Interpolation and Approximation: Lagrange Interpolation PDF

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What is Lagrange interpolation?

different way of interpolating a function is known as Lagrange interpolation. of some functionf that is otherwise unknown. We want to reconstruct apolynomial that attains the same function values asf. For the sake ofoverview, we put this into a table: x0 x1 : : : xm y0 y1 : : : ym 1 0 1 6 2 4

What is multivariate interpolation?

kind of interpolation, in which most development has been made, is interpolation by means of univariate polynomials. Multiple formulae for polynomial interpolation have been given, notably those of Newton and Lagrange [1]. Multivariate interpolation has applications in computer graphics, numerical quadrature, cubature,

Why is high-degree interpolation a problem?

It is interesting to note that the error closely resembles the Taylor remainderRn(x). If the number of data points is large, then polynomial interpolation becomes problematic sincehigh-degree interpolation yields oscillatory polynomials, when the data may t a smooth function.

What is the basic principle of polynomial interpolation?

whereI is the interval betweenaandx. The basic principle of polynomial interpolation is that we “takemeasurements” off by looking at the values of the function (and itsderivatives) at certain points. We then construct a polynomial that satisesthe same measurements. different way of interpolating a function is known as Lagrange interpolation.

Lagrange Polynomials

Interpolation and Approximation: Lagrange Interpolation

Martin Licht

UC San Diego

Winter Quarter ????

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Lagrange Interpolation

Given a functionf:[a;b]!Rover some interval[a;b], we would like to approximatefby a polynomial.

How do we find a good polynomial?

We have already one example, namely the Taylor polynomial around a pointa: T maf(x) =mX k=?f (k)(a)k!(xa)k

Note that this can be written as

T maf(x) =mX k=?c k(xa)k; where c k=f(k)(a)k!: Evidently, we construct the Taylor polynomial by evaluatingfand its derivatives at a particular pointa2R.

Lagrange Interpolation

We recall some representations of the error:

Theorem

Let f :R!Rhave continuous derivatives up to order m+?. Then I

We have

R maf(x) =Z x af (m+?)(t)m!(ta)mdt: I For every x2Rthere existsxin the closed interval between a and x with R maf(x) =f(m+?)(x)(m+?)!(xa)m+?: I For every x2Rthere existsxin the closed interval between a and x with R maf(x) =f(m+?)(x)m!(x)m(xa):

Lagrange Interpolation

From each of those representations of the error we can derive f(x)Tmaf(x)=Rmaf(x)?m!max2I f(m+?)() jxajm+?: or even f(x)Tmaf(x)=Rmaf(x)?(m+?)!max2I f(m+?)() jxajm+?: whereIis the interval betweenaandx.

Lagrange Interpolation

The basic principle of polynomial interpolation is that we "take measurements" offby looking at the values of the function (and its derivatives) at certain points. We then construct a polynomial that satisfies the same measurements.In the case of the Taylor polynomial, we have a single numberx?2Rand take the derivatives up to orderm, to construct a degreempolynomialp(x)with p(x?) =f(x?);p0(x?) =f0(x?);p00(x?) =f00(x?); :::p(m)(x?) =f(m)(x?): A different way of interpolating a function is known as Lagrange interpolation. In the case of Lagrange interpolation, we havemdifferent numbers x ?;x?;:::;xm2Rand take function evaluations up to orderm, to construct a degreempolynomialp(x)with p(x?) =f(x?);p(x?) =f(x?);p(x?) =f(x?); :::p(xm) =f(xm):

Lagrange Interpolation

Example

Suppose we have got pointsx?;x?;:::;xmand values

y ?=f(x?);y?=f(x?); :::ym=f(xm) of some functionfthat is otherwise unknown. We want to reconstruct a polynomial that attains the same function values asf. For the sake of overview, we put this into a table: xx ?x ?:::x myy ?y ?:::y m

For this example, let us consider the casem=? and

x ?=?;x?=?;x?=?; y ?=6;y?=?;y?=?:

Lagrange Interpolation

Example

The table is

x??? y6??

