Search found 19 matches

by c.adryan
Mon May 25, 2009 4:47 pm
Forum: BACALAUREAT
Topic: Sir admitere
Replies: 0
Views: 542

Sir admitere

Fie sirul \( a_n=\sum_{k=1}^n{\frac{k(k+1)}{2x^{k-1}}} \) si \( |x|>1 \) Calculati \( \lim_{n \to \infty} a_n \)
by c.adryan
Sat May 16, 2009 1:26 pm
Forum: Analiza matematica
Topic: Integrala interesanta
Replies: 2
Views: 582

Integrala interesanta

Se considera o functie continua \( f:[0,1]\rightarrow \mathbb{R} \) cu proprietatea :
\( \int_0^1{(x-1)^2f(x)dx}=0 \)
Sa se arate ca exista \( \alpha\in (0,1) \) astfel incat sa avem : \( \int_0^{\alpha}{(\alpha-x)f(x)dx}=0 \)

I.V.Maftei, Concursul Laurentiu Duican 2009
by c.adryan
Thu Apr 23, 2009 5:35 pm
Forum: Algebra
Topic: Numar prim care divide o valoare a indicatorului lui Euler
Replies: 2
Views: 653

\mathbb{Z}_{a^p+1} =\{\widehat{0},\widehat{1},....\widehat{a^2-1},...\widehat{a^p-1},\widehat{a^p}\} inel. a^{2p}-1=(a^p-1)(a^p+1) \Rightarrow\widehat{a^{2p}-1} =\widehat{0} Avem \hat{a^{2p}}=1, de unde rezulta \hat a inversabil si ord(\hat a)\in\{2,p,2p\} . Ordinul lui \hat a nu poate sa fie 2 sau...
by c.adryan
Sat Apr 04, 2009 3:29 pm
Forum: Algebra
Topic: Inel cu 2p elemente
Replies: 3
Views: 742

Inel cu 2p elemente

Fie \( (A,+,\cdot) \) un inel cu \( 2p \) elemente, unde \( p\ge 3 \) e un numar prim. Demonstrati ca \( A \) e izomorf cu \( \mathbb{Z}_{2p} \).
by c.adryan
Tue Mar 17, 2009 10:31 pm
Forum: Algebra
Topic: Polinom ireductibil
Replies: 1
Views: 547

Polinom ireductibil

Fie f=X^p+a_{p-1}X^{p-1}+\ldots+a_1X+p un polinom cu coeficienti intregi unde p\geq3 este un numar prim. Stiind ca radaciniile lui f reprezinta in planul complex afixele varfurilor unui poligon regulat cu p laturi, sa se demostreze ca f este ireductibil in inelul \mathbb{Q}[X] Ioan Baetu, GM
by c.adryan
Wed Mar 04, 2009 7:27 pm
Forum: Analiza matematica
Topic: Inegalitate integrala 2
Replies: 1
Views: 460

Inegalitate integrala 2

Fie \( f:\mathbb{R}\rightarrow\mathbb{R} \) o functie continua cu proprietatea ca \( f^3(x)+f(x)\geq x, \forall x \in \mathbb{R}. \)
Demonstrati ca \( \int_0^2f(x)dx\geq\frac{5}{4} \)

Cristinel Mortici, OJM Constanta 1997
by c.adryan
Mon Mar 02, 2009 4:15 pm
Forum: Analiza matematica
Topic: Inegalitate integrala 1
Replies: 2
Views: 761

f crescatoare \Rightarrow f^\prime \geq0 Din inegalitatea mediilor avem: \frac{f^\prime(x)}{xf^\prime(x)+f(x)}= \frac{2}{x+\frac{f(x)}{f^\prime(x)}} \leq\frac{1}{2}(\frac{1}{x}+\frac{f^\prime(x)}{f(x)}) Integrand inegalitatea obtinem : \int_1^2\frac{f^\prime(x)}{xf^\prime(x)+f(x)}dx\leq\frac{1}{2}\...
by c.adryan
Tue Jan 20, 2009 6:06 pm
Forum: Analiza matematica
Topic: Sir cu integrale
Replies: 0
Views: 367

Sir cu integrale

Fie \( 0<a<b \) si \( (x_n)_n_\geq_0,\ (y_n)_n_\geq_0 \) doua siruri cu \( x_0=a,\ y_0=0 \)
\( x_{n+1}=\int_{x_n}^{y_n} e^{-\frac{a^2}{t^2}} dt \),
\( y_{n+1}=\int_{y_n}^{x_n} e^{-\frac{b^2}{t^2}} dt \).
Demonstrati ca cele doua siruri sunt convergente si aflati limita lor.
by c.adryan
Sun May 11, 2008 9:58 pm
Forum: Analiza matematica
Topic: Functii total discontinue cu proprietatea lui Darboux
Replies: 4
Views: 1163

