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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 147<br />

2 ◦ Punând x = cosθ avem<br />

π<br />

Ai =<br />

0<br />

cosnθ<br />

cosθ −cosθi<br />

căci cosnθi = 0, i = 1,n. Din relat¸ia<br />

rezultă pentruj ≥ 2 că<br />

T ′ n<br />

1 δn<br />

= ,<br />

(xi) (xi)<br />

T ′ n<br />

cos(j +1)θ+cos(j −1)θ = 2cosθcosjθ<br />

δj+1 +δj−1 = 2<br />

= 2<br />

π<br />

0 π<br />

0<br />

cosθcosjθ−cosθicosjθi<br />

dθ<br />

cosθ−cosθi<br />

cosjθdθ+2cosθiδj<br />

s¸i δ0 = 0 s¸i δ1 = π. Relat¸ia <strong>de</strong> recurent¸ă δj+1 −2cosθiδj + δj−1 = 0 are<br />

solut¸ia generalăδj = Acosjθi +Bsinjθi; se obt¸ine<br />

s¸i cum<br />

se <strong>de</strong>duce căAi = π<br />

,i = 1,n. n<br />

3 ◦ Din expresia restului se obt¸ine<br />

Rn(f) = f(2n) (ξ)<br />

(2n)!<br />

δn = πsinnθi<br />

sinθi<br />

T ′ n (xi) = nsinnθi<br />

sinθi<br />

1<br />

−1<br />

T2 n(x)<br />

22n−2√ π<br />

=<br />

1−x 2dx 22n−1 Problema 9.5.6 Deducet¸i o formulă <strong>de</strong> cuadratură <strong>de</strong> forma<br />

1<br />

−1<br />

f (2n) (ξ)<br />

.<br />

(2n)!<br />

√ 1−x 2 f(x)dx = A1f(x1)+A2f(x2)+A3f(x3)+R3(f)<br />

Solut¸ie. Formula va fi <strong>de</strong> tip Gauss; polinoamele ortogonale care dau nodurile<br />

vor fi polinoamele Cebâs¸ev <strong>de</strong> spet¸a a II-a.<br />

Qn(t) = sin[(n+1)arccost]<br />

√ , t ∈ [−1,1]<br />

1−t 2

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