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Culegere de probleme de Analiz˘a numeric˘a

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130 Aproximarea funct¸ionalelor liniare<br />

9.2.2 Formule Newton-Cotes <strong>de</strong>schise<br />

La aceste formule nodurile sunt echidistante<br />

xi = x0 +ih, i = 0,m, h = b−a<br />

m+2<br />

x0 = ah, xm = b−h<br />

x−1 = a, xm+1 = b<br />

Coeficient¸ii au expresia<br />

b<br />

Ai =<br />

a<br />

li(x)dx = (−1) m−i h<br />

i!(m−i)!<br />

m+1<br />

−1<br />

t [m+1]<br />

t−i dt<br />

Problema 9.2.5 Deducet¸i formula Newton-Cotes <strong>de</strong>schisă pentrum = 1.<br />

căci<br />

Solut¸ie.<br />

b−a<br />

2<br />

b<br />

a<br />

f(x)dx = A0f(x0)+A1f(x1)+R1(f)<br />

A0 = A1 = −h<br />

2<br />

R1(f) =<br />

⎧<br />

⎪⎨<br />

K1(t) =<br />

⎪⎩<br />

t(t−1)<br />

dt =<br />

t<br />

3h<br />

2<br />

−1<br />

b<br />

a<br />

(a−t) 2<br />

2<br />

(a−t) 2<br />

2<br />

(b−t) 2<br />

2<br />

K1(t)f ′′ (t)dt<br />

+ b−a<br />

2<br />

2a+b<br />

3 −t<br />

<br />

2a+b<br />

3 −t<br />

<br />

a+2b<br />

+<br />

3 −t<br />

<br />

=<br />

Se verifică că pentru oricet ∈ [a,b], K1(t) ≥ 0.<br />

Aplicând corolarul la teorema lui Peano obt¸inem<br />

= b−a<br />

2<br />

b<br />

a<br />

(x−t)dx<br />

R1(f) = 1<br />

2! f′′ (ξ)R(e2) =<br />

= 1<br />

2 f′′ b<br />

(ξ) x<br />

a<br />

3 dx− b−a<br />

2a+b 2 <br />

2<br />

a+2b<br />

+ =<br />

2 3 3<br />

= 1<br />

2 f′′ (ξ) b−a<br />

<br />

b<br />

3<br />

2 +ab+a 2 − 5a2 +8ab+5b 2<br />

=<br />

6<br />

= (b−a)3<br />

f<br />

36<br />

′′ (ξ) = 3h3<br />

4 f′′ (ξ).

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