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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature) Numerical Analysis and Computing^ Lecture Notes #07— Numerical Differentiation and Integration —
Differentiation; Richardson’s Extrapolation; Integration
Peter Blomgren, 〈[email protected]
Department of Mathematics and Statistics
Dynamical Systems GroupComputational Sciences Research Center San Diego State UniversitySan Diego, CA 92182-7720 http://terminus.sdsu.edu/^ Fall 2009 Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (1/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Outline^1 Numerical Differentiation
Ideas and Fundamental Tools Moving Along... 2 Richardson’s Extrapolation A Nice Piece of “Algebra Magic” Homework #6 – Preliminary Version 3 Numerical Integration (Quadrature) The “Why?” and Introduction Trapezoidal & Simpson’s Rules Newton-Cotes Formulas Homework #6 – Final Version Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (2/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
Numerical Differentiation: The Big Picture^ The goal of numerical differentiation is to compute an accurateapproximation to the derivative(s) of a function.^ Given
measurements
n {f}i i=^ of the underlying function
f^ (x) at the
node values
n {x}i i= , our task is to estimate
′ f(x) (and, later,
higher derivatives) in the same nodes. The strategy:
Fit a polynomial to a cleverly selected subset of thenodes, and use the derivative of that polynomial asthe approximation of the derivative. Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (3/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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Numerical Differentiation^ Definition (Derivative as a limit)^ The derivative of
f^ at^ x^0
is ′ f (x) = lim^0
f^ (x^0 h→ 0 +^ h)^ −^ f^ (x)^0. h
The obvious approximation is to fix
h^ “small” and compute ′ f (x)^ ≈^0 f^ (x+^0 h)^ −^ f^ (x
) 0. h
Problems:
Cancellation and roundoff errors. — For small valuesof^ h,^ f
(x+h)^0 ≈^ f^ (x) so the difference may have very^0 few^ significant digits
in finite precision arithmetic. ⇒^ smaller
h^ not necessarily better numerically. Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (4/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
Main Tools for Numerical Differentiation
1 of 2
In the discussion on Numerical Differentiation (and laterIntegration) we will rely on our old friend (nemesis?) — the Taylorexpansions... Theorem (Taylor’s Theorem) Suppose f
n ∈ C [a
(n+1), b], f ∃^ on^ [a,^ b], and x
∈^ [a,^ b 0 ]. Then
∀x^ ∈^ (a,
b),^ ∃ξ(x
)^ ∈^ (min(
x,^ x),^ max(^0
x,^ x))^ with^0
f^ (x) =^ P
(x) +^ Rn
(x)^ wheren nX P(x) =n k= (k) f (x)^0 (x^ k!^
k^ − x),^ R 0
(nf (^) (x) = (^) n +1)(ξ(x))( (n^ + 1)!^
(n+1)x − x) 0 .
P(x)^ is then
Taylor polynomial of degree
n, and
R(x)^ is then
remainder term
(truncation error).
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (5/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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Main Tools for Numerical Differentiation
2 of 2
Our second tool for building Differentiation and Integrationschemes are the
Lagrange Coefficients^ Ln,
(x) =k
n∏^ x^ − j=0,j^6 =k
xj x−^ xk j
Recall:^
L(x) is then,k^
nth degree polynomial which is 1 in
xandk^
0 in the other nodes (
x,^ j^6 =^ kj^
Previously we have used the family
L(x),n,^0
L(x),n,^1
...^ ,^ Ln,
(x) ton
build the
Lagrange interpolating polynomial.
— A good tool for
discussing polynomial behavior, but not necessarily for computingpolynomial values (
c.f.^ Newton’s divided differences). Now, lets combine our tools and look at differentiation.^ Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (6/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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Getting an Error Estimate — Taylor Expansion^ f^ (x
+^ h)^ − 0 f^ (x)^0 h
[ (^1) = f^ h (x) +^ hf^0
′ (x) +^0
(^2) h′′f^ (ξ(x 2 ))^ −^ f^ (x
] ) 0
′ = f (x^0 h′′) + f( 2 ξ(x))
′′If f (ξ(x )) is bounded,
i.e. ′′ |f (ξ(x))
|^ <^ M,^
∀ξ(x)^ ∈^
(x,^ x+^00
h)
then we have′^ f^ (x)^0
f^ (x^0 ≈ +^ h)^ −^ f
(x)^0 ,^ h
with an error less than
M|h|.^2
This is the
approximation error
(Roundoff error,
∼^ ǫmach^
−^16 ≈ 10
, not taken into account).
