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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