DesignMat uge 1 F2009Preben Alsholm, 9/1 2009Pakkerrestart;with(plots): #Meget brugtwith(plottools): #Sj\303\246ldent brugtDe hyperbolske funktionerGraferneplot([sinh(x),cosh(x),tanh(x)],x=-4..4,-3..3,color=[red,blue,maroon],legend=["sinh","cosh","tanh"],thickness=2);Nogle formler:cosh(x)^2-sinh(x)^2: % = simplify(%);cosh(x)^2+sinh(x)^2: % = combine(%);sinh(2*x): % = expand(%);Differentialkvotienterne:Diff(sinh(x),x): % = value(%);Diff(cosh(x),x): % = value(%);Diff(tanh(x),x): % = value(%);Til sammenligning tager vi den trigonometriske tangens:Diff(tan(x),x): % = value(%);Stamfunktioner:Int(sinh(x),x): % = value(%);Int(cosh(x),x): % = value(%);Int(tanh(x),x): % = value(%);Hvorfor kaldes de hyperbolske funktioner 'hyperbolske'? Fordi (x, y) = (cosh(t), sinh(t)) ligger p\303\245 hyperblen 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.Nedenfor har vi lavet et animeret plot af kurverne givet ved parameterfremstillingerne (x, y) = (cosh(t), sinh(t)) og (x, y) = (-cosh(t), sinh(t)):animate(plot,[[x,-x,[cosh(t),sinh(t),t=-2..T],[-cosh(t),sinh(t),t=-2..T]],x=-4..4,scaling=constrained,color=[green,green,red,blue],thickness=[1,1,2,2]],T=-2..2);Omvendt funktionEksempel p\303\245 en ikke-monoton, men dog enentydig funktionLad f v\303\246re defineret p\303\245 intervallet [0, 1] ved forskriftenf:=x->piecewise(x<1/2,x,3/2-x);plot(f(x),x=0..1,discont=true);f ses at v\303\246re enentydig, og har derfor en invers, der er defineret p\303\245 v\303\246rdim\303\246ngden for f, som ogs\303\245 er [0, 1].Eksempel: Eksponentialfunktionen og logaritmefunktionenf:=exp:plot(f,-2..2,thickness=2,title="Eksponentialfunktionen");Den inverse er jo logaritmefunktionen:solve(y=f(x),x);finv:=unapply(%,y);At logaritmefunktionen er den omvendte funktion til eksponentialfunktionen betyder, at (x,y) ligger p\303\245 grafen for exp hvis og kun hvis (y,x) ligger p\303\245 grafen for ln. Alts\303\245 er graferne for exp og ln spejlbilleder af hinden i linien y = x.plot([f,finv,x->x],-2..2,-2..2,color=[red,blue,black],legend=["f","Den inverse til f","y = x"],thickness=[2,2,1]);Differentialkvotienten af den inverse i yo er den reciprokke af differentialkvotienten af f i xo. Begrundelsen er egentlig simpel:x0:=-1/2:
pf:=plot([f(x),f(x0)+D(f)(x0)*(x-x0),x],x=-2..2,y=-2..2,thickness=[2,1,1],color=[red,green,black]):pkt:=plot([[x0,f(x0)]],style=point,symbol=circle,symbolsize=20,color=black):display(pf,pkt);p1:=%:roed:=op(select(x->round(op(2,x))=1,indets(p1,specfunc(anything,COLOUR)))):groen:=op(select(x->round(op(3,x))=1,indets(p1,specfunc(anything,COLOUR)))):display(p1,subs(roed=COLOUR(RGB,0,0,1),groen=COLOUR(RGB,1,0,1),reflect(p1,[[0,0],[1,1]])),view=[-2..2,-2..2]);Kontrol af p\303\245standen for vores funktion og i punktet xo. Vi udregner begge sider af'D(f)'(x0)=1/'D(finv)'(f(x0));og f\303\245r%;Den inverse funktion skal omg\303\270re, hvad f g\303\270r:simplify(finv(f(x))) assuming x::real;og omvendt:simplify(f(finv(x)));Eksempel 2 p\303\245 en ikke-monoton, men dog enentydig funktionf:=x->x/(1-x);plot(f(x),x=-4..4,y=-4..4,discont=true);solve(y=f(x),x);finv:=unapply(%,y);plot([f,finv],-4..4,-4..4,discont=true,color=[red,blue],legend=["f","Den inverse til f"]);Den inverse funktion skal omg\303\270re, hvad f g\303\270r:simplify(finv(f(x)));og omvendt:simplify(f(finv(x)));Omvendte trigonometriske funktionersin og arcsinplot(sin,-3*Pi..3*Pi,scaling=constrained,tickmarks=[5,0],caption="Sinusfunktionens graf");En ad gangen:p1:=plot(sin,-Pi/2..Pi/2,color=blue,thickness=2):p3:=plot(arcsin,-1..1,thickness=2):p2:=plot(x,x=-1.5..1.5,color=green):
p12:=display(p1,p2):
