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Material Type: Notes; Class: INTRO NUM ANALYS & COMPUT; Subject: Mathematics; University: San Diego State University; Term: Fall 2002;
Typology: Study notes
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Richardson’s Extrapolation Numerical Integration (Quadrature)
Lecture Notes # — Numerical Differentiation and Integration — Differentiation; Richardson’s Extrapolation; Integration
Peter Blomgren, 〈[email protected]〉
Department of Mathematics and Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA 92182- http://terminus.sdsu.edu/
Fall 2009
Richardson’s Extrapolation Numerical 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
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Numerical Differentiation
Definition (Derivative as a limit) The derivative of f at x 0 is
f ′(x 0 ) = lim h→ 0
f (x 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 values of h, f (x 0 +h) ≈ f (x 0 ) so the difference may have very few significant digits in finite precision arithmetic. ⇒ smaller h not necessarily better numerically.
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Main Tools for Numerical Differentiation 1 of 2
In the discussion on Numerical Differentiation (and later Integration) we will rely on our old friend (nemesis?) — the Taylor expansions...
Theorem (Taylor’s Theorem) Suppose f ∈ C n[a, b], f (n+1)∃ on [a, b], and x 0 ∈ [a, b]. Then ∀x ∈ (a, b), ∃ξ(x) ∈ (min(x 0 , x), max(x 0 , x)) with f (x) = Pn(x) + Rn(x) where
Pn(x) =
X^ n k=
f (k)(x 0 ) k! (x^ −^ x^0 )
k (^) , Rn(x) = f^ (n+1)(ξ(x)) (n + 1)! (x^ −^ x^0 )
(n+1).
Pn(x) is the Taylor polynomial of degree n, and Rn(x) is the remainder term (truncation error).
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Getting an Error Estimate — Taylor Expansion
f (x 0 + h) − f (x 0 ) h
= (^1) h
f (x 0 ) + hf ′(x 0 ) + h
2 2 f^
′′(ξ(x)) − f (x 0 )
= f ′(x 0 ) + h 2 f′′(ξ(x))
If f ′′(ξ(x)) is bounded, i.e.
|f ′′(ξ(x))| < M, ∀ξ(x) ∈ (x 0 , x 0 + 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 ≈ 10 −^16 , not taken into account).
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Using Higher Degree Polynomials to get Better Accuracy
Suppose {x 0 , x 1 ,... , xn} are distinct points in an interval I, and f ∈ C n+1(I), we can write
f (x) =
∑^ n
k=
f (xk )Ln,k (x) ︸ ︷︷ ︸ Lagrange Interp. Poly.
∏n k=0(x^ −^ xk^ ) (n + 1)!
f (n+1)(ξ(x)) ︸ ︷︷ ︸ Error Term
Formal differentiation of this expression gives:
f ′(x) =
∑^ n
k=
f (xk )L′ n,k (x) +
d dx
[ ∏n k=0(x^ −^ xk^ ) (n + 1)!
f (n+1)(ξ(x))
∏n k=0(x^ −^ xk^ ) (n + 1)!
d dx
f (n+1)(ξ(x))
Note: When we evaluate f ′(xj ) at the node points (xj ) the last term gives no contribution. (⇒ we don’t have to worry about it...)
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
The (n + 1) point formula for approximating f ′(xj )
Putting it all together yields what is known as the (n + 1) point formula for approximating f ′(xj ):
f ′(xj ) =
∑^ n
k=
f (xk )L′ n,k (xj ) +
f (n+1)(ξ) (n + 1)!
∏^ n
k = 0 k 6 = j
(xj − xk )
Note: The formula is most useful when the node points are equally spaced (it can be computed once and stored), i.e. xk = x 0 + kh.
Now, we have to compute the derivatives of the Lagrange coefficients, i.e. Ln,k (x)... [We can no longer dodge this task!]
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Example: 3-point Formulas, I/III
Building blocks:
L 2 , 0 (x) =
(x − x 1 )(x − x 2 ) (x 0 − x 1 )(x 0 − x 2 )
, L′ 2 , 0 (x) =
2 x − x 1 − x 2 (x 0 − x 1 )(x 0 − x 2 )
L 2 , 1 (x) =
(x − x 0 )(x − x 2 ) (x 1 − x 0 )(x 1 − x 2 )
, L′ 2 , 1 (x) =
2 x − x 0 − x 2 (x 1 − x 0 )(x 1 − x 2 )
L 2 , 2 (x) =
(x − x 0 )(x − x 1 ) (x 2 − x 0 )(x 2 − x 1 )
, L′ 2 , 2 (x) =
2 x − x 0 − x 1 (x 2 − x 0 )(x 2 − x 1 )
Formulas:
f ′(xj ) = f (x 0 )
2 xj − x 1 − x 2 (x 0 − x 1 )(x 0 − x 2 )
2 xj − x 0 − x 2 (x 1 − x 0 )(x 1 − x 2 )
2 xj − x 0 − x 1 (x 2 − x 0 )(x 2 − x 1 )
f (3)(ξj ) 6
k = 0 k 6 = j
(xj − xk ).