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (7/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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Using Higher Degree Polynomials to get Better Accuracy^ Suppose
{x,^ x,... ,^01
x}^ are distinct points in an intervaln
I, and
n+1 f ∈ C (I), we can write^ n∑ f (x) =^ k=
f^ (x)Lk^ n,
(x)k ︸^ ︷︷
Lagrange Interp. Poly.
∏nk=0 + (x^ −^ x)k^ (n + 1)!^
(n+1)f (ξ (x)) ︸^
︷︷^
Error Term
Formal differentiation of this expression gives:′^ f^ (x)^
n∑=^ f k=
′ (x)Lk n,k
d(x) + dx [ ∏n(k=
]x − x)k (^) (n + 1)! (n+1) f (ξ (x))
∏n(k=0+ x^ −^ x)k^ (n + 1)!
[ d(n+1)^ f^ dx
] (ξ(x))^.
Note:^ When we evaluate
′ f (x)^ at the node pointsj^
(x) the last termj^
gives no contribution. (
⇒^ we don’t have to worry about it...)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (8/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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Example: 3-point Formulas, III/III^
′ f (x) =^0 1 [−^3 f^ (x 2 h^
) + 4f^ (x 0
+^ h)^ −^0 f^ (x+ 2h^0
(^2) h(3))] + f^3 (ξ)^0
′∗ f(x) =^0
1 ∗[−f(x^0 2h^
−^ h) +^ f
∗(x+^ h)] 0
(^2) h(^3 ) − f 6 (ξ)^1
′+ f (x) =^0
[ 1 + f^ (x^02 h −^2 h)^ −
]^ h+x)+^0 2 (3)f^ (ξ)^23
After the substitution
x+^ h^ →^0
∗ xin the second equation, and 0
x+ 2h^ →^0
- xin the third equation. 0 Note#1:^ The third equation can be obtained from the first one by setting
h^ → −h.
Note#2:^ The error is smallest in the second equation. Note#3:^ The second equation is a two-sided approximation, the first and third one-sided approximations. Note#4:^ We can drop the superscripts
∗+ ,...
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (13/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
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3-point Formulas: Illustration
Centered Formula
−2^ −1.^
−1^ −0.^
0 0.^
1 1.^
2
(^10) −1 −2 −3 −4 −5 −6 −7^1 ′ f (x) =^02 h [−f^ (x^0 −^ h) +^ f (x+^ h)]^0
(^2) h(3) − f^6 (ξ)^1
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (14/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
3-point Formulas: Illustration
Forward Formula
−2^ −1.^
−1^ −0.^
0 0.^
1 1.^
2
(^10) −1 −2 −3 −4 −5 −6 − ′ f (x) =^0
1 [−^3 f 2 h^ (x) + 4^0
f^ (x+^ h^0
)^ −^ f^ (x^0
(^2) h(3) f^ (ξ 3
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (15/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
3-point Formulas: Illustration
Backward Formula
−2^ −1.^
−1^ −0.^
0 0.^
1 1.^
2
(^10) −1 −2 −3 −4 −5 −6 − ′ f (x) =^0
1 [f^ (x^02 h^ −^2 h)^ −
4 f^ (x−^0
h) + 3f^
h(x)] + (^0) 2 (3)f^ (ξ^23
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (16/51)
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5-point Formulas^ If we want even better approximations we can go to 4-point,5-point, 6-point, etc
...^ formulas. The most accurate (smallest error term) 5-point formula is:′^ f^ (
f^ (x) = (^0) x−^2 h)−^80 f^ (x−h)+8^0
f^ (x+h)−f^0
(x+2h)^0 12 h^
(^4) h(5)+ f^30 (ξ)
Sometimes (
e.g^ for end-point approximations like the clamped splines), we need one-sided formulas′ f^ (x) =^0
−^25 f^ (x)+48^0
f^ (x+h)−^0 36 f^ (x+2h^0
)+16f^ (x+3^0
h)−^3 f^ (x+4^0
h) 12 h^
(^4) h(5)+ f^ ( 5 ξ).
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (17/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
5-point Formulas: Illustration
Centered Formula
−2^ −1.^
−1^ −0.^
0 0.^
1 1.^
2
(^10) −1 −2 −3 −4 −5 −6 −7 f^ (x−′^0 f (x) =^0 2 h)−^8 f^ (x^0
−h)+8f^ (x
+h)−f^ (x 00 +2h) 12 h^
(^4) h(5)+ f^30 (ξ)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (18/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
Higher Order Derivatives^ We can derive approximations for higher order derivatives in thesame way. — Fit a
kth degree polynomial to a cluster of points {x,^ f^ (xi^ i^
n+k+1)} i=n^ , and compute the appropriate derivative of the polynomial in the point of interest.The standard centered approximation of the second derivative isgiven by
′′ f (x) =^0
f^ (x+^0 h)^ −^2 f^ (
x) +^ f^ (^0
x−^ h)^0 (^2) h
(^2) + O(h)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (19/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
Ideas and Fundamental ToolsMoving Along...
