Escuela de Ingeniería Informática de Oviedo...Introduction Motivation Often, when solving...
Transcript of Escuela de Ingeniería Informática de Oviedo...Introduction Motivation Often, when solving...
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Interpolation
Escuela de Ingeniería Informática de Oviedo
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Outline
1 Introduction
2 Lagrange interpolation
3 Piecewise polynomial interpolation
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Introduction
Outline
1 Introduction
2 Lagrange interpolation
3 Piecewise polynomial interpolation
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Introduction
Motivation
Often, when solving mathematical problems, we need to get the value of afunction in several points.
But:
it can be expensive in terms of processor use or machine time (forinstance, a complicated function which we need to evaluate many times).possibly, we only have the function values in few points (for instance, ifthey come from data trials).
A possible solution is to replace the complicated function by another similarbut simpler to evaluate.
These simpler functions are polynomials, trigonometric functions, ...
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Introduction
Interpolation
Data:n + 1 different points x0,x1, . . . , xn
n + 1 values ω0,ω1, . . . , ωn
Problem:
Find a (simple) function f̃ verifying f̃ (xi ) = yi with i = 0,1, . . . ,n.
Remarks:x0,x1, . . . , xn are called interpolation nodesω0,ω1, . . . , ωn can be the values of a function f in the nodes:
ωi = f (xi ) i = 0,1, . . . ,n
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Introduction
ExampleHere, f̃ ≡ P5 is a polynomial of degree 5:
(x0,ω
0)
(x1,ω
1)
(x2,ω
2)
(x3,ω
3)
(x4,ω
4)
(x5,ω
5)
(xi,ω
i)
P5
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Introduction
Polynomial interpolation
If f̃ is a polynomial or a piecewise polynomial function we speak ofpolynomic interpolation.
Main examples:1 Lagrange interpolation.2 Piecewise polynomial interpolation.
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Lagrange interpolation
Outline
1 Introduction
2 Lagrange interpolation
3 Piecewise polynomial interpolation
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Lagrange interpolation
Lagrange interpolation
Data:n + 1 different points x0,x1, . . . , xn
n + 1 values ω0,ω1, . . . , ωn
Problem: We seek for a polynomial Pn of degree at most n verifying
Pn (x0) = ω0 Pn (x1) = ω1 Pn (x2) = ω2 . . . P(n)n (xn) = ωn
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Lagrange interpolation
Lagrange linear interpolation (n = 1)
Data:2 different points x0,x1
2 values ω0,ω1
Problem: Find a polynomial Pn of degree at most 1 verifying
P1 (x0) = ω0 P1 (x1) = ω1
y = P1 (x) is the straight line joining (x0, ω0) and (x1, ω1)
P1 (x) = ω0 +ω1 − ω0
x1 − x0(x1 − x0)
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Lagrange interpolation
Lagrange interpolation (for any n)
Problem: Find a polynomial
Pn (x) = a0 + a1x + a2x2 + · · ·+ anxn
satisfying
Pn (x0) = ω0 Pn (x1) = ω1 Pn (x2) = ω2 . . . Pn (xn) = ωn (1)
Conditions (1) imply that coefficients solve1 x0 x2
0 · · · xn0
1 x1 x21 · · · xn
1...
......
. . ....
1 xn x2n · · · xn
n
a0a1...
an
=
ω0ω1...ωn
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Lagrange interpolation
The coefficient matrix is of Vandermonde type:
A =
1 x0 x2
0 · · · xn0
1 x1 x21 · · · xn
1...
......
. . ....
1 xn x2n · · · xn
n
,
withdet (A) =
∏0≤l≤k≤n
(xk − xl ) 6= 0.
Therefore, the system has a unique solution which determines Pn.
Pn is the Lagrange interpolation polynomial at x0,x1, . . . , xn relative to thevalues ω0,ω1, . . . , ωn.
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Lagrange interpolation
An example of Lagrange interpolation
(xi,ω
i)
P1(x)
(xi,ω
i)
P2(x)
(xi,ω
i)
P3(x)
(xi,ω
i)
P4(x)
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Computing the Lagrange’s interpolationpolynomial
Motivation:Solving the Vandermonde system of equations is not efficient and large errorsmay arise due to bad conditioning.
Two ways of computing:
Using fundamental Lagrange polynomials.Using divided differences.
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Fundamental Lagrange polynomialsFundamental polynomials:For each i = 0,1, . . . ,n there exists an unique polynomial `i of degree at mostn such that `i (xk ) = δik (zero if i 6= k , one if i = k ):
`i (x) =n∏
j = 0j 6= i
x − xj
xi − xj,
`0, `1, . . . , `n are the fundamental Lagrange polynomials of degree n.
