1 Interpolation. 2 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1 ), …… (x n,y n ), find the value of ‘y’ at a

1

Interpolation

http://numericalmethods.eng.usf.edu

2

What is Interpolation ?Given (x0,y0), (x1,y1), …… (xn,yn), find the value of ‘y’ at a value of ‘x’ that is not given.

Figure 1 Interpolation of discrete.


3

Interpolants

Polynomials are the most common choice of interpolants because they are easy to:

Evaluate Differentiate, and Integrate

1. Direct method

4


5

Direct MethodGiven ‘n+1’ data points (x0,y0), (x1,y1),…………..

(xn,yn),pass a polynomial of order ‘n’ through the data as

given below:

where a0, a1,………………. an are real constants. Set up ‘n+1’ equations to find ‘n+1’ constants. To find the value ‘y’ at a given value of ‘x’,

simply substitute the value of ‘x’ in the above polynomial.

.....................10n

nxaxaay


6

Example 1 The upward velocity of a rocket is given as

a function of time in Table 1.

Find the velocity at t=16 seconds using the direct method for linear interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1 Velocity as a function of time.

Figure 2 Velocity vs. time data for the rocket example

s ,t m/s ,tv


7

Linear Interpolation taatv 10

78.3621515 10 aav

35.5172020 10 aav

Solving the above two equations gives,

93.1000 a 914.301 a

Hence .2015,914.3093.100 tttv

m/s 7.39316914.3093.10016 v

00 , yx

xf1

11, yx

x

y

Figure 3 Linear interpolation.


8



Find the velocity at t=16 seconds using the direct method for quadratic interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



s ,t m/s ,tv


9

Quadratic Interpolation

2210 tataatv

04.227101010 2210 aaav

78.362151515 2210 aaav

35.517202020 2210 aaav

Solving the above three equations gives

05.120 a 733.171 a 3766.02 a

Quadratic Interpolation

00 , yx

11, yx

22 , yx

xf2

y

x

Figure 6 Quadratic interpolation.


10

Quadratic Interpolation (cont.)

10 12 14 16 18 20200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired

2010 x s range x desired

2010,3766.0733.1705.12 2 ttttv

2163766.016733.1705.1216 v

m/s 19.392

The absolute relative approximate error obtained between the results from the first and second order polynomial is

%38410.0

10019.392

70.39319.392

a

a


11



Find the velocity at t=16 seconds using the direct method for cubic interpolation.

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



s ,t m/s ,tv


12

Cubic Interpolation

33

2210 tatataatv

33

2210 10101004.22710 aaaav

33

2210 15151578.36215 aaaav

33

2210 20202035.51720 aaaav

33

2210 5.225.225.2297.6025.22 aaaav

2540.40 a 266.211 a 13204.02 a 0054347.03 a

y

x

xf3

33 , yx

22 , yx

11, yx

00 , yx

Figure 7 Cubic interpolation.


13

Cubic Interpolation (contd)

10 12 14 16 18 20 22 24200

300

400

500

600

700602.97

227.04

y s

f range( )

f x desired

22.510 x s range x desired

5.2210,0054347.013204.0266.212540.4 32 tttttv

m/s 06.392

160054347.01613204.016266.212540.416 32

v

The absolute percentage relative approximate error between second and third order polynomial is

a

%033269.0

10006.392

19.39206.392

a


14

Comparison Table

Order of Polynomial

1 2 3

m/s 16tv 393.7 392.19 392.06

Absolute Relative Approximate Error

---------- 0.38410 % 0.033269 %

Table 4 Comparison of different orders of the polynomial.


15

Distance from Velocity ProfileFind the distance covered by the rocket from t=11s to t=16s ?

5.2210,0054606.013064.0289.213810.4 32 tttttv

m 1605

40054347.0

313204.0

2266.212540.4

0054347.013204.0266.212540.4

1116

16

11

432

16

11

32

16

11

tttt

dtttt

dttvss


16

Acceleration from Velocity Profile

5.2210,0054347.013204.0266.212540.4 32 ttttFind the acceleration of the rocket at t=16s given that

5.2210 ,016382.026130.0289.21

0054347.013204.0266.212540.4

2

32

ttt

tttdt

d

tvdt

dta

2

2

m/s 665.29

16016304.01626408.0266.2116

a

2. Spline Method

17


18

Why Splines ?2251

1)(

xxf

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


19

Why Splines ?

