Recorte 15: Fórmulas de Newton-Cotes
Introdução ¶ No texto a seguir, vamos construir uma fórmula geral de Newton-Cotes para integração numérica fechada . Assumimos que os nós de integração são igualmente espaçados e que os pontos extremos também são incluídos na fórmula de quadratura. Isto significa que para n + 1 n+1 n + 1
a = x 0 < x 1 < x 2 < … < x n − 1 < x n = b , com h = x k + 1 − x k , k = 0 , 1 , … , n − 1. a = x_0 < x_1 < x_2 < \ldots < x_{n-1} < x_n = b, \ \ \text{com} \ \ h = x_{k+1} - x_k, \, k = 0,1,\ldots,n-1. a = x 0 < x 1 < x 2 < … < x n − 1 < x n = b , com h = x k + 1 − x k , k = 0 , 1 , … , n − 1. Integrar f ( x ) f(x) f ( x ) [ a , b ] [a,b] [ a , b ] ∫ a b f ( x ) d x \int_a^b f(x) \, dx ∫ a b f ( x ) d x ∫ a b ω ( x ) f ( x ) d x \int_a^b \omega(x) f(x) \, dx ∫ a b ω ( x ) f ( x ) d x ω ( x ) ≥ 0 \omega(x) \geq 0 ω ( x ) ≥ 0 função-peso que pode anular-se em um número finito de pontos. Entretanto, usaremos ω ( x ) ≡ 1 \omega(x) \equiv 1 ω ( x ) ≡ 1
Para cada nó x k x_k x k f k = f ( x k ) f_k = f(x_k) f k = f ( x k ) n n n
∫ a b f ( x ) d x ≈ ∑ k = 0 n f k A k = ∑ k = 0 n f k ∫ x 0 x n L n , k ( x ) d x , \int_a^b f(x) \, dx \approx \displaystyle \sum_{k=0}^n f_k A_k = \displaystyle \sum_{k=0}^n f_k \int_{x_0}^{x_n} \mathcal{L}_{n,k}(x) \, dx, ∫ a b f ( x ) d x ≈ k = 0 ∑ n f k A k = k = 0 ∑ n f k ∫ x 0 x n L n , k ( x ) d x , em que, evidentemente, A k = ∫ x 0 x n L n , k ( x ) d x A_k = \int_{x_0}^{x_n} \mathcal{L}_{n,k}(x) \, dx A k = ∫ x 0 x n L n , k ( x ) d x L n , k ( x ) \mathcal{L}_{n,k}(x) L n , k ( x )
Antes de prosseguirmos, vale a pena escrever as funções de base de Lagrange e, consequentemente, o próprio polinômio de Lagrange em uma forma especial. Para isso, vamos utilizar a seguinte mudança de variável
u = x − x 0 h . u = \dfrac{x-x_0}{h}. u = h x − x 0 . Esta mudança está embasada em dois resultados (teoremas que podem ser provados por indução):
Para r ∈ Z + r \in \mathbb{Z}_{+} r ∈ Z + x − x r = ( u − r ) h x - x_r = (u-r)h x − x r = ( u − r ) h
Para r , s ∈ Z + r,s \in \mathbb{Z}_{+} r , s ∈ Z + x r − x s = ( r − s ) h x_r - x_s = (r-s)h x r − x s = ( r − s ) h
