01037 Uge 6Preben Alsholm, 24/10 2006Opgave 110Sum((n+2)/(n+1)*x^n,n=0..infinity);Konvergensradius findes ved kvotientkriteriet til 1.Maple finder v\303\246rdien tilvalue(%);Ved h\303\245ndkraft deler vi op i to r\303\246kker, der begge har konvergensradius 1:Sum(x^n,n=0..infinity)+Sum(1/(n+1)*x^n,n=0..infinity);Den f\303\270rste er en kvotientr\303\246kke, den anden:Diff(x*Sum(1/(n+1)*x^n,n=0..infinity),x)=simplify(diff(Sum(1/(n+1)*x^(n+1),n=0..infinity),x));Opgave 112S:=Sum(x^n,n=0..infinity);simplify(applyop(diff,1,S,x))=diff(value(S),x);Af denne f\303\245s ved omordning:Sum(n*x^n,n=1..infinity): res1:=%=value(%);simplify(applyop(diff,1,lhs(res1),x))=normal(diff(rhs(res1),x));s\303\245 ved multiplikation med x f\303\245s:Sum(n^2*x^n,n=1..infinity): res2:=%=value(%);eval(res1,x=1/2);eval(res2,x=1/2);Opgave 115Sum(x^n,n=0..infinity): %=value(%);Sum((-x)^n,n=0..infinity): %=value(%);Sum((-1)^n*x^(2*n),n=0..infinity): %=value(%);Opgave 116Sum((-1)^n*(2*n+2)/(2*n+1)!*x^(2*n+1),n=0..infinity);value(%);Kvotientkriteriet:a:=n->n*(2*n+2)/(2*n+1)!*x^(2*n+1);a(n+1)/a(n);simplify(%);limit(%,n=infinity);S\303\245 konvergensradius er uendelig.Opgave 121S:=Sum((n+1)/n/n!*x^n,n=1..infinity);value(%);Kvotientkriteriet:a:=unapply(op(1,S),n);a(n+1)/a(n);simplify(%);limit(%,n=infinity);Konvergensradius er uendelig.simplify(applyop(diff,1,S,x));value(%);Opgave 134S:=Sum(x^n/n/2^n,n=1..infinity);value(%);Kvotientkriteriet:a:=unapply(op(1,S),n);a(n+1)/a(n);simplify(%);limit(%,n=infinity);Konvergensradius er 2.Opgave 135Sum(sin(n)/n^3*(-1)^n,n=1..infinity);Absolut konvergent, daSum(1/n^3,n=1..infinity);er konvergent.Opgave 136Sum(2^n/n*x^n,n=1..infinity);value(%);a:=unapply(op(1,%%),n);a(n+1)/a(n);simplify(%);limit(%,n=infinity);Konvergensradius 1/2.Opgave 137Sum(x^n/n/(n+1),n=1..infinity);value(%);a:=unapply(op(1,%%),n);a(n+1)/a(n);simplify(%);limit(%,n=infinity);Konvergensradius 1.