Figure : Higher order polynomial interpolation is a bad idea

-0.8

-0.4

0

0.4

0.8

1.2

-1 -0.5 0 0.5 1

x

y

19th Order Polynomial f (x) 5th Order Polynomial


20

Linear InterpolationGiven nnnn yxyxyxyx ,,,......,,,, 111100 , fit linear splines to the data. This simply involves

forming the consecutive data through straight lines. So if the above data is given in an ascending

order, the linear splines are given by )( ii xfy

Figure : Linear splines


21

Linear Interpolation (contd)

),()()(

)()( 001

010 xx

xx

xfxfxfxf

10 xxx

),()()(

)( 112

121 xx

xx

xfxfxf

21 xxx

.

.

.

),()()(

)( 11

11

nnn

nnn xx

xx

xfxfxf nn xxx 1

Note the terms of

1

1)()(

ii

ii

xx

xfxf

in the above function are simply slopes between 1ix and ix .


22

Example The upward velocity of a rocket is given as

a function of time in Table 1. Find the velocity at t=16 seconds using linear splines.Table Velocity as a

function of time

Figure. Velocity vs. time data for the rocket example

(s) (m/s)0 0

10 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv


23

Linear Interpolation

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


,150 t 78.362)( 0 tv

,201 t 35.517)( 1 tv

)()()(

)()( 001

010 tt

tt

tvtvtvtv

)15(1520

78.36235.51778.362

t

)15(913.3078.362)( ttv

At ,16t

)1516(913.3078.362)16( v

7.393 m/s


24

Quadratic InterpolationGiven nnnn yxyxyxyx ,,,,......,,,, 111100 , fit quadratic splines through the data. The splines

are given by

,)( 112

1 cxbxaxf 10 xxx

,222

2 cxbxa 21 xxx

.

.

.

,2nnn cxbxa nn xxx 1

Find ,ia ,ib ,ic i 1, 2, …, n


25

Quadratic Interpolation (contd)

Each quadratic spline goes through two consecutive data points

)( 0101

2

01 xfcxbxa

)( 11112

11 xfcxbxa .

.

.

)( 11

2

1 iiiiii xfcxbxa

)(2

iiiiii xfcxbxa .

.

.

)( 11

2

1 nnnnnn xfcxbxa

)(2

nnnnnn xfcxbxa

This condition gives 2n equations


26

Quadratic Splines (contd)The first derivatives of two quadratic splines are continuous at the interior points.

For example, the derivative of the first spline

112

1 cxbxa is 112 bxa

The derivative of the second spline

222

2 cxbxa is 222 bxa

and the two are equal at 1xx giving

212111 22 bxabxa

022 212111 bxabxa


27

Quadratic Splines (contd)Similarly at the other interior points,

022 323222 bxabxa

.

.

.

022 11 iiiiii bxabxa

.

.

.

022 1111 nnnnnn bxabxa

We have (n-1) such equations. The total number of equations is )13()1()2( nnn .

We can assume that the first spline is linear, that is 01 a


28

Quadratic Splines (contd)This gives us ‘3n’ equations and ‘3n’ unknowns. Once we find the ‘3n’ constants,

we can find the function at any value of ‘x’ using the splines,

,)( 112

1 cxbxaxf 10 xxx

,222

2 cxbxa 21 xxx

.

.

.

,2nnn cxbxa nn xxx 1


29

Quadratic Spline ExampleThe upward velocity of a rocket is given as a function of time. Using quadratic splinesa) Find the velocity at t=16 secondsb) Find the acceleration at t=16 secondsc) Find the distance covered between t=11 and t=16 seconds

Table Velocity as a function of time

Figure. Velocity vs. time data for the rocket example

(s) (m/s)0 0

10 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv


30

Solution,)( 11

21 ctbtatv 100 t

,222

2 ctbta 1510 t

,332

3 ctbta 2015 t,44

24 ctbta 5.2220 t

,552

5 ctbta 305.22 t

Let us set up the equations


31

Each Spline Goes Through Two Consecutive Data

Points,)( 11

21 ctbtatv 100 t

0)0()0( 112

1 cba

04.227)10()10( 112

1 cba


32

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Each Spline Goes Through Two Consecutive Data

