Theoretical-Numerical-Analysis-3-rd--部分习题答案

homeworkforthescienticcomputationDanyiHe31101013001Approximationofcontinuousfunctionsbypoly-nomialsExercise1.1.3.1.1ProveCorollary3.1.3byapplyingtheorem3.1.2Proof.onlyneedtoverifytheconditionoftheorem3.1.2(a)constantfunctionsarepolynomials.(b)ifu,varepolynomial,thenu
vispolynomial.(c)foreachpairofpointsx,y2D,x̸=y,considerpolinomialv(x)=x,thenv(x)̸=v(y).sothesetofallpolynomialsisdenseinC(D).Exercise1.2.3.1.2ProveCorollary3.1.4byapplyingTheorem3.1.2Proof.thebasistrigonometricfunctionssin(t);cos(t)arecontinuousin[;],bythepropertiesofcontinuousfunction,thetrigonometricpolynomialsarecontinuousin[;].onlytoverifythethreeconditionsintheorem3.1.2(a)(b)issimple,(c)considersin(t);cos(t),atleastoneofthetwofunctionwillsatisfyconditionthree.forifsin(x)=sin(y);cos(x)=cos(y),wehavetan(x)=tan(y),thenx=y+=2,whichcontradictswithsin(x)=sin(y);cos(x)=cos(y).Exercise1.3.3.1.3ProveCorollary3.1.4byapplyingCorollary3.1.3Proof.applycorollary3.1.forunitcircleD,forDiscompactsetinR2,wehavethesetofpolynomialsinDisdenseinC(D).butinD,Pn(x;y)=∑nm=0∑i+j=maijxiyjistrigonometricpolynomialsifwechangex=sin(t);y=cos(t).andthe\mapisbijection.sothesetofalltrigonometricpolynomialsisdenseinC(D).Exercise1.4.3.1.4LetDbeacompactsetinRd.Assumef2C(D)issuchthat∫Df(x)xdx=0foranymuti-index=(1;


;d).thenf(x)=0forx2D.Proof.bythelinearityofintegral,wehavef;p=0;8polynomialsinD,since:;:hereisdenedfromtheintegraloverD,andthesetofpolynomialsoverDisdenseinC(D),bythecontinuityofintegral,wehavef;g=0;8g2C(D).takeg=f,wehavef=0.1Exercise1.5.3.1.5Showthateverycontinuousfunctionfdenedon[0;1)withthepropertylimx!1f(x)=0canbeapproximatedbyasequenceoffunc-tionsoftheformqn(x)=n∑j=1cn;jejaxwherea0isanyxednumber,andcn;jareconstants.Proof.usehint.considerg(t)=f(logt=a);0t1,usethepropertylimx!1f(x)=0,wemaydefg(0)=0,itiswell-denedtokeepcontinuity.forg(t),bythm3.1.1,wehaveapproximationfunctionspn(x)=∑nj=1cn;jxj,transferfunctiongandpn(x)=qn(logt=a),wewillgettheconclusion.Exercise1.6.3.1.6Assumef2C[1;1]isanevenfunctionisanevenfunc-tion.Showthatbeuniformlyapproximatedon[1;1]byasequenceofpolyno-mialsoftheformpn(x2).Proof.usehint.considerg(x)=f(px);x2[0;1],forfisanevenfunction,giswell-denedandcontinuous.forg,wehaveuniformlyapproximationfunctionspn(x).wetransferbothg;qn,wherepn(x)=qn(sqrtx),soqn(x)=pn(x2)isaapproximationoff.Exercise1.7.3.1.7Letf2Cm[a;b];m0.Showthatthereisasequenceofpolynomialsfpngn1suchthat∥fpn∥Cm[a;b]!0asn!1Proof.wemayapproximatef(m)rst.saywehave∥p(m)nf(m)∥ϵ,wemayintegralp(m)nandaddascalartogetp(m1)nsuchthatp(m1)n(a)=f(m1)(a).wehave∥p(m1)nf(m1)∥ϵ
(ba).wedoittilltogetpn(a)=f(a),andwehave∥pnf∥ϵ
(ba)m,so∥fpn∥Cm[a;b]ϵ
M,M=(ba)+:::+(ba)misaconstant.sowegettheapproximation.Exercise1.8.3.1.8LetΩRdbeadomain,andf2Cm(Ω);m0.Showthatthereisasequenceofpolynomialsfpngn1suchthat∥fpn∥Cm(overlineΩ)!0asn!1Proof.similartoexercise3.1.7.fortherststep,wemayapplycorollary3.1.3togettheapproximationofderivativef(m),andthenintegralandconstructasbefore.2InterpolationtheoryExercise2.1.3.2.1Showthatthereisauniquequadraticfunctionp2satisfyingthecongditionp2(0)=a0;p2(1)=a1;∫10P2(x)dx=a2withgivena0;a1;anda:Proof.supposep2(x)=ax2+bx+c,thenwehavec=a0a+b+c=a11=3
a+1=2
b+c=athedeterminationofthematrixisnonzero.Exercise2.2.3.2.2GivenafunctionfonC[a;b],themomentproblemistondpn2Pn(a;b)suchthat∫abxipn(x)dx=∫abxif(x)dx;0in:showthattheproblemhasauniquesolutionandthesolutionpnsatises∫ab[f(x)pn(x)]2dx∫ab[f(x)q(x)]2dx8q2Pn(a;b):Proof.supposepn(x)=a0+a1x+


