Ch18 Interpolation

The Islamic University of Gaza

Faculty of Engineering

Civil Engineering Department

Numerical AnalysisECIV 3306

Chapter 18

Interpolation

IntroductionThe most common method for Estimation ofintermediate values between precise data pointsis polynomial interpolation. :

Polynomial interpolation is used when the pointdetermined are very precise. The curverepresenting the behavior has to pass throughevery point.There is one and only one nth-order polynomialthat fits n+1 points

nn xaxaxaaxf 2

210)(

Introduction

First order (linear) 3rd order (cubic)2nd order (quadratic)

n = 2 n = 3 n = 4

Introduction

There are a variety of mathematical formats inwhich this polynomial can be expressed:

The Newton polynomial (sec. 18.1)

The Lagrange polynomial (sec. 18.2)

Lagrange Interpolating Polynomials

• The general form for n+1 data points is:

n

ijj ji

ji

n

iiin

xxxx

xL

xfxLxf

0

0

)(

)()()(

designates the “product of”


)()()( 101

00

10

11 xf

xxxxxf

xxxxxf

• Linear version (n = 1):Used for 2 points of data: (xo,f(xo)) and (x1,f(x1)),

)(xLo )(1 xL


1,)(1 jxL

)(

)(

)()(

21202

10

12101

20

02010

212

xfxxxx

xxxx

xfxxxxxxxx

xfxxxx

xxxxxf

2,)(2 jxL

• Second order version (n = 2):

0,)( jxLo

n

ijj ji

ji xx

xxxL

0

)(

Lagrange Interpolating Polynomials - Example

Use a Lagrange interpolating polynomial of thefirst and second order to evaluate ln(2) on thebasis of the data:

10x 0)1ln()( 0xf

41x62x

386294.1)4ln()( 1xf791760.1)6ln()( 2xf

Lagrange Interpolating Polynomials – Example(cont’d)

• First order polynomial:

)()()( 101

00

10

11 xf

xxxxxf

xxxxxf

4620981.0386294.114120

4142)2(1f

Lagrange Interpolating Polynomials –Example (cont’d)

• Second order polynomial:

606x

404x

xxxx

xxxxxL

2o

2

1o

1o )(

646x

040x

xxxx

xxxxxL

21

2

o1

o1 )(

464x

060x

xxxx

xxxxxL

12

1

o2

o2 )(


5658444.0791760.1)46)(16()42)(12(

386294.1)64)(14()62)(12(

0)61)(41()62)(42()2(2f

n

0iiin xfxLxf )()()( )()( ij

xxxx

xLn

0j ji

ji


Pseudocode – Lagrange interpolation

Coefficients of an Interpolating Polynomial

• Although “Lagrange” polynomials are well suited fordetermining intermediate values between points, theydo not provide a polynomial in conventional form:

• Since n+1 data points are required to determine n+1coefficients, simultaneous linear systems of equationscan be used to calculate “a”s.

nx xaxaxaaxf 2

210)(

Coefficients of an Interpolating Polynomial(cont’d)

nnnnnn

nn

nn

xaxaxaaxf

xaxaxaaxfxaxaxaaxf

2210

12121101

02020100

)(

)(

)(

Where “x”s are the knowns and “a”s are theunknowns.

Possible divergence of an extrapolatedproduction

Spline Interpolation

• Polynomials are the most common choice ofinterpolants.

• There are cases where polynomials can lead toerroneous results because of round off error andovershoot.

• Alternative approach is to apply lower-orderpolynomials to subsets of data points. Suchconnecting polynomials are called spline functions.

18

Why Spline Interpolation?

Apply lower-order polynomials to subsets of data points.

Spline provides a superior approximation of the behavior of functionsthat have local, abrupt changes.

Spline InterpolationThe concept of spline is using a thin , flexible strip(called a spline) to draw smooth curves through a setof points….natural spline (cubic)

Linear Spline

The first order splines for a group of ordered datapoints can be defined as a set of linear functions:

ii

iii xx

xfxfm1

1 )()(

)()()( 000 xxmxfxf10 xxx

)()()( 111 xxmxfxf 21 xxx

)()()( 111 nnn xxmxfxf nn xxx 1

Linear spline - Example

Fit the following data with first order splines.Evaluate the function at x = 5.

x f(x)

3.0 2.54.5 1.07.0 2.59.0 0.5

6.05.4715.2m

3.15.06.00.1

)5.45()5.4()5( mff

Linear Spline

• The main disadvantage of linear spline is that they arenot smooth. At the data points where 2 splines meetscalled (a knot), the slope changes abruptly.