Points04.227)10()10( 22

22 cba

78.362)15()15( 222

2 cba

78.362)15()15( 332

3 cba

35.517)20()20( 332

3 cba

67.901)30()30( 552

5 cba

35.517)20()20( 442

4 cba

97.602)5.22()5.22( 442

4 cba

97.602)5.22()5.22( 552

5 cba


33

Derivatives are Continuous at Interior Data Points

,)( 112

1 ctbtatv 100 t

,222

2 ctbta 1510 t

10

222

210

112

1

tt

ctbtadt

dctbta

dt

d

10221011 22

ttbtabta

2211 102102 baba

02020 2211 baba


34

Derivatives are continuous at Interior Data Points

0)10(2)10(2 2211 baba

0)15(2)15(2 3322 baba

0)20(2)20(2 4433 baba

0)5.22(2)5.22(2 5544 baba

At t=10

At t=15

At t=20

At t=22.5


35

Last Equation

01 a


36

Final Set of Equations

0

0

0

0

0

67.901

97.602

97.602

35.517

35.517

78.362

78.362

04.227

04.227

0

000000000000001

01450145000000000

00001400140000000

00000001300130000

00000000001200120

130900000000000000

15.2225.506000000000000

00015.2225.506000000000

000120400000000000

000000120400000000

000000115225000000

000000000115225000

000000000110100000

000000000000110100

000000000000100

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

c

b

a

c

b

a

c

b

a

c

b

a

c

b

a


37

Coefficients of Splinei ai bi ci1 0 22.704 0

2 0.8888 4.928 88.88

3 −0.1356 35.66 −141.61

4 1.6048 −33.956

554.55

5 0.20889 28.86 −152.13


38

Final Solution,704.22)( ttv 100 t

,88.88928.48888.0 2 tt 1510 t,61.14166.351356.0 2 tt 2015 t,55.554956.336048.1 2 tt 5.2220 t,13.15286.2820889.0 2 tt 305.22 t


39

Velocity at a Particular Pointa) Velocity at t=16

,704.22)( ttv 100 t,88.88928.48888.0 2 tt 1510 t

,61.14166.351356.0 2 tt 2015 t,55.554956.336048.1 2 tt 5.2220 t,13.15286.2820889.0 2 tt 305.22 t

m/s24.394

61.1411666.35161356.016 2

v


40


16)()16(

ttv

dt

da

b) The quadratic spline valid at t=16 is given by

)61.14166.351356.0()( 2 ttdt

dta

,66.352712.0 t 2015 t

66.35)16(2712.0)16( a 2m/s321.31

,61.14166.351356.0 2 tttv 2015 t


41

Distance from Velocity Profile

c) Find the distance covered by the rocket from t=11s to t=16s.

16

11

)(1116 dttvSS

2015,61.14166.351356.0

1510,88.88928.48888.02

2

ttt

ttttv

m9.1595

61.14166.351356.088.88928.48888.0

1116

16

15

215

11

2

16

11

15

11

16

15

dtttdttt

dttvdttvdttvSS

3. Newton’s Divided Differences

42

43

Newton’s Divided Difference Method

Linear interpolation: Given pass a linear interpolant through the data

where

),,( 00 yx ),,( 11 yx

)()( 0101 xxbbxf

)( 00 xfb

01

011

)()(

xx

xfxfb

44

Example The upward velocity of a rocket is given as a

function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for linear interpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1: Velocity as a function of time

Figure 2: Velocity vs. time data for the rocket example

45

Linear Interpolation

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


,150 t 78.362)( 0 tv

,201 t 35.517)( 1 tv

)( 00 tvb 78.362

01

011

)()(

tt

tvtvb

914.30

)()( 010 ttbbtv

46

Linear Interpolation (contd)

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


)()( 010 ttbbtv

),15(914.3078.362 t 2015 t

At 16t

)1516(914.3078.362)16( v

69.393 m/s

47

Quadratic InterpolationGiven ),,( 00 yx ),,( 11 yx and ),,( 22 yx fit a quadratic interpolant through the data.

