[Fourier-series傅里叶级数]例题01

Math462:HW5SolutionsDueonAugust15,2014JackyChong1JackyChongRemark:WeareworkinginthecontextofRiemannIntegrals.Problem15.1.4FindtheFouriercosineseriesofthefunctionsjsinxjintheinterval(;).Useittondthesums1Xn=114n21and1Xn=1(1)n4n21:Solution:Sincejsinxjisanevenfunctionon(;),thenithasaFouriercosineseriesgivenbyjsinxj12A0+1Xn=1AncosnxwheretheAnscanbereadilycomputedbytheintegralformulaAn=1Zjsinxjcosnxdx=1Z0sinxcosnxdx1Z0sinxcosnxdx=2Z0sinxcosnxdxConsiderthetrigonometricidentitysin[(n+1)x]sin[(n1)x]2=sinxcosnx;thenitfollowsAn=(41n21ifniseven0ifnisoddforalln2N.Moreover,itistrivialtocheckthatA0=4.HencetheFouriercosineseriesisgivenbyjsinxj24Xevenn1n21cosnx=241Xn=114n21cos2nx:AssumetheFouriercosineseriesconvergespointwisetojsinxjon(;),thenwehavethatjsinxj=241Xn=114n21cos2nx:Setx=0,wegetthat1Xn=114n21=12:Setx=2,wegetthat1=241Xn=1(1)n4n21or1Xn=1(1)n4n21=124:Page2of9JackyChongProblem1Problem25.2.11FindthefullFourierseriesofexon(l;l)initsrealandcomplexforms.Solution:ThecomplexformofthefullFourierseriesofexon(l;l)isgivenbyex1Xn=1cneinx=lwherecncanbereadilycomputeviathefollowingformulawhichyieldscn=12lZllexp1inlxdx=elinel+in2(lin):Henceitfollowsex1Xn=1elinel+in2(lin)einx=l=1Xn=1(1)nsinhll+inl2+n22einx=l:Therealformisgivenbyex1Xn=1(1)nsinhll+inl2+n22einx=l=sinhll+1Xn=1(1)nsinhll+inl2+n22einx=l+1Xn=1(1)nsinhll+inl2+n22einx=l=sinhll+1Xn=1(1)nsinhllinl2+n22einx=l+1Xn=1(1)nsinhll+inl2+n22einx=l=sinhll+1Xn=1(1)nsinhll2+n22l(einx=l+einx=l)neinx=leinx=li=sinhll+21Xn=1(1)nsinhll2+n22hlcosnxlnsinnxli:Problem35.3.10(TheGram-Schmidtorthogonalizationprocedure)IfX1;X2;:::isanysequence(niteorinnite)oflinearlyindependentvectorsinanyvectorspacewithaninnerproduct,itcanbereplacedbyasequenceoflinearcombinationsthataremutuallyorthogonal.Theideaisthatateachsteponesubtractsothecomponentsparalleltothepreviousvectors.Theprocedureisasfollows.First,weletZ1=X1=kX1k.Second,wedeneY2=X2(X2;Z1)Z1andZ2=Y2kY2k:Third,wedeneY3=X3(X3;Z2)Z2(X3;Z1)Z1andZ3=Y3kY3k;andsoon.(a)ShowthatallthevectorsZ1;Z2;Z3;:::areorthogonaltoeachother.(b)Applytheproceduretothepairoffunctionscosx+cos2xand3cosx4cos2xintheinterval(0;)toProblem3continuedonnextpage...Page3of9JackyChongProblem3(continued)getanorthogonalpair.Solution:(a)Weshallprovethestatementbystronginduction.Thebasecaseistriviallytrue.Now,supposeZ1;:::;Zk1aremutuallyorthogonal,i.e.(Zi;Zj)=(1ifi=j0otherwisefori=1;:::;k1andj=1;:::;k1.ConsiderZk=XkPk1i=1(Xk;Zi)ZikXkPk1i=1(Xk;Zi)Zikthenforax1lk1wehavethat(Zk;Zl)=(Xk;Zl)Pk1i=1(Xk;Zi)(Zi;Zl)kXkPk1i=1(Xk;Zi)Zik=(Xk;Zl)Pk1i=1(Xk;Zi)ilkXkPk1i=1(Xk;Zi)Zik=(Xk;Zl)(Xk;Zl)kXkPk1i=1(Xk;Zi)Zik=0:HenceZkisorthogonaltoallZlwherel=1;:::;k1.Moreover,wealsohavethat(Zk;Zk)=(XkPk1i=1(Xk;Zi)Zi;XkPk1i=1(Xk;Zi)Zi)kXkPk1i=1(Xk;Zi)Zik2=1:(b)Observekcosx+cos2xk22=Z0jcosx+cos2xj2dx=Z0cos2x+2cosxcos2x+cos22xdx=Z01+cos2x2+cos3x+cosx+1+cos4x2dx=Z01+cosx+12cos2x+cos3x+12cos4xdx=;thenitfollowsZ1=cosx+cos2xkcosx+cos2xk2=1p(cosx+cos2x):Next,observe(X2;Z1)=1pZ0(cosx+cos2x)(3cosx4cos2x)dx=1pZ03cos2xcosxcos2x4cos22xdx=12pZ01cosx+3cos2xcos3x4cos4xdx=p2Page4of9JackyChongProblem3whichmeansY2=X2(X2;Z1)Z1=72(cosxcos2x):ComputingthenormofY2yieldskY2k22=494Z0(cosxcos2x)2dx=494thenitfollowsZ2=Y2kY2k2=1p(cosxcos2x):Problem45.3.12ProveGreen'srstidentity:Foreverypairoffunctionsf(x);g(x)on(a;b),Zbaf00(x)g(x)dx=Zbaf0(x)g0(x)dx+f0gba:Solution:Assumef2C2[a;b]andg2C1[a;b],thenf0g2C1[a;b].Inparticular,wemayapplytheFundamentalTheoremofCalculusforRiemannIntegraltogetf0(x)g(x)ba=Zbadds(f0(s)g(s))0ds=Zbaf00(s)g(s)+f0(s)g0(s)ds:HenceitfollowsZbaf00(x)g(x)dx=f0(x)g(x)baZbaf0(x)g0(x)dx:Problem55.4.5Let(x)=0for0x1and(x)=1for1x3.(a)FindtherstfournonzerotermsofitsFouriercosineseriesexplicitly.(b)Foreachx(0x3),whatisthesumofthisseries?(c)Doesitconvergeto(x)intheL2sense?Why?(d)Putx=0tondthesum1+121415+17+18110111+


