newton’s divided difference polynomial method of interpolation

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Newton’s Divided Difference Polynomial

Method of InterpolationChemical Engineering Majors

Authors: Autar Kaw, Jai Paul

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

Undergraduates

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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.


Interpolants Polynomials are the most

common choice of interpolants because they are easy to:

EvaluateDifferentiate, and Integrate.


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

To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table 1. Use Newton’s divided difference method with a first order and then a second order polynomial to determine the value of the specific heat at T = 61°C.


Example

Table 1 Specific heat of water as a function of temperature.

Figure 2 Specific heat of water vs. temperature.

Temperature, Specific heat,

22425282100

41814179418641994217

CT

Ckg

JpC


Linear Interpolation

50 60 70 80 904185

4190

4195

42004.199 103

4.186 103

y s

f range( )

f x desired

x s110x s0

10 x s range x desired

)()( 010 TTbbTC p

,520 T 4186)( 0 TC p

,821 T 4199)( 1 TC p

)( 00 TCb p 4186

01

011

)()(TT

TCTCb pp

528241864199

43333.0


Linear Interpolation (contd)

)()( 010 TTbbTC p

),52(43333.04186 T 8252 T

At 61T )5261(43333.04186)61( pC

Ckg

J

9.4189

50 60 70 80 904185

4190

4195

42004.199 103

4.186 103

y s

f range( )

f x desired

x s110x s0



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

)()()()(

xxxx

xfxfxx

xfxf

b


Quadratic Interpolation (contd)

40 45 50 55 60 65 70 75 80 854175

4180

4185

4190

4195

42004.199 103

4.179 103

y s

f range( )

f x desired


))(()()( 102010 TTTTbTTbbTC p

,420 T 4179)( 0 TC p

,521 T 4186)( 1 TC p

,822 T 4199)( 2 TC p



)( 00 TCb p 4179

01

011

)()(TT

TCTCb pp

425241794186

7.0

02

01

01

12

12

2

)()()()(

TTTT

TCTCTT

TCTC

b

pppp

4282

425241794186

528241864199

40

7.043333.0

3106667.6



))(()()( 102010 TTTTbTTbbTC p

),52)(42(106667.6)42(7.04179 3 TTT 8242 T

At ,61T )5261)(4261(106667.6)4261(7.04179)61( 3

pC

Ckg

J

2.4191

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

second order polynomial is

1002.4191

9.41892.4191

a

%030063.0


General Form))(()()( 1020102 xxxxbxxbbxf

where

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

)(][ 000 xfxfb

01

01011

)()(],[

xxxfxf

xxfb

02

01

01

12

12

02

01120122

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

],,[xx

xxxfxf

xxxfxf

xxxxfxxf

xxxfb


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


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

))()(](,,,[

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

2100123

1001200103

xxxxxxxxxxfxxxxxxxfxxxxfxfxf

0b

0x )( 0xf 1b

],[ 01 xxf 2b

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

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

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

],[ 23 xxf

3x )( 3xf

To find how much heat is required to bring a kettle of water to its boiling point, you are asked to calculate the specific heat of water at 61°C. The specific heat of water is given as a function of time in Table 1. Use Newton’s divided difference method with a third order polynomial to determine the value of the specific heat at T = 61°C.


Example

Table 1 Specific heat of water as a function of temperature.

Figure 2 Specific heat of water vs. temperature.

Temperature, Specific heat,

22425282100

41814179418641994217

CT

Ckg

JpC


Example The specific heat profile is chosen as

))()(())(()()( 2103102010 TTTTTTbTTTTbTTbbTC p

We need to choose four data points that are closest to CT 61 . ,420 T 4179)( 0 TC p

,521 T 4186)( 1 TC p

,822 T 4199)( 2 TC p

,1003 T 4217)( 3 TC p


Example 0b

,420 T 4179 1b

7.0 2b ,521 T 4186 3106667.6 3b

43333.0 4101849.3 ,822 T 4199 011806.0

1 ,1003 T 4217

The values of the constants are found to be

0b 4179 1b 7.0 2b 3106667.6 3b 41018493 .


Example))()(())(()()( 2103102010 TTTTTTbTTTTbTTbbTC p

10042825242101849.3

)52)(42(106667.6)42(7.041794

3

TTTT

TTT

At ,61T

CkgJ

C p

0.4190

826152614261101849.3

)5261)(4261(106667.6)4261(7.04179)61(4

3

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

1000.4190

2.41910.4190

a

%027295.0


Comparison Table

Order of Polynomial 1 2 3

TCp CkgJ

9.4189 2.4191 0.4190

Absolute Relative Approximate Error ---------- %030063.0 %027295.0

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/newton_divided_difference_method.html



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newton’s divided difference polynomial method of interpolation

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