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Culegere de probleme de Analiz˘a numeric˘a

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9.5. Formule <strong>de</strong> cuadratură <strong>de</strong> tip Gauss 143<br />

iar coeficient¸ii s¸i restul au expresiile<br />

s¸i respectiv<br />

(m!)<br />

Ak =<br />

4 (b−a) 2m+1<br />

[(2m)!] 2 (xk −a)(b−xk)[v ′ k<br />

Rm(f) = (m!)4<br />

[(2m)!] 3<br />

(xk)] 2, k = 1,m<br />

(b−a) 2m+1<br />

f<br />

2m+1<br />

(2m) (ξ), ξ ∈ [a,b]<br />

Problema 9.5.1 Stabilit¸i o formulă <strong>de</strong> cuadratură <strong>de</strong> tip Gauss în cazulw(x) ≡ 1<br />

s¸im = 3.<br />

Solut¸ie. Polinomul Legendre <strong>de</strong> grad 3 corespunzând intervalului[−1,1] este<br />

P3(t) = 1<br />

2 (5t3 −3t)<br />

cu rădăcinile <br />

3<br />

t1 = −<br />

5 , t2 = 0, t3 =<br />

Coeficient¸ii sunt solut¸iile sistemului<br />

⎧<br />

⎪⎨ A1 +A2 +A3 <br />

= 2<br />

3 −<br />

⎪⎩<br />

5A1 <br />

3 + 5A3 = 0<br />

Pentru rest se obt¸ine<br />

3<br />

5 A1 + 3<br />

5 A2 = 2<br />

3<br />

A1 = A3 = 5<br />

9 A2 = 8<br />

9<br />

R3(f) = (3!)4<br />

(6!) 3<br />

3<br />

5<br />

(b−a) 7<br />

f<br />

7<br />

(6) (ξ)<br />

Trecerea <strong>de</strong> la[−1,1] la[a,b] se poate face prin schimbarea <strong>de</strong> variabilă<br />

b<br />

a<br />

x = b+a b−a<br />

+<br />

2 2 t<br />

f(x)dx = b−a<br />

1 <br />

b+a b−a<br />

f +<br />

2 −1 2 2 t<br />

<br />

dt<br />

b<br />

f(x)dx ≈ b−a<br />

m<br />

Aif(xi)<br />

2<br />

a<br />

un<strong>de</strong> xi = b+a b−a<br />

+<br />

2 2 t2, ti fiind rădăcinile polinomului Legendre corespunzător<br />

intervalului[−1,1].<br />

i=1

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