Uma vez que o polinômio de Lagrange é dado por P n ( x ) = ∑ k = 0 n f k L n , k ( x ) P_n(x) = \sum_{k=0}^n \, f_k \mathcal{L}_{n,k}(x) P n ( x ) = ∑ k = 0 n f k L n , k ( x )
L n , k ( x ) = ( x − x 0 ) ( x − x 1 ) … ( x − x k − 1 ) ( x − x k + 1 ) … ( x − x n ) ( x k − x 0 ) ( x − x 1 ) … ( x k − x k − 1 ) ( x k − x k + 1 ) … ( x k − x n ) , \mathcal{L}_{n,k}(x) = \dfrac{ (x-x_0)(x-x_1) \ldots (x-x_{k-1})(x-x_{k+1}) \ldots (x-x_n) }{ (x_k-x_0)(x-x_1) \ldots (x_k-x_{k-1})(x_k-x_{k+1}) \ldots (x_k-x_n) }, L n , k ( x ) = ( x k − x 0 ) ( x − x 1 ) … ( x k − x k − 1 ) ( x k − x k + 1 ) … ( x k − x n ) ( x − x 0 ) ( x − x 1 ) … ( x − x k − 1 ) ( x − x k + 1 ) … ( x − x n ) , a mudança de variável anterior permite-nos escrevê-lo como
P n ( x 0 + u h ) = ∑ k = 0 n f k λ n , k ( u ) , P_n(x_0 + uh) = \sum_{k=0}^n \, f_k \lambda_{n,k}(u), P n ( x 0 + u h ) = k = 0 ∑ n f k λ n , k ( u ) , para
λ n , k ( u ) = u ( u − 1 ) … ( u − ( k − 1 ) ) ( u − ( k + 1 ) ) … ( u − n ) k ( k − 1 ) … ( k − ( k − 1 ) ) ( k − ( k + 1 ) ) … ( k − n ) . \lambda_{n,k}(u) = \dfrac{ u(u-1) \ldots (u-(k-1))(u-(k+1)) \ldots (u-n) }{ k(k-1) \ldots (k-(k-1))(k-(k+1)) \ldots (k-n) }. λ n , k ( u ) = k ( k − 1 ) … ( k − ( k − 1 )) ( k − ( k + 1 )) … ( k − n ) u ( u − 1 ) … ( u − ( k − 1 )) ( u − ( k + 1 )) … ( u − n ) . Logo, P n ( x 0 + u h ) P_n(x_0 + uh) P n ( x 0 + u h ) forma de Lagrange para pontos igualmente espaçados .
Fórmulas gerais ¶ A mudança de variável nos permite ter d x = h d u dx = hdu d x = h d u
{ x = x 0 , → u = 0 x = x n , → u = n \begin{cases}
x=x_0,& \rightarrow u = 0 \\
x=x_n,& \rightarrow u = n \\
\end{cases} { x = x 0 , x = x n , → u = 0 → u = n Assim, a aproximação de nossa integral converte-se em:
∫ x 0 x n f ( x ) d x ≈ ∑ k = 0 n f k h C k n ≈ ∑ k = 0 n f k h ∫ 0 n λ n , k ( u ) d u , \int_{x_0}^{x_n} f(x) \, dx \approx \displaystyle \sum_{k=0}^n f_k h C_k^n \approx \displaystyle \sum_{k=0}^n f_k h \int_{0}^{n} \lambda_{n,k}(u) \, du, ∫ x 0 x n f ( x ) d x ≈ k = 0 ∑ n f k h C k n ≈ k = 0 ∑ n f k h ∫ 0 n λ n , k ( u ) d u , para C k n = ∫ 0 n λ n , k ( u ) d u C_k^n = \int_{0}^{n} \lambda_{n,k}(u) \, du C k n = ∫ 0 n λ n , k ( u ) d u
A fórmula anterior, totalmente independente dos limites de integração fornece-nos as mais diversas (e conhecidas) fórmulas de Newton-Cotes.
