20.07.2013 Views

Culegere de probleme de Analiz˘a numeric˘a

Culegere de probleme de Analiz˘a numeric˘a

Culegere de probleme de Analiz˘a numeric˘a

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

9.4. Cuadraturi repetate. Metoda lui Romberg 141<br />

Astfel<br />

Rf = Kf (m+1) (ξ ′ )+Lf (m+1) (ξ ′′ )<br />

Deoarece K < 0 s¸i L < 0, Rf = (K + L)f (n+1) (ξ) pentru ξ ∈ (ξ ′ ,ξ ′′ ).<br />

Deoarece<br />

ϕn+1(x) = d<br />

dx φn(x)(x−b)<br />

integrarea prin părt¸i ne dă<br />

K +L = In.<br />

9.4 Cuadraturi repetate. Metoda lui Romberg<br />

Se vor utiliza formulele<br />

Rk,1 = 1<br />

⎡<br />

2<br />

⎣Rk−1,1 +hk−1<br />

2k−2 <br />

i=1<br />

f<br />

<br />

a+ i− 1<br />

<br />

hk−1<br />

2<br />

⎤<br />

⎦, k = 2,n<br />

Rk,j = 4j−1Rk,j−1−Rk−1,j−1 4j−1 , k = 2,n<br />

−1<br />

R1,1 = h1 b−a<br />

[f(a)+f(b)] =<br />

2 2 [f(a)+f(b)]<br />

hk = hk−1<br />

2<br />

= b−a<br />

2 k−1<br />

Problema 9.4.1 Aproximat¸i π<br />

0 sinxdx prin metoda lui Romberg, ε = 10−2 .<br />

Solut¸ie.<br />

I =<br />

π<br />

0<br />

sinxdx = 2<br />

R1,1 = π<br />

(0+0) = 0<br />

2<br />

R2,1 = 1<br />

<br />

R1,1 +πsin<br />

2<br />

π<br />

<br />

= 1.571<br />

2<br />

R2,2 = 1.571+(1,571−0)/3 = 2.094<br />

R3,1 = 1<br />

<br />

R2,1 +<br />

2<br />

π<br />

2<br />

(R2,2 −R1,1) > 0.01<br />

<br />

sin π 3π<br />

+sin<br />

4 4<br />

<br />

= 1.895

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!