We search for a polynomialpof degreem=? such that

p(?) =6;p(?) =?;p(?) =?:

The solution is the polynomial

p(x) =?x+?x?: In these notes, we describe different ways to computing and representing such polynomials.

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Monomial Basis

Suppose we have pairwise different pointsx?;x?;:::;xmand that we search for the coefficientsa?;a?;:::;amof a polynomial p(x) =a?+a?x++amxm such that for some given valuesy?;y?;:::;ymwe have p(x?) =y?;p(x?) =y?; :::p(xm) =ym: That is, we search formunknown variablesa?;a?;:::;am2Rsuch that them constraints given by the point evaluations are satisfied. This translates into a linear system of equations a ?+a?x?+a?x??++amxm?=y?; a ?+a?x?+a?x??++amxm?=y?; a ?+a?x?+a?x??++amxm?=y?; a ?+a?xm+a?x?m++amxmm=ym:

Monomial Basis

We can rewrite this in matrix notation as

0 B

BBBBB@?x?x??:::xm?

?x?x??:::xm? ?x?x??:::xm?............... ?xmx?m:::xmm1 C

CCCCCA0

BBBBB@a

a a a m1 C

CCCCCA=0

BBBBB@y

y y y m1 C

CCCCCA:

The matrix in that system called theVandermonde matrixassociated to the pointsx?;x?;:::;xm. We would like to understand the linear system of equations has got a solution, and for that purpose the Vandermonde matrix.Theorem

The determinant of the Vandermonde matrix V is

det(V) =Y ?iMonomial Basis

Proof.

For the proof we use elementary properties of determinants. Let x ?;x?;:::;xm2Rbe pairwise different. Since the determinant is invariant under row additions and subtractions, we get the identity det 0 B

BBBBB@?x?x??:::xm?

?x?x??:::xm? ?x?x??:::xm?............... ?xmx?m:::xmm1 C

CCCCCA= det0

BBBBB@?x?x??:::xm?

?x?x?x??x??:::xm?xm? ?x?x?x??x??:::xm?xm?............... ?xmx?x?mx??:::xmmxm?1 C

CCCCCA

Similarly, the determinat is invariant under additions of columns. We perform a number of column substractions: we subtractx?-times them-th column from the(m+?)-th column, subtractx?-times the(m?)-th column from the m-th column, subtractx?-times the(m?)-th column from the(m?)-th column, and and so on, until we have subtractedx?-times the first column from the second column.

Monomial Basis

Proof.

Consequently, we end up with the determinant

det 0 B

BBBBB@? ? ?:::?

?x?x?(x?x?)x?:::(x?x?)xm? ??x?x?(x?x?)x?:::(x?x?)xm? ?xmx?(xmx?)xm:::(xmx?)xm?m1 C

CCCCCA

The rows of this determinant have the common factors (x?x?);(x?x?); :::(xmx?):

Monomial Basis

Proof.

We can extract these common factors from the determinant and get the value m Y i=?(xix?)det0 B

BBBBB@? ? ?:::?

? ?x?:::xm? ?? ?x?:::xm? ? ?xm:::xm?m1 C

CCCCCA

mY i=?(xix?)det0 B

BBB@?x?:::xm?

??x?:::xm? ?xm:::xm?m1 C CCCA The last term is the determinat of the Vandermonde matrix for the points x ?;:::;xm.

Monomial Basis

Proof.

We can repeat this calculation recursively until we only need to compute the determinant of the Vandermonde matrix for the single pointx?, which is just equals ?. Working up from there, the determinant becomes m Y i=?(xix?)Y ?iThis completes the proof.

