NTERPOLATION AND CURVE FITTING · Chapter 3 Interpolation & Curve Fitting / 2 3.1 Introduction • Interpolation is a technique to estimate the value between a set of data. • A

3 INTERPOLATION AND CURVE FITTING

Introduction

Newton Interpolation: Finite Divided Difference

Lagrange Interpolation

Spline Interpolation

Polynomial Regression

Multivariable Interpolation

Chapter 3 Interpolation & Curve Fitting / 2

3.1 Introduction

• Interpolation is a technique to estimate the value between a set of data.

• A general approach is to map the data into an n-th order polynomial:

(3.1) ∑=

=+++=n

i

ii

nnn xaxaxaaxf

010)( L

• This chapter covers three types of techniques, i.e. the Newton interpolation, the Lagrange interpolation and the Spline interpolation.

• The resulting equation can be used for curve fitting.


3.2 Newton Interpolation: Finite Divided Difference

• For the linear interpolation, consider Fig. 3.1 and the following relation:

( ) ( ) ( ) ( )01

01

0

01

xxxfxf

xxxfxf

−−

=−−

x

f(x)

x x1

f(x1)

f(x0)

x0

f(x)

FIGURE 3.1 Linear interpolation

Rearranging the above relation leads to

( ) ( )( ) ( ) ( 0

01

0101 xxxx

xfxfxfxf −⋅−

)−+= (3.2)

The term ( ) ( )[ ] ( )0101 xxxfxf −− is referred to as the first-order finite divided difference.

Example 3.1

Calculate an approximate value of ln 2 using the linear interpolation using the following ranges: (1,4) and (1,6). Use ln 1 = 0, ln 4 = 1.3862944 and ln 6 = 1.7917595 as the source data. Estimate the relative error if the actual value is ln 2 = 0.69314718. Solution

For the linear interpolation between x0 = 1 and x1 = 6:

( ) ( ) 35835190.01216

07917595.1021 =−⋅−−

+=f

This produces a relative error of εt = 48.3%.


For the linear interpolation between x0 = 1 dan x1 = 4:

( ) ( ) 46209812.01214

03862944.1021 =−⋅−−

+=f

This produces a relative error of εt = 33.3%, which smaller than the previous case.

• For the quadratic interpolation, consider Fig. 3.2 and the following relation:

0

1

2

0 2 4 6

yf(x) = ln x

x

f1(x)

FIGURE 3.2 The result for the linear interpolation for ln 2

( ) ( ) ( )( )1020102 xxxxbxxbbxf −−+−+= (3.3)

By taking the values of x equal to x0, x1 dan x2 in Eq. (3.3), the coefficients bi can be calculated as followed:

( )00 xfb = (3.4a)

( ) ( )

01

011 xx

xfxfb−−

= (3.4b)

( ) ( ) ( ) ( )

02

01

01

12

12

2 xxxx

xfxfxx

xfxf

b−

−−

−−−

= (3.4c)


Example 3.2

Repeat Example 3.1 using the quadratic interpolation. Solution

From Example 3.1, the data can be rewritten as followed:

7917595.1)(:63862944.1)(:4

0)(:1

22

11

00

======

xfxxfxxfx

Using Eq. (3.4):

051873113.016

1403862944.1

463862944.17917595.1

46209812.014

03862944.10

2

1

0

−=−

−−

−−−

=

=−

−=

=

b

b

b

Hence, Eq. (3.3) yields ( ) ( )( )41051873113.0146209812.00)(2 −−−−+= xxxxf

For x = 2:

( ) ( ) ( )( ).56584436.0

,4212051873113.01246209812.0022=

−−−−+=f

This produces the relative error εt = 18.4% as shown below.


