Culegere de probleme de Analiz˘a numeric˘a
Culegere de probleme de Analiz˘a numeric˘a
Culegere de probleme de Analiz˘a numeric˘a
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8.3. Alt¸i operatori liniari s¸i pozitivi 119<br />
8.3 Alt¸i operatori liniari s¸i pozitivi<br />
Problema 8.3.1 (operatorul lui Fejer) Se obt¸ine din polinomul <strong>de</strong> interpolare<br />
Hermite cu noduri duble rădăcini ale polinomului Cebâs¸ev <strong>de</strong> spet¸a I, Tm+1.<br />
xk = cos<br />
(H2m+1)(x) =<br />
2k +1<br />
π k = 0,m<br />
2(m+1)<br />
m<br />
hk0(x)f(x)+<br />
k=0<br />
m<br />
hk1(x)f ′ (x)<br />
omit¸ând a doua sumă sau consi<strong>de</strong>rând echivalentf ′ (xk) = 0, k = 0,n<br />
Solut¸ie.<br />
(F2m+1)(x) =<br />
k=0<br />
m<br />
hk(x)f(xk)<br />
k=0<br />
<br />
Tm+1(x)<br />
hk(x) = hk0(x) = (1−xkx)<br />
(m+1)(x−xk)<br />
F2m+1((t−x) 2 ;x) =<br />
=<br />
F2m+1f ⇉ f pe [−1,1]<br />
F2m+1(1;x) = 1 x ∈ [−1,1]<br />
m<br />
<br />
Tm+1(x)<br />
(1−xkx)<br />
(m+1)(x−xk)<br />
n=0<br />
1<br />
(m+1) 2T2 m+1(x)<br />
m<br />
k=0<br />
2<br />
2<br />
(xk −x) 2 =<br />
(1−xkx) = 1<br />
m+1 T2 m+1(x) ≤ 1<br />
m+1<br />
căci m<br />
k=0 xk = 0.<br />
Deci,<br />
lim<br />
m→∞ F2m+1((t−x) 2 ;x) = 0 uniform pe [−1,1]<br />
Problema 8.3.2 (Operatorul lui Meyer-König s¸i Zeller) Fie B[0,1) spat¸iul liniar<br />
al funct¸iilor reale <strong>de</strong>finite s¸i mărginite pe[0,1).<br />
Se <strong>de</strong>fines¸te operatorul lui Meyer-König s¸i Zeller Mm : B[0,1) → C[0,1)<br />
pentru oricex ∈ [0,1] prin egalitatea<br />
(Mmf)(x) =<br />
m<br />
<br />
m+k<br />
x<br />
k<br />
k (1−x) m+1 <br />
k<br />
f<br />
m+k<br />
k=0