Wrapping Up Numerical Differentiation^ We now have the tools to build high-order accurate approximationsto the derivative.We will use these tools and similar techniques in buildingintegration schemes in the following lectures.Also, these approximations are the backbone of finite differencemethods for numerical solution of differential equations (
see
Math 542
, and^ Math 693b
Next, we develop a general tool for combining low-order accurateapproximations (to derivatives, integrals, anything! (almost))... inorder to hierarchically constructing higher order approximations.^ Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (20/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
A Nice Piece of “Algebra Magic”Homework #6 – Preliminary Version
Building High Accuracy Approximations
IV/V
Now,^ M
−^ N(j+
h) =^ Ehj^
[ j^ α+ (1j^
(^1) − α) j j 2 ](^ +^ O^ h
)j+
We want to select
αso that the expression in the bracket is zero.j^ This gives^ α=^ j^
−^1 ,^ j^2 −^1
1 −^ α=j^
j (^2) j 2 −^1
j^ (2−^ =
(^1 1) + j 2 −
Therefore,
N(h) =j+
N(h/2) +j^
N(h/2)j^ −^ N(h)j^ j 2 −^1
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (25/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
A Nice Piece of “Algebra Magic”Homework #6 – Preliminary Version
Building High Accuracy Approximations
V/V
The following table illustrates how we can use Richardson’sextrapolation to build a 5th order approximation, using five 1storder approximations:^ O^ (
h)^
()^2 O h
()^3 O h
()^4 O h
()^5 O h
N(h)^1 N(h/2)^1
N(h)^2 N(h/4)^1
N(h/2)^2
N(h)^3
N(h/8)^1
N(h/4)^2
N(h/^3
2)^ N(^4
h)
N(h/16)^1
N
(h/8)^2 N(h/4)^3
N(h/^4
2)^ N(^5
h)
↑^ Measurements
↑^
Extrapolations
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (26/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
A Nice Piece of “Algebra Magic”Homework #6 – Preliminary Version
Example (
c.f.^ slide#13, and slide#17) The centered difference formula approximating
′ f (x) can be expressed:^0
′ f (x) =^0 f^ (x^ +^ h)^
−^ f^ (x^ −^ h) 2 h ︸^
︷︷^
︸ N(h)^2
(^2) h′′′ − f^ ( 6 ξ) +^ O(h
4 ) ︸^ ︷︷
︸ error term
In order to eliminate the
(^2) hpart of the error, we let our new approximation be
N(h) =^3 N(h/2) +^2
N(h/2)^ −^ N(h). 3
N(2h)^3
f^ (x+h= )−f^ (x−h)^2 h^
f^ (x+h)−f^ + (x−h)f^ (x+2−^2 h
h)−f^ (x−^2 h)^4 h 3
8 f^ (x+ = h)−^8 f^ (x−h
)f^ (x+2−^6 h h)−f^ (x−^2 h
) 6 h
(^1) = [f 12h^ (x^ −^ 2h)^
−^ 8f(x^ −
h) +^ 8f(
x^ +^ h)^ −^
f(x^ +^ 2h)]
.
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (27/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
A Nice Piece of “Algebra Magic”Homework #6 – Preliminary Version
Example,
f^ (x) =^
2 x^ xe. x^ f(x) 1.70^ 15.8197 1.80^ 19.6009 1.90^ 24.1361 2.00^ 29.5562 2.10^ 36.0128 2.20^ 43.6811 2.30^ 52.
′ f (x) = (
(^2) x + x)e x^ , ′ f (2) = 8
(^2) e= 59.
f^ (2.1)−f^ (
.566. (Fwd Difference, 2pt) f^ (2.1)−f^ (
.384. (Ctr Difference, 3pt) f^ (2.2)−f^ (
.201. (Ctr Difference) (4∗^59.^384
−^60 .201)
- (Richardson)
f^ (1.8)−^8 f^ (1.9)+8f^ (
.1)−f^ (2.2) 1. 2
= 59.111. (5pt)
Peter Blomgren,
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∂〉 ; Richardson’s Extrapolation; ∂x^
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— (28/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
A Nice Piece of “Algebra Magic”Homework #6 – Preliminary Version
Wrap-up / Homework #6 — Due Friday 11/6/2009, 12-noon^ We are going to use Richardson extrapolation in combination withsome of the simpler integration schemes we will develop in order togenerate general schemes for numerically computing integrals up tohigh order.^ Note:
In^ order
to^ use
Richardson
extrapolation,
we^ must
know the form of the error — hence finding error termsin our approximations turns out to have a very practical use. (Part-1) BF-4.1.5BF-4.1.27BF-4.2.9^ Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (29/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Integration: Introduction — The
“Why?”