• The first derivative of the function is discontinuous atthese points.

• Using higher order polynomial splines ensuresmoothness at the knots by equating derivatives atthese points.

Quadratic Splines

iiii cxbxaxf 2)(• Objective: to derive a second order polynomial for eachinterval between data points.• Terms: Interior knots and end points

For n+1 data points:• i = (0, 1, 2, …n),• n intervals,• 3n unknownconstants (a’s, b’s andc’s)

Quadratic Splines

1. The function values of adjacent polynomialmust be equal at the interior knots 2(n-1).

2. The first and last functions must pass throughthe end points (2).

nixfcxbxa

nixfcxbxa

iiiiiii

iiiiiii

,...,4,3,2)(

,...,4,3,2)(

112

1

11112

11

)(

)(2

01012

01

nnnnnn xfcxbxa

xfcxbxa

Quadratic Splines

• The first derivatives at the interior knotsmust be equal (n-1).

• Assume that the second derivate is zero atthe first point (1)

(The first two points will be connected by a straight line)

iiiiii

iii

bxabxabxaxf

1111

'

222)(

01a

Quadratic Splines - Example

Fit the following data with quadratic splines.Estimate the value at x = 5.

Solutions:There are 3 intervals (n=3), 9 unknowns.

x 3.0 4.5 7.0 9.0f(x) 2.5 1.0 2.5 0.5


1. Equal interior points:

For first interior point (4.5, 1.0)

The 1st equation:

The 2nd equation:)( 12212

21 xfcbxax

0.15.425.20 111 cba

)( 1111121 xfcbxax

)5.4(5.4)5.4( 1112 fcba

0.15.425.20 222 cba)5.4(5.4)5.4( 2222 fcba


For second interior point (7.0, 2.5)

The 3rd equation:

The 4th equation:

5.2749 222 cba

5.2749 333 cba

)( 2222222 xfcbxax

)7(7)7( 2222 fcba

)( 2332322 xfcbxax

)7(7)7( 3332 fcba


First and last functions pass the end points

For the start point (3.0, 2.5)

For the end point (9, 0.5)

5.239 111 cba

5.0981 333 cba

)( 0110120 xfcbxax

)( 3333323 xfcbxax


Equal derivatives at the interior knots.

For first interior point (4.5, 1.0)

For second interior point (7.0, 2.5)

Second derivative at the first point is 00)( 10

'' axf

2211 99 baba221111 22 baxbax

3322 1414 baba333222 22 baxbax


00

5.05.25.25.2

11

01140114000000190119810000000000013174900000

00017490000015.425.200000000015.4

3

3

3

2

2

2

1

1

cbacbacb


5.511

5.21

1315.4 1

1

1

cb

cb

46.1876.664.0

15.2

1

019174915.425.20

2

2

2

2

2

2

cba

cba


Solving these 8 equations with 8 unknowns

3.91,6.24,6.146.18,76.6,64.0

5.5,1,0

333

222

111

cbacbacba

,5.5)(1 xxf 5.40.3 x

,46.1876.664.0)( 22 xxxf 0.75.4 x

,3.916.246.1)( 23 xxxf 0.90.7 x

f1(x)f2(x)

f3(x)

Cubic Splines

iiiii dxcxbxaxf 23)(

Objective: to derive a third order polynomial foreach interval between data points.Terms: Interior knots and end points

For n+1 data points:• i = (0, 1, 2, …n),• n intervals,• 4n unknown constants (a’s, b’s ,c’s and d’s)

Cubic Splines

• The function values must be equal at the interior knots(2n-2).

• The first and last functions must pass through the endpoints (2).

• The first derivatives at the interior knots must be equal(n-1).

• The second derivatives at the interior knots must beequal (n-1).

• The second derivatives at the end knots are zero (2), (the 2nd

derivative function becomes a straight line at the end points)

Alternative technique to get Cubic Splines

• The second derivative within each interval [xi-1, xi ] is a straight line. (the 2nd

derivatives can be represented by first order Lagrange interpolatingpolynomials.