Regra do Trapézio Simples ¶ Quando n = 1 n=1 n = 1
∫ x 0 x 1 f ( x ) d x ≈ f 0 h C 0 1 + f 1 h C 1 1 . \int_{x_0}^{x_1} f(x) \, dx \approx f_0 h C_0^1 + f_1 h C_1^1. ∫ x 0 x 1 f ( x ) d x ≈ f 0 h C 0 1 + f 1 h C 1 1 . Porém, notemos que:
C 0 1 = ∫ 0 1 λ 1 , 0 ( u ) d u = ∫ 0 1 u − 1 0 − 1 d u = ∫ 0 1 ( 1 − u ) d u = 1 2 C 1 1 = ∫ 0 1 λ 1 , 1 ( u ) d u = ∫ 0 1 u − 0 1 − 0 d u = ∫ 0 1 u d u = 1 2 \begin{aligned}
C_0^1 &= \int_{0}^{1} \lambda_{1,0}(u) \, du = \int_{0}^{1} \dfrac{u-1}{0-1} \, du = \int_{0}^{1} (1-u) \, du = \dfrac{1}{2} \\
C_1^1 &= \int_{0}^{1} \lambda_{1,1}(u) \, du = \int_{0}^{1} \dfrac{u-0}{1-0} \, du = \int_{0}^{1} u \, du = \dfrac{1}{2} \end{aligned} C 0 1 C 1 1 = ∫ 0 1 λ 1 , 0 ( u ) d u = ∫ 0 1 0 − 1 u − 1 d u = ∫ 0 1 ( 1 − u ) d u = 2 1 = ∫ 0 1 λ 1 , 1 ( u ) d u = ∫ 0 1 1 − 0 u − 0 d u = ∫ 0 1 u d u = 2 1 Logo,
∫ x 0 x 1 f ( x ) d x ≈ f 0 h 1 2 + f 1 h 1 2 = h 2 ( f 0 + f 1 ) = h 2 [ f ( x 0 ) + f ( x 1 ) ] , \int_{x_0}^{x_1} f(x) \, dx \approx f_0 h \dfrac{1}{2} + f_1 h \dfrac{1}{2} = \dfrac{h}{2}(f_0 + f_1) = \dfrac{h}{2}[f(x_0) + f(x_1)], ∫ x 0 x 1 f ( x ) d x ≈ f 0 h 2 1 + f 1 h 2 1 = 2 h ( f 0 + f 1 ) = 2 h [ f ( x 0 ) + f ( x 1 )] , que é a tradicional regra do trapézio simples .
Regra do Trapézio Composta ¶ A regra do trapézio composta equivale a aplicar a regra simples para cada subintervalo de [ a , b ] [a,b] [ a , b ]
∫ x 0 x n f ( x ) d x ≈ h 2 [ f ( x 0 ) + 2 ( f ( x 1 ) + f ( x 2 ) + … + f ( x n − 1 ) ) + f ( x n ) ] , \int_{x_0}^{x_n} f(x) \, dx \approx \dfrac{h}{2}[f(x_0) + 2( f(x_1) + f(x_2) + \ldots + f(x_{n-1})) + f(x_n)], ∫ x 0 x n f ( x ) d x ≈ 2 h [ f ( x 0 ) + 2 ( f ( x 1 ) + f ( x 2 ) + … + f ( x n − 1 )) + f ( x n )] , Regra 1/3 de Simpson ¶ Quando n = 2 n=2 n = 2
∫ x 0 x 2 f ( x ) d x ≈ f 0 h C 0 2 + f 1 h C 1 2 + f 2 h C 2 2 . \int_{x_0}^{x_2} f(x) \, dx \approx f_0 h C_0^2 + f_1 h C_1^2 + f_2 h C_2^2. ∫ x 0 x 2 f ( x ) d x ≈ f 0 h C 0 2 + f 1 h C 1 2 + f 2 h C 2 2 . Porém, notemos que:
C 0 2 = ∫ 0 2 λ 2 , 0 ( u ) d u = ∫ 0 2 ( u − 1 ) ( u − 2 ) ( 0 − 1 ) ( 0 − 2 ) d u = 1 2 ∫ 0 2 u 2 − 3 u + 2 d u = 1 3 C 1 2 = ∫ 0 2 λ 2 , 1 ( u ) d u = ∫ 0 2 ( u − 0 ) ( u − 2 ) ( 1 − 0 ) ( 1 − 2 ) d u = − ∫ 0 2 u 2 − 2 u d u = 4 3 C 2 2 = 1 3 \begin{aligned}
