CE30125 - Lecture 3

p. 3.1

LECTURE 3

LAGRANGE INTERPOLATION

• Fit points with an degree polynomial

• = exact function of which only discrete values are known and used to estab-lish an interpolating or approximating function

• = approximating or interpolating function. This function will pass through allspecified interpolation points (also referred to as data points or nodes).

N 1+ Nth

f1

x0

g(x)f(x)

f0

f2

f3 f4fN

x1 x2 x3 x4 xN...

f x N 1+g x

g x N 1+

CE30125 - Lecture 3

p. 3.2

• The interpolation points or nodes are given as:

:

• There exists only one degree polynomial that passes through a given set of points. It’s form is (expressed as a power series):

where = unknown coefficients, ( coefficients).

• No matter how we derive the degree polynomial,

• Fitting power series

• Lagrange interpolating functions

• Newton forward or backward interpolation

The resulting polynomial will always be the same!

xo f xo fo

x1 f x1 f1

x2 f x2 f2

xN f xN fN

Nth

N 1+

g x ao a1x a2x2 a3x3 aNxN+ + + + +=

ai i 0 N= N 1+

Nth

CE30125 - Lecture 3

p. 3.3

Power Series Fitting to Define Lagrange Interpolation

• must match at the selected data points

: :

• Solve set of simultaneous equations

• It is relatively computationally costly to solve the coefficients of the interpolating func-tion (i.e. you need to program a solution to these equations).

g x f x

g xo fo = ao a1xo a2xo2 aNxo

N+ + + + fo=

g x1 f1 = ao a1x1 a2x12

+ + aNx1N

+ + f1=

g xN fN = ao a1xN+ a2xN2 aNxN

N+ + + fN=

1 xo xo2 xo

N

1 x1 x12 x1

N

1 xN xN2 xN

N

ao

a1

:

aN

fo

f1

:

fN

=

g x

CE30125 - Lecture 3

p. 3.4

Lagrange Interpolation Using Basis Functions

• We note that in general

• Let

where = polynomial of degree associated with each node such that

• For example if we have 5 interpolation points (or nodes)

Using the definition for : ; ; ; ;

,we have:

g xi fi=

g x fi Vi x i 0=

N

=

Vi x N i

Vi xj 0 i j1 i = j

g x3 foVo x3 f1V1 x3 f2V2 x3 f3V3 x3 f4V4 x3 + + + +=

Vi xj V0 x3 0= V1 x3 0= V2 x3 0= V3 x3 1=

V4 x3 0=

g x3 f3=

CE30125 - Lecture 3

p. 3.5

• How do we construct ?

• Degree

• Roots at (at all nodes except )

•

• Let

• The function is such that we do have the required roots, i.e. it equals zero at nodes except at node

• Degree of is

• However in the form presented will not equal to unity at

• We normalize and define the Lagrange basis functions

Vi x

N

xo x1 x2 xi 1– xi 1+ xN xi

Vi xi 1=

Wi x x xo– x x1– x x2– x xi 1–– x xi 1+– x xN– =

Wi

xo x1 x2 ... , xN xi

Wi x N

Wi x xi

Wi x Vi x

Vi x x xo– x x1– x x2– x xi 1–– x xi 1+– x xN–

xi xo– xi x1– xi x2– xi xi 1–– xi xi 1+– xi xN– --------------------------------------------------------------------------------------------------------------------------------------------------------=

CE30125 - Lecture 3

p. 3.6

• Now we have such that equals:

• We also satisfy

e.g.

• The general form of the interpolating function with the specified form of is:

• The sum of polynomials of degree is also polynomial of degree

• is equivalent to fitting the power series and computing coefficients .

Vi x Vi xi

Vi xi xi xo– xi x1– xi x2– xi xi 1–– 1 xi xi 1+– xi xN–

xi xo– xi x1– xi x2– xi xi 1–– xi xi 1+– xi xN– ------------------------------------------------------------------------------------------------------------------------------------------------------------=

Vi xi 1=

Vi xj 0 for i j=

V1 x2 x2 xo– 1 x2 x2– x2 x3– x2 xN– x1 xo– 1 x1 x2– x1 x3– x1 xN–

--------------------------------------------------------------------------------------------------------- 0= =

g x Vi x

g x fiVi x i 0=

N

=

N N

g x ao aN

CE30125 - Lecture 3

p. 3.7

Lagrange Linear Interpolation Using Basis Functions

• Linear Lagrange is the simplest form of Lagrange Interpolation

where

and

N 1=

g x fiVi x i 0=

1

=

g x foVo x f1V1 x +=

Vo x x x1– xo x1–

---------------------x1 x– x1 xo–

--------------------- = = V1 x x xo– x1 xo–

---------------------=

x0

(x)

