Culegere de probleme de Analiz˘a numeric˘a

9.5. Formule de cuadratură de tip Gauss 145
=
Momentele funct¸iei pondere sunt
∞
µ0 = e
0
−x dx = 1 µ1 = 1 µ2 = 2
A1 +A2 = 1
A1x1 +A2x2 = 1 ⇒ A1 = 2+√2 , A2 =
4
2−√2 4
R2(f) = f(4) (ξ)
4!
b
w(x)u 2 (x)dx
b
a
w(x)u
a
2 ∞
(x) = (x
0
2 −4x+2) 2 e −x dx =
∞
(x
0
4 +16x 2 +4−8x 3 +4x 2 −16x)e −x dx = 4+32+4−24+8−16 = 8
Problema 9.5.4 Aceeas¸i problemă pentru gradul de exactitater = 3 s¸i
∞
−∞
e −x2
f(x)dx
Solut¸ie. Nodurile formulei gaussiene căutate vor fi rădăcinile polinoamelor
Hermite ortogonale pe(−∞,∞) relativ la pondereaw(t) = e−t2. hn(t) = (−1) n dn
t2 e
dtn(e−t2 ) t ∈ R
h0(t) = 1, h1(t) = 2t
hn+1(t) = 2thn(t)−2nhn−1(t)
h2(t) = 2(2t 2 −1) = 2th1(t)−2 = 4t 2 −2
h3(t) = 2th2(t)−2h1(t) = 2t(4t 2 −2)−8t = 4t(2t 2 −3)
3
t1 = −
2 , t2 = 0, t3 =
∞
µ0 =
−∞ ∞
µ1 =
−∞ ∞
µ2 =
−∞
e −t2
dt = √ π
te −t2
dt = 0
t 2 e −t2
dt = 1
⎧
⎪⎨
⎪⎩
4
∞
−∞
3
2
(2t)(2t)e −t2
dt = 1
4 22 ·2! √ π = 2 √ π
A1 +A2 +A3 = √ π
−A1 +A3 = 0
A1 +A3 = 2
3 ·2√π = 4√
π
3