Interpolation - IntroductionEstimation of intermediate values betweenprecise data points. The most common method ispolynomial interpolation:

Polynomial interpolation is used when the pointdetermined are very precise. The curverepresenting the behavior has to pass throughevery point (has to touch).

There is one and only one nth-order polynomialthat fits n+1 points

f ( x )=a0+a1 x+a2 x2+⋯ +an x n

Introduction

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

n = 2 n = 3 n = 4

3

Interpolation

Polynomials are the most commonchoice of interpolation because theyare easy to:

EvaluateDifferentiate, andIntegrate.

Introduction

There are a variety of mathematical formats in which thispolynomial can be expressed:

The Newton polynomial (sec. 18.1)

The Lagrange polynomial (sec. 18.2)

5

Newton’s Divided-DifferenceInterpolating PolynomialsLinear Interpolation/

Is the simplest form of interpolation, connecting two data pointswith a straight line.

f1(x) designates that this is a first-order interpolating polynomial.

)()()(

)()(

)()()()(

001

0101

01

01

0

01

xxxx

xfxfxfxf

xx

xfxf

xx

xfxf

−−−+=

−−=

−−

Linear-interpolationformula

Slope and afinite divideddifferenceapproximation to1st derivative

6

Figure

18.2

7

Quadratic Interpolation/

If three data points are available, the estimate is improvedby introducing some curvature into the line connectingthe points.

A simple procedure can be used to determine the valuesof the coefficients.

))(()()( 1020102 xxxxbxxbbxf −−+−+=

02

01

01

12

12

22

01

0111

000

)()()()(

)()(

)(

xx

xx

xfxf

xx

xfxf

bxx

xx

xfxfbxx

xfbxx

−−−−

−−

==

−−==

==

8

General Form of Newton’s Interpolating Polynomials/

0

02111011

011

0122

011

00

01110

012100100

],,,[],,,[],,,,[

],[],[],,[

)()(],[

],,,,[

],,[

],[

)(

],,,[)())((

],,[))((],[)()()(

xx

xxxfxxxfxxxxf

xx

xxfxxfxxxf

xx

xfxfxxf

xxxxfb

xxxfb

xxfb

xfb

xxxfxxxxxx

xxxfxxxxxxfxxxfxf

n

nnnnnn

ki

kjjikji

ji

jiji

nnn

nnn

n

−−=

−−

=

−−

=

=

===

−−−++−−+−+=

−−−−

−

−−

Bracketed functionevaluations are finitedivided differences

Lagrange Interpolating Polynomials

The general form for n+1 data points is:

∏

∑

≠=

=

−−

=

=

n

ijj ji

ji

n

iiin

xx

xxxL

xfxLxf

0

0

)(

)()()(

designates the “product of”

Lagrange Interpolating Polynomials

f 1( x )=x− x1

x0− x1

f ( x0 )+x− x0

x1− x0

f ( x1 )

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

Lo( x ) L1 ( x )

Lagrange InterpolatingPolynomials

L1 ( x ) , j≠ 1

f 2 ( x )=( x− x1)( x− x2)( x0− x1 )( x0− x 2 )

f ( x0)

+( x− x0) (x− x2)( x1− x0 )( x1− x 2 )

f ( x1 )

+( x− x0)( x− x1)( x2− x0 )( x2− x 1 )

f ( x 2) L2( x ) , j≠ 2

• Second order version (n = 2):

Lo( x ) , j≠ 0

Lagrange Interpolating Polynomials -Example

Use a Lagrange interpolating polynomial of the first andsecond order to evaluate ln(2) on the basis of the data:

x 0=1 f ( x 0 )=ln (1 )=0

x 1=4x 2=6

f ( x 1 )=ln (4 )=1.386294f ( x 2)=ln (6)=1.791760

Lagrange Interpolating Polynomials –Example (cont’d)

● First order polynomial:

f 1( x )=x− x 1

x 0− x 1f ( x 0)+

x− x 0

x 1− x 0f ( x 1 )

f 1( 2)=2− 41− 4

⋅ 0+ 2− 14− 1

⋅ 1 . 386294=0 . 4620981

Lagrange Interpolating Polynomials –Example (cont’d)

