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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

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