01037 Uge 10Preben Alsholm, 13/11 2006restart;libname:="g:/SigLinSys/lib01037",libname;with(SigLinSys);Version();Opgave 202 (vii)For god ordens skyld bringer vi besvarelsen af hele opgaven.f1:=x->x^2/4-Pi*x/2;f:=PeriodiskUdvidelse(f1,0..2*Pi);plot(f(x),x=-2*Pi..4*Pi,scaling=constrained);f(-x) assuming x<2*Pi,x>0;simplify(%);f(x) assuming x>=0,x<2*Pi;index = n sikrer brug af globalt n.S:=FourierR\303\246kke(f1(x),x=0..2*Pi,index=n);f1(0)=eval(S,x=0);BeskrivProc(Vurdering);Vurdering(x->x^(-2),N,N+1) assuming N::posint;value(%);FourierR\303\246kke(f1(x),x=0..2*Pi,antal=10,output=animation,plotoptions=[x=-2*Pi..2*Pi]);F:=FourierR\303\246kke(f1(x),x=0..2*Pi,antal=100):plot(F-f1(x),x=0..2*Pi);Sp\303\270rgsm\303\245l (vii): Parsevals s\303\246tning.Fourierr\303\246kken:S;a0:=op(1,S)*2;an:=op([2,1,1],S);Vi har if\303\270lge Parsevals s\303\246tning1/2/Pi*Int(f1(x)^2,x=0..2*Pi)=1/4*a0^2+1/2*Sum(an^2,n=1..infinity);applyop(value,1,%);2*(rhs-lhs)(%)=0;Maple-kontrol:Sum(an^2,n=1..infinity): %=value(%);Sp\303\270rgsm\303\245l (viii) ved ledvis integration:Int(S,x=0..X);expand(%);Ombyt(%,[Int,Sum]);value(%);s,r:=selectremove(has,%,sin);Int(f1(x),x=0..X): %=value(%);f2:=rhs(%)-r;Kontrol:FourierR\303\246kke(f2,X=0..2*Pi);Opgave 214 (iv)For god ordens skyld bringer vi besvarelsen af hele opgaven.f:=t->abs(sin(t));F:=FourierR\303\246kke(f(x),x=0..Pi,index=n);Da f er kontinuert og stykkevist differentiabel, er Fourierr\303\246kken uniformt konvergent med sum f(x). f(x)=F;eval(%,x=0);expand(-Pi/4*%);%+(1/2=1/2);Sp\303\270rgsm\303\245l (iv): Parsevals s\303\246tning.Fourierr\303\246kken:F;a0:=op(1,F)*2;an:=eval(op([2,1],F),x=0);Vi har if\303\270lge Parsevals s\303\246tning1/Pi*Int(f(x)^2,x=0..Pi)=1/4*a0^2+1/2*Sum(an^2,n=1..infinity);applyop(value,1,%);expand(Pi^2/8*(rhs-lhs)(%))=0;Maple-kontrol:GangInd(Pi^2/16*Sum(an^2,n=1..infinity)): %=value(%);Fourierr\303\246kken p\303\245 kompleks form:convert(F,exp);combine(%);res1:=subs(exp(-2*I*n*x)=0,%);res2:=subs(exp(2*I*n*x)=0,2/Pi=0,%%);res3:=SkiftVariabel(res2,n=-n);res1+res3;Dette kan samlet skrivessubsop(2=(n=-infinity..infinity),res3);simplify(%);F;abs(f(x)-applyop(Afsnit,2,F,N));Dette er mindre end eller ligR:=Sum(4/Pi/(4*n^2-1),n=N+1..infinity);som er lig med Sum(4/Pi/(4*n^2-1),n=1..infinity)-Sum(4/Pi/(4*n^2-1),n=1..N);Men f\303\270rste led er fundet i sp\303\270rgsm\303\245l (iii):R:=applyop(value,1,%);seq(evalf(eval(R,N=k)),k=1..5);Vi ser, at N = 3 er nok.F;F3:=FourierR\303\246kke(f(x),x=0..Pi,antal=3);plot(f(x)-F3,x=0..Pi);BeskrivProc(FunkAnimateSum);p1,p2:=FunkAnimateSum(-4*cos(2*n*x)/Pi/(4*n^2-1),n=1..10,x=0..Pi,eps=.1,sumfunk=f(x)-2/Pi):p1;p2;Opgave HermiteL:=seq(simplify(HermiteH(k,x)),k=0..5);plot(expand(exp(-x^2/2)*[L]),x=-1..1);HermiteH(4,x);simplify(%);Int(exp(-x^2)*HermiteH(7,x)*HermiteH(3,x),x=-infinity..infinity);value(%);printlevel:=2:for k from 0 to 5 do