Lagrange formula:Lagrange’s polynomial at x0,x1, . . . , xn relative to ω0,ω1, . . . , ωn is
Pn (x) = ω0`0 (x) + ω1`1 (x) + · · ·+ ωn`n (x) .
Remark: If we want to add new nodes, we need to re-compute all thefundamental polynomials.
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Example
0 1 2 3 4−2
−1
0
1
2
3
l0(x)
0 1 2 3 4−2
−1
0
1
2
3
l1(x)
0 1 2 3 4−2
−1
0
1
2
3
l2(x)
0 1 2 3 4−2
−1
0
1
2
3
ω0 l
0(x)
0 1 2 3 4−2
−1
0
1
2
3
ω1 l
1(x)
0 1 2 3 4−2
−1
0
1
2
3
ω2 l
2(x)
0 1 2 3 4−2
−1
0
1
2
3
P2(x)
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Divided differences
Divided differences of 0-order
[ωi ] = ωi , i = 0,1, . . . ,n.
Divided differences of order 1
[ωi , ωi+1] =ωi+1 − ωi
xi+1 − xi, i = 0,1, . . . ,n − 1.
Divided differences of order 2
[ωi , ωi+1, ωi+2] =[ωi+1, ωi+2]− [ωi , ωi+1]
xi+2 − xi, i = 0,1, . . . ,n − 2.
Divided differences of order k (k = 1, . . . ,n)
[ωi , ωi+1, . . . , ωi+k ] =[ωi+1, . . . , ωi+k ]− [ωi , ωi+1, . . . , ωi+k−1]
xi+k − xi,
i = 0,1, . . . ,n − k .
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Newton’s formula:Lagrange polynomial at x0,x1, . . . , xn relative to ω0,ω1, . . . , ωn is
Pn (x) = [ω0] + [ω0, ω1] (x − x0) + [ω0, ω1, ω2] (x − x0) (x − x1) + · · ·++ [ω0, ω1, . . . , ωn] (x − x0) (x − x1) · · · (x − xn−1) .
Consequence:P0,P1, . . . ,Pn may be computed recursively, adding in each iteration a newterm,
Pn (x) = Pn−1 (x) + [ω0, ω1, . . . , ωn] (x − x0) (x − x1) · · · (x − xn−1) .
Notation:If ωi = f (xi ), with i = 0,1, . . . ,n,
f [xi , . . . , xi+k ] = [ωi , . . . , ωi+k ] .
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Lagrange interpolation Computing the Lagrange’s interpolation polynomial
Example
P1(x) P
2(x)
P3(x) P
4(x)
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Lagrange interpolation Error
Error
Hypothesis:f : [a,b]→ R is n + 1 times differentiable in [a,b] with continuousderivatives.x0, x1, . . . , xn ∈ [a,b] .
ωi = f (xi ) for i = 0,1, . . . ,n.
Error estimate: The absolute error satisfies
|f (x)− Pn (x)| ≤ supy∈[a,b]
∣∣∣f (n+1) (y)∣∣∣ |(x − x0) (x − x1) · · · (x − xn)|
(n + 1)!.
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Piecewise polynomial interpolation
Outline
1 Introduction
2 Lagrange interpolation
3 Piecewise polynomial interpolation
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Piecewise polynomial interpolation
Piecewise polynomial interpolation
Motivation:If we increase the number of nodes
The Lagrange’s interpolation polynomial degree increases as well,generating oscillations.
Often, when increasing the number of nodes the error also increases (notintuitive).
Estrategy:Use low degree piecewise polynomials.
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Piecewise polynomial interpolation
Example: oscillations of Lagrange polynomials
e−x
2
(xi,ω
i)
P10
(x)
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Piecewise polynomial interpolation
Example: piecewise linear interpolation
e−x
2
(xi,ω
i)
s
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Piecewise polynomial interpolation
Constantwise polynomial interpolation
Interpolation by degree zero piecewise polynomials is that in which thepolynomials are constant between nodes. For instance,
f̃ (x) =
ω0 if x ∈ [x0, x1),ω1 if x ∈ [x1, x2),. . .ωn−1 if x ∈ [xn−1, xn),ω0 if x = xn.
Observe that if ωi 6= ωi+1 then f̃ is discontinuous at xi+1.
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Piecewise polynomial interpolation
Linearwise polynomial interpolation
The degree one piecewise polynomial interpolation is that in which thepolynomials are straight lines joining two consecutive nodes,
f̃ (x) = ωi + (ωi+1 − ωi )x − xi
xi+1 − xiif x ∈ [xi , xi+1],
for i = 0, . . . ,n − 1.