● Second order polynomial:

60

6x

40

4x

xx

xx

xx

xxxL

2o

2

1o

1o −

−⋅−−=

−−⋅

−−=)(

64

6x

04

0x

xx

xx

xx

xxxL

21

2

o1

o1 −

−⋅−−=

−−⋅

−−=)(

46

4x

06

0x

xx

xx

xx

xxxL

12

1

o2

o2 −

−⋅−−=

−−⋅

−−=)(

Lagrange Interpolating Polynomials –Example (cont’d)

f 2 (2)=(2− 4 )(2− 6)(1− 4 )(1− 6)

⋅ 0

+( 2− 1)( 2− 6 )( 4− 1 )(4− 6)

⋅ 1.386294

+(2− 1)(2− 4 )(6− 1)(6− 4 )

1.791760=0 . 5658444

∑=

=n

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

xx

xxxL

n

0j ji

ji ≠

−−

= ∏=

Lagrange Interpolating Polynomials –Example (cont’d)

Coefficients of an Interpolating Polynomial

● Although “Lagrange” polynomials are well suitedfor determining intermediate values betweenpoints, they do not provide a polynomial inconventional form:

● Since n+1 data points are required to determinen+1 coefficients, simultaneous linear systems ofequations can be used to calculate “a”s.

f ( x )=a0+a1 x+a2 x2+⋯ +a x xn

Coefficients of an InterpolatingPolynomial (cont’d)

f ( x0 )=a0+a1 x0+a2 x02⋯ +an x0

n

f ( x1 )=a0+a1 x1+a2 x12⋯ +an x1

n

⋮f ( xn )=a0+a1 xn+a2 x n

2⋯ +an xnn

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

Possible divergence of an extrapolatedproduction

20

Why Spline Interpolation?

Apply lower-order polynomials to subsets of data points. Splineprovides a superior approximation of the behavior of functions thathave local, abrupt changes.

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 splinefunctions.

22

Why Splines ?

f ( x )=1

1+25 x2

Table : Six equidistantly spaced points in [-1, 1]

Figure : 5th order polynomial vs. exact function

x 2251

1

xy

+=

-1.0 0.038461

-0.6 0.1

-0.2 0.5

0.2 0.5

0.6 0.1

1.0 0.038461

23

Why Splines ?

Figure : Higher order polynomial interpolation is a bad idea

OriginalFunction

17th OrderPolynomial

9th OrderPolynomial

5th OrderPolynomial

Spline InterpolationThe concept of spline is using a thin , flexible strip(called a spline) to draw smooth curves through aset of 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:

mi=f ( xi+1 )− f ( x i )

x i+1− x i

f ( x )= f ( x0)+m0( x− x0 ) x0≤ x≤ x1

f ( x )= f ( x1 )+m1 ( x− x1 ) x1≤ x≤ x2

f ( x )= f ( xn− 1 )+mn− 1 ( x− x n− 1 ) xn− 1≤ x≤ xn

⋮

Linear spline - ExampleFit the following data with first order splines. Evaluatethe function at x = 5.

x f(x)

3.0 2.54.5 1.07.0 2.59.0 0.5

m=2. 5− 17− 4 . 5

=0 . 6

f (5)= f (4 .5 )+m(5− 4 .5)=1 . 0+0 . 6×0. 5=1 . 3

Linear Spline● The main disadvantage of linear spline is that

they are not smooth. The data points where 2splines meets called (a knot), the changesabruptly.

● The first derivative of the function is discontinuousat these points.

● Using higher order polynomial splines ensuresmoothness at the knots by equating derivativesat these points.

Quadric Splines

f i( x )=a i x2+bi x+ci

• 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)

Quadric Splines (3n conditions)

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

● The first and last functions must passthrough the end points (2).

a i− 1 xi− 1

2+bi− 1 xi− 1+ci− 1= f i ( xi− 1 ) i=2, 3, 4, . .. , n

a i xi− 1

2+b i x i− 1+ci= f i( xi− 1 ) i=2, 3, 4, . .. , n

a1 x0

2+b1 x0+c1= f ( x0 )

an xn

2+bn xn+cn= f ( xn )

Quadric Splines (3n conditions)● The first derivatives at the interior knots

must be equal (n-1).

