Integrazione numerica – Parabole

Metodo di Cavalieri-Simpson

Con n pari ogni due intervalli si utilizza la parabola passante per 3 punti (di seguito trovi le formule finali)
Osserva lo sviluppo delle formule al crescere di n.

h = $\displaystyle \frac{b-a}{n}$
x = a, a+h, a+2h, …, b
y = f(a), f(a+h), f(a+2h), …, f(b)

n = 2

$\displaystyle h=\frac{b-a}{2}$

una parabola

$\displaystyle \frac{h}{3}\big[f(a)+4 f(a+h)+f(b)\big]$

$\displaystyle (b-a)\frac{f(a)+4 f(a+h)+f(b)}{6}$

n = 4

$\displaystyle h=\frac{b-a}{4}$

due parabole

$\displaystyle \frac{h}{3}\big[f(a)+4 f(a+h)+f(a+2h)\big]$

$\displaystyle+\frac{h}{3}\big[f(a+2h)+4 f(a+3h)+f(b)\big]$

$\displaystyle \frac{h}{3}\big[f(a)+4 f(a+h)+2 f(a+2h)+4 f(a+3h)+f(b)\big]$

$\displaystyle (b-a)\frac{f(a)+4 f(a+h)+2 f(a+2h)+ 4 f(a+3h)+f(b)}{12}$

n = 6

$\displaystyle h=\frac{b-a}{6}$

tre parabole

$\displaystyle \frac{h}{3}\big[ f(a)+4 f(a+h)+f(a+2h)\big]$

$\displaystyle+\frac{h}{3}\big[ f(a+2h)+4 f(a+3h)+f(a+4h)\big]$

$\displaystyle+\frac{h}{3}\big[ f(a+4h)+4 f(a+5h)+f(b)\big]$

$\displaystyle \frac{h}{3}\left[f(a)+4 f(a+h)+2 f(a+2h)+4 f(a+3h)+2\,f(a+4h)+4 f(a+5h)+f(b)\right]$

$\displaystyle (b-a)\frac{f(a)+4 f(a+h)+2 f(a+2h)+4 f(a+3h)+ 2 f(a+4h)+4 f(a+5h)+f(b)}{18}$

…

n pari

$\displaystyle h=\frac{b-a}{n}$

n/2 parabole

$\displaystyle \frac{h}{3}\left[f(a)+4 f(a+h)+f(a+2h)\right]$

$\displaystyle+\frac{h}{3}\big[ f(a+2h)+4 f(a+3h)+f(a+4h)\big]$

$+\dots$

$\displaystyle+ \frac{h}{3}\left[f(a+(n-2)h)+4 f(a+(n-1)h)+f(b)\right]$

$\displaystyle \frac{h}{3}\left[f(a)+4 f(a+h)+2 f(a+2h)+\dots + 2 f(a+(n-2)h)+ 4 f(a+(n-1)h)+f(b)\right]$

Un unico rettangolo con base h/3 e altezza data dalla somma pesata delle n+1 altezze.

$\displaystyle (b-a)\frac{f(a)+4 f(a+h)+2 f(a+2h)+\dots + 2 f(a+(n-2)h)+4 f(a+(n-1)h)+f(b)}{3n}$

Un unico rettangolo con base (b-a) e altezza data dalla media pesata delle n+1 altezze.

Formulazioni con le sommatorie

$\displaystyle \frac{h}{3}\, \left[f(x_0)+4\cdot\sum _{i\ dispari}f(x_i)+2\cdot\sum _{i\ pari} f(x_i)+f(x_{n})\right]$	$\displaystyle \frac{h}{3}\, \left[f(a)+4\cdot\sum _{i\ dispari}f(a+(2i-1)h)+2\cdot\sum _{i\ pari} f(a+2i\cdot h)+f(b)\right]$
$\displaystyle \frac{h}{3}\, \left[f(x_0)+4\cdot\sum _{i=1}^{n/2}f(x_{2i-1})+2\cdot\sum _{i=1}^{n/2-1} f(x_{2i})+f(x_{n})\right]$	$\displaystyle \frac{h}{3}\, \left[f(a)+4\cdot\sum _{i=1}^{n/2}f(a+(2i-1)h)+2\cdot\sum _{i=1}^{n/2-1} f(a+2i\cdot h)+f(b)\right]$