In this case, f̃ is continuous, but its first derivative is, in general, discontinuousat the nodes.
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Piecewise polynomial interpolation
Spline interpolation
Data:n + 1 points x0 < x1 < . . . < xn.
n + 1 values ω0, ω1, . . . , ωn.
Problem: Interpolating by splines of order p (or degree p) consists on findinga function f̃ such that:
1 f̃ is p − 1 times continuously differentiable in [x0, xn].2 f̃ is a piecewise function given by the polynomials f̃0, f̃1, . . . , f̃n−1 defined,
respectively, in [x0, x1] , [x1, x2] , . . . , [xn−1, xn], and of degree lower orequal to p.
3 Interpolation condition: f̃0 (x0) = ω0, . . . , f̃n (xn) = ωn.
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Piecewise polynomial interpolation
The problem has solution!Each solution f̃ is called spline interpolator of order p at x0,x1, . . . , xnrelative to ω0,ω1, . . . , ωn.
The most used spline is that ot third order, known as cubic spline.
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Piecewise polynomial interpolation Cubic spline
Example
sin(x)
Lineal
Spline
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Piecewise polynomial interpolation Cubic spline
Cubic spline
Definition:A Cubic spline is a function f̃ such that
1 f̃ is twice continuously differentiable in [x0, xn].2 f̃0, f̃1, . . . , f̃n−1 are of degree at most 3.3 f̃ (x0) = ω0 f̃ (x1) = ω1 . . . f̃ (xn) = ωn
Remarks:
In each sub-interval [xi , xi+1], the interpolant f̃ is determined from thevalues of f̃ ′′ at xi and xi+1.Such values (f̃ ′′(xi ) and f̃ ′′(xi+1)) are obtained by solving a linear systemof equations.
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Piecewise polynomial interpolation Cubic spline
Computing the cubic spline
Task: Deduce an algorithm to compute the cubic spline.
By definition, f̃ ′′ is continuous in [x0, xn]. Therefore
ω′′0 = f̃ ′′0 (x0)
ω′′1 = f̃ ′′0 (x1) = f̃ ′′1 (x1)
ω′′2 = f̃ ′′1 (x2) = f̃ ′′2 (x2)· · · · · · · · ·ω′′n−1 = f̃ ′′n−2 (xn−1) = f̃ ′′n−1 (xn−1)
ω′′n = f̃ ′′n−1 (xn)
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Piecewise polynomial interpolation Cubic spline
Integrating twice, we get
f̃i (x) = ω′′i(xi+1 − x)3
6hi+ ω′′i+1
(x − xi )3
6hi+ ai (xi+1 − x) + bi (x − xi ) ,
whereai =
ωi
hi− ω′′i
hi
6, bi =
ωi+1
hi− ω′′i+1
hi
6.
Therefore, once ω′′i are known, the cubic spline is fully determined.
Using that f̃ is twice continuously differentiable gives the n−1 linear equations
hi
6ω′′i +
hi+1 + hi
3ω′′i+1 +
hi+1
6ω′′i+2 =
ωi
hi−( 1
hi+1+
1hi
)ωi+1 +
ωi+2
hi+1.
For full determination of the n + 1 values ω′′i we need two additional equations.
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Piecewise polynomial interpolation Cubic spline
Natural cubic spline:There are several options to determine uniquely the cubic spline, for instance
adding two equations to close the system,fix the value of two unknowns.
If the values of ω′′0 and ω′′n are fixed, we talk about natural splines. In this case,ω′′in = (ω1, . . . , ωn) are solution of Hω′′in = 6d, where
d = (∆1 −∆0, . . . ,∆n−1 −∆n−2), with ∆i = (ωi+1 − ωi )/hi ,
H =
2 (h0 + h1) h1 0 · · · 0 0
h1 2 (h1 + h2) h2 · · · 0 0...
......
. . ....
...0 0 0 · · · 2 (hn−3 + hn−2) hn−20 0 0 · · · hn−2 2 (hn−2 + hn−1)
.
(Dpto. de Matemáticas-UniOvi) Numerical Computation Interpolation 33 / 34
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Piecewise polynomial interpolation Error
Error
Hypothesis:f : [a,b]→ R is four times continuously differentiable in [a,b] .
x0, x1, . . . , xn ∈ [a,b] .
Error estimation:Let h̃ = max
i=0,...,nhi . Then, for all x ∈ [a,b] we have
∣∣∣f (x)− f̃ (x)∣∣∣ ≤ ch̃p+1 max
y∈[a,b]
∣∣∣f (p+1) (y)∣∣∣ ,
where c is a constant independent of f , x and h̃.
(Dpto. de Matemáticas-UniOvi) Numerical Computation Interpolation 34 / 34