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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24 Elemente <strong>de</strong> Analiză funct¸ională s¸i teoria aproximării<br />
Solut¸ie. Avem arccosξk = 2k−1π,<br />
k = 1,n+1. Să calculăm acum produsul<br />
2n+2<br />
scalar:<br />
(Ti,Tj) T = (cosiarccost,cosjarccost) T =<br />
n+1<br />
= cos(iarccosξk)cos(jarccosξk) =<br />
k=1<br />
n+1<br />
<br />
<br />
2k −1<br />
= cos i<br />
2(n+1)<br />
k=1<br />
π<br />
<br />
2k −1<br />
cos j<br />
2(n+1) π<br />
<br />
=<br />
= 1 n+1<br />
<br />
2k −1 2k −1<br />
cos(i+j) π +cos(i−j)<br />
2 2(n+1) 2(n+1) π<br />
<br />
=<br />
k=1<br />
= 1 n+1<br />
i+j 1 n+1<br />
i−j<br />
cos(2k−1) π + cos(2k −1)<br />
2 2(n+1) 2 2(n+1) π.<br />
k=1<br />
Notămα = i+j i−j<br />
π, β = π s¸i<br />
2(n+1) 2(n+1)<br />
Deoarece<br />
S1 = 1<br />
n+1<br />
2<br />
k=1<br />
S2 = 1 n+1<br />
2<br />
k=1<br />
k=1<br />
<br />
cos(2k −1)α,<br />
cos(2k −1)β.<br />
2sinαS1 = sin2(n+1)α,<br />
2sinβS2 = sin2(n+1)β,<br />
se obt¸ineS1 = 0 s¸iS2 = 0. Cealaltă proprietate se <strong>de</strong>monstrează analog.<br />
Problema 2.4.4 Polinoame Cebîs¸ev <strong>de</strong> spet¸a a II-a.<br />
Definit¸ia 2.4.5 Qn ∈ Pn dat <strong>de</strong><br />
Qn(t) = sin[(n+1)arccost]<br />
√ , t ∈ [−1,1]<br />
1−t 2<br />
se numes¸te polinomul lui Cebîs¸ev <strong>de</strong> spet¸a a II-a.<br />
Qn = 1<br />
n+1 T′ n+1 (t), t ∈ [−1,1]