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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88 Interpolare<br />
Problema 6.3.5 Se consi<strong>de</strong>ră f : [−1,1] → R. Se notează cu F2n+1f polinomul<br />
Hermite cu noduri duble <strong>de</strong>terminat <strong>de</strong> condit¸iile<br />
(F2m+1f)(xk) = f(xk), k = 0,m<br />
(F2m+1f) ′ (xk) = 0.<br />
Să se arate că dacăx0,x1,...,xm sunt rădăcinile polinomului lui Cebâs¸ev <strong>de</strong><br />
spet¸a I avem:<br />
-<br />
Solut¸ie.<br />
(F2m+1f)(x) =<br />
1−(x−xk) u′ k (xk)<br />
uk(xk)<br />
1<br />
(m+1) 2<br />
m<br />
2 Tm+1(x)<br />
(1−xkx) f(xk).<br />
x−xk<br />
k=0<br />
hk0(x) = uk(x)<br />
<br />
1−(x−xk)<br />
uk(xk)<br />
u′ <br />
(xk)<br />
uk(xk)<br />
w(x) = (x−x0)(x−x1)...(x−xm)<br />
uk(x) = w2 <br />
(x) 1<br />
=<br />
(x−xk) 2 2m 2 Tm+1(x)<br />
x−xk<br />
<br />
1 1<br />
= (x−xk) + +···+<br />
x−xk x0 −xk<br />
uk(xk) = w ′2 (xk)<br />
u ′ k (xk) = w ′ (xk)w ′′ (xk)<br />
(−1) k<br />
<br />
2 1−x k<br />
w ′′ (x) = m+1<br />
2m <br />
xsin[(m+1)arccosx]−<br />
w ′ (xk) = m+1<br />
2 m<br />
(m+1) √ 1−x 2 √ 3 cos[(m+1)arccosx] / 1−x 2<br />
w ′′ (xk) = m+1<br />
2 m<br />
hk0(x) =<br />
1<br />
2 m<br />
Tm+1(x)<br />
x−xk<br />
(−1) kxk 3 1−x 2 k<br />
2<br />
<br />
1−(x−xk) w′ (xk)w ′′ (xk)<br />
w ′2 (xk)<br />
1<br />
w ′2 (xk)<br />
<br />
=<br />
1<br />
xn −xk