Exemplul lui Lebesgue: f:[0,1]\rightarrow R are P.D. si este discontinua in fiecare pct din [0,1]. Orice x\in [0,1] se scrie in baza 10 astfel: x=\overline{0,a_1a_2...a_n} . Definim f(x)=\overline{0,a_{2n}a_{2n+2}...} daca sirul a_1,a_3,a_5 este periodic de perioada 2n-1, si f(x)=0 daca sirul preced...
by c.adryan
Sun May 04, 2008 9:01 pm
Forum: Algebra
Topic: Problema 4 ONM 2008
Replies: 4
Views: 1798

A e matrice antisimetrica, deci are valoriile proprii pur imaginare. Fie ele ia_k, \overline{k=1,n} . Avem \det(A+xI_n)=\prod(ia_k+x) \det(A+yI_n)=\prod(ia_k+y) \det(A+\sqrt{xy}I_n)=\prod(ia_k+\sqrt{xy}) Determinantii sunt reali iar produsul primilor doi e pozitiv. Aplicam modul si inegalitatea devi...
by c.adryan
Tue Apr 22, 2008 3:33 pm
Forum: Intrebari teoretice
Topic: Valori proprii nenule
Replies: 2
Views: 1398

am intrebat pentru ca am gasit urmatorul post
http://www.mathlinks.ro/Forum/viewtopic ... 84&t=13989
by c.adryan
Tue Apr 22, 2008 12:25 pm
Forum: Intrebari teoretice
Topic: Valori proprii nenule
Replies: 2
Views: 1398

Valori proprii nenule

Cate valori proprii nenule are o matrice de rang \( k \)?
by c.adryan
Sun Apr 13, 2008 10:56 pm
Forum: Algebra
Topic: Concursul "Al. Myller" problema 2
Replies: 2
Views: 824

Fie b_i,a_i,s_i , i=\overline{1,3} valorile proprii pt B ,A si S . \det(S)\neq0\Rightarrow s_i\neq0 Din B=S^{-1}AS rezulta, trecand la valori proprii b_i=\frac{1}{s_i} a_i s_i\Rightarrow b_i=a_i Avem \tr(B^2)=b_1^2+b_2^2+b_3^3 \tr(B^*)=b_1b_2+b_2b_3 +b_3b_1 Rezulta \tr(B^2)+2\tr(B^*)=(b_1+b_2+b_3)^2...
by c.adryan
Wed Apr 09, 2008 10:46 pm
Forum: Algebra
Topic: AB+BA=O si AX+XA=B implica B nilpotenta
Replies: 5
Views: 1079

Bogdan Cebere wrote:Eu nu inteleg in baza carui rezultat se poate trece la valori proprii o relatie ca \( AX+XA=B \)
vezi teorema 2 din urmatorul link
http://cnmv.ploiesti.roedu.net/mambo/Ex ... 2001.3.pdf
by c.adryan
Wed Apr 09, 2008 8:11 pm
Forum: Algebra
Topic: AB+BA=O si AX+XA=B implica B nilpotenta
Replies: 5
Views: 1079

fie \( x_i , \) \( y_i, \) \( z_i, \) \( i=\overline{1,n} \) valorile proprii pt \( A, B, \) respectiv \( X \).
Avem \( y_i=2x_iz_i \) iar din prima relatie avem \( 2x_i^2z_i=0 \) de unde rezulta \( x_iz_i=0 \), ceea ce implica \( B^n=O_n \).

PS Sper sa fie bine :D
by c.adryan
Fri Mar 28, 2008 7:49 pm
Forum: Analiza matematica
Topic: Functie Darboux
Replies: 1
Views: 724

Fie \( g:[0,1]\rightarrow[0,1],\ g(x)=f(x) \) pt orice \( x\in [0,1] \).
\( g(g(x))=cos^2x \) injectiva \( \Rightarrow \) g injectiva.
g are P.D. \( \Rightarrow \) g strict monotona.
g(g(x)) strict descrescatoare ceea ce e o contradictie.
by c.adryan
Tue Mar 25, 2008 8:33 pm
Forum: Algebra
Topic: Rolle si determinanti
Replies: 2
Views: 814

si eu am facut la fel, dar am folosit g^\prime (x)=\sum_{j=0}^{n}\left| \begin{array}{clr} f_{0}(x) & f_{1}(x) & .. .&f_{n}(x) \\ f_{0}(\alpha_{1}) & f_{1}(\alpha_{1}) & .. .&f_{n}(\alpha_{1}) \\ ...&...&...&...\\...&...&...&...\\f_{0}^\prime(\alpha_{j...

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