+anxn.letbi=∫baxif(x)dx;Aij=∫baxixjdx=xi;xj,thenweneedonlytosolveanequation264x0;x0


x0;xn.........xn;x0


xn;xn3750B@a0...an1CA=0B@b0...bn1CAweneedtoprovethatthedeterminantofthematrixisnonzero.supposethereisC=(c0;


;cn)TsuchthatAC=0,thenbythelinearityofinnerproduct,wehave∑ni=0ci
xi;xi=0;j=0;1;:::;n,whichimplies∑ni=0ci
xi;∑ni=0ci
xi=0.soci=0.thedeterminantofAisnonzero.wehaveuniquesolutionfortheequation.thesecondpartisthroughthe(innerproduct)norminequality.bylinearityofintegral,wehavepnf;q=0;8q2Pn(a;b),especiallyforpnf;pnq=0.then∥fq∥=∥fpn+pnq∥=∥fpn∥+∥pnq∥thenormisnonnegative,sowegettheinequality.Exercise2.3.3.2.3Letx0x1x2bethreerealnumbers.Considerndingapolynomialp(x)ofdegree3forwhichp(x0)=y0;p(x2)=y2p′(x1)=y′1;p′′(x1)=y′′1withgivendatafy0;y2;y′1;y′′1g.Showthereexistsauniquesuchpolynomial.3Proof.thepolynomialisuniquecanbeobtainedfromthefundamentaltheoremfoalgebra.weprovetheexistencebyconstructingone.Taylorexpansethepolynomialatx1,wemaywritep(x)=a0+y′1(xx1)+1=2
y′′(xx1)2+a3x3.usep(x0)=y0;p(x2)=y2,wegettheequationofa0;a3a0+a3
x30=b1;a0+a3
x32=b2b1;b2isconstant.andthedeterminantofthematrixisnonzeroforx0̸=x2.sotheexistenceisproved,actuallytheuniquenesscanalsobeobtained.Exercise2.4.3.2.4Derivetheformula(3.2.2)fortheVandermondedetermi-nantofordern+1.Proof.usethehinttodoso.IntroduceVn(x)=det26666641x0x20


xn01x1x21


xn1...............1xn1x2n1


xnn11xx2


xn3777775wehavethethingsbelow:det26666641x0x20


xn01x1x21


xn1...............1xn1x2n1


xnn11xx2


xn3777775=d