C_0^2 &= \int_{0}^{2} \lambda_{2,0}(u) \, du = \int_{0}^{2} \dfrac{(u-1)(u-2)}{(0-1)(0-2)} \, du = \dfrac{1}{2}\int_{0}^{2} u^2 - 3u + 2 \, du = \dfrac{1}{3} \\
C_1^2 &= \int_{0}^{2} \lambda_{2,1}(u) \, du = \int_{0}^{2} \dfrac{(u-0)(u-2)}{(1-0)(1-2)} \, du = -\int_{0}^{2} u^2 - 2u \, du = \dfrac{4}{3} \\
C_2^2 &= \dfrac{1}{3} \end{aligned} C 0 2 C 1 2 C 2 2 = ∫ 0 2 λ 2 , 0 ( u ) d u = ∫ 0 2 ( 0 − 1 ) ( 0 − 2 ) ( u − 1 ) ( u − 2 ) d u = 2 1 ∫ 0 2 u 2 − 3 u + 2 d u = 3 1 = ∫ 0 2 λ 2 , 1 ( u ) d u = ∫ 0 2 ( 1 − 0 ) ( 1 − 2 ) ( u − 0 ) ( u − 2 ) d u = − ∫ 0 2 u 2 − 2 u d u = 3 4 = 3 1 Logo,
∫ x 0 x 2 f ( x ) d x ≈ h 3 ( f 0 + 4 f 1 + f 2 ) = h 3 [ f ( x 0 ) + 4 f ( x 1 ) + f ( x 2 ) ] , \int_{x_0}^{x_2} f(x) \, dx \approx \dfrac{h}{3}(f_0 + 4f_1 + f_2) = \dfrac{h}{3}[f(x_0) + 4f(x_1) + f(x_2)], ∫ x 0 x 2 f ( x ) d x ≈ 3 h ( f 0 + 4 f 1 + f 2 ) = 3 h [ f ( x 0 ) + 4 f ( x 1 ) + f ( x 2 )] , que é a tradicional regra 1/3 de Simpson simples .
Regra 1/3 de Simpson composta ¶ A regra 1/3 de Simpson composta é obtida de forma similar. Dividimos o intervalo de integração em subintervalos de comprimento h = ( b − a ) / 2 n h = (b-a)/2n h = ( b − a ) /2 n
∫ x 0 x 2 n f ( x ) d x ≈ h 3 [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + … + 2 f ( x 2 n − 2 ) + 4 f ( x 2 n − 1 ) + f ( x 2 n ) ] , \int_{x_0}^{x_{2n}} f(x) \, dx \approx \dfrac{h}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + \ldots + 2f(x_{2n-2}) + 4f(x_{2n-1}) + f(x_{2n})], ∫ x 0 x 2 n f ( x ) d x ≈ 3 h [ f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + … + 2 f ( x 2 n − 2 ) + 4 f ( x 2 n − 1 ) + f ( x 2 n )] , Regra 3/8 de Simpson ¶ Quando n = 3 n=3 n = 3
∫ x 0 x 3 f ( x ) d x ≈ f 0 h C 0 3 + f 1 h C 1 3 + f 2 h C 2 3 + f 2 h C 3 3 . \int_{x_0}^{x_3} f(x) \, dx \approx f_0 h C_0^3 + f_1 h C_1^3 + f_2 h C_2^3 + f_2 h C_3^3. ∫ x 0 x 3 f ( x ) d x ≈ f 0 h C 0 3 + f 1 h C 1 3 + f 2 h C 2 3 + f 2 h C 3 3 . Porém, notemos que:
