(190 SlideShares)
, STATISTICAL CONSULTANT/ANALYST/TUTOR/CIVIL ENGINEER /MATHEMATICIAN/SUBJECT MATTER EXPERT
at CEO @ TG TUTORIALS / S.M.E @ TUTOR VISTA AND STUDENT OF FORTUNE ,
1. Gaussian Quadratures
• Newton-Cotes Formulae
– use evenly-spaced functional values
– Did not use the flexibility we have to select the quadrature points
• In fact a quadrature point has several degrees of freedom.
Q(f)=∑i=1m ci f(xi)
A formula with m function evaluations requires specification of
2m numbers ci and xi
• Gaussian Quadratures
– select both these weights and locations so that a higher order
polynomial can be integrated (alternatively the error is proportional
to a higher derivatives)
• Price: functional values must now be evaluated at nonuniformly distributed points to achieve higher accuracy
• Weights are no longer simple numbers
• Usually derived for an interval such as [-1,1]
• Other intervals [a,b] determined by mapping to [-1,1]
2. Gaussian Quadrature on [-1, 1]
∫
1
−1
n
f ( x )dx ≈ ∑ c i f ( x i ) = c 1 f ( x 1 ) + c 2 f ( x 2 ) + L + c n f ( x n )
i =1
• Two function evaluations:
– Choose (c1, c2, x1, x2) such that the method yields “exact
integral” for f(x) = x0, x1, x2, x3
n=2:
∫
1
−1
f(x)dx
= c 1 f(x 1 ) + c 2 f(x 2 )
-1
x1
x2
1
3. Finding quadrature nodes and weights
• One way is through the theory of orthogonal
polynomials.
• Here we will do it via brute force
• Set up equations by requiring that the 2m points
guarantee that a polynomial of degree 2m-1 is integrated
exactly.
• In general process is non-linear
– (involves a polynomial function involving the unknown point
and its product with unknown weight)
– Can be solved by using a multidimensional nonlinear solver
– Alternatively can sometimes be done step by step
4. Gaussian Quadrature on [-1, 1]
n=2:
∫
1
−1
f(x)dx = c 1 f(x 1 ) + c 2 f(x 2 )
Exact integral for f = x0, x1, x2, x3
– Four equations for four unknowns
⎧f
⎪
⎪
⎪f
⎨
⎪f
⎪
⎪f
⎩
1
= 1 ⇒ ∫ 1dx = 2 = c 1 + c 2
−1
1
= x ⇒ ∫ xdx = 0 = c 1 x 1 + c 2 x 2
−1
2
2
2
= x ⇒ ∫ x dx = = c 1 x 1 + c 2 x 2
−1
3
2
1
2
1
3
3
= x ⇒ ∫ x 3 dx = 0 = c 1 x 1 + c 2 x 2
3
−1
1
I = ∫ f ( x )dx = f ( −
−1
⎧c 1 = 1
⎪c = 1
⎪ 2
−1
⎪
⇒ ⎨ x1 =
3
⎪
1
⎪
⎪ x2 = 3
⎩
1
3
)+ f (
1
3
)
5. Error
• If we approximate a function with a Gaussian quadrature
formula we cause an error proportional to 2n th
derivative
6. Gaussian Quadrature on [-1, 1]
n=3:
∫
1
−1
f ( x )dx = c 1 f ( x 1 ) + c 2 f ( x 2 ) + c 3 f ( x 3 )
x2
x3
-1 x1
1
• Choose (c1, c2, c3, x1, x2, x3) such that the method
yields “exact integral” for f(x) = x0, x1, x2, x3,x4, x5
8. Gaussian Quadrature on [-1, 1]
Exact integral for f = x0, x1, x2, x3, x4, x5
I=∫
1
−1
3
5
3
8
5
)
f ( x )dx = f ( −
)+ f (0 )+ f (
9
5
9
5
9
9. Gaussian Quadrature on [a, b]
Coordinate transformation from [a,b] to [-1,1]
b−a
b+a
t=
x+
2
2
⎧ x = −1 ⇒ t = a
⎨
⎩x = 1 ⇒ t = b
a
∫
b
a
t1
f ( t )dt = ∫
1
−1
t2
b
1
b−a
b+a b−a
f(
x+
)(
)dx = ∫ g ( x )dx
−1
2
2
2
10. Example: Gaussian Quadrature
4
I = ∫ te dt = 5216.926477
Evaluate
2t
0
Coordinate transformation
b−a
b+a
t=
x+
= 2 x + 2; dt = 2dx
2
2
4
1
I = ∫ te dt = ∫ (4 x + 4)e
2t
4 x+4
−1
0
1
dx = ∫ f ( x)dx
−1
Two-point formula
1
I = ∫ f ( x )dx = f (
−1
−1
)+ f (
1
)= (4 −
4
)e
3
3
3
= 9.167657324 + 3468.376279 = 3477.543936
4−
4
3
+(4 +
4
)e
3
( ε = 33.34%)
4+
4
3
11. Example: Gaussian Quadrature
Three-point formula
1
5
8
5
I = ∫ f ( x )dx = f ( − 0.6 ) + f ( 0 ) + f ( 0.6 )
−1
9
9
9
5
8
5
4 − 0 .6
4
= ( 4 − 4 0.6 )e
+ ( 4 )e + ( 4 + 4 0.6 )e 4 + 0.6
9
9
9
5
8
5
= ( 2.221191545 ) + ( 218.3926001 ) + ( 8589.142689 )
9
9
9
= 4967.106689
( ε = 4.79%)
Four-point formula
I = ∫ f ( x )dx = 0.34785[ f ( −0.861136 ) + f ( 0.861136 )]
1
−1
+ 0.652145 [ f ( −0.339981 ) + f ( 0.339981 )]
= 5197.54375
( ε = 0.37%)
12. Other rules
• Gauss-Lobatto:
– requiring end points be included in the formula
• Gauss-Radau
– Require one end point be in the formula
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