lagrangian method power point interpolation · the upward velocity of a rocket is given as a...

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Lagrangian Interpolation

Major: All Engineering Majors

Authors: Autar Kaw, Jai Paul

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Lagrange Method of Interpolation

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


Interpolants

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

EvaluateDifferentiate, and Integrate.


Lagrangian InterpolationLagrangian interpolating polynomial is given by

∑=

=n

iiin xfxLxf

0

)()()(

where ‘ n ’ in )(xfn stands for the thn order polynomial that approximates the function )(xfy =

given at )1( +n data points as ( ) ( ) ( ) ( )nnnn yxyxyxyx ,,,,......,,,, 111100 −− , and

∏≠= −

−=

n

ijj ji

ji xx

xxxL

0

)(

)(xLi is a weighting function that includes a product of )1( −n terms with terms of ij = omitted.


ExampleThe upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for linear interpolation.

Table Velocity as a function of time

Figure. Velocity vs. time data for the rocket example

(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv


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 s110+x s0

10− x s range, x desired,

)()()(1

0ii

itvtLtv

=∑=

)()()()( 1100 tvtLtvtL +=

( ) 78.362,15 00 == tt ν

( ) 35.517,20 11 == tt ν


Linear Interpolation (contd)∏≠= −

−=

1

00 0

0 )(jj j

j

tttt

tL 10

1

tttt−−

=

∏≠= −

−=

1

10 1

1 )(jj j

j

tttt

tL 01

0

tttt

−−

=

)()()( 101

00

10

1 tvtttt

tvtttttv

−−

+−−

= )35.517(1520

15)78.362(2015

20−−

+−−

=tt

)35.517(15201516)78.362(

20152016)16(

−−

+−−

=v

)35.517(2.0)78.362(8.0 +=

7.393= m/s.


Quadratic InterpolationFor the second order polynomial interpolation (also called quadratic interpolation), we

choose the veloc ity given by

∑=

=2

0

)()()(i

ii tvtLtv

)()()()()()( 221100 tvtLtvtLtvtL ++=


ExampleThe upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for quadratic interpolation.



(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv


Quadratic Interpolation (contd)

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,

,100 =t 04.227)( 0 =tv

,151 =t 78.362)( 1 =tv

,202 =t 35.517)( 2 =tv

∏≠= −

−=

2

00 0

0 )(jj j

j

tttt

tL

−−

−−

=20

2

10

1

tttt

tttt

∏≠= −

−=

2

10 1

1 )(jj j

j

tttt

tL

−−

−−

=21

2

01

0

tttt

tttt

∏≠= −

−=

2

20 2

2 )(jj j

j

tttt

tL

−−

−−

=12

1

02

0

tttt

tttt


Quadratic Interpolation (contd)( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( )( ) ( )( ) ( )( )m/s19.392

35.52712.078.36296.004.22708.0

35.51715201516

1020101678.362

20152016

1015101604.227

20102016

1510151616

212

1

02

01

21

2

01

00

20

2

10

1

=++−=

−−

−−

+

−−

−−

+

−−

−−

=

−−

−−

+

−−

−−

+

−−

−−

=

v

tvtttt

tttttv

tttt

tttttv

tttt

tttttv

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

a∈

%38410.0

10019.392

70.39319.392

=

×−

=∈a


Cubic Interpolation

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,

For the third order polynomial (also called cubic interpolation), we choose the velocity given by

∑=

=3

0

)()()(i

ii tvtLtv

)()()()()()()()( 33221100 tvtLtvtLtvtLtvtL +++=


ExampleThe upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds using the Lagrangian method for cubic interpolation.



(s) (m/s)

0 010 227.0415 362.7820 517.35

22.5 602.9730 901.67

t )(tv


Cubic Interpolation (contd)( ) 04.227,10 == oo tvt ( ) 78.362,15 11 == tvt

( ) 35.517,20 22 == tvt ( ) 97.602,5.22 33 == tvt

∏≠= −

−=

3

00 0

0 )(jj j

j

tttt

tL

−−

−−

−−

=30

3

20

2

10

1

tttt

tttt

tttt

;

∏≠= −

−=

3

10 1

1 )(jj j

j

tttt

tL

−−

−−

−−

=31

3

21

2

01

0

tttt

tttt

tttt

∏≠= −

−=

3

20 2

2 )(jj j

j

tttt

tL

−−

−−

−−

=32

3

12

1

02

0

tttt

tttt

tttt

;

∏≠= −

−=

3

30 3

3 )(jj j

j

tttt

tL

−−

−−

−−

=23

2

13

1

03

0

tttt

tttt

tttt

10 12 14 16 18 20 22 24

200

300

400

500

600

700602.97

227.04

y s

f range( )

f x desired( )

22.510 x s range, x desired,


Cubic Interpolation (contd)( ) ( ) ( )

( ) ( )

( ) ( ) ( )

( ) ( )

( )( ) ( )( ) ( )( ) ( )( )m/s06.392

97.6021024.035.517312.078.362832.004.2270416.0

97.602205.22

2016155.22

1516105.22

101635.5175.22205.2216

15201516

10201016

78.3625.22155.2216

20152016

1015101604.227

5.22105.2216

20102016

1510151616

323

2

13

1

13

12

32

3

12

1

02

0

231

3

21

2

01

01

30

3

20

2

10

1

=−+++−=

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

=

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

=

v

tvtttt

tttt

tttttv

tttt

tttt

tttt

tvtttt

tttt

tttttv

tttt

tttt

tttttv

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

a∈

%033269.0

10006.392

19.39206.392

=

×−

=∈a


Comparison Table

Order of Polynomial 1 2 3

v(t=16) m/s 393.69 392.19 392.06

Absolute Relative Approximate Error -------- 0.38410% 0.033269%


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

,00544.013195.0265.21245.4)( 32 ttttv +++−= 5.2210 ≤≤ t

∫=−16

11

)()11()16( dttvss

∫ +++−≈16

11

32 )00544.013195.0265.21245.4( dtttt

1611

432

]4

00544.03

13195.02

265.21245.4[ tttt +++−=

1605= m

)5727.2)(300065045()1388.4)(33755.7125.47()9348.1)(45008755.52()36326.0)(67505.10875.57()(

2323

2323

−+−+−−+−+

−+−+−−+−=

tttttt tttttttv


Acceleration from Velocity Profile

,

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

( ) ( ) ( )32 00544.013195.0265.21245.4 tttdtdtv

dtdta +++−==

201632.026390.0265.21 tt ++=

2)16(01632.0)16(26390.0265.21)16( ++=a

665.29= 2/ sm

,00544.013195.0265.21245.4)( 32 ttttv +++−= 5.2210 ≤≤ t

Additional ResourcesFor all resources on this topic such as digital audiovisual lectures, primers, textbook chapters, multiple-choice tests, worksheets in MATLAB, MATHEMATICA, MathCad and MAPLE, blogs, related physical problems, please visit

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