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LhAncosnc LtBnsinnc LtiThereforethenaturalfrequenciesintimearencLforn123ForpartbweshowedinclassthatthewaveequationwithoneendxedandoneendfreeyieldedthefollowingsolutionSeethec. page40110SOR201(2002)(ii)Bernoullir.v.{ifp1=p;p0=1p=q;pk=0;k6=0or1,thenGX(s)=E(sX)=q+ps:(3:4)(iii)Geometricr.v.{ifpk=pqk1;k=1;2;:::;q=1p;thenGX(s)=ps1qsifjsjq1(seeHWSheet4.):(3:5)(iv)Binomialr.v. x2eR2x=(1+x2)dxdx=xZ1 x21 1+x2dx=(1+xtan1x)Thus,twoLIsolutionsareY1=xandY2=1+xtan1x)Herep(x)=2x=(1+x2)andq(x)=2=(1+x2)areanalyticatx=0withcommonradiusofconvergenceR=1.Lety(x)=1Xn=0cnxn:Nowusin 2ntana 2nwhereaisnotanintegermultipleof.7.Prove:1Xn=13n1sin3a 3n=1 4(asina):2NoTrig(notentirelystolenfromTitu97)1.GetaniceformulaforPnk=1k!(k2+k+1).Solution:Summandis(k+1)(k+1)!(k)(k)!.Soweget(n+1 {z }b=0BBBBBB@x2x1x3x2......xN1xN21CCCCCCA| {z }A0@121A| {z }:Unlikethepreviousp=1case,tryingtoexpressthesolution^=ATA1ATbanalyticallyisnottrivial.WestartwithATA1=266640@x2x3


xN1x1x2


andthemapF:`2(ZNZN)!`2(ZNZN)givenbyF(I(n;m))0n;m=bI(n;m)0n;misthesocalleddiscreteFouriertransform(DFT).ThismapisrealizedbytheformulabI(k;l)=hI;Expk;li kExpk;lk2=1 N2N1Xn=0N1Xm=0I(n;m)e2i(k 5.Sincejrj1,thisseriesdiverges.6.nn! (n+2)!oWerstsimplify:n! (n+2)!=1 (n+1)(n+2)sothelimitasn!1is0.7.1Xn=1lnn n+1Alittletricky...First,notethatthiscanbewrittenasln(n)ln(n+1).Now,let'swriteoutthen 5.(Thecomplexexponential.)Denef:C!Cbyf(z)=ez=1Xn=0zn n!:Here,ifz=x+iy,thenez=exeiy=ex(cosy+isiny):Thefactthatfisahomomorphismfollowsfromtheidentityez1+z2=ez1ez2:Thecomplexexponentialissurjective:eve u(32;t)t642320Figure2:Sketchofu(32;t)for0t6.Notethatuiscontinuous,bututisnot.Forpart(a)x=,y=,f()=e2Forpart(b)x=,y=,f()=1.Integrateupw.r.tsusingtheI.C.'swheres=0istheinitialcurve:Inbo Mechanisms to Mitigate Unfairness. Krishna P. . Gummadi. Joint work with . Muhammad . Bilal . Zafar. , Isabel Valera, . Manuel . Gomez-Rodriguez. Max Planck Institute for Software Systems. Context: Machine decision . The Epistles of John. Prepared by Bro. Ted Hodge Jr.. (Shippensburg Bible School 2017). Class 2 Agenda. “God is light. . and in him is no darkness. . at all”. The message of light and darkness. 2+1 31 4+1 51 6:::=1Xn=1(1)n+11 n=ln2;(8.1)convergeveryslowly.Thesameistrueforabsolutelyconvergentseries,suchas1Xn=11 n2=(2)=2 6:(8.2)IfwecallthepartialsumforthelatterNXn=11 n2=SN;(8.3)thedieren 1 2FIRSTTHINGSFIRST(5)Duringclasstheinstructorhasthenaldecisionondeterminingwhetheranar-gumentmaystandornot.Hisverdictmaystillbechallengedafteraproofis\published"(seerule(6)).(6)Ifsomeoneothertha ngettinglargewhenngettinggreater.So,limx!1sin(2n) 1+p n=0.Thesequenceconvergeto0.2.Accordingtothequestion,wegetlimx!1p x2�1=Landlimx!�1p x2�1=L.Thenwecansolvethem.Wegetlimx!1p x2�1=xandlim f2xn0xn01f3xnfioforder204Invariant110K111x2n1x2n120K121xn1xnxn1xn0K1x2n1x2n150K151xn1xn160K161xn1xn220K2210KintegrationconstantSolutionuptohomographyellipticfunctionssampledoverequidistantpoints2xnxn1