Lecture5 Newton Divided DifferenceBUE

8/14/2019 Lecture5 Newton Divided DifferenceBUE

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Week Date Topic Classification of Topic

1 9 Feb. 2010 Introduction toNumerical Methods

and Type of Errors

Measuring errors, Binary representation,Propagation of errors and Taylor series

2 14 Feb. 2010 Nonlinear Equations Bisection Method

3 21 Feb. 2010 Newton-Raphson Method

4 28 Feb. 2010 Interpolation Lagrange Interpolation

5 7 March 2010 Newton's Divided Difference Method

6 14 March 2010 Differentiation Newton's Forward and BackwardDivided Difference

7 21 March 2010 Regression Least squares

8 28 March 2010 Systems of Linear

Equations

Gaussian Jordan

9 11 April 2010 Gaussian Seidel

10 18 April 2010 Integration Composite Trapezoidal and SimpsonRules

11 25 April 2010 Ordinary Differential

Equations

Euler's Method

12 2 May2010 Runge-Kutta 2nd and4th order Method

Schedule



Interpolation

Newton’s Divided Difference



What is Interpolation ?

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



Interpolants

Polynomials are the most commonchoice of interpolants because they

are easy to:

Evaluate

Differentiate, andIntegrate.



Newton’s Divided Difference

Method: Given pass aLinear interpolation

linear interpolant through the data

where

),,( 00 y x ),,( 11 y x

)()( 0101 x xbb x f

)( 00 x f b

01

011

)()(

x x

x f x f b



ExampleThe upward velocity of a rocket is given as a function of

time in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for linearinterpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67

Table 1: Velocity as afunction of time

Figure 2: Velocity vs. time data

for the rocket example



Linear Interpolation

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f ran g e( )

f x desired

x s1

10x s0

10 x s range x desired

1520

78.36235.517

,150 t 78.362)( 0 t v

,201 t 35.517)( 1 t v

)( 00 t vb 78.362

01

011

)()(

t t

t vt vb

)tt( b b)t(v 0101

914.30



Linear Interpolation (contd)

10 12 14 16 18 20 22 24350

400

450

500

550517.35

362.78

y s

f range( )

f x desired

x s1

10x s0


)tt( b b)t(v 0101

),15(914.3078.362 t 2015 t

At 16t

)1516(914.3078.362)16( v

69.393 m/s



Quadratic Interpolation

Given ),,( 00 y x ),,( 11 y x and ),,( 22 y x fit a quadratic interpolant through the data.

))(()()( 1020102 x x x xb x xbb x f

)( 00 x f b

01

01

1

)()(

x x

x f x f b

02

01

01

12

12

2

)()()()(

x x

x x

x f x f

x x

x f x f

b



ExampleThe upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for quadraticinterpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67






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


,100 t 04.227)( 0 t v

,151 t 78.362)( 1 t v

,202 t 35.517)( 2 t v




)( 00 t vb

04.227

01

01

1

)()(

t t

t vt vb

1015

04.22778.362

148.27

02

01

01

12

12

2

)()()()(

t t

t t

t vt v

t t

t vt v

b

1020

1015

04.22778.362

1520

78.36235.517

10

148.27914.30

37660.0




)tt)(tt( b)tt( b b)t(v 1020102

),15)(10(37660.0)10(148.2704.227 t t t 2010 t

At ,16t

)16(v2 )1516)(1016(37660.0)1016(148.2704.227

19.392 m/s

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

order and second order polynomial is

a 100x19.392

69.39319.392

= 0.38502 %



General Form

))(()()( 1020102 x x x xb x xbb x f

where

Rewriting

))(](,,[)](,[][)( 1001200102 x x x x x x x f x x x x f x f x f

)(][ 000 x f x f b

01

01

011

)()(],[

x x

x f x f x x f b

02

01

01

12

12

02

01120122

)()()()(

],[],[],,[

x x

x x

x f x f

x x

x f x f

x x

x x f x x f x x x f b



General Form

Given )1( n data points, nnnn y x y x y x y x ,,,,......,,,, 111100 as

))...()((....)()( 110010 nnn x x x x x xb x xbb x f

where

][ 00 x f b

],[ 011 x x f b

],,[ 0122 x x x f b

],....,,[ 0211 x x x f b nnn

],....,,[ 01 x x x f b nnn



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

)xx)(xx)(xx](x,x,x,x[f

)xx)(xx](x,x,x[f )xx](x,x[f ]x[f )x(f

2100123

1001200103

0b

0 x )( 0 x f 1

b

],[ 01 x x f 2b

1 x )( 1 x f ],,[ 012 x x x f 3b

],[ 12 x x f ],,,[ 0123 x x x x f

2 x )( 2 x f ],,[ 123 x x x f

],[ 23 x x f

3

x )(3

x f



Example

The upward velocity of a rocket is given as a function of time in Table 1. Find the velocity at t=16 seconds usingthe Newton Divided Difference method for cubicinterpolation.

t v(t)

s m/s

0 0

10 227.04

15 362.78

20 517.35

22.5 602.97

30 901.67






Example

The velocity profile is chosen as

)tt)(tt)(tt( b)tt)(tt( b)tt( b b)t(v 21031020103

we need to choose four data points that are closest to 16

t ,100 t 04.227)( 0 t v

,151 t 78.362)( 1 t v

,202 t 35.517)( 2 t v

,5.223 t 97.602)( 3 t v



Example

b0 = 227.04; b1 = 27.148; b2 = 0.37660; b3 = 5.4347*10-3

0b

100 t 04.227 1b

148.27 2b

,151 t 78.362 37660.0 3b

914.30 3104347.5 x

,202 t 35.517 44453.0

248.34

,5.223 t 97.602



Example

Hence))()(())(()()(

2103102010 t t t t t t bt t t t bt t bbt v

)20)(15)(10(10*4347.5

)15)(10(37660.0)10(148.2704.227

3

t t t

t t t

At ,16

t

)2016)(1516)(1016(10*4347.5

)1516)(1016(37660.0)1016(148.2704.227)16(

3

v

06.392 m/s

The absolute relative approximate error a obtained is

a 100x06.392

19.39206.392

= 0.033427 %



Comparison Table

Order of

Polynomial

1 2 3

v(t=16)

m/s

393.69 392.19 392.06

Absolute Relative

Approximate Error

---------- 0.38502 % 0.033427 %

Lecture5 Newton Divided DifferenceBUE

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