● Assume that the second derivate is zeroat the first point (1)

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

fi' ( x )=2ai x+bi

2ai− 1 x i− 1+bi− 1=2a i xi− 1+bi

a 1=0

Quadric Splines - Example

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

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

x 3.0 4.5 7.0 9.0

f(x) 2.5 1.0 2.5 0.5

Quadric Splines - Example1. Equal interior points:

For first interior point (4.5, 1.0)

The 1st equation:

The 2nd equation:

Quadric Splines - Example

For second interior point (7.0, 2.5)

The 3rd equation:

The 4th equation:

49 a2+7b2+c2=2. 5

49 a3+7b3+c3=2 .5

x22 a2+ x2 b2+c2= f ( x2 )

(7)2 a2+7b2+c2= f (7)

x22 a3+x2 b3+c3= f ( x2 )

(7)2 a3+7b3+c3= f (7 )

Quadric Splines - Example

First and last functions pass the endpoints

For the start point (3.0, 2.5)

For the end point (9, 0.5)

9a1+3b1+c1=2. 5

81a3+9b3+c3=0 .5

x02 a1+x0 b1+c1= f ( x0 )

x32 a1+x3 b3+c3= f ( x3 )

Quadric Splines - ExampleEqual 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 0

Quadric Splines - Example

[ ] [ ] [ ] [ ] [ ] [ ] [ ]

righ

righ

c

b

a

c

b

a

c

b

[]

righ

0

0

0.5

2.5

2.5

2.5

1

1

0114011400

00001901

198100000

00000013

174900000

000174900

00014.520.2500

00000014.5

3

3

3

2

2

2

1

1

−−−−

Quadric Splines - ExampleSolving these 8 equations with 8 unknowns

a1=0, b1=− 1, c1=5 . 5

a2=0 . 64 , b2=− 6. 76 , c2=18. 46a3=− 1. 6, b3=24. 6, c3=− 91. 3

f 1( x )=− x+5. 5, 3 .0≤ x≤ 4 . 5

f 2 ( x )=0. 46 x2− 6 .76 x+18 . 46 , 4 . 5≤ x≤ 7 . 0

f 3( x )=− 1. 6x2+24 .6x− 91. 3, 7 . 0≤ x≤ 9 .0

Cubic Splines

f i( x )=a i x3+bi x2+c i x+d i

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 (4n conditions)● The function values must be equal at the interior

knots (2n-2).● The first and last functions must pass through the

end points (2).● The first derivatives at the interior knots must be

equal (n-1).● The second derivatives at the interior knots must

be equal (n-1).● The second derivatives at the end knots are zero (2),

(the 2nd derivative function becomes a straight line atthe end points)

Alternative technique to get CubicSplines● The second derivative within each interval [xi-1, xi ] is a straight line.

(the 2nd derivatives can be represented by first order Lagrangeinterpolating polynomials.

fi''( x )= f

i''( xi− 1)

x− x i

x i− 1− x i

+ fi''( xi )

x− x i− 1

xi− xi− 1

A 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 evaluatedby applying the boundary conditions:1. f(x) = f (xi-1) at xi-1

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

Unknowns:

i = 0, 1,…, n

Cubic Splines

( xi− x i− 1) f ''( x i− 1)+2( xi+1− x i− 1 ) f ''( xi )

+( x i+1− xi ) f ''( xi+1 )=6x i+1− xi

[ f ( x i+1 )− f ( xi )]

+6xi− xi− 1

[ f ( x i− 1)− f ( x i)]

• For each interior point xi (n-1):

This equation result with n-1 unknown secondderivatives where, for boundary points:f˝(xo) = f˝(xn) = 0

fi− 1

' ( x i )= f i' ( x 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 9.0

f(x) 2.5 1.0 2.5 0.5

f ''( x0)= f ''(3 )=0, f ''( x3 )= f ''(9 )=0

Cubic Splines - Example For 1st interior point (x1 = 4.5)