0

1

2

0 2 4 6

y

f(x) = ln x

x

f2(x)

f1(x)

FIGURE 3.3 The result for the quadratic interpolation for ln 2

• The general form representing an n-th order interpolation function (requires n+1 pairs of data) is:

( ) ( ) ( )( ) ( )110010 −−−−++−+= nnn xxxxxxbxxbbxf LL (3.5)

where the coefficients bi can be obtained via

( )[ ][ ]

[ ]011

0122

011

00

,,,,

,,,

xxxxfb

xxxfbxxfb

xfb

nnn K

M

−=

===

where the terms […] are finite divided differences and are defined as:

[ ] ( ) ( )ji

jiji xx

xfxfxxf

−

−=, (3.6a)

[ ] [ ] [ ]ki

kjjikji xx

xxfxxfxxxf

−

−=

,,,, (3.6b)

[ ] [ ] [ ]0

0121121011

,,,,,,,,,,,,xx

xxxxfxxxxfxxxxfn

nnnnnn −

−= −−−−

KKK (3.6c)

Hence, Eq. (3.5) forms the Newton interpolation polynomial:

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

( )( ) [ 01110012100100

,,,)(,,,

xxxfxxxxxxxxxfxxxxxxfxxxfxf

nnn

n

KLL −−−−−++ ]−−+−+=

(3.7)


TABLE 3.1 Newton finite divided difference interpolation polynomial

i xi f(xi ) I II III0 x0 f (x0) f [x1,x0] f [x2,x1,x0] f [x3,x2,x1,x0]

1 x1 f (x1) f [x2,x1] f [x3,x2,x1]

2 x2 f (x2) f [x3,x2]

3 x3 f (x3)

• The relation for error for the n-th order interpolation function can be estimated using:

[ ]( )( ) ( )nnnnn xxxxxxxxxxfR −−−≈ −+ LK 10011 ,,,, (3.8)

Example 3.3

Repeat Example 3.1 with an additional fourth point of x3 = 5, f(x3) = 1.6094379 using the third order Newton interpolation polynomial. Solution

The third order Newton interpolation polynomial can be written as

( ) ( ) ( )( ) ( )( )( )21031020103 xxxxxxbxxxxbxxbbxf −−−+−−+−+=

i xi f(xi ) I II III0 1 0 0.46209813 −0.0597386 0.00786553

1 4 1.3862944 0.22314355 −0.0204110

2 5 1.6094379 0.18232156

3 6 1.7917595

Thus,

( ) ( ) ( )( )( )( )( ),54100786553.0

410597386.0146209812.003−−−+

−−−−+=xxx

xxxxf

( ) ( ) ( )( )( )( )( )

.6287691.0,52421200786553.0

42120597386.01246209812.0023

=−−−+

−−−−+=f

and producing a relative error of εt = 9.29%.


0

1

2

0 2 4 6

y

f(x) = ln x

x

f3(x)

TABLE 3.4 The result for the cubic interpolation for ln 2


3.3 Lagrange Interpolation

• The formula for the Lagrange interpolation is:

(3.9) ( ) ( ) ( )∑=

=n

iiin xfxLxf

0

where the parameter Li(x) is defined as

( ) ∏≠= −

−=

n

ijj ji

ji xx

xxxL

0 (3.10)

• For n = 1, the first order Lagrange interpolation function can be written as:

( ) ( ) ( )101

00

10

11 xfxx

xxxfxxxxxf

−−

+−−

=

• For n = 2, the second order Lagrange interpolation can be written as:

( ) ( )( )( )( ) ( )( )( )

( )( ) ( )

( )( )( )( ) ( )21202

10

12101

200

2010

212

xfxxxx

xxxx

xfxxxx

xxxxxfxxxx

xxxxxf

−−−−

+

−−−−

+−−

−−=

• The error term for the Lagrange interpolation can be estimated using:

(3.11) [ ] (∏=

−+ −=n

iinnnn xxxxxxfR

0011 ,,,, K )

Example 3.5

Repeat Examples 3.1 and 3.2 using the first and second order Lagrange interpolation functions, respectively. Solution