))(()()( 1020102 xxxxbxxbbxf

)( 00 xfb

01

011

)()(

xx

xfxfb

02

01

01

12

12

2

)()()()(

xx

xx

xfxf

xx

xfxf

b

48


function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for quadratic interpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



49


0104.203.0 623 xxxxf10 12 14 16 18 20

200

250

300

350

400

450

500

550517.35

227.04

y s

f range( )

f x desired


,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv

,202 t 35.517)( 2 tv

50


)( 00 tvb

04.227

01

011

)()(

tt

tvtvb

1015

04.22778.362

148.27

02

01

01

12

12

2

)()()()(

tt

tt

tvtv

tt

tvtv

b

1020

1015

04.22778.362

1520

78.36235.517

10

148.27914.30

37660.0

51


))(()()( 102010 ttttbttbbtv

),15)(10(37660.0)10(148.2704.227 ttt 2010 t

At ,16t

)1516)(1016(37660.0)1016(148.2704.227)16( v 19.392 m/s

The absolute relative approximate error a obtained between the results from the first

order and second order polynomial is

a 100x19.392

69.39319.392

= 0.38502 %

52

General Form

))(()()( 1020102 xxxxbxxbbxf

where

Rewriting))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf

)(][ 000 xfxfb

01

01011

)()(],[

xx

xfxfxxfb

02

01

01

12

12

02

01120122

)()()()(

],[],[],,[

xx

xx

xfxf

xx

xfxf

xx

xxfxxfxxxfb

53

General FormGiven )1( n data points, nnnn yxyxyxyx ,,,,......,,,, 111100 as

))...()((....)()( 110010 nnn xxxxxxbxxbbxf

where

][ 00 xfb

],[ 011 xxfb

],,[ 0122 xxxfb

],....,,[ 0211 xxxfb nnn

],....,,[ 01 xxxfb nnn

54

General formThe third order polynomial, given ),,( 00 yx ),,( 11 yx ),,( 22 yx and ),,( 33 yx is

))()(](,,,[

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

2100123

1001200103

xxxxxxxxxxf

xxxxxxxfxxxxfxfxf

0b

0x )( 0xf 1b

],[ 01 xxf 2b

1x )( 1xf ],,[ 012 xxxf 3b

],[ 12 xxf ],,,[ 0123 xxxxf

2x )( 2xf ],,[ 123 xxxf

],[ 23 xxf

3x )( 3xf

55


function of time in Table 1. Find the velocity at t=16 seconds using the Newton Divided Difference method for cubic interpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67



56

Example

The velocity profile is chosen as))()(())(()()( 2103102010 ttttttbttttbttbbtv

we need to choose four data points that are closest to 16t

,100 t 04.227)( 0 tv

,151 t 78.362)( 1 tv ,202 t 35.517)( 2 tv ,5.223 t 97.602)( 3 tv

The values of the constants are found as : b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347*10-3

57

Example

b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347*10-3

0b

100 t 04.227 1b

148.27 2b

,151 t 78.362 37660.0 3b

914.30 3104347.5 x

,202 t 35.517 44453.0

248.34

,5.223 t 97.602

58

ExampleHence

))()(())(()()( 2103102010 ttttttbttttbttbbtv

)20)(15)(10(10*4347.5

)15)(10(37660.0)10(148.2704.2273

ttt

ttt

At ,16t

)2016)(1516)(1016(10*4347.5

)1516)(1016(37660.0)1016(148.2704.227)16(3

v

06.392 m/s

The absolute relative approximate error a obtained is

a 100x06.392

19.39206.392

= 0.033427 %

59

Comparison Table

Order of Polynomial

1 2 3

v(t=16) m/s

393.69 392.19 392.06

Absolute Relative Approximate Error

---------- 0.38502 %

0.033427 %

60

Distance from Velocity Profile

Find the distance covered by the rocket from t=11s tot=16s ?

)20)(15)(10(10*4347.5

)15)(10(37660.0)10(148.2704.227)(3

ttt

ttttv 5.2210 t

32 0054347.013204.0265.212541.4 ttt 5.2210 t So

16

11

1116 dttvss

dtttt )0054347.013204.0265.212541.4( 3216

11

16

11

432

40054347.0

313204.0

2265.212541.4

tttt

m 1605

61


Find the acceleration of the rocket at t=16s given that

32 0054347.013204.0265.212541.4)( ttttv

32 0054347.013204.0265.212541.4)()( tttdt

dtv

dt

dta

2016304.026408.0265.21 tt

2)16(016304.0)16(26408.0265.21)16( a

2/ 664.29 sm

Documents

1 Interpolation. 2 What is Interpolation ? Given (x 0,y 0 ), (x 1,y 1 ), …… (x n,y n ), find the value of ‘y’ at a