---Apply the following equation:

( xi− x i− 1) f ''( x i− 1)+2( xi+1− x i− 1 ) f ''( xi )+( x i+1− x i) f ''( xi+1 )

¿6xi+1− x i

[ f ( xi+1 )− f ( x i )] +6x i− xi− 1

[ f ( xi− 1 )− f ( xi )]

x 3.0 4.5 7.0 9.0

f(x) 2.5 1.0 2.5 0.5x i− x i− 1=x1− x0=4 . 5− 3 . 0=1. 5

x i+1− xi= x2− x1=7− 4.5=2 . 5

x i+1− xi− 1= x2− x0=7− 3. 0=4

Cubic Splines - Example

1. 5f ''(3 )+2×4f ''(4 .5)+2 . 5f ''(7 )=6

2 . 5(2 . 5− 1 )+

61 .5(2 . 5− 1 )

f ''(3 )=0

8f ''( 4. 5 )+2 .5f ''(7)=9. 6 .. . .. .. . .. .. . . (eq .1 )

x 3.0 4.5 7.0 9.0

f(x) 2.5 1.0 2.5 0.5

Since

For 2nd interior point (x2 = 7 )

x i− x i− 1=x2− x1=7− 4 . 5=2. 5

x i+1− xi− 1= x3− x1=9− 4 . 5=4 . 5

x i+1− xi= x3− x2=9− 7=2

Cubic Splines - Example

Apply the following equation:

( xi− x i− 1) f ''( x i− 1)+2( xi+1− x i− 1 ) f ''( xi )+( x i+1− x i) f ''( xi+1 )

¿6xi+1− x i

[ f ( xi+1 )− f ( x i )] +6x i− xi− 1

[ f ( xi− 1 )− f ( xi )]

2 . 5f ''( 4. 5)+2×4 .5f ''(7)+2f ''(9 )=62(0. 5− 2 . 5)+

62.5(1− 2. 5)

Since f ''(9 )=0

2 . 5f ''(4 . 5)+9f ''(7 )=− 9 .6 . .. .. . .. .. . .. ( equ 2 )

Cubic Splines - ExampleSolve the two equations:

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

8f i''( 4. 5 )+2 .5f i

''(7)=9.6

2. 5f i''( 4. 5 )+9f i

''(7)=− 9 . 6

¿ }¿¿ yeild f ''(4 . 5)=1. 67909 , f ''(7)=− 1 .53308¿

f i( x )=f

i''( xi− 1 )

6( x i− x i− 1)(xi− x )3+

fi''( xi )

6( xi− xi− 1 )( x− xi− 1)

3

+[ f i ( xi− 1)xi− xi− 1

−f

i''( xi− 1 )(xi− x i− 1)

6 ]( x i− x)+[ f i( x i )xi− x i− 1

−f

i''( xi )(x i− x i− 1)

6 ]( x− x i− 1)

f 1( x )=0 .186566 ( x− 3 )3+1 .6667 (4 . 5− x )+0 . 24689( x− 3 )

f 1( x )=0 ( xi− 3 )3+1. 679096(1. 5)

( x− 3 )3+[ 2 . 51 . 5

− 0(1.5 )6 ](4 . 5− x )+[ 1

1. 5− 1. 67909(1. 5)

6 ]( x− 3)

Cubic Splines - Example

f 2 ( x )=1 . 679096 (2 .5 )

(7− x )3+− 1 .533086(2 .5 )

( x− 4 .5 )3+[12 .5− − 1. 67909(2 .5)

6 ](7− x)

+[2.52.5

− − 1 . 53308(2 . 5)6 ]( x− 4 . 5)

f 2 ( x )=0. 111939(7− x )3− 0 . 102205 ( x− 4 .5 )3− 0 . 29962(7− x )+1. 638783 ( x− 4 . 5)

f 3( x )=− 0 .127757 (9− x )3+1.761027 (9− x )+0. 25 ( x− 7 )

f 2 ( x )= f 2(5)=1. 102886

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

The 3rd interval (i =3),

For x = 5:

Credits:● Chapra, Canale● The Islamic University of Gaza, Civil Engineering Department