From Examples 3.1:

7917595.1)(:63862944.1)(:4

0)(:1

22

11

00

======

xfxxfxxfx


For the first order Lagrange interpolation:

( ) ( ) ( )

( ) ( ) ( )

4620981.0

3862944.114120

414221

101

00

10

11

=−−

+−−

=∴

−−

+−−

=

f

xfxxxxxf

xxxxxf

For the second order Lagrange interpolation:

( ) ( )( )( )( ) ( )( )( )

( )( ) ( )

( )( )( )( ) ( )21202

10

12101

200

2010

212

xfxxxx

xxxx

xfxxxx

xxxxxfxxxx

xxxxxf

−−−−

+

−−−−

+−−

−−=

( )( )( )( ) ( )

( )( )( )( ) ( )

( )( )( )( ) ( )

5658444.0

7917595.146164212

3862944.1641462120

61416242)2(2

=−−−−

+

−−−−

+−−−−

=f


3.4 Spline Interpolation

• The polynomial interpolation can cause oscillation to the function and this can be remedied by using the spline interpolation.

x

f(x)

x

f(x)

(a) polynomial (b) Linear spline FIGURE 3.5 Comparison between polynomial and spline interpolation

• The formula for the linear spline interpolation for n+1 data between points x0 and xn (n intervals) is

( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( ) nnnnnn xxx

xxxxxx

xxmxfxf

xxmxfxfxxmxfxf

≤≤

≤≤≤≤

−+=

−+=−+=

−−−− 1

21

10

1111

11121

00011

untuk

untuk untuk

MM (3.12)

dengan mi is the gradient for the (i+1)-th interval:

( ) ( )

ii

iii xx

xfxfm−−

=+

+

1

1 (3.13)

Example 3.6

Use the following data to estimate the function f(x) at x = 5 using the linear spline interpolation.

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


Solution

The value of x = 5 is in the range 75.4 ≤≤ x , where the gradient is

60.05.40.70.15.2

1 =−−

=m

From Eq. (3.12): ( ) ( ) ( )( ) ( ) .3.15.4560.00.15

,

21

11121

=−+=

−+=

fxxmxfxf

0

2

2 4 6 8 x

f(x)

FIGURE 3.6 Linear spline interpolation for Example 3.6

• For the quadratic spline, the general polynomial for each interval is

( ) iiii cxbxaxf ++= 22 (3.14)

The coefficient ai, bi and ci for each interval can be evaluated using:

1. Continuity at vertices (2n−2 equations):

( )11112 11 −−−−−− =++ iiiiii xfcxbxa (3.15) ( )112 1 −−− =++ iiiiii xfcxbxa (3.16)

2. Conditions at end points (2 equations):

( )0101201 xfcxbxa =++ (3.17) ( )nnnnnn xfcxbxa =++2 (3.18)


3. Differentiability at vertices (n−1 equations):

( ) baxxf +=′ 2 iiiiii bxabxa +=+ −−−− 1111 22 (3.19)

4. Assumption at the first interval (1 persamaan):

01 =a (3.20)

x

f(x)

0

2

2 4 6 8

FIGURE 3.7 The quadratic spline interpolasi for example 3.7

• For the cubic spline, the general polynomial for each interval is

(3.21) ( ) iiiii dxcxbxaxf +++= 233

• The formulation for the cubic spline can be derived as followed:

For interval (xi−1, xi), the second derivative can be written as:

( ) ( ) ( )1

1

11

−

−

−− −

−′′+−

−′′=′′ii

iii

ii

iiii xx

xxxfxx

xxxfxf (3.22)

The integration of Eq. (3.22) twice produces two integration constants, which can be evaluated using ( ) ( )1−= ixfxf at xi−1 and at x( ) ( )ixfxf = i, thus produces

( ) ( )( ) ( )( )

( ) ( )

( )( )

( )( ) ( )

( )( )