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Example: 3-point Formulas, II/III
When the points are equally spaced...
f ′(x 0 ) =
1 2 h
[− 3 f (x 0 ) + 4f (x 1 ) − f (x 2 )] +
h^2 3
f (3)(ξ 0 )
f ′(x 1 ) =
1 2 h
[−f (x 0 ) + f (x 2 )] −
h^2 6
f (3)(ξ 1 )
f ′(x 2 ) =
1 2 h
[f (x 0 ) − 4 f (x 1 ) + 3f (x 2 )] +
h^2 3
f (3)(ξ 2 )
Use x 0 as the reference point — xk = x 0 + kh:
f ′(x 0 ) = 1 2 h
[− 3 f (x 0 ) + 4f (x 0 + h) − f (x 0 + 2h)] + h
2 3
f (3)(ξ 0 )
f ′(x 0 + h) =
1 2 h
[−f (x 0 ) + f (x 0 + 2h)] −
h^2 6
f (3)(ξ 1 )
f ′(x 0 + 2h) = 1 2 h
[f (x 0 ) − 4 f (x 0 + h) + 3f (x 0 + 2h)] + h
2 3
f (3)(ξ 2 )
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Example: 3-point Formulas, III/III
f ′(x 0 ) = 1 2 h
[− 3 f (x 0 ) + 4f (x 0 + h) − f (x 0 + 2h)] + h
2 3
f (3)(ξ 0 )
f ′(x 0 + h) =
1 2 h
[−f (x 0 ) + f (x 0 + 2h)] −
h^2 6
f (3)(ξ 1 )
f ′(x 0 + 2h) = 1 2 h
[f (x 0 ) − 4 f (x 0 + h) + 3f (x 0 + 2h)] + h
2 3
f (3)(ξ 2 )
Make the substitution x 0 + h → x 0 ∗ in the second equation.
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Example: 3-point Formulas, III/III
f ′(x 0 ) =
1 2 h
[− 3 f (x 0 ) + 4f (x 0 + h) − f (x 0 + 2h)] +
h^2 3
f (3)(ξ 0 )
f ′(x 0 ∗ ) = 1 2 h
[−f (x 0 ∗ − h) + f (x 0 ∗ + h)] − h
2 6
f (3)(ξ 1 )
f ′(x 0 + ) =
1 2 h
[ f (x 0 + − 2 h) − 4 f (x 0 + − h) + 3f (x 0 + )
]
h^2 3
f (3)(ξ 2 )
After the substitution x 0 + h → x 0 ∗ in the second equation, and x 0 + 2h → x 0 + in the third equation.
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
Example: 3-point Formulas, III/III
f ′(x 0 ) =
1 2 h
[− 3 f (x 0 ) + 4f (x 0 + h) − f (x 0 + 2h)] +
h^2 3
f (3)(ξ 0 )
f′(x∗ 0 ) = 1 2h
[−f(x∗ 0 − h) + f(x∗ 0 + h)] − h
2 6
f(^3 )(ξ 1 )
f ′(x 0 + ) =
1 2 h
[ f (x 0 + − 2 h) − 4 f (x 0 + − h) + 3f (x 0 + )
]
h^2 3
f (3)(ξ 2 )
After the substitution x 0 + h → x 0 ∗ in the second equation, and x 0 + 2h → x 0 + in the third equation. 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 ∗,+...
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
3-point Formulas: Illustration Forward Formula
−7 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−
−
−
−
−
−
0
1
f ′(x 0 ) =
2 h
[− 3 f (x 0 ) + 4f (x 0 + h) − f (x 0 + 2h)] +
h^2 3
f (3)(ξ 0 )
Richardson’s Extrapolation Numerical Integration (Quadrature)
Ideas and Fundamental Tools Moving Along...
3-point Formulas: Illustration Backward Formula
−7 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2
−
−
−
−
−
−
0
1
f ′(x 0 ) =
2 h
[f (x 0 − 2 h) − 4 f (x 0 − h) + 3f (x 0 )] +
h^2 3
f (3)(ξ 2 )
