P1.  Reprezentaţi pe acelaşi grafic funcţia </COMMENT>Browser's Java Plug-in not enabled. şi aproximantele </COMMENT>Browser's Java Plug-in not enabled. , </COMMENT>Browser's Java Plug-in not enabled. şi </COMMENT>Browser's Java Plug-in not enabled. pe intervalul </COMMENT>Browser's Java Plug-in not enabled. . 

Soluţie. 

 

> restart;
 

> taylor(sin(x),x=0,12);
 

</COMMENT>Browser's Java Plug-in not enabled. (1)
 

> Tn:=(n,x)->convert(taylor(sin(x),x=0,n+1),polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (2)
 

> Tn(1,x),Tn(3,x),Tn(5,x);
 

</COMMENT>Browser's Java Plug-in not enabled. (3)
 

> plot([sin(x),Tn(1,x),Tn(3,x),Tn(5,x)],x=-Pi..Pi,view=[-Pi..Pi,-2..2],legend=["f",typeset(T[1]),typeset(T[3]),typeset(T[5])],color=["red","green","blue","black"]);
 

Plot_2d
 

animaţie 

> for k from 1 to 15 do p[k]:=plot([sin(x),Tn(k,x)],x=-2*Pi..2*Pi,view=[-2*Pi..2*Pi,-5..5]): end do:
 

> plots[display](seq(p[2*i-1],i=1..8),insequence=true);
 

Plot_2d
 

>
 

P2. La fel pentru </COMMENT>Browser's Java Plug-in not enabled. , aproximantele </COMMENT>Browser's Java Plug-in not enabled. , ..., </COMMENT>Browser's Java Plug-in not enabled. şi intervalul [-2,2]. 

Soluţie.  

 

> restart;
 

> taylor(exp(x),x=0,10);
 

</COMMENT>Browser's Java Plug-in not enabled. (4)
 

> Te:=(n,x)->convert(taylor(exp(x),x=0,n+1),polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (5)
 

> plot([exp(x),seq(Te(k,x),k=1..5)],x=-2..5,view=[-2..2,-1..7]);
 

Plot_2d
 

> for k from 1 to 15 do p[k]:=plot([exp(x),Te(k,x)],x=-2..2,view=[-2..2,-1..8]): end do:
 

> plots[display](seq(p[2*i-1],i=1..8),insequence=true);
 

Plot_2d
 

P4. Dezvoltaţi funcţia eroare 

</COMMENT>Browser's Java Plug-in not enabled. , 

īn serie utilizānd seria pentru exponenţială şi integrānd. Calculaţi seria Taylor a lui erf(x) īn jurul lui zero direct. Sunt cele două serii identice? Evaluaţi erf(1) adunānd patru termeni ai seriei şi comparaţi cu valoarea erf(1) = 0.8427, care este dată cu patru zecimale corecte. Indicaţie: Din teorema fundamentală a calculului integral rezultă că 

</COMMENT>Browser's Java Plug-in not enabled. . 

 

</COMMENT>Browser's Java Plug-in not enabled. (6)
 

> restart;
 

> f:=t->2/sqrt(Pi)*exp(-t^2);
 

</COMMENT>Browser's Java Plug-in not enabled. (7)
 

> taylor(f(t),t=0,12);
 

</COMMENT>Browser's Java Plug-in not enabled. (8)
 

> dt1:=convert(%,polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (9)
 

> serf1:=int(dt1,t=0..x);
 

</COMMENT>Browser's Java Plug-in not enabled. (10)
 

> f2:=x->2/sqrt(Pi)*Int(exp(-t^2),t=0..x);
 

</COMMENT>Browser's Java Plug-in not enabled. (11)
 

> taylor(f2(x),x=0,12);
 

</COMMENT>Browser's Java Plug-in not enabled. (12)
 

> serf2:=convert(%,polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (13)
 

> evalf(eval(serf1,x=1));
 

</COMMENT>Browser's Java Plug-in not enabled. (14)
 

> evalf(eval(serf2,x=1));
 

</COMMENT>Browser's Java Plug-in not enabled. (15)
 

> err:=coeftayl(erf(x),x=0,13);
 

</COMMENT>Browser's Java Plug-in not enabled. (16)
 

> evalf(%);
 

</COMMENT>Browser's Java Plug-in not enabled. (17)
 

P3. Deduceţi seria Taylor pentru </COMMENT>Browser's Java Plug-in not enabled. şi aproximaţi ln2 folosind primii 8 termeni. Cāţi termeni sunt necesari pentru a obţine ln2 cu 5 zecimale corecte. Aceeaşi problemă pentru </COMMENT>Browser's Java Plug-in not enabled. . 

Soluţie. 

 

> restart;
 

> taylor(ln(1+x),x=0,9);
 

</COMMENT>Browser's Java Plug-in not enabled. (18)
 

> apln:=convert(%,polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (19)
 

> eval(apln,x=1);
 

</COMMENT>Browser's Java Plug-in not enabled. (20)
 

> evalf(%);
 

</COMMENT>Browser's Java Plug-in not enabled. (21)
 

> evalf(ln(2));
 

</COMMENT>Browser's Java Plug-in not enabled. (22)
 

Cāţi termeni sunt necesari? 

> solve(1/(n+1)<1e-5);
 

</COMMENT>Browser's Java Plug-in not enabled. (23)
 

>
 

Totuşi, lucrurile merg bine īn vecinătatea originii (de exemplu pentru ln 1.1) 

 

> eval(apln,x=0.1);
 

</COMMENT>Browser's Java Plug-in not enabled. (24)
 

> solve((1/10)^(n+1)/(n+1)<1e-5,n);
 

</COMMENT>Browser's Java Plug-in not enabled. (25)
 

Remediu: să se utilizeze dezvoltarea Taylor a lui </COMMENT>Browser's Java Plug-in not enabled.  

> taylor(ln((1+x)/(1-x)),x=0,13);
 

</COMMENT>Browser's Java Plug-in not enabled. (26)
 

> apln2:=convert(%,polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (27)
 

> solve((1+x)/(1-x)=2,x);
 

</COMMENT>Browser's Java Plug-in not enabled. (28)
 

> evalf(eval(apln2,x=1/3));
 

</COMMENT>Browser's Java Plug-in not enabled. (29)
 

> evalf(ln(2));
 

</COMMENT>Browser's Java Plug-in not enabled. (30)
 

> abs(%-%%);
 

</COMMENT>Browser's Java Plug-in not enabled. (31)
 

Numărul de termeni necesar 

> R:=2/(2*p+1)*sum(1/3^(2*k+1),k=p..infinity);
 

</COMMENT>Browser's Java Plug-in not enabled. (32)
 

> R:=p->1/(4*(2*p+1)*3^(2*p-1));
 

</COMMENT>Browser's Java Plug-in not enabled. (33)
 

> solve(R(p)<1e-5,p);
 

</COMMENT>Browser's Java Plug-in not enabled. (34)
 

> Tlog:=convert(taylor(log((1+x)/(1-x)),x=0,11),polynom);
 

</COMMENT>Browser's Java Plug-in not enabled. (35)
 

> eval(Tlog,x=1/3);
 

</COMMENT>Browser's Java Plug-in not enabled. (36)
 

> evalf(%)-evalf(ln(2));
 

</COMMENT>Browser's Java Plug-in not enabled. (37)
 

>
 

>