( )( ) ( )111

11

1

1

31

1

3

1

13

6

6

66

−−

−

−−

−

−

−−−

−

−⎥⎦

⎤⎢⎣

⎡ −′′−

−+

−⎥⎦

⎤⎢⎣

⎡ −′′−

−+

−−′′

+−−

′′=

iiii

ii

i

iiii

ii

i

iii

ii

ii

ii

xxxxxfxxxf

xxxxxfxx

xf

xxxx

xfxxxx

xfxf

(3.23)


The differentiability can be maintained troughout the function via

( ) ( )iiii xfxf ′=′−1 (3.24)

Eq. (3.23) can be differentiated for both the (i−1)-th and i-th intervals and following Eq. (3.24):

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )[ ] ( ) ( ) ( )[ iiiiiiii

iiiiiiiii

xfxfxx

xfxfxx

xfxxxfxxxfxx

−−

+−−

=

′′

]

−+′′−+′′−

−−

++

++−+−−

11

11

111111

662

(3.25)

Example 3.8

Repeat Example 3.6 using the cubic spline interpolation. Solution

Using Eq. (3.25) for the first interval (i = 1):

( ) ( ) ( ) ( ) ( ) (

( )

)

( ) ( ) ( )15.235.4615.2

5.476

75.475.4372335.4

−−

+−−

=

′′−+′′−+′′− fff

It is known that . Hence ( ) 03 =′′f( ) ( ) 6.975.25.48 =′′+′′ ff

For the second interval (i = 2): ( ) ( ) 6.9795.45.2 −=′′+′′ ff

Both equations is solved simultaneously to yield ( )

( ) 53308.1767909.15.4

=′′=′′

ff

Using Eq. (3.23) for the first interval:

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ).3246894.05.4666667.13186566.0

,36

35.467909.135.4

1

5.435.4

5.2335.46

67909.1

3

313

−+−+−=

−⎥⎦⎤

⎢⎣⎡ −−

−+

−−

+−−

=

xxx

x

xxxf


For the second and third intervals:

( ) ( ) ( ) (( )

( ) ( ) ( ) ( )725.09761027.19127757.05.4638783.1

7299621.05.4102205.07111939.0

333

3323

−+−+−−=

−+

−−−−−=

xxxxf

xxxxxf )

At x = 5, use the function for the second interval:

( ) ( ) ( ) (( )

.102886.1,5.45638783.1

57299621.05.45102205.057111939.05 3323

=−+

−−−−−=f )

0

2

2 4 6 8 x

f(x)

Cubic Spline

Cubic Newton/ Lagrange Interpolation

FIGURE 3.8 The cubic spline interpolation for Example 3.8


3.5 Polynomial Regression

• A “best fit” polynomial of an order lower than the number of intervals is sometimes required to represent the data and can be evaluated via polynomial regression.

• Consider the following form of polynomial:

(3.26) ( ) nn xaxaxaaxf ++++= L2210where the error for the i-th data (xi,yi) is

( ) niniiiiii xaxaxaayxfye −−−−−=−= L2210

The sum of the squares of error for N data is

(3.27) ( )∑∑==

−−−−−==N

i

niniii

N

ii xaxaxaayeS

1

22210

1

2 L

These errors can be minimised as followed:

( )( )

( )( )

( )∑

∑

∑

=

=

=

−−−−−−==∂∂

−−−−−−==∂∂

−−−−−−==∂∂

N

i

ni

niniii

n

N

ii

niniii

N

i

niniii

xxaxaxaayaS

xxaxaxaayaS

xaxaxaayaS

1

2210

1

2210

1

1

2210

0

20

20

120

L

M

L

L

( )

producing the following system of equations:

(3.28)

ini

nin

ni

ni

ni

iininiii

iininiii

ininii

yxxaxaxaxa

yxxaxaxaxa

yxxaxaxaxa

yxaxaxaNa

∑=∑++∑+∑+∑

∑=∑++∑+∑+∑

∑=∑++∑+∑+∑

∑=∑++∑+∑+

++

+

+

222

110

2242

31

20

132

210

2210

L

M

L

L

L


or, in a form of matrix equation:

(3.29)

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

∑

∑∑∑

=

⎪⎪⎪

⎭

⎪⎪⎪

⎬

⎫

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

⋅

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

∑∑∑∑

∑∑∑∑∑∑∑∑∑∑∑

++

+

+

ini

ii

ii

i

nn

ini

ni

ni

niiii

niiii

niii

yx

yxyx

y

a

aaa

xxxx

xxxxxxxxxxxN

MM

L

MOMMM

L

L

L

22

1

0

221

2432

132

2

• The standard deviation σ for this case can be evaluated as followed:

111

2

+−=

+−=

∑=

nN

e

nNS

N

ii

σ (3.30)

• The case for n = 1 is referred to as the linear regression:

xaay 10 +=

(3.31) ⎭⎬⎫

⎩⎨⎧

∑∑

=⎭⎬⎫

⎩⎨⎧

⋅⎥⎦

⎤⎢⎣

⎡∑∑∑

ii

i

ii

i

yxy

aa

xxxN

1

02

Example 3.9

Determine a best-fit quadratic equation which can represent the following experimental data:

xi 0.05 0.11 0.15 0.31 0.46 0.52 yi 0.956 0.890 0.832 0.717 0.571 0.539

xi 0.70 0.74 0.82 0.98 1.17 yi 0.378 0.370 0.306 0.242 0.104

Solution

For n = 2, the quadratic polynomial can be written as

( ) 2210 xaxaaxf ++=


From Eq. (3.28):

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

∑∑∑

=⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧⋅

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∑∑∑∑∑∑∑∑

ii

ii

i

iii

iii

ii

yxyx

y

aaa

xxxxxxxxN

22

1

0

432

32

2

From the following table:

3357.19161.31839.21150.4905.56545.4

1101.6

24

3

2

=∑=∑=∑=∑=∑=∑==∑

iii

iii

ii

i

yxxyxxyxNx

Hence the matrix equation becomes:

225.0018.1998.03357.11839.2905.5

9161.31150.46545.41150.46545.401.66545.401.611

210

2

1

0

=−==

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧⋅

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

aaaaaa

Hence, the quadratic function which can represent the data is

( ) 2225.0018.1998.0 xxxf +−=

0

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1 1.2

y

x

f(x)

FIGURE 3.9 The fitting of data using a quadratic function


3.6 Multivariable Interpolation

• The Newton, Lagrange and spline interpolation method can be applied for multivariable cases, e.g. for two and three variables:

(3.32) ( ) ( ) ( ) (∑∑= =

⋅=p

i

q

jjiyxpq yxfyLxLyxf ji

0 0,, )

) (3.33) ( ) ( ) ( ) ( ) (∑∑∑= = =

⋅⋅=p

i

q

j

r

kkjizyxpqr zyxfzLyLxLzyxf kji

0 0 0

,,,,

where

( ) ( ) ( ) ∏∏∏≠=

≠=

≠= −

−=

−−

=−−

=r

kll lk

lz

q

jll lj

ly

p

ill li

lx zz

zzzLyyyyyL

xxxxxL

kji000

• For example, Eq. (3.32) can be expanded as

( ) ( )( )( )( ) ( )( )( )

( )( ) ( )

( )( )( )( ) ( )

( )( )( )( ) ( )110101

0010

0110

01

011001

1000

1010

111,1

,,

,,,

yxfyyxx

yyxxyxfxyxx

yyxx

yxfyyxx

yyxxyxfyyxx

yyxxyxf

−−−−

+−−

−−+

−−−−

+−−

−−=

Example 3.10

The following table show the topology of a 3-D non-planar surface:

y z = f(x,y)

0.2 0.3 0.4 0.5

1.0 0.640 1.003 1.359 1.703

1.5 0.990 1.524 2.045 2.549 x

2.0 1.568 2.384 3.177 3.943

Use the Newton interpolation to estimate the value of z at the coordinat (1.6, 0.33). As a comparison, the above data follows the function

( ) 1.0sin, −+= yyeyxf x , thus f(1.6, 0.33) = 1.8350.