V0 (x)

x1

V1(x)1.0

CE30125 - Lecture 3

p. 3.8

Example

• Given the following data:

Find the linear interpolating function

• Lagrange basis functions are:

and

• Interpolating function g(x) is:

xo 2 = fo 1.5=

x1 5 = f1 4.0=

g x

Vo x 5 x–3

----------- = V1 x x 2–3

-----------=

g x 1.5Vo x 4.0V1 x +=

CE30125 - Lecture 3

p. 3.9

x0 = 2

1.5 V0 (x)4

2

4

2

x1 = 5x

4.0 V1(x)

x0 = 2 x1 = 5x

x0 = 2 x1 = 5

g(x) = 1.5 V0(x) + 4.0V1(x)

CE30125 - Lecture 3

p. 3.10

Lagrange Quadratic Interpolation Using Basis Functions

• For quadratic Lagrange interpolation, N=2

where

g x fi Vi x i 0=

2

=

g x foVo x f1V1 x f2V2 x + +=

Vo x x x1– x x2–

xo x1– xo x2– ------------------------------------------=

V1 x x xo– x x2–

x1 xo– x1 x2– ------------------------------------------=

V2 x x xo– x x1–

x2 xo– x2 x1– ------------------------------------------=

CE30125 - Lecture 3

p. 3.11

• Note that the location of the roots of , and are defined such that the

basic premise of interpolation is satisfied, namely that . Thus:

x0

V0 (x)

1.0

x1

xx2

V1(x) V2(x)

V0 x V1 x V2 x g xi fi=

g xo Vo xo fo V1 xo f1 V2 xo f2+ + f0= =

g x1 Vo x1 fo V1 x1 f1 V2 x1 f2+ + f1= =

g x2 Vo x2 fo V1 x2 f1 V2 x2 f2+ + f2= =

CE30125 - Lecture 3

p. 3.12

Example

• Given the following data:

Find the quadratic interpolating function

• Lagrange basis functions are

• Interpolating function g(x) is:

xo 3= fo 1=

x1 4 = f1 2=

x2 5 = f2 4=

g x

Vo x x 4– x 5– 3 4– 3 5–

----------------------------------=

V1 x x 3– x 5– 4 3– 4 5–

----------------------------------=

V2 x x 3– x 4– 5 3– 5 4–

----------------------------------=

g x 1.0Vo x 2.0V1 x 4.0V2 x + +=

CE30125 - Lecture 3

p. 3.13

1.0 V0 (x)1.0x

4.0 V2(x)

x1 = 4x0 = 3 x2 = 5

2.0

4.0

x1 = 4x0 = 3 x2 = 5

x1 = 4x0 = 3 x2 = 5

x

x

2.0 V1(x)

x1 = 4x0 = 3 x2 = 5

4.0g(x) = 1.0 V0(x) + 2.0V1(x) + 4.0V2(x)

CE30125 - Lecture 3

p. 3.14

Lagrange Cubic Interpolation Using Basis Functions

• For Cubic Lagrange interpolation, N=3

Example

• Consider the following table of functional values (generated with )

• Find as:

0 0.40 -0.916291

1 0.50 -0.693147

2 0.70 -0.356675

3 0.80 -0.223144

f x xln=

i xi fi

g 0.60

g x fo

x x1– x x2– x x3– xo x1– xo x2– xo x3–

---------------------------------------------------------------- f1

x xo– x x2– x x3– x1 xo– x1 x2– x1 x3–

----------------------------------------------------------------+=

f2

x xo– x x1– x x3– x2 xo– x2 x1– x2 x3–

---------------------------------------------------------------- f3

x xo– x x1– x x2– x3 xo– x3 x1– x3 x2–

----------------------------------------------------------------+ +

CE30125 - Lecture 3

p. 3.15

g 0.60 0.9162910.60 0.50– 0.60 0.70– 0.60 0.80– 0.40 0.50– 0.40 0.70– 0.40 0.80–

------------------------------------------------------------------------------------------------–=

0.6931470.60 0.40– 0.60 0.70– 0.60 0.80– 0.50 0.40– 0.50 0.70– 0.50 0.80–

------------------------------------------------------------------------------------------------–

0.3566750.60 0.40– 0.60 0.50– 0.60 0.80– 0.70 0.40– 0.70 0.50– 0.70 0.80–

------------------------------------------------------------------------------------------------–