After taking calculus, I thought I could differentiate and/orintegrate every function...Then came physics, mechanical engineering, etc...The need for numerical integration was painfully obvious!Sometimes (most of the time?), the anti-derivative is not availablein closed form.
∫^ f^ (x)^ dx
=^ F^
(x)^ + ︸ ︷︷^ ︸ Anti-Derivative
C
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (30/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Numerical Quadrature^ The basic idea is to replace integration by clever summation:
∫^ b^ f^ (x)a
dx^ →
n∑^ af,i^ i^ i=
where^ a^
≤^ x<^ x^0
<^ · · ·^ < 1
x≤^ b,^ n^
f=^ f^ (xi^
).i
The coefficients
aand the nodesi^
xare to be selected.i^
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (31/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Building Integration Schemes with Lagrange Polynomials^ Given the nodes
{x,^ x,... ,^01
x}^ we can use then
Lagrange
interpolating polynomial P(x) =n
n∑^ fL(i^ n,i^ i=
x),^ with error
E(x) =n
(n+1)f (
n∏ξ(x)) (n + 1)! i= (x−x)i^
to obtain
∫^ b^ f^ (x)a
∫ dx = b^ P(x)na
dx ︸^ ︷︷^
The Approximation
∫^ b +^ Ea (x)^ dxn ︸ ︷︷^ ︸ The Error Estimate
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〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
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— (32/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Example #2a: Simpson’s Rule (sub-optimal error bound)^ Let^ x
=^ a,^ x 0 1
a+b= ,^2 x=^ b, let^2
b−a h = 2 and use the
quadratic
interpolating polynomial^ ∫^ b^ f^ (xa
∫^ )dx = [x 2 f^ (x)^0 x 0 (x^ −^ x)(^1
x^ −^ x)^2 (x−^ x^0 )(x−^ x^0
+^ f^ (x^1 )
(x^ −^ x) )(x^ −^ x 0
(x−^ x^1 )(x−^ x^1
+^ f^ (x)^2
(x^ −^ x)(^0
x^ −^ x)^1 (x−^ x^2 )(x−^ x^2
]^ dx) ∫^ x^2 ( +^ x^0 x^ −^ x)(^0 x^ −^ x)(^1
x^ −^ x)^2 f 6
(3) (ξ(x))
dx^ ...
∫^ b^ f(x)^ a
[ dx = h f(x) +^0
4f(x) +^1
] f(x) 2 + 3
4 (^3 O(hf )(ξ)).
Peter Blomgren,
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— (37/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Example #2b: Simpson’s Rule (optimal error bound)^ The optimal error bound for Simpson’s rule can be obtained byTaylor expanding
f^ (x) about the mid-point
x:^1
f^ (x) =^ f^ (x) +^1
′ f (x)(x^ −^ x^11
′′f (x)^1 ) + (x^2
′′′f (^2) − x)+ (^1) (x)^13 (x^ −^ x)^16
(4)f (ξ(x)) + 24
(^4) (x − x) 1
Then formally integrating this expression^ "^ Z^ b^ f^ (x^1 a
′) + f (x)(x^ −^1
′′f (x)^1 x) + 12
(^2) (x − x)+^1 ′′′f (x)^1 (x^ −^6
(4)f (ξ (^3) x)+ (^1) (x))^4 (x^ −^ x)^124
#^ dx
After use of the weighted mean value theorem, and the theapproximation
′′ f (x) =^1
1 [f^ (x^2 0 h )^ −^2 f^ (x
) +^ f^ (x 1
(^2) h)] − f 212 (4) (ξ),
and a whole lot of algebra (see BF pp 189–190) we end up with^ ∫^ x^2 f^ (x)^ dx x 0
[^ f( = h x) +^ 4f^0
(x) +^ f^1
](x) 2 −^3 (^5) h(^4 )f(ξ 90
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
— (38/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Example #2: Simpson’s Rule^ ∫
b^ f(x)^ dxa
[^ f( = h x) +^ 4f^0
(x) +^ f(^1
x)^2 3
]^ +^ O(h
5 (^4 )f(ξ))
f(x) p(x) − Simpson’s Rule Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
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— (39/51)
Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Integration Examples^ f^ (x)
[a,^ b]^
R^ bf^ (x)dx^ a^
Trapezoidal
Error
Simpson
Error
x^ [0,^ 1]^
1/^
0.^
0 0.
