Culegere de probleme de Analiz˘a numeric˘a
Culegere de probleme de Analiz˘a numeric˘a
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).