0.2231440.60 0.40– 0.60 0.50– 0.60 0.70– 0.80 0.40– 0.80 0.50– 0.80 0.70–

------------------------------------------------------------------------------------------------–

g 0.60 0.509976–=

CE30125 - Lecture 3

p. 3.16

Errors Associated with Lagrange Interpolation

• Using Taylor series analysis, the error can be shown to be given by:

where

derivative of w.r.t. evaluated at

• Notes

• If = polynomial of degree where , then

for all x

Therefore will be an exact representation of

e x f x g x –=

e x L x f N 1+ = xo xN

fN 1+ N 1

th+= f x

L x x xo– x x1– x xN–

N 1+ !--------------------------------------------------------------- an N 1

th degree polynomial+= =

f x M M N

fN 1+

x 0 = e x 0=

g x f x

CE30125 - Lecture 3

p. 3.17

• Since in general is not known, if the interval is small and if does not change rapidly in the interval

where .

• can be estimated by using Finite Difference (F.D.) formulae

• will significantly effect the distribution of the error

• is a minimum at the center of and a maximum near the edges

• e.g. using 6 point interpolation looks like:

• at all data points

• largest . becomes very large outside of the interval.

xo xN fN 1+

x

e x L x f N 1+ xm xm

xo xN+

2------------------=

fN 1+

L x

L x xo xN

L x

0 1 2 3 4 5

L x 0=

L x 0 x 1 4 x 5 L x

CE30125 - Lecture 3

p. 3.18

• As the size of the interpolating domain increases, so does the maximum error withinthe interval

• As increases from a small value,

• However as for a given and thus

• Therefore convergence as does not necessarily occur!!

• Properties of will also influence error as and vary

D xN xo–= Lmax x0 x xN

emax x0 x xN

N Lmax x0 x xN

emax x0 x xN

N NCRIT Lmax x0 x xN

xo xN emax x0 x xN

N

fN 1+ D N

CE30125 - Lecture 3

p. 3.19

Example

• Estimate the error made in the previous example knowing that (usuallywe do not have this information).

f x x ln=

e x L x f N 1+ xm

e x x xo– x x1– x x2– x x3–

3 1+ !--------------------------------------------------------------------------- f

3 1+ xm

e 0.60 0.60 0.40– 0.60 0.50– 0.60 0.70– 0.60 0.80– 3 1+ !

--------------------------------------------------------------------------------------------------------------------------------f3 1+

0.6

e 0.60 0.000017 f4

0.6 =

CE30125 - Lecture 3

p. 3.20

• We estimate the fourth derivative of f(x) using the analytical function itself

• Therefore

• Exact error is computed as:

Therefore error estimate is excellent

• Typically we would also have to estimate using a Finite Difference (F.D.)

approximation (a discrete differentiation formula).

f x xln=

f1

x x1–

=

f2

x x2–

–=

f3

x 2x3–

=

f4

x 6x4–

–=

f4

0.6 46.29–=

e 0.60 0.00079–=

E x 0.60 g 0.60 –ln 0.00085–= =

fN 1+

xm

CE30125 - Lecture 3

p. 3.21

SUMMARY OF LECTURES 2 AND 3

• Linear interpolation passes a straight line through 2 data points.

• Power series data points degree polynomial find coefficients bysolving a matrix

• Lagrange Interpolation passes an degree polynomial through data points Use specialized nodal functions to make finding easier.

where

= the interpolating function approximating f(x)

fi = the value of the function at the data (or interpolation) point i

= the Lagrange basis function

• Each Lagrange polynomial or basis function is set up such that it equals unity at thedata point with which it is associated, zero at all other data points and nonzero in-between.

N 1+ Nth

Nth

N 1+g x

g x fiVi x i 0=

N

=

g x

Vi x

CE30125 - Lecture 3

p. 3.22

• For example when N 2 = 3 data points

V0 V1 V2

0 1 2

g x foVo x f1V1 x f2V2 x + +=

f0

f1

f2 g(x)

CE30125 - Lecture 3

p. 3.23

• Linear interpolation is the same as Lagrange Interpolation with

• Error estimates can be derived but depend on knowing (or at some point inthe interval).

where

derivative of w.r.t. evaluated at

N 1=

fN 1+

xm

e x L x f N 1+ = xo xN

fN 1+ N 1

th+= f x

L x x xo– x x1– x xN–

N 1+ !--------------------------------------------------------------- an N 1

th degree polynomial+= =