C 0 3 = ∫ 0 3 λ 3 , 0 ( u ) d u = ∫ 0 3 ( u − 1 ) ( u − 2 ) ( u − 3 ) ( 0 − 1 ) ( 0 − 2 ) ( 0 − 3 ) d u = − 1 6 ∫ 0 3 u 3 − 6 u 2 + 11 u − 6 d u = 3 8 C 1 3 = ∫ 0 3 λ 3 , 1 ( u ) d u = ∫ 0 3 u ( u − 2 ) ( u − 3 ) ( 1 − 0 ) ( 1 − 2 ) ( 1 − 3 ) d u = 1 2 ∫ 0 3 u 3 − 5 u 2 + 6 u d u = 9 8 C 2 3 = 9 8 C 3 3 = 3 8 \begin{aligned}
C_0^3 &= \int_{0}^{3} \lambda_{3,0}(u) \, du = \int_{0}^{3} \dfrac{(u-1)(u-2)(u-3)}{(0-1)(0-2)(0-3)} \, du = -\dfrac{1}{6}\int_{0}^{3} u^3 - 6u^2 + 11u - 6 \, du = \dfrac{3}{8} \\
C_1^3 &= \int_{0}^{3} \lambda_{3,1}(u) \, du = \int_{0}^{3} \dfrac{u(u-2)(u-3)}{(1-0)(1-2)(1-3)} \, du = \dfrac{1}{2}\int_{0}^{3} u^3 - 5u^2 + 6u \, du = \dfrac{9}{8} \\
C_2^3 &= \dfrac{9}{8} \\
C_3^3 &= \dfrac{3}{8}\end{aligned} C 0 3 C 1 3 C 2 3 C 3 3 = ∫ 0 3 λ 3 , 0 ( u ) d u = ∫ 0 3 ( 0 − 1 ) ( 0 − 2 ) ( 0 − 3 ) ( u − 1 ) ( u − 2 ) ( u − 3 ) d u = − 6 1 ∫ 0 3 u 3 − 6 u 2 + 11 u − 6 d u = 8 3 = ∫ 0 3 λ 3 , 1 ( u ) d u = ∫ 0 3 ( 1 − 0 ) ( 1 − 2 ) ( 1 − 3 ) u ( u − 2 ) ( u − 3 ) d u = 2 1 ∫ 0 3 u 3 − 5 u 2 + 6 u d u = 8 9 = 8 9 = 8 3 Logo,
∫ x 0 x 3 f ( x ) d x ≈ 3 h 8 ( f 0 + 3 f 1 + 3 f 2 + f 3 ) = 3 h 8 [ f ( x 0 ) + 3 ( f ( x 1 ) + f ( x 2 ) ) + f ( x 3 ) ] , \int_{x_0}^{x_3} f(x) \, dx \approx \dfrac{3h}{8}(f_0 + 3f_1 + 3f_2 + f_3) = \dfrac{3h}{8}[f(x_0) + 3(f(x_1) + f(x_2)) + f(x_3)], ∫ x 0 x 3 f ( x ) d x ≈ 8 3 h ( f 0 + 3 f 1 + 3 f 2 + f 3 ) = 8 3 h [ f ( x 0 ) + 3 ( f ( x 1 ) + f ( x 2 )) + f ( x 3 )] , que é regra 3/8 de Simpson simples .
Regra 3/8 de Simpson composta ¶ A regra 3/8 de Simpson composta é obtida de forma similar. Dividimos o intervalo de integração em subintervalos de comprimento h = ( b − a ) / 3 n h = (b-a)/3n h = ( b − a ) /3 n
∫ x 0 x 3 n f ( x ) d x ≈ 3 h 8 [ f ( x 0 ) + 3 ( f ( x 1 ) + f ( x 2 ) ) + 2 f ( x 3 ) + … + + 2 f ( x 3 n − 3 ) + 3 ( f ( x 3 n − 2 ) + f ( x 3 n − 1 ) ) + f ( x 3 n ) ] , \int_{x_0}^{x_{3n}} f(x) \, dx \approx \dfrac{3h}{8}[f(x_0) + 3(f(x_1) + f(x_2)) + 2f(x_3) + \ldots + \\
+ \ \ 2f(x_{3n-3}) + 3(f(x_{3n-2}) + f(x_{3n-1})) + f(x_{3n})], ∫ x 0 x 3 n f ( x ) d x ≈ 8 3 h [ f ( x 0 ) + 3 ( f ( x 1 ) + f ( x 2 )) + 2 f ( x 3 ) + … + + 2 f ( x 3 n − 3 ) + 3 ( f ( x 3 n − 2 ) + f ( x 3 n − 1 )) + f ( x 3 n )] ,