Monomial Basis

In particular, sincex?;x?;:::;xmare pairwise different, the determinant of the Vandermonde matrix is non-zero, and hence that the matrix is invertible. We conclude that the interpolation problem has a got a unique solution.Theorem

Given pairwise distinct points x

?;x?;:::;xm2Rand values y?;y?;:::;ym2R, there exists a unique polynomial p of degree m such that

p(x?) =f(x?);p(x?) =f(x?);p(x?) =f(x?); :::p(xm) =f(xm):The polynomials of degreemare a vector space of dimensionm+?, with a

basis being the monomials up to orderm: ?;x;x?; :::xm; In particular, if we express the interpolation problem using the monomial basis, then the basis does not depend on the interpolation points x ?;x?;:::;xm. However, the Vandermonde matrix in the formulation has several disadvantageous properties, e.g., it is very dense.

Monomial Basis

Example

Consider again the quadratic interpolation problem with the following table: x??? y6??

The solution is

0 B ? ? ?1 C A0 B @a a a ?1 C A=0 B @6 ?1 C A: We check that the determinant of Vandermonde matrix is det 0 B ? ? ?1 C

A= (?)(?)(?) =?:

Monomial Basis

Example

The inverse of that Vandermonde matrix is

0 B ? ? ?1 C A? 0 B ?? ?1 C A; and we readily check that 0 B ?? ?1 C A0 B @6 ?1 C A=0 B ?1 C A; which is precisely the coefficients of the solutionp(x) =?x+?x?.

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Newton Polynomials

We pose the same interpolation but with a different basis. This time, the basis incorporates the interpolation pointsx?;x?;:::;xm2R. We define the

Newton polynomials

p ?(x) =? p ?(x) = (xx?) p ?(x) = (xx?)(xx?) p ?(x) = (xx?)(xx?)(xx?) p m(x) = (xx?)(xx?)(xx?) (xxm?)

So we have the form

p k(x) =k?Y i=?(xxk) = (xx?)(xx?) (xxk?):

Consequently,

p k(x?) ==pk(xk?) =?:

Newton Polynomials

Using this basis lets us formulate the interpolation problem in a simplified manner. Using the Newton polynomials, we search coefficientsquotesdbs_dbs22.pdfusesText_28

[PDF] Interpolation and Approximation: Lagrange Interpolation

What is Lagrange interpolation?

What is multivariate interpolation?

Why is high-degree interpolation a problem?

What is the basic principle of polynomial interpolation?

Lagrange Polynomials

Martin Licht

UC San Diego

Winter Quarter ????

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Lagrange Interpolation

How do we find a good polynomial?

Note that this can be written as

Lagrange Interpolation

We recall some representations of the error:

Theorem

We have

Lagrange Interpolation

Lagrange Interpolation

Lagrange Interpolation

Example

Suppose we have got pointsx?;x?;:::;xmand values

For this example, let us consider the casem=? and

Lagrange Interpolation

Example

The table is

We search for a polynomialpof degreem=? such that

The solution is the polynomial

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Monomial Basis

Monomial Basis

We can rewrite this in matrix notation as

BBBBB@?x?x??:::xm?

CCCCCA0

BBBBB@a

CCCCCA=0

BBBBB@y

CCCCCA:

The determinant of the Vandermonde matrix V is

Proof.

BBBBB@?x?x??:::xm?

CCCCCA= det0

BBBBB@?x?x??:::xm?

CCCCCA

Monomial Basis

Proof.

Consequently, we end up with the determinant

BBBBB@? ? ?:::?

CCCCCA

Monomial Basis

Proof.

BBBBB@? ? ?:::?

CCCCCA

BBB@?x?:::xm?

Monomial Basis

Proof.

Monomial Basis

Given pairwise distinct points x

Monomial Basis

Example

The solution is

A= (?)(?)(?) =?:

Monomial Basis

Example

The inverse of that Vandermonde matrix is

Lagrange Interpolation

Monomial Basis

Newton Polynomials

Lagrange Polynomials

Error Analysis for Lagrange Polynomials

Newton Polynomials

Newton polynomials