45134583 quadrature

7/27/2019 45134583 Quadrature

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Numerical Integration/Quadrature

We just finished with differentiation, so let’s look now to integration. Remember that an

integral is associated with the area under a curve. We may approach the computation

of an integral as a sum:

Z b

a

f ( x)dx ≈ Q[ f ] = M

∑k =0

wk f ( xk ) = w0 f ( x0) + w1 f ( x1) + w2 f ( x2) + . . . + w M f ( x M )

where a = x0 < x1 < . . . < x M = b and

Z b

a f ( x)dx = Q[ f ] + E [ f ]

If we integrate by representing f ( x) as a polynomial:

f ( x) ≈ P N ( x) = a N x N + a N −1 x

N −1 + . . . + a1 x + a0

then the truncation error E [ f ] = K f ( N +1)(c).

This process leads to the Newton-Cotes formulas.

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0 2 4 6 8 10 120

5

10

Interpolation for Trapezoidal Rule

0 2 4 6 8 10 120

5

10

Interpolation for Simpson′s Rule

0 2 4 6 8 10 120

5

10

Interpolation for Simpson′s 3/8 Rule

5

10

Interpolation for the n=4 Newton−Cotes Formula

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Newton-Cotes FormulasAssume xk = x0 + h ·k and f k = f ( xk ). Then:

I.

Z

x1

x0

f ( x) dt =h

2( f 0 + f 1)−

h3

12 f (2)(c) Trapezoidal rule

II.

Z x2

x0

f ( x) dt =h

3( f 0 +4 f 1 + f 2)−

h5

90 f (4)(c) Simpson’s rule

III.

Z x3

x0 f ( x) dt =

3h

8 ( f 0 +3 f 1 +3 f 2 + f 3)−

3h5

80 f (4)

(c) Simpson’s 3/8 rule

IV.

Z x4

x0

f ( x) dt =2h

45(7 f 0 +32 f 1 +12 f 2 +32 f 3 +7 f 4)−

8h7

945 f (6)(c) n = 4

How do we get these formulas? Lagrange interpolation:

f ( x) ≈ P M ( x) = M

∑k =0

f k L M ,k ( x)

Z x M

x0

f ( x) dx ≈

Z x M

x0

P M ( x) dx =Z x M

x0

M

∑k =0

f k L M ,k ( x)

dx =

M

∑k =0

f k

Z x M

x0

L M ,k ( x) dx

As a result,R x M

x0f

( x

)dx ≈ ∑ M

k =0w

k f

k where the weights w

k =

R x M

x0L

M ,k ( x

)dx.

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Example: Simpson’s Rule

P2( x) = f 0( x− x1)( x− x2)

( x0− x1)( x0− x2)+ f 1

( x− x0)( x− x2)

( x1− x0)( x1− x2)+ f 2

( x− x0)( x− x1)

( x2− x0)( x2− x1)

Let’s integrate our interpolant to find an approximation of the integral:

Z x2

x0 f ( x) dx ≈ f 0

Z x2

x0( x−

x1)( x−

x2)( x0− x1)( x0− x2)dx + f 1

Z x2

x0( x−

x0)( x−

x2)( x1− x0)( x1− x2)dx

+ f 2

Z x2

x0

( x− x0)( x− x1)

( x2− x0)( x2− x1)dx

Let’s evaluate the integrals using even spacing and a change of variables: t = ( x−

x0)/h, dt = dx/h, x1 = x0 + h and x2 = x0 +2h:Z x2

x0

f ( x) dx =Z 2

0 f (t ) h dt ≈ f 0

Z 2

0

h2(t −1)(t −2)

(−h)(−2h)h dt + f 1

Z 2

0

h2(t )(t −2)

(h)(−h)h dt

+ f 2

Z 2

0

h2(t )(t −1)

(2h)(h)h dt

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Z x2

x0

f ( x) dx ≈

h f 0

2

Z 2

0

(t 2−3t +2) dt −h f 1

Z 2

0

(t 2−2t ) dt +h f 2

2

Z 2

0

(t 2− t ) dt

=h f 0

2

t 3

3−

3t 2

2+2t

2

0

−h f 1

t 3

3− t 2

2

0

+

h f 2

2

t 3

3−

t 2

2

2

0

=h f 0

2 8

3

−

12

2

+4−h f 18

3

−4+h f 2

2 8

3

−

4

2=

h

3( f 0 +4 f 1 + f 2)

which gives us Simpson’s rule.