p123:=display(p12,p3):
display(p1,p12,p123,insequence=true,caption="sin og arcsin");p1:=plot(sin,-Pi/2..Pi/2,color=blue,legend="sinus",thickness=2):p3:=plot(arcsin,-1..1,legend="arcussinus",thickness=2):p2:=plot(x,x=-1.5..1.5,color=green,legend="Linien y = x"):
display(p1,p2,p3);V\303\246rdier.Vi har at'arcsin(1/2)'=arcsin(1/2);da'sin(Pi/6)'=sin(Pi/6);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiNkYnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 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.Vi har at'arcsin(1)'=arcsin(1);da'sin(Pi/2)'=sin(Pi/2);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiMkYnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 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.Vi har at'arcsin(sqrt(2)/2)'=arcsin(sqrt(2)/2);da'sin(Pi/4)'=sin(Pi/4);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiNEYnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 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.x0:='x0':Differentialkvotient:Diff(arcsin(x),x): %=value(%);Og med D:D(arcsin);Da vi \303\245benbart harInt(1/sqrt(1-x^2),x): %=value(%);f\303\270lger eksempelvisA:=Int(1/sqrt(1-x^2),x=0..1): A=value(A);cos og arccosplot(cos,-3*Pi..3*Pi,scaling=constrained,tickmarks=[5,0]);p3:=plot(arccos,-1..1,thickness=2):p1:=plot(cos,0..Pi,color=blue,thickness=2):p2:=plot(x,x=-1..3,color=green):p12:=display(p1,p2):
p123:=display(p12,p3):
display(p1,p12,p123,insequence=true,caption="cos og arccos");p2:=plot(arccos,-1..1,legend="arcuscosinus",thickness=2):p1:=plot(cos,0..Pi,color=blue,legend="cosinus",thickness=2):p3:=plot(x,x=-1..3,color=green,legend="y = x"):display(p1,p2,p3);V\303\246rdier.Vi har at'arccos(1/2)'=arccos(1/2);da'cos(Pi/3)'=cos(Pi/3);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiM0YnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkobWZlbmNlZEdGJDYmLUYjNiYtSSNtbkdGJDYkUSIwRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVEiLEYnRjQvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHUSV0cnVlRicvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdRLDAuMzMzMzMzM2VtRictSSNtaUdGJDYlUSUmcGk7RicvJSdpdGFsaWNHRj1GNEY0RjQvJSVvcGVuR1EiW0YnLyUmY2xvc2VHUSJdRidGNA==.Vi har at'arccos(1)'=arccos(1);da'cos(0)'=cos(0);og da i \303\270vrigt 0 ligger i det rigtige interval, alts\303\245 LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkobWZlbmNlZEdGJDYmLUYjNiYtSSNtbkdGJDYkUSIwRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVEiLEYnRjQvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHUSV0cnVlRicvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdRLDAuMzMzMzMzM2VtRictSSNtaUdGJDYlUSUmcGk7RicvJSdpdGFsaWNHRj1GNEY0RjQvJSVvcGVuR1EiW0YnLyUmY2xvc2VHUSJdRidGNA==.Vi har at'arccos(sqrt(3)/2)'=arccos(sqrt(3)/2);da'cos(Pi/6)'=cos(Pi/6);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiNkYnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkobWZlbmNlZEdGJDYmLUYjNiYtSSNtbkdGJDYkUSIwRicvJSxtYXRodmFyaWFudEdRJ25vcm1hbEYnLUkjbW9HRiQ2LVEiLEYnRjQvJSZmZW5jZUdRJmZhbHNlRicvJSpzZXBhcmF0b3JHUSV0cnVlRicvJSlzdHJldGNoeUdGPS8lKnN5bW1ldHJpY0dGPS8lKGxhcmdlb3BHRj0vJS5tb3ZhYmxlbGltaXRzR0Y9LyUnYWNjZW50R0Y9LyUnbHNwYWNlR1EmMC4wZW1GJy8lJ3JzcGFjZUdRLDAuMzMzMzMzM2VtRictSSNtaUdGJDYlUSUmcGk7RicvJSdpdGFsaWNHRj1GNEY0RjQvJSVvcGVuR1EiW0YnLyUmY2xvc2VHUSJdRidGNA==Differentialkvotient:Diff(arccos(x),x): %=value(%);Og med D:D(arccos);tan og arctanplot(tan,-3*Pi..3*Pi,-10..10,scaling=constrained,discont=true);p3:=plot(arctan,-5..5,-5..5,thickness=2):p1:=plot(tan,-Pi/2..Pi/2,-5..5,color=blue,thickness=2):p2:=plot(x,x=-5..5,color=green):
p12:=display(p1,p2):