Solution

At x = 1.0:

j yj f(1.0, yj) I II III0 0.2 0.640 0.363 −0.007 0.005

1 0.3 1.003 0.356 −0.012

2 0.4 1.359 0.344

3 0.5 1.703

( ) ( ) ( )( )( )( )( )

( ) .1108.133.0,0.1,4.03.02.0005.0

3.02.0007.02.0363.0640.0,0.1

1

1

=−−−+

−−−−+=

fyyy

yyyyf

At x = 1.5:

j yj f(1.5, yj) I II III0 0.2 0.990 0.534 −0.013 −0.004

1 0.3 1.524 0.521 −0.017

2 0.4 2.045 0.504

3 0.5 2.549

( ) ( ) ( )( )( )( )( )

( ) .6818.133.0,5.1,4.03.02.0004.0

3.02.0013.02.0534.0990.0,5.1

1

1

=−−−−

−−−−+=

fyyy

yyyyf

At x = 2.0:

j yj f(2.0, yj) I II III0 0.2 1.568 0.816 −0.023 −0.004

1 0.3 2.384 0.793 −0.027

2 0.4 3.177 0.766

3 0.5 3.943

( ) ( ) ( )( )( )( )( )

( ) .6245.233.0,0.2,4.03.02.0004.0

3.02.0023.02.0816.0568.1,0.2

1

1

=−−−−

−−−−+=

fyyy

yyyyf


Next, at y = 0.33:

i xi f(xi, 0.33) I II0 1.0 1.1108 0.5710 0.3717

1 1.5 1.6818 0.9427

2 2.0 2.6245

( ) ( ) ( )( )( ) .8406.133.0,6.1

,5.10.13717.00.15710.01108.133.0,

1,1

1,1

=

−−+−+=

fxxxxf

This produces a relative error of εt = 0.305%.


Exercise

1. The fuel consumption of an engine has been recorded as shown in the following table.

Time, hour Fuel, liter

1.2 0.33201 1.7 0.54739 1.8 0.60496 2.0 0.73891

If a user runs the engine for 1.55 hours, determine the estimated fuel consumption using the Newton and Lagrange interpolation methods.

2. The following data shows the height function of a hill a a distance x from a reference. Form a cubic polynomial via regression.

xi 0 1 2 3 4 5 6 7 8

hi 4 5 10 17 21 16 11 3 1

Also, calculate the corresponding standard deviation.

3. The following data represents temperature distribution T(x,y) in a metal plate:

y coordinate Temperature T(x,y) 0.5 1.0 1.5 2.0

0.5 7.51 10.05 12.70 15.67

1.0 10.00 10.00 10.00 10.00

1.5 12.51 9.95 7.32 4.33

x co

ordi

nate

2.0 15.00 10.00 5.00 0.00

Estimate the temperature at the coordinate (x, y) = (1.15, 1.42) using:

a. the Newton linear interpolation, b. the Newton cubic interpolation.

3.1 Introduction3.2 Newton Interpolation: Finite Divided DifferenceExample 3.1SolutionExample 3.2Solution

Example 3.3Solution

3.3 Lagrange InterpolationExample 3.5Solution

3.4 Spline InterpolationExample 3.6SolutionExample 3.8Solution

3.5 Polynomial RegressionExample 3.9Solution

3.6 Multivariable InterpolationExample 3.10Solution

Exercise

Documents

NTERPOLATION AND CURVE FITTING · Chapter 3 Interpolation & Curve Fitting / 2 3.1 Introduction • Interpolation is a technique to estimate the value between a set of data. • A