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Culegere de probleme de Analiz˘a numeric˘a

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148 Aproximarea funct¸ionalelor liniare<br />

Ele au rădăciniletk = cos kπ<br />

, k = 1,n<br />

n+1<br />

În cazul nostru avem<br />

Q3(t) = 8t 3 −4t Q3(t) = 1<br />

8 (8t3 −4t)<br />

Rădăcinile vor fi<br />

t1 = −<br />

√ 2<br />

2 , t0 = 0, t2 =<br />

Pentru coeficient¸i, t¸inând cont că formula are gradul <strong>de</strong> exactitate2m−1 = 5<br />

obt¸inem sistemul ⎧<br />

⎨ A1 +A2 +A)3 = µ0<br />

un<strong>de</strong><br />

1<br />

µ2 =<br />

−1<br />

⎩<br />

Se observă că µ2k+1 =<br />

este impară.<br />

Sistemul are solut¸iile<br />

Restul va fi<br />

A1t1 +A2t2 +A3t3 = µ1<br />

A1t2 1 +A2t2 2 +A3t2 3 = µ2<br />

1<br />

µk =<br />

−1<br />

1<br />

µ0 = π<br />

2 , µ1 =<br />

t 2√ 1−t 2dt = 1<br />

4<br />

t k√ 1−t 2 dt<br />

−1<br />

1<br />

√ 2<br />

2<br />

t √ 1−t 2 dt = 0<br />

(2t)(2t) √ 1−t 2dt = π<br />

8<br />

−1<br />

1<br />

t<br />

−1<br />

2k+1√ 1−t 2dt = 0, <strong>de</strong>oarece funct¸ia <strong>de</strong> integrat<br />

A1 = A3 = π<br />

8 , A2 = π<br />

4<br />

Rm(f) = f(2m) (ξ)<br />

(2m)!<br />

= f(2m) (ξ)<br />

(2m)!<br />

1<br />

· 1 π<br />

·<br />

2m 2 =<br />

−1<br />

w(x)u 2 (x)dx =<br />

π<br />

2 m+1 (2m)! f(2m) (ξ)<br />

Am obt¸inut formula<br />

1<br />

f(x)dx = π<br />

√ √ <br />

2 2<br />

f − +2f(0)+f<br />

8 2 2<br />

−1<br />

+ π<br />

2 4 6! f(6) (ξ)

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