for n from 0 to k do
s[k,n]:=int(exp(-x^2)*HermiteH(k,x)*HermiteH(n,x),x=-infinity..infinity)
end do
end do;seq(s[k,k],k=0..5);map(ifactor,expand([%]/sqrt(Pi)));seq(sqrt(Pi)*2^n*n!,n=0..5);Ekstraf:=x->exp(x);an:=n->if not type(n,integer)
then 'procname'(n)
else
evalf(simplify(Int(f(t)*HermiteH(n,t)*exp(-t^2),t=-infinity..infinity)/(sqrt(Pi)*2^n*n!)*HermiteH(n,x)))
end if;n:='n':p1,p2:=FunkAnimateSum(an(n),n=0..5,x=-2..2,eps=.1,sumfunk=f(x));p1;p2;Opgave LegendreL:=seq(simplify(LegendreP(k,x)),k=0..5);plot([L],x=-1..1);LegendreP(4,x);simplify(%);Int(LegendreP(7,x)*LegendreP(3,x),x=-1..1);value(%);printlevel:=2:for k from 0 to 5 do
for n from 0 to k do
s[k,n]:=int(LegendreP(k,x)*LegendreP(n,x),x=-1..1)
end do
end do;seq(s[k,k],k=0..5);seq(2/(2*n+1),n=0..5);n:='n':S:=Sum(Int(f(t)*LegendreP(n,t),t=-1..1)/Int(LegendreP(n,t)^2,t=-1..1)*LegendreP(n,x),n=0..infinity);SN:=eval(S,infinity=N);f:=x->sin(Pi*x);an:=n->if not type(n,integer)
then 'procname'(n)
else
evalf(simplify(Int(f(t)*LegendreP(n,t),t=-1..1)/(2/(2*n+1))*LegendreP(n,x)))
end if;int((f(x)-add(an(k),k=1..N+1))^2,x=-1..1);p1,p2:=FunkAnimateSum(an(n),n=0..5,x=-1..1,eps=.1,sumfunk=f(x));p1;p2;Opgave 226f:=t->piecewise(t<=Pi,sin(t),0);F:=FourierR\303\246kke(f(x),x=0..2*Pi);fu:=PeriodiskUdvidelse(f,0..2*Pi);plot(fu(t),t=-2*Pi..4*Pi);N:='N': n:='n': k:='k':Sp\303\270rgsm\303\245l (ii). Vi bruger (6.9) p. 124:abs('f'(x)-'SN'(x))<=1/sqrt(N*Pi)*sqrt(Int('D(f)'(t)^2,t=0..2*Pi));1/sqrt(N*Pi)*sqrt(Int(D(f)(t)^2,t=0..2*Pi)): %=value(%);solve(rhs(%)=.1,N);Da f er kontinuert og stykkevist differentiabel, er Fourierr\303\246kken uniformt konvergent med sum f(x). Fourierr\303\246kken erFourierR\303\246kke(f(t),t=0..2*Pi,checkantal=1);SkiftVariabel(%,n=2*k);F:=map(simplify,%);SN:=subs(infinity=N,F);Resultatet (iv) f\303\270lger af at1/(2*n-1)-1/(2*n+1): %=normal(%);Summen teleskoperer, s\303\245Sum(1/(2*n-1)/(2*n+1),n=N+1..infinity): %=value(%);abs('f'(x)-'SN'(x))<=Sum(1/(2*n-1)/(2*n+1),n=N+1..infinity);solve(value(rhs(%))=.1,N);I animationen b\303\270r bem\303\246rkes, at alle led medatges ogs\303\245 de lige (der er 0).FourierR\303\246kke(f(x),x=0..2*Pi,antal=50,output=animation);