p123:=display(p12,p3):display(p1,p12,p123,insequence=true,caption="tan og arctan");p2:=plot(arctan,-5..5,-5..5,legend="arcustangens",thickness=2):p1:=plot(tan,-Pi/2..Pi/2,-5..5,color=blue,legend="tangens",thickness=2):p3:=plot(x,x=-5..5,color=green,legend="y = x"):display(p1,p2,p3);plot([arctan(x),-Pi/2,Pi/2],x=-5..5,
color=[red,blue,blue],caption=typeset("arctan med asymptoterne ",y=Pi/2," og ",y=-Pi/2),scaling=constrained,tickmarks=[5,3]);V\303\246rdier.Vi har at'arctan(1)'=arctan(1);da'tan(Pi/4)'=tan(Pi/4);og da i \303\270vrigt LUklbXJvd0c2Iy9JK21vZHVsZW5hbWVHNiJJLFR5cGVzZXR0aW5nR0koX3N5c2xpYkdGJzYkLUkmbWZyYWNHRiQ2KC1GIzYkLUkjbWlHRiQ2JVElJnBpO0YnLyUnaXRhbGljR1EmZmFsc2VGJy8lLG1hdGh2YXJpYW50R1Enbm9ybWFsRidGNy1GIzYkLUkjbW5HRiQ2JFEiNEYnRjdGNy8lLmxpbmV0aGlja25lc3NHUSIxRicvJStkZW5vbWFsaWduR1EnY2VudGVyRicvJSludW1hbGlnbkdGRS8lKWJldmVsbGVkR0Y2Rjc= ligger i det rigtige interval, alts\303\245 det \303\245bne interval mellem tallene 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 .Differentialkvotienten.Diff(arctan(x),x): %=value(%);Og med D:D(arctan);Da vi \303\245benbart harInt(1/(1+x^2),x): %=value(%);f\303\270lger eksempelvisA:=Int(1/(1+x^2),x=0..1): A=value(A);og Int(1/(1+x^2),x=0..R): %=value(%);Int(1/(1+x^2),x=0..infinity): %=value(%);Diverse formlerDer g\303\246lder formlenarcsin(x)+arccos(x)=Pi/2;Den kan vises s\303\245ledes, vi differentierer udtrykketu:=arcsin(x)+arccos(x);diff(u,x);s\303\245 u er konstant p\303\245 det \303\245bne interval ]-1,1[, men da udtrykket u er kontinuert i x p\303\245 hele det lukkede interval [-1,1], s\303\245 er u konstant der.Konstantenarcsin(0)+arccos(0);sin(arccos(x));cos(arcsin(x));StamfunktionInt(arcsin(x),x): % = value(%);Kontrol:diff(%,x);Int(arccos(x),x): % = value(%);Int(arctan(x),x): % = value(%);f:='f':
Diff(Int(f(t),t=a..x),x): %=value(%);Delvis integration og integration ved substitutionMaple er god til at integrere, s\303\245 denne pakke har man sj\303\246ldent brug for.Man kan dog undertiden \303\270nske at tvinge Maple til delvis integration eller til at lave en bestemt substitution.with(IntegrationTools);A:=Int(f(x)*g(x),x);A=Parts(A,g(x));subs(int(f(x),x)=F(x),%);A:=Int(f(g(x))*diff(g(x),x),x);A=Change(A,x=(g@@(-1))(t));simplify(%);Eksempel p\303\245 delvis integrationA:=Int(x*sin(x),x);value(A);Den faktor, der skal differentieres, skal st\303\245 som andet argument: i dette tilf\303\246lde x:Parts(A,x);value(%);Eksempel p\303\245 substitutionA:=Int(sin(x^2)*x,x);Change(A,t=x^2);value(%): %=subs(t=x^2,%);Eksempel p\303\245 omvendt substitutionA:=Int(sin(sqrt(x)),x);value(A);Change(A,x=t^2);simplify(%) assuming t>0;value(%);subs(t=sqrt(x),%);Bestemt integral: Illustration af definitionen som gr\303\246nsev\303\246rdi for summerwith(Student:-Calculus1):ApproximateInt(sin(x),x=0..Pi,output=plot,method=random);ApproximateInt(sin(x),x=0..Pi,output=plot,method=random,partition=random[.4],
pointoptions=[symbolsize=15]);ApproximateInt(sin(x),x=0..Pi,output=animation,method=random,partition=2,refinement=random,iterations=8);Numerisk kontrol af udregning af bestemt integralEksempel 1Int(x/sqrt(1-x^4),x=0..1): %=value(%);Venstre side udregnes nu ved numerisk integration og der findes en decimalbr\303\270kstiln\303\246rmelse til det eksakte resultat til h\303\270jre:evalf(%);P\303\245 n\303\246r afrundingsfejl stemmer de to resultater overens.Int(1/sqrt(1+sin(x)),x=0..Pi): %=value(%);Man kan evt. s\303\246tte n\303\270jagtigheden op: evalf[20](%);Ogs\303\245 OK!Eksempel 2Dette g\303\245r ikke godt:Int(exp(-x)/(1-exp(-x)),x=0..infinity): %=value(%);evalf(%);Der er jo un\303\246gtelig forskel p\303\245 de to sider!