iωt − e −iωtdv(ω) =2i==Z π0Z π0Z π0⇐(49)e iωt − e −iωt à !2dz(ω) − du(ω)2iie −iωt − e iωt(2dz(ω) − du(ω))2Z π(e −iωt − e iωt e iωt − e −iωt)dz(ω)+du(ω).0 2Substitute this last result back into (2.106) and cancel terms to get30 This is in fact not true because E(u i v i ) 6= 0, but as we let N → ∞, theimportance of these terms become negligible.⇐(50)