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Newton’s Divided Difference Polynomial

Method of Interpolation

Civil 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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ExampleTo maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the thermocline. The characteristic feature of this area is the sudden change in temperature. We are given the temperature vs. depth plot for a lake. Determine the value of the temperature at z = −7.5 using Newton’s Divided Difference method for linear interpolation.

Temperature vs. depth of a lake

Temperature Depth

T (oC) z (m)19.1 019.1 -119 -2

18.8 -318.7 -418.3 -518.2 -617.6 -711.7 -89.9 -99.1 -10

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Linear Interpolation

15 10 5 011

12

13

14

15

16

17

1817.6

11.7

y s

f range( )

f x desired

x s1

10x s0

10 x s range x desired

)()( 010 zzbbzT

,80 z 7.11)( 0 zT

,71 z 6.17)( 1 zT

)( 00 zTb

7.11

01

011

)()(

zz

zTzTb

87

7.116.17

9.5

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Linear Interpolation (contd)

15 10 5 011

12

13

14

15

16

17

1817.6

11.7

y s

f range( )

f x desired

x s1

10x s0

10 x s range x desired

)()( 010 zzbbzT

),8(9.57.11 z 78 z

At 5.7z

)85.7(9.57.11)5.7( T

C 65.14

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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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ExampleTo maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the thermocline. The characteristic feature of this area is the sudden change in temperature. We are given the temperature vs. depth plot for a lake. Determine the value of the temperature at z = −7.5 using Newton’s Divided Difference method for quadratic interpolation.

Temperature vs. depth of a lake

Temperature Depth

T (oC) z (m)19.1 019.1 -119 -2

18.8 -318.7 -418.3 -518.2 -617.6 -711.7 -89.9 -99.1 -10

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Quadratic Interpolation (contd)

9 8.5 8 7.5 78

10

12

14

16

1817.6

9.89238

y s

f range( )

f x desired

79 x s range x desired

))(()()( 102010 zzzzbzzbbzT

,90 z 9.9)( 0 zT

,81 z 7.11)( 1 zT

,72 z 6.17)( 2 zT

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Quadratic Interpolation (contd)

9.9)( 00 zTb

8.178

9.97.11)()(

01

011

zz

zTzTb

02

01

01

12

12

2

)()()()(

zz

zz

zTzT

zz

zTzT

b

97

98

9.97.11

87

7.116.17

2

8.19.5

05.2

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Quadratic Interpolation (contd)

))(()()( 102010 zzzzbzzbbzT

),8)(9(05.2)9(8.19.9 zzz 79 z

At ,5.7z

)85.7)(95.7(05.2)95.7(8.19.9)5.7( T

C 138.14 The absolute relative approximate error a obtained between the results

from the first and second order polynomial is

100138.14

65.14138.14

a

%6251.3

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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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ExampleTo maximize a catch of bass in a lake, it is suggested to throw the line to the depth of the thermocline. The characteristic feature of this area is the sudden change in temperature. We are given the temperature vs. depth plot for a lake. Determine the value of the temperature at z = −7.5 using Newton’s Divided Difference method for cubic interpolation.

Temperature vs. depth of a lake

Temperature Depth

T (oC) z (m)19.1 019.1 -119 -2

18.8 -318.7 -418.3 -518.2 -617.6 -711.7 -89.9 -99.1 -10

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Example

The temperature profile is chosen as

))()(())(()()( 2103102010 zzzzzzbzzzzbzzbbzT

We need to choose four data points that are closest to 5.7z

,90 z 9.9)( 0 zT

,81 z 7.11)( 1 zT

,72 z 6.17)( 2 zT

,63 z 2.18)( 3 zT

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Example

The values of the constants are obtained as

0b 9.9 1b 8.1 2b 05.2 3b 5667.1

0b

,90 z 9.9 1b

8.1 2b

,81 z 7.11 05.2 3b

9.5 5667.1

,72 z 6.17 65.2

6.0

,63 z 2.18

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Example

))()(())(()()( 2103102010 zzzzzzbzzzzbzzbbzT

)7)(8)(9(5667.1)8)(9(05.2)9(8.19.9 zzzzzz , −9 ≤ z ≤ −6

At ,5.7z

)75.7)(85.7)(95.7(5667.1

)85.7)(95.7(05.2)95.7(8.19.9)5.7(

T

C 725.14 The absolute relative approximate error a obtained between the

results from the second and third order polynomial is

100725.14

138.14725.14

a

%9898.3

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Comparison Table

Order of Polynomial

1 2 3

Temperature (oC) 65.14 138.14 725.14

Absolute Relative Approximate Error

---------- %6251.3 %9898.3

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ThermoclineWhat is the value of depth at which the thermocline exists?

The position where the thermocline exists is given where 02

2

dz

Td

.

69,5667.155.3558.2629.615

69,7895667.18905.298.19.932

zzzz

zzzzzzzzT

69,7.41.7158.262 2 zzzdz

dT

69,4.91.712

2

zzdz

Td

Simply setting this expression equal to zero, we get

69,4.910.710 zz

m5638.7z

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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