:Solution:(a)It'sclearthatA0=43.Next,observeAn=23Z31cosnx3dx=2nsinn3Page5of9JackyChongProblem5wheresinn3=8:0ifn0;3mod6p32ifn1;2mod6p32ifn4;5mod6:Hence,itfollowsAn=8:0ifn0;3mod6p3nifn1;2mod6p3nifn4;5mod6:ThisgivesusthefollowingFouriercosineseries(x)23p3cosx3p32cos2x3+p34cos4x3+p35cos5x3:::(b)ByTheorem4(ii),theFouriercosineseriesconvergespointwiseeverywhereonR.Moreover,letusextendtoanevenperiodicfunctionthenforeachxedx2[0;3]wehavethat12A0+1Xn=1Ancosnx3=12[ext(x+)+ext(x)]where12[ext(x+)+ext(x)]=8:0if0x112ifx=1;31if1x3(c)Since(x)isL2integrable,thenbyTheorem3weknowtheFouriercosineseriesdoesindeedconvergeintheL2senseto.(d)Setx=0yields0=23p3p32+p34+p35p37p38+:::or23p3=1+121415+17+18110111+


:Problem65.4.15Let(x)1for0x.Expand1=1Xn=0Bncosn+12x:(a)FindBn.(b)Let2x2.Forwhichsuchxdoesthisseriesconverges?Foreachsuchx,whatisthesumoftheseries?(c)ApplyParseval'sequalitytothisseries.Useittocalculatethesum1+132+152+


:Problem6conti