Escuela de Ingeniería Informática de Oviedo...Introduction Motivation Often, when solving mathematical problems, we need to get the value of a function in several points. But: it

Interpolation

Escuela de Ingeniería Informática de Oviedo

(Dpto. de Matemáticas-UniOvi) Numerical Computation Interpolation 1 / 34

Outline

1 Introduction

2 Lagrange interpolation

3 Piecewise polynomial interpolation


Introduction

Outline

1 Introduction




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, ...


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


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


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.


Lagrange interpolation

Outline

1 Introduction






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



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)



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



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.



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)


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.



Fundamental Lagrange polynomialsFundamental polynomials:For each i = 0,1, . . . ,n there exists an unique polynomial ì of degree at mostn such that ì (xk ) = δik (zero if i 6= k , one if i = k ):

ì (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.



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)



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 .



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 ] .



Example

P1(x) P

2(x)

P3(x) P

4(x)


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)!.


Piecewise polynomial interpolation

Outline

1 Introduction






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.



Example: oscillations of Lagrange polynomials

e−x

2

(xi,ω

i)

P10

(x)



Example: piecewise linear interpolation

e−x

2

(xi,ω

i)

s



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.



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.



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.



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.


Piecewise polynomial interpolation Cubic spline

Example

sin(x)

Lineal

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.



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)



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.



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)

.


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.


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Escuela de Ingeniería Informática de Oviedo...Introduction Motivation Often, when solving mathematical problems, we need to get the value of a function in several points. But: it