Numerical Integration

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Numerical Integration

Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

June 7, 2009

T.M. Huang (Nat. Taiwan Normal Univ.) Numerical Integration June 7, 2009 1 / 7

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If we choose x0= a, x1= ¹₂(a + b), x2 = b, h = (b − a)/2, and the second order Lagrange polynomial to interpolate f (x), then

Z b a

f (x)dx = Z x2

x0

 (x − x₁)(x − x₂)

(x0− x₁)(x0− x₂)f (x₀) + (x − x₀)(x − x₂) (x1− x₀)(x1− x₂)f (x₁) + (x − x₀)(x − x₁)

(x2− x₀)(x2− x₁)f (x2)

 dx +

Z x2

x0

(x − x₀)(x − x₁)(x − x₂)

6 f⁽³⁾(ζ(x))dx.

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It holds that Z b

a

f (x)dx = h 1

3f (x0) +4

3f (x1) +1 3f (x2)



+ Z x2

x0

(x − x0)(x − x1)(x − x2)

6 f⁽³⁾(ζ(x))dx.

It can show that there exists ξ ∈ (a, b) such that Z b

a

f (x) dx = h

3 [f (x₀) + 4f (x₁) + f (x₂)] − f⁽⁴⁾(ξ) 90 h⁵. This gives theSimpson’s rule formulation.

T.M. Huang (Nat. Taiwan Normal Univ.) Numerical Integration June 7, 2009 3 / 7

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To illustrate the procedure, we choose an even integer n and partition the interval [a, b] into n subintervals by nodes a = x₀< x₁ < · · · < x_n= b, and apply Simpson’s rule on each consecutive pair of subintervals. With

h = b − a

n and xj = a + jh, j = 0, 1, . . . , n, we have on each interval [x2j−2, x2j],

Z x2j

x2j−2

f (x) dx = h

3 [f (x2j−2) + 4f (x2j−1) + f (x2j)] − h⁵

90f⁽⁴⁾(ξj), for some ξ_j ∈ (x_2j−2, x_2j), provided that f ∈ C⁴[a, b].

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The composite rule is obtained by summing up over the entire interval, that is,

Z b a

f (x) dx =

n/2

X

j=1

Z x2j

x2j−2

f (x) dx

=

n/2

X

j=1

 h

3(f (x2j−2) + 4f (x2j−1) + f (x2j)) − h⁵

90f⁽⁴⁾(ξj)



= h

3[f (x0) + 4f (x1) + f (x2) +f (x₂) + 4f (x₃) + f (x₄) +f (x4) + 4f (x5) + f (x6)

...

+f (xn−2) + 4f (xn−1) + f (xn)] − h⁵ 90

n/2

X

j=1

f⁽⁴⁾(ξj)

T.M. Huang (Nat. Taiwan Normal Univ.) Numerical Integration June 7, 2009 5 / 7

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Hence Z b

a

f (x) dx = h

3[f (x₀) + 4f (x₁) + 2f (x₂) + 4f (x₃) + 2f (x₄) + 4f (x₅) + · · · + 2f (xn−2) + 4f (xn−1) + f (xn)] − h⁵

90

n/2

X

j=1

f⁽⁴⁾(ξj)

= h

3



f (x0) + 4

n/2

X

j=1

f (x2j−1) + 2

(n/2)−1

X

j=1

f (x2j) + f (xn)





−b − a

180 f⁽⁴⁾(µ)h⁴, for some µ ∈ (a, b).

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Composite Simpson’s Rule

Z _b

a

f (x) dx = h 3



f (a) + 4

n/2

X

j=1

f (x_2j−1) + 2

(n/2)−1

X

j=1

f (x_2j) + f (b)





−b − a

180 f⁽⁴⁾(µ)h⁴,

where n is an even integer, h = (b − a)/n, x_j = a + jh, for j = 0, 1, . . . , n, and some µ ∈ (a, b).

T.M. Huang (Nat. Taiwan Normal Univ.) Numerical Integration June 7, 2009 7 / 7