If we wish to solve over a large interval, we can combine the trapezoidal rule orSimpson’s rule into composite formulas:

Z x4

x0

f ( x) dx =

Z x1

x0

f ( x) dx +

Z x2

x1

f ( x) dx +

Z x3

x2

f ( x) dx +

Z x4

x3

f ( x) dxComposite

Trapezoidal ruleZ x4

x0

f ( x) dx =Z x2

x0

f ( x) dx +Z x4

x2

f ( x) dx Composite Simpson’s rule

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Composite FormulasSuppose we want to integrate

R ba f ( x) dx. Divide the interval [a, b] into M subintervals

[ xk , xk +1] such that xk = a + h · k . Integrate using trapezoidal rule:

Z b

a f ( x) dx =

M

∑k =1

Z xk

xk −1

f ( x) dx ≈ M

∑k =1

h

2[ f ( xk −1) + f ( xk )]

=h

2[ f 0 +2 f 1 +2 f 2 + . . . +2 f M −1 + f M ] =

h

2( f 0 + f M ) + h

M −1

∑k =1

f k

Composite Trapezoidal Rule

We can use the same approach with Simpson’s rule as long as the number of intervals

is a multiple of 2 (say, 2 M with h = (b−a)/(2 M ).):

Z b

a f ( x) dx =

M

∑k =1

Z x2k

x2k −

2

f ( x) dx ≈ M

∑k =1

h

3[ f ( x2k −2) +4 f ( x2k −1) + f ( x2k )]

=h

3[ f 0 +4 f 1 +2 f 2 +4 f 3 +2 f 4 + . . . +2 f M −2 +4 f M −1 + f M ]

=h

3( f 0 + f M ) +

2h

3

M −1

∑k =1

f 2k +4h

3

M

∑k =1

f 2k −1

Composite Simpson’s Rule

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Error AnalysisWith the trapezoidal rule, we approximate the function f ( x) using linear interpolation

between successive points and then integrating the interpolant. The true function

may be written:

f ( x) = P( x) + E ( X ) = f 0 + xf 1− f 0

x1− x0+

( x− x0)( x− x1)

2f (2)(c)

Integrating and changing variables to t = x− x0:

Z

x1

x0

f ( x) dx =h

2( f 0 + f 1) +

Z

x1

x0

( x− x0)( x− x1)2

f (2)(c) dx

=h

2( f 0 + f 1) +

f (2)(c)

2

Z h

0t (t −h) dt

=h

2( f 0 + f 1) +f (2)(c)

2 t 3

3 −ht 2

2

h

0

=

h

2( f 0 + f 1)−

h3 f (2)(c)

12

As a result, the composite trapezoidal rule is accurate locally to O(h3) even though

the approximation to the function is just linear.

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However, we are really interested in the global error over the whole interval [a, b], not

just a single sub-interval [ xk −1, xk ]. (In going from the 2nd to the 3rd step, we use the

identity Mh = b−a.)

Z b

a f ( x) dx =

M

∑k =1

Z xk

xk −1

f ( x) dx = M

∑k =1

h

2( f k −1 + f k )−

h3 f (2)(c)

12

=h

2( f 0 +2 f 1 + . . . +2 f M −1 + f M )−

h3 f (2)(c)12

M

∑k =1

1

=h

2( f 0 +2 f 1 + . . . +2 f M −1 + f M )−

h2(b−a)

12f (2)(c)

Therefore, the global error for composite trapezoidal rule is O(h2) while the local error

is O(h3).

Similarly, the global error for composite Simpson’s rule is:−h4(b−a)

180f (4)(c).

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Example

Let’s put the composite formulas to work:

Z ln(2)

0exp( x) dx = 1

N Trapezoidal – O(h2) Simpson – O(h4) n = 4 – O(h6)

4 1.00250110798165 1.00000499159078 1.00000005641290

8 1.00062551160333 1.00000031281055 1.00000000089187

16 1.00015639257365 1.00000001956376 1.00000000001398

32 1.00003909906062 1.00000000122294 1.00000000000022

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Numerical Integration and Recursion

We would like to improve our the quality of our numerical integration without the com-

plication of deriving higher order schemes.