0
(^2) x[0,^ 1] 1/
0.5^ 0.
0
(^3) x[0,^ 1] 1/
0.5^ 0.
0
(^4) x[0,^ 1] 1/
0.5^ 0.
x^ e[0,^ 1] e^ −
1 1.
The Trapezoidal rule gives exact solutions for linear functions. —The error terms contains a second derivative.Simpson’s rule gives exact solutions for polynomials of degree lessthan 4. — The error term contains a fourth derivative.^ Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Degree of Accuracy (Precision)^ Definition (Degree of Accuracy)^ The^ Degree of Accuracy
, or^ precision
, of a quadrature formula is
the largest positive integer
n^ such that the formula is exact for
k x
∀k^ = 0,^ 1 ,... ,^ n. With this definition:
Scheme^
Degree of Accuracy Trapezoidal
Simpson’s
Trapezoidal and Simpson’s are examples of a class of methodsknown as
Newton-Cotes formulas
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Newton-Cotes Formulas — Two Types
Closed
Closed^
The (n^ + 1) point closed NCF uses nodes
x=^ xi^0
+^ ih,
i^ = 0,^1 ,... ,
n, where
x=^ a,^0
x=^ b^ andn^
h^ = (b−
a)/n. It
is called closed since the endpoints are included as nodes. Peter Blomgren,^ 〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Newton-Cotes Formulas — Two Types
Open
Open^
The (n^ + 1) point open NCF uses nodes
x=^ x^0 i^
+^ ih,
i^ = 0,^1 ,... ,
n^ where h^ = (b^ −
a)/(n^ + 2) and
x=^ a^ +^0
h,
x=^ b^ −n^
h. (We label
x=^ a−^1
,^ x=n+^
b.)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Closed Newton-Cotes Formulas^ The approximation is
∫^ b^ f^ (x) a
n∑ dx ≈ i= af^ (x),i^ i^
where
∫^ x a=i n^ L(x)n,i^ x 0
∫^ dx = nxn∏^ ( x 0 j = 0 j 6 =^ i x^ −^ x)j^ (x−^ x)^ i^ j^ dx.
Note:^
The Lagrange polynomial
L(x) models a function whichn,i^
takes the value 0 at all
x(j^6 =^ j^
i), and 1 at
x. Hence, thei^
coefficient
acaptures the integral of a function which is 1i^ in^ xand zero in the other node points.i^ Peter Blomgren,^ 〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
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The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Open Newton-Cotes Formulas — Examples^ n^ =
0 :^
2 hf^ (x) +^0
(^3) h′′ f^ (ξ) 3
n^ =^1 :^
[ 3 h^ f^ (x^02 ) +^ f^ (x^1
]^33 h)^ +^4
′′f (ξ)
n^ =^2 :^
[ 4 h^2 f^ (x 3 )^ −^ f^ (x 0
) + 2f^ ( 1
]^14 x)^ + 2 (^5) h(4)f^ (ξ 45
n^ =^3 :^
[ 5 h^11 f^ ( 24 x) +^ f^ (^0
x) +^ f^ (^1 x) + 11^2
] f (x)^ + 3
(^595) h(4)f^144 (ξ)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Divide and Conquer!^ Say you want to compute:
∫^100 f^ (x^0
)^ dx.
Is it a Good Idea
TM^ to directly apply your favorite Newton-Cotes formula to this integral?!? No! With the closed 5-point NCF, we have
h^ = 25 and
(^5) h/^90 ∼
5 10 so
even with a bound on
(6) f (ξ) the error will be large. Better: Apply the closed 5-point NCF to the integrals
∫^ 4(i+1)^4 i
f^ (x)^ dx,^
i^ = 0,^1 ,... ,
then sum.
“Composite Numerical Integration.”
(next time)
Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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Numerical DifferentiationRichardson’s ExtrapolationNumerical Integration (Quadrature)
The “Why?” and IntroductionTrapezoidal & Simpson’s RulesNewton-Cotes FormulasHomework #6 – Final Version
Homework #6 — Due Friday 11/6/2009, 12-noon^ (Part-1)^ BF-4.1.5BF-4.1.27BF-4.2.9^ (Part-2)^ BF-4.3.1-a,b.BF-4.3.5-a,b.^ Peter Blomgren,
〈[email protected]
∂〉 ; Richardson’s Extrapolation; ∂x^
R^ f^ (x)^ dx^
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