PDF-ekiXkkfddfe Xkfi fkeldiXkfi XeefdeXkfi fkkeldiXkfi XeefdeXkfipk NXk VV e dgck
Author : cheryl-pisano | Published Date : 2014-10-19
ekiXkkfddfe Xkfi fkeldiXkfi XeefdeXkfi fkkeldiXkfi XeefdeXkfipk NXk VV e dgckfid Xkfif1 Xkfif1 fXe 1 INIATHESSON filnkkilXip
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ekiXkkfddfe Xkfi fkeldiXkfi XeefdeXkfi fkkeldiXkfi XeefdeXkfipk NXk VV e dgck: Transcript
ekiXkkfddfe Xkfi fkeldiXkfi XeefdeXkfi fkkeldiXkfi XeefdeXkfipk NXk VV e dgckfid Xkfif1 Xkfif1 fXe 1 INIATHESSON filnkkilXip. \Z`[`e^n_\k_\ikfi\Yl`c[fii\jli]XZ\X^i\\ej_flc[Y\YXj\[fej\m\iXcjk\gj+Ie\`jkf\mXclXk\k_\`ek\^i`kpf]k_\[iX`eX^\jpjk\d+B\i\(Xk\jk_fc\nXj[l^fek_\_`^_j`[\f]k_\glkk`e^^i\\eXe[nXk\inXjX[[\[[`i\Zkcpkf `ekXZk( 2ntana 2nwhereaisnotanintegermultipleof.7.Prove:1Xn=13n 1sin3a 3n=1 4(a sina):2NoTrig(notentirelystolenfromTitu97)1.GetaniceformulaforPnk=1k!(k2+k+1).Solution:Summandis(k+1)(k+1)! (k)(k)!.Soweget(n+1 ('( Mfij\ Ski\\k# NXk\ikfne MA ')+.) nnn%[\ fi -(. 0)-$'+0( for YOYO Man Nemo Concepcion e YoYo Man I fell under the yoyo’s spell on the rst really hot day of Spring in BinomialFormulas nXk=0nkxkyn k=(x+y)n (BinomialTheorem)nXk=0nk=2n(Lowersummation)nXm=kmk=n+1k+1(Uppersummation)nXk=0m+kk=n+m+1n(Parallelsummation)nXk=0nk2=2nn(Squaresummation)rXk=0mk k(k 1)=1+nXk=21 k 1 1 k=1+1 1 n2:So,thesequence(sn)ofpartialsumsisboundedinRandsinceitisalsoincreasing,itconverges.Example6.5.TheharmonicseriesP1=kdiverges.Indeed,ifsn=Pnk=11=k,thenjs2n snj=1 n+1+:: 2FXf=1NXk=1Pfkhck(DXi=1Cifvi)2 NXk=1bckhck(1)whereP2RFNisamatrixwithnon-positiveentries,Nisthenumberofhiddenunitsandbcisavectorofbi-ases.Eachtermintherstsumconsistsofatripletofvari- Figure2.Toyillus n=1 1 2+1 3 iseasilyseentobetheTaylorseriesexpansionoflog(1+x)evaluatedatx=1.Thatis,itssumislog2.Asweknow,thisseriesisconditionallyconvergent.Ifwetrytherearrangementabove(twooddtermsfollowedbyaneve 2FXf=1NXk=1Pfkhck(DXi=1Cifvi)2 NXk=1bckhck(1)whereP2RFNisamatrixwithnon-positiveentries,Nisthenumberofhiddenunitsandbcisavectorofbi-ases.Eachtermintherstsumconsistsofatripletofvari- Figure2.Toyillus 1�pforjpj1;wewouldget1Xk=0kpk�1=1 (1�p)2;1 and1Xk=0k2pk�1=p (1�p)20=1+p (1�p)3forjpj1.Finally,wehave1Xk=0k2pk=p(1+p) q3:Plugginginthisexpression,itfollowsthata0=�1&# (4)Problem14onpage60ofAxler.(Hint:youmightneedproblem3onpage59ofAxler.)Solution.FirstsupposethatTisinjective.DeneS0:rangeT VbyS0(Tv)=v.BecauseTisinjective,eachelementofrangeTcanberepre-sentedinth 1IntroductionWeightedessentiallynon-oscillatory(WENO)schemeshavebeenusedextensivelyinnu-mericalsolutionsofhyperbolicpartialdi erentialequations(PDEs)andotherconvectiondominatedproblems.The rstWENOsche Thm.[B] LetX1;X2;;Xkbeeigenvectorscorrespondingto distinct eigenvalues1;2;;kofA.ThenfX1;X2;;Xkgis linearlyindependent . Proof.AssumethatfX1;X2;&#
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