Culegere de probleme de Analiz˘a numeric˘a

132 Aproximarea funct¸ionalelor liniare
9.3 Alte formule de tip interpolator
Problema 9.3.1 Obt¸inet¸i o formulă de cuadratură de forma
b
f(x)dx = A00f(a)+A10f(b)+A01f
a
′ (a)+A11f ′ (b)+R(f)
b b
(x−b)
Solut¸ie.A00 = h00(x)dx =
a a
2
(a−b) 3[3a−b−2x]dx
b b
(x−a)
A10 = h10(x)dx =
a a
2
(b−a) 3[3b−a−2x]dx
A00 = A10 = b−a
2 b
A01 = −A10 = (x−a) (x−b)2 (b−a)2
(a−b) 2dx
=
12
b
a
R(f) = K3(t)f
a
(4) (t)dt
K3(t) = 1
(b−t)
3!
4
−
4
b−a
2 (a−t)3 b−a
+ −
2 (b−t)3 + −
− (b−a)2
·
12
3(a−t)2 +
+
0
(b−a)2
3(b−t)
122
2
+ =
= 1
4 (b−t)
−
3! 4
b−a
2 (b−t)3 + (b−a)2
(b−t)
4
2
=
= (b−t)2
[b
4!
2 −2bt+t 2 −2(b−a)(b−t)+(b−a) 2 ] =
= (b−t)2
4!
(b−t) 2 (a−t) 2
4!
R3(f) =
[b 2 − 2bt + t 2 − 2b 2 + 2bt + 2ab − 2at + b 2 − 2ab + a 2 ] =
5 2! (b−a)
f
4! 5
(4) (ξ), ξ ∈ [a,b]
Problema 9.3.2 Generalizare pentrum = 1 s¸i r0 = r1 = s−1.
Solut¸ie.
b
A0j =
a
b
a
s−1
f(x)dx =
h0j(x)dx =
b
a
j=0
[A0jf (j) (a)+A1jf (j) (b)]+R2s−1(f)
s j x−b (x−a)
a−b j!
n−j
n+ν x−a
dx =
ν b−a
ν=0