One approach is to recursively refine the intervals over which we integrate, i.e. from

2h to h to h/2 to h/4, etc. To start, let xk = a + k ·h with

a = x0 < x1 < x2 < . . . < x2 M −2 < x2 M −1 < x2 M = b

and consider composite trapezoidal rule with a spacing of 2h:

Z b

a f

( x

)dx

≈T

( f ,2h

) =

2h

2

M

∑k =1 [ f

(2k −2

) +f

(2k

)]

=2h

2[( f 0 + f 2) + ( f 2 + f 4) + . . . + ( f 2 M −2 + f 2 M )]

=2h

2[( f 0 +2 f 2 +2 f 4 + . . . +2 f 2 M −2 + f 2 M ]

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Now, consider composite trapezoidal with a spacing of h:Z b

a f ( x) dx ≈ T ( f , h) =

h

2

2 M

∑k =1

[ f (k −1) + f (k )]

=h

2[( f 0 +2 f 1 +2 f 2 + . . . +2 f 2 M −1 + f 2 M ]

Notice that T ( f , h) may be rearranged to give:

=h

2[ f 0 +2 f 2 +2 f 4 + . . . +2 f 2 M −2 + f 2 M ] +

h

2[2 f 1 +2 f 3 + . . . +2 f 2 M −1]

T ( f , h) =T ( f ,2h)

2+ h

M

∑

k =1

f 2k −1

This method allows you to go from a spacing of h to h/2 without having to repeat

the work that you have already done. Let’s look next to combining successive O(h2)

approximations to give an O(h4) approximation.

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Recursive Simpson’s Rule

If [a, b] is broken up into intervals of length h, then

S ( f , h) =4T ( f , h)−T ( f ,2h)

3

so that you can compute Simpson’s rule using two trapezoidal rule computations.

Why?:

Z b

a f ( x) dx≈ T ( f , h) = h

2( f 0 +2 f 1 +2 f 2 + . . . +2 f 2 M −1 + f 2 M )

4T ( f , h) = h (2 f 0 +4 f 1 +4 f 2 + . . . +4 f 2 M −1 +2 f 2 M )Z b

a f ( x) dx≈ T ( f ,2h) =

2h

2( f 0 +2 f 2 +2 f 4 + . . . +2 f 2 M −2 + f 2 M )

3

Z b

a f ( x) dx≈ 4T ( f , h)−T ( f ,2h) = h ( f 0 +4 f 1 +2 f 2 + . . . +2 f 2 M −2 +4 f 2 M −1 + f 2 M ) Simpson’s Rule

Z b

a f ( x) dx≈ S ( f , h) =

4T ( f , h)−T ( f ,2h)

3

Therefore, we can compute a numerical approximation that is O(h4) by combining

successive approximations using trapezoidal rule, an O(h2) method!

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Romberg Integration

Romberg integration allows us to recursively reduce the error in integration from

O(h2k ) to O(h2k +2) without deriving higher order quadrature formulas.

It can be shown that:Z b

a f ( x) dx = T ( f , h) + E T ( f , h) where E T ( f , h) = a1h2 + a2h4 + a3h6 + . . .

Then consider two Trapezoidal rule approximations with spacings 2h and h:

2h step

Z b

a f ( x) dx = T ( f ,2h) + a1 4h2 + a2 16h4 + a3 64h6 + . . .

h step

Z b

a f ( x) dx = T ( f , h) + a1 h2 + a2 h4 + a3 h6 + . . .

Multiplying second equation by four and subtracting the first equation yields:

3

Z b

a f ( x) dx = 4T ( f , h)−T ( f ,2h) + a2 12h4 + a3 60h6 + . . .

Z b

a f ( x) dx =

4T ( f , h)−T ( f ,2h)

3+ a2 4h4 + a3 20h6 + . . .

= S ( f , h) + O(h

4

)

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Can we continue to improve accuracy in this way?

Suppose we are given a quadrature scheme R(

h,

k −1)

of accuracy O(

h2k

):

R(2h, k −1) ≈ Q =Z b

a f ( x) dx

R(h, k −1) ≈ Q

Then, the error can be expanded in a series as follows:

Q = R(h, k −1) + c1h2k + c2h2k +2 + . . .

Q = R(2h, k −1) + c1 4k h2k + c2 4

k +1h2k +2 + . . .

Using Richardson extrapolation, we can find:

Z b

a f ( x) dx = Q =

4k R(h, k −1)−

R(2h, k −1)

4k −1 R(h,k )

+O(h2k +2)

Therefore, with two successive approximations of O(h2k ) accuracy, we can generate

an approximation of order O(h2k +2).

