Bine ai venit guest
 
User:
Pass:

[Creare cont]
[Am uitat parola]
iBac = materialul ULTRACOMPLET de pregătire pentru bac la mate. Dacă vrei poţi.
Forum pro-didactica.ro  [Căutare în forum]

Forum » Examene de admitere » UTC 428
[Subiect nou]   [Răspunde]
[1]
Autor Mesaj
xStefx
Grup: membru
Mesaje: 110
18 Jun 2013, 21:52

[Trimite mesaj privat]

UTC 428    [Editează]  [Citează] 


gauss
Grup: Administrator
Mesaje: 6933
18 Jun 2013, 21:19

[Trimite mesaj privat]


Inainte de a spune mai mult, iata ce spune computerul despre primele valori ale lui x(n):


(20:20) gp > for( n=10, 20, x=solve( x = (n+.0001)*Pi, (n+0.4999)*Pi, tan(x) - x ); print( "(",n,"+1/2)*Pi - x(",n,") = ", (n+1/2)*Pi - x ) )
(10+1/2)*Pi - x(10) = 0.03033382287035227855869904239
(11+1/2)*Pi - x(11) = 0.02769327190701154561396494663
(12+1/2)*Pi - x(12) = 0.02547580870822263850285863291
(13+1/2)*Pi - x(13) = 0.02358725533028886347211155435
(14+1/2)*Pi - x(14) = 0.02195946306072213281428111391
(15+1/2)*Pi - x(15) = 0.02054189868740805738392981772
(16+1/2)*Pi - x(16) = 0.01929629695191892224923335008
(17+1/2)*Pi - x(17) = 0.01819314993244580238773401254
(18+1/2)*Pi - x(18) = 0.01720933691558231692133615272
(19+1/2)*Pi - x(19) = 0.01632648462656777726852961028
(20+1/2)*Pi - x(20) = 0.01552980803334767659859565970

(20:20) gp > for( n=110, 120, x=solve( x = (n+.0001)*Pi, (n+0.4999)*Pi, tan(x) - x ); print( "(",n,"+1/2)*Pi - x(",n,") = ", (n+1/2)*Pi - x ) )
(110+1/2)*Pi - x(110) = 0.002880648390065819782379667198
(111+1/2)*Pi - x(111) = 0.002854812696480711747228345682
(112+1/2)*Pi - x(112) = 0.002829436311535373607649466285
(113+1/2)*Pi - x(113) = 0.002804507094655265960738397281
(114+1/2)*Pi - x(114) = 0.002780013329401846366562094451
(115+1/2)*Pi - x(115) = 0.002755943705111133384171051769
(116+1/2)*Pi - x(116) = 0.002732287299477957788488004993
(117+1/2)*Pi - x(117) = 0.002709033562029559883553334610
(118+1/2)*Pi - x(118) = 0.002686172298435995615634636741
(119+1/2)*Pi - x(119) = 0.002663693655608331485886838848
(120+1/2)*Pi - x(120) = 0.002641588107538862843107229040

(20:20) gp > for( n=1110, 1120, x=solve( x = (n+.0001)*Pi, (n+0.4999)*Pi, tan(x) - x ); print( "(",n,"+1/2)*Pi - x(",n,") = ", (n+1/2)*Pi - x ) )
*** solve: roots must be bracketed in solve.

(20:20) gp > for( n=1110, 1120, x=solve( x = (n+.0001)*Pi, (n+0.499999)*Pi, tan(x) - x ); print( "(",n,"+1/2)*Pi - x(",n,") = ", (n+1/2)*Pi - x ) )
(1110+1/2)*Pi - x(1110) = 0.0002866365633667666416848771911
(1111+1/2)*Pi - x(1111) = 0.0002863786806904511365705512466
(1112+1/2)*Pi - x(1112) = 0.0002861212616235173426362214774
(1113+1/2)*Pi - x(1113) = 0.0002858643049169051353858072797
(1114+1/2)*Pi - x(1114) = 0.0002856078093260373350154860045
(1115+1/2)*Pi - x(1115) = 0.0002853517736107996125261227981
(1116+1/2)*Pi - x(1116) = 0.0002850961965355205038196316158
(1117+1/2)*Pi - x(1117) = 0.0002848410768689515311032540549
(1118+1/2)*Pi - x(1118) = 0.0002845864133842474309271385177
(1119+1/2)*Pi - x(1119) = 0.0002843322048589464881923895401
(1120+1/2)*Pi - x(1120) = 0.0002840784500749509754647408687



---
df (gauss)
gauss
Grup: Administrator
Mesaje: 6933
18 Jun 2013, 21:24

[Trimite mesaj privat]


Se pare ca ne aflam in cazul "zero x infinit"...
Este natural in acest secol sa batem cod care sa ne dea evidenta numrica pentru lucrurile de calculat.

Desigur ca daca ne manjim mai departe pe maini si cerem
  • valoarea aproximativa din care trebuie sa calculam limita pe de o parte,
    apoi
  • valorile numerice oferite ca posibilitati de raspuns pe cealalta parte...

    (Eu tot incerc sa conving lumea sa invete rudimente de programare.)


  • ---
    df (gauss)
    RazzvY
    Grup: membru
    Mesaje: 329
    18 Jun 2013, 21:52

    [Trimite mesaj privat]



    [1]


    Legendă:  Access general  Conţine mesaje necitite  47510 membri, 58526 mesaje.
    © 2007, 2008, 2009, 2010 Pro-Didactica.ρ