1

1''

11

'''' )()()(ii

iii

ii

iiii xx

xxxfxx

xxxfxfA straight lineconnecting the firstknot f’’(xi-1) and thesecond knot f’’(xi)

The second derivative at any point x within the interval

Cubic Splines• The last equation can be integrated twice

2 unknown constants of integration can be evaluated byapplying the boundary conditions:1. f(x) = f (xi-1) at xi-1

2. f(x) = f (xi) at xi

11

''

1

11''

1

1

31

1

''3

1

1''

6)()(

6)()(

6)(

6)()(

iiiii

ii

ii

iiiii

ii

ii

iii

iii

ii

iii

xxxxxfxxxf

xxxxxfxx

xf

xxxx

xfxxxx

xfxf

)('' ixf

Unknowns:

)('' 1ixfi = 0, 1,…, n

Cubic Splines

)()(6

)()(6)()(

)()(2)()(

11

11

1''

1

''111

''1

iiii

iiii

iii

iiiiii

xfxfxx

xfxfxx

xfxx

xfxxxfxx

• For each interior point xi (n-1):

This equation result with n-1 unknown secondderivatives where, for boundary points:

(xo) = f (xn) = 0

)()( ''1 iii xfxf

i

Cubic Splines - Example

Fit the following data with cubic splinesUse the results to estimate the value at x=5.

Solution:

Natural Spline:

x 3.0 4.5 7.0 9.0f(x) 2.5 1.0 2.5 0.5

0)9()(,0)3()( ''3

''''0

'' fxffxf


For 1st interior point (x1 = 4.5)

-

-

-Apply the following equation:

)()(6)()(6)()()()(2)()(

11

11

1''

1''

111''

1

iiii

iiii

iiiiiiiii

xfxfxx

xfxfxx

xfxxxfxxxfxx

x 3.0 4.5 7.0 9.0f(x) 2.5 1.0 2.5 0.5

5.10.35.4011 xxxx ii

5.25.47121 xxxx ii

40.370211 xxxx ii


)15.2(5.1

6)15.2(5.2

6)7(5.2)5.4(42)3(5.1 '''''' fff

0)3(''f

)1.(..............6.9)7(5.2)5.4(8 '''' eqff

x 3.0 4.5 7.0 9.0f(x) 2.5 1.0 2.5 0.5

Since

For 2nd interior point (x2 = 7 )

5.25.47121 xxxx ii

5.45.491311 xxxx ii

279231 xxxx ii


Apply the following equation:

)()(6)()(6)()()()(2)()(

11

11

1''

1''

111''

1

iiii

iiii

iiiiiiiii

xfxfxx

xfxfxx

xfxxxfxxxfxx

)5.21(5.2

6)5.25.0(26)9(2)7(5.42)5.4(5.2 '''''' fff

Since 0)9(''f

)2(.............6.9)7(9)5.4(5.2 '''' equff

Cubic Splines - ExampleSolve the two equations:

The first interval (i=1), apply for the equation:

53308.1)7(,67909.1)5.4(6.9)7(9)5.4(5.2

6.9)7(5.2)5.4(8 ''''''''

''''

ffyeildffff

i

ii

i

11

''

1

11''

1

1

31

1

''3

1

1''

6)()(

6)()(

6)(

6)()(

iiiii

ii

iii

iiii

ii

ii

iii

iii

ii

iii

xxxxxfxxxfxxxxxf

xxxf

xxxx

xfxxxx

xfxf

)3(24689.0)5.4(6667.1)3(186566.0)( 31 xxxxf

)3(6

)5.1(67909.15.1

15.46

)5.1(05.15.2)3(

)5.1(667909.1)3(0)( 33

1 xxxxxf i


)5.4(6

)5.2(53308.15.25.2

76

)5.2(67909.15.2

1)5.4()5.2(6

53308.1)7()5.2(6

67909.1)( 332

x

xxxxf

)5.4(638783.1)7(29962.0)5.4(102205.0)7(111939.0)( 332 xxxxxf

)7(25.0)9(761027.1)9(127757.0)( 33 xxxxf

102886.1)5()( 22 fxf

The 2nd interval (i =2), apply for the equation:

The 3rd interval (i =3),

For x = 5:

Documents

Ch18 Interpolation