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We can compute successively better approximations using:

R(h, k ) =4k R(h, k −1)− R(2h, k −1)

4k −1

= R(h, k −1) + R(h, k −1)− R(2h, k −1)

4k −1

Romberg integration leads to the following tables:

R(h,0)

R(h/2,0) → R(h/2,1)

R(h/4,0) → R(h/4,1) → R(h/4,2)

R(h/8,0)

→R(h/8,1)

→R(h/8,2)

→R(h/8,3)

R(h/16,0) → R(h/16,1) → R(h/16,2) → R(h/16,3) → R(h/16,4)... ... ... ... . . .

Trapezoid Simpson’s O(h6) O(h8) O(h10)

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Example

Let’s put the Romberg to work, starting with h = b−a.Z ln(2)

0exp( x) dx = 1

N R(ln(2)/ N ,0) R(ln(2)/ N ,1) R(ln(2)/ N ,2) R(ln(2)/ N ,3)

1 1.03972077083992

2 1.00998945715423 1.00007901925900

4 1.00250110798165 1.00000499159078 1.00000005641290

8 1.00062551160333 1.00000031281055 1.00000000089187 1.00000000001058

Note that:

R(h,1) =4 R(h,0)− R(2h,0)

3R(h,2) =

16 R(h,1)− R(2h,1)

15

R(h,3) =64 R(h,2)− R(2h,2)

63

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Gaussian Quadrature

Instead of evaluating a function at equally spaced points and combining those values

to approximate an integral (as in the Newton-Cotes formulas), try to choose the points

optimally to maximize the order of accuracy of the scheme.

Z b

a

f ( x) dx

≈

N

∑k =1

ck f ( xk )

We would like our formula to be exact for polynomials up to as high as degree as

possible. We have N coefficients ci and N points xi ∈ [a, b]. This gives us 2 N degrees

of freedom which can be used to exactly interpolate and integrate polynomials of

degree up to (2 N −1).

Let’s try the case with N = 2 to demonstrate the derivation and power of the method.

Consider [a, b] = [−1,1] for simplicity:

Z 1

−1

f ( x) dx≈ c1 f ( x1) + c2 f ( x2)

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We would like this to be an exact representation for polynomials of degree as high as

three. Assume the function is a cubic polynomial:

f ( x) = a0 + a1 x + a2 x2 + a3 x3

This implies that:

Z 1

−1 f ( x) dx =

Z 1

−1

a0 + a1 x + a2 x

2 + a3 x3

dx

= c1a0 + a1 x1 + a2 x

2

1 + a3 x

3

1+ c2a0 + a1 x2 + a2 x

2

2 + a3 x

3

2a0

c1 + c2−

Z 1

−1dx

+ a1

c1 x1 + c2 x2−

Z 1

−1 x dx

+ a2

c1 x

21 + c2 x

22−

Z 1

−1 x2 dx

+a3

c1 x

31 + c2 x

32−

Z 1

−1 x3 dx

= 0

Since a0, a1, a2 and a3 are arbitrary, then their coefficients must all be zero. Thisimplies:

c1 + c2 =Z 1

−1dx = 2 c1 x1 + c2 x2 =

Z 1

−1 x dx = 0

c1 x2

1+ c

2 x2

2=

Z 1

−1 x2 dx =

2

3c1 x3

1+ c

2 x3

2=

Z 1

−1 x3 dx = 0

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Some algebra leads to:

c1 = 1 c2 = 1 x1 =−

√ 3

3 x2 =

√ 3

3

Therefore:

Z 1

−1 f ( x) dx≈ f

−√ 3

3

+ f

√ 3

3

These points and coefficients can be derived using the Legendre polynomials andare readily available in tables in our book (up to N=5) and elsewhere. These formulas

can be very powerful. They can allow you to exact integrate a polynomial in less time

than it would take to compute the analytical/exact solution.

Trick: If you are integrating over an interval other than [−1,1], change variables.Example: x ∈ [a, b]. Let t =

2( x−a)(b−a)

−1 so that x ∈ [a, b]←→ t ∈ [−1,1].

Z b

a f ( x) dx =

Z 1

−1 f

(b−a)(t +1)

2+ a

b−a

2dt

45134583 quadrature

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