Newton’s Divided Difference Polynomial Method of Interpolation

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Newton’s Divided Difference Polynomial

Method of Interpolation

Mechanical Engineering Majors

Authors: Autar Kaw, Jai Paul

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Undergraduates

Newton’s Divided Difference 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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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

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Example A 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 direct 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 s1

10x s0

10 x s range x desired

)()( 010 TTbbT

,00 T 60 10006 .Tα

,601 T 61 10585 .Tα

)( 00 Tb

610006 .

01

011

)()(

TT

TTb

060

1000610585 66

..

6100070 .

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Linear Interpolation (contd)

5045403530252015105.5

5.6

5.7

5.8

5.9

6

5.58

y s

f range( )

f x desired

x s1

10x s0

10 x s range x desired

010 TTbbTα

,T.. 010007010006 66

060 T

Fin/in/109025

01410007010006146

66

.

..α

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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

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Example A 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 direct 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

102010 TTTTbTTbbTα

,800 T 60 10476 .Tα

,01 T 61 10006 .Tα

,602 T 62 10585 .Tα

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Quadratic Interpolation (contd)

600 10476 .Tαb

9

66

01

011 108755

800

1047610006

...

TT

TαTαb

12

66

6666

02

01

01

12

12

2

1003578

140

100058750100070

8060800

1047610006060

1000610585

.

..

....

TT

TTTαTα

TTTαTα

b

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Quadratic Interpolation (contd)

102010 TTTTbTTbbTα

,TT.T.. 08010035788010875510476 1296 8060 T

At ,14T

Fin/in/1090725

01480141003578801410875510476146

1296

.

...α ---

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

first and second order polynomial is

%087605.0

1001090725

10902510907256

66

-

--

a .

..

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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

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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

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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

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Example A 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 direct 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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Example

The coefficient of thermal expansion profile is chosen as

2103102010 TTTTTTbTTTTbTTbbTα

,800 T 60 10476 .Tα

,01 T 61 10006 .Tα

,602 T 62 10585 .Tα

,1603 T 63 10724 .Tα

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Example 0b

,800 T 610476 . 1b

9108755 . 2b

,01 T 610006 . 121003578 . 3b

6100070 . 151018458 .

,602 T 610585 . 1110

61000860 .

,1603 T 610724 .

The values of the constants are 6

0 10476 .b 91 108755 .b 12

2 1003578 .b 153 1018458 .b

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Example

600801018458

0801003578801087551047615

1296

2103102010

TT-T-.

TT.T..

TTTTTTbTTTTbTTbbTα

At ,14T

Fin/in/1090775

601401480141018458

0148014100357880141087551047614

6

15

1296

.

.

...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

Tf

Tr

dTDD

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Reduction in DiameterWe know from interpolation that

80160

,101845.8101944.8104786.61000.6 31521296

T

TTTT

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

dTD

D f

r

T

T

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

6

6

106.1110

80108109077.5

rf TT

TD

D

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

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THE END

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