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Lagrangian Interpolation

Mechanical Engineering Majors

Authors: Autar Kaw, Jai Paul

http://numericalmethods.eng.usf.eduTransforming Numerical Methods Education for STEM

Undergraduates

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.

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Interpolants

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

EvaluateDifferentiate, and Integrate.

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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.

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ExampleA trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the Lagrangian method for linear interpolation.

Temperature (oF)

Thermal Expansion Coefficient (in/in/oF)

80 6.47 × 10−6

0 6.00 × 10−6

−60 5.58 × 10−6

−160 4.72 × 10−6

−260 3.58 × 10−6

−340 2.45 × 10−6

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Linear Interpolation

5045403530252015105.5

5.6

5.7

5.8

5.9

6

5.58

y s

f range( )

f x desired( )

x s110+x s0

10− x s range, x desired,

∑=

=1

0)()()(

iii TTLT αα

)()()()( 1100 TTLTTL αα +=

( ) 600 100060 −×== .Tα,T

( ) 611 1058560 −×=−= .Tα,T

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Linear Interpolation (contd)∏≠= −

−=

1

00 0

0 )(jj j

j

TTTT

TL10

1

TTTT−−

=

∏≠= −

−=

1

10 1

1 )(jj j

j

TTTT

TL01

0

TTTT

−−

=

( ) ( ) ( )

( ) ( ) 060,10585060

01000660060 66

101

00

10

1

≤≤−×−−−

+×++

=

−−

+−−

=

−− T.T.T

TαTTTTTα

TTTTTα

( ) ( ) ( )( ) ( )

Fin/in/ 1090251058523333010006766670

1058506001410006

600601414

6

66

66

°×=

×+×=

×−−−−

+×++−

=−

−

−−

−−

.....

..α

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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 ++=

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ExampleA trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the Lagrangian method for quadratic interpolation.

Temperature (oF)

Thermal Expansion Coefficient (in/in/oF)

80 6.47 × 10−6

0 6.00 × 10−6

−60 5.58 × 10−6

−160 4.72 × 10−6

−260 3.58 × 10−6

−340 2.45 × 10−6

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Quadratic Interpolation (contd)

60 40 20 0 20 40 60 805.4

5.6

5.8

6

6.2

6.4

6.66.47

5.58

y s

f range( )

f x desired( )

8060− x s range, x desired,

( ) 61047680 −×== .Tα,T oo

( ) 611 100060 −×== .Tα,T

( ) 622 1058560 −×=−= .Tα,T

∏≠= −

−=

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

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Quadratic Interpolation (contd))()()()( 2

12

1

02

01

21

2

01

00

20

2

10

1 TTTTT

TTTT

TTTTT

TTTT

TTTTT

TTTTT αααα

−−

−−

+

−−

−−

+

−−

−−

=

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

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

Fin/in/109072.510585156670100069008301047605750

1058506080600148014

10006600800

60148014104766080080

601401414

6

666

6

66

°×=

×+×+×−=

×−−−−−−−−

+

×+−

+−−−+×

+−+−−−

=−

−

−−−

−

−−

......

.

..α

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

1001090725

10902510907256

66

××

×−×=∈ −

−−

...

a

%087605.0=

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Cubic Interpolation

200 150 100 50 0 50 1004.5

5

5.5

6

6.56.47

4.72

y s

f range( )

f x desired( )

80160− x s range, x desired,

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

the coefficient of thermal expansion given by

∑=

=3

0)()()(

iii TTLT αα

)()()()()()()()( 33221100 TTLTTLTTLTTL αααα +++=

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ExampleA trunnion is cooled 80°F to − 108°F. Given below is the table of the coefficient of thermal expansion vs. temperature. Determine the value of the coefficient of thermal expansion at T=−14°F using the Lagrangian method for cubic interpolation.

Temperature (oF)

Thermal Expansion Coefficient (in/in/oF)

80 6.47 × 10−6

0 6.00 × 10−6

−60 5.58 × 10−6

−160 4.72 × 10−6

−260 3.58 × 10−6

−340 2.45 × 10−6

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Cubic Interpolation (contd)

200 150 100 50 0 50 1004.5

5

5.5

6

6.56.47

4.72

y s

f range( )

f x desired( )

80160− x s range, x desired,

( ) 61047680 −×== .Tα,T oo ( ) 611 100060 −×== .Tα,T

( ) 622 1058560 −×=−= .Tα,T ( ) 6

33 10724160 −×=−= .Tα,T

∏≠= −

−=

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

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Cubic Interpolation (contd)( ) ( ) ( )

( ) ( )123

2

13

1

03

00

32

3

12

1

02

0

131

3

21

2

01

00

30

3

20

2

10

1

TαTTTT

TTTT

TTTTTα

TTTT

TTTT

TTTT

TαTTTT

TTTT

TTTTTα

TTTT

TTTT

TTTTTα

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

+

−−

−−

−−

=

, 30 TTT ≤≤

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

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

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

Fin/in/ 109077.51072401576501058522873010006822010104760349790

1072460160016080160

6014014801410585160600608060160140148014

100061600600800

160146014801410476160806080080

16014601401414

6

6666

66

66

°×=

×−+×+×+×−=

×+−−−−−

+−−−−−+×

+−−−−−+−−−−−

+

×++−

+−+−−−+×

++−+−+−−−

=−

−

−−−−

−−

−−

........

. .

..α

The absolute relative approximate error a∈ obtained between the results from the second and third order polynomial is

%0083867.0

1001090775

109072510907756

66

=

××

×−×=∈ −

−−

...

a

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Comparison Table

Order of Polynomial 1 2 3 Thermal Expansion

Coefficient (in/in/oF) 5.902 × 10−6 5.9072 × 10−6 5.9077 × 10−6

Absolute Relative Approximate Error ---------- 0.087605% 0.0083867%

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Reduction in DiameterThe actual reduction in diameter is given by

where Tr = room temperature (°F)Tf = temperature of cooling medium (°F)

Since Tr = 80 °F and Tr = −108 °F,

Find out the percentage difference in the reduction in the diameter by the above integral formula and the result using the thermal expansion coefficient from the cubic interpolation.

∫−

=∆108

80

dTDD α

∫=∆Tf

Tr

dTDD α

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Reduction in DiameterWe know from interpolation that

( )80160

,101845.8101944.8104786.61000.6 31521296

≤≤−×+×−×+×= −−−−

TTTTTα

Therefore,

( )

6

180

80

415

312

296

108

80

31521296

109.1105

4101845.8

3101944.8

2104786.61000.6

101845.8101944.8104786.61000.6

−

−

−−−−

−−−−−

×−=

×+×−×+×=

×+×−×+×=

=∆

∫

∫

TTTT

dTTTT

dTDD f

r

T

T

α

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Reduction in diameterUsing the average value for the coefficient of thermal expansion from cubic interpolation

( )( )6

6

106.111080108109077.5

−

−

×−=

−−×=

−=

∆=∆

rf TT

TDD

α

α

The percentage difference would be

( )

%42775.0

100109.1105

106.1110109.11056

66

=

××−

×−−×−=∈ −

−−

a

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

http://numericalmethods.eng.usf.edu/topics/lagrange_method.html

THE END

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