8/14/2019 Lec 6BUE

1/21

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 LinearEquations

Gaussian Jordan

9 11 April 2010 Gaussian Seidel

10 18 April 2010 Integration Composite Trapezoidal and SimpsonRules

11 25 April 2010 Ordinary DifferentialEquations

Euler's Method

12 2 May 2010 Runge-Kutta 2nd and4th order Method

Schedule

8/14/2019 Lec 6BUE

2/21

Interpolation

Newton's Forward DividedDifference

8/14/2019 Lec 6BUE

3/21

What is Interpolation ?

Given (x0,y0), (x1,y1), (xn,yn), find the valueof y at a value of x that is not given such that

the differences are now constant h.

x

0 0( )y f x( )x

0x

1 0( )y f x h 2 0( 2 )..y f x h 0( )ny f x nh

1 0x x h 2 0 2 ..x x h 0nx x nh

1( )i ix x

8/14/2019 Lec 6BUE

4/21

Interpolants

Polynomials are the most commonchoice of interpolants because they

are easy to:

Evaluate

Differentiate, andIntegrate.

8/14/2019 Lec 6BUE

5/21

The equation of a straight line

1

1

1

( ) ( )

( ) ( ) ( )( )

( ) ( ) ( )

o

o o o

o o

f x f x

f x f x x sh xh

f x f x f x

: Given pass a linearLinear interpolation

interpolant through the data

),,( 00 yx ),,( 11 yx

0x xsh

Such that ,and represented the first difference

and it is called delta.

8/14/2019 Lec 6BUE

6/21

Example

To maximise a catch of bath in lake, it is suggested to throw the

line to the depth of the thermocline. The characteristic featureof this area is the sudden change in temperature (T ). We aregiven the temperature , depth z(m) data for a lake in table1

Depth z

m

18.3 -5

18.5 -4

18.8 -3

19 -2

19.4 -1

19.4 0

Table 1 Temperature vs. depthfor a lake.

Figure 1 Temperature vs. depth of a lake.

T

0c

8/14/2019 Lec 6BUE

7/21

Using the given data, we see the largest change in temperatureis between z =-3 m and z =-4 m. Determine the value of the

temperature at z =-3.5 m. using Newton's forward dividedmethod of interpolation and a first order polynomial.

z T(z)

-4 18.5

0.3

-3 18.8

T

0z

1z

1 0 0

1

3.5 ( 4)

( ) ( ) ( ) 18.5 0.3 0.51

( 3.5) 18.8 0.3(0.5) 18.5 0.15 18.65

T z T z s T z s s

T

8/14/2019 Lec 6BUE

8/21

Quadratic Interpolation (contd)Using the given data, we see the largest change in temperatureis between z =-4 m, z =-3 m and z=-2. Determine the value ofthe temperature at z =-3.5 m using Newton's forward dividedmethod of interpolation of a second order polynomial.

z T(z)

-4 18.5

0.3

-3 18.8 -0.1

0.2

-2 19

0z

1z

2z

T 2

T

8/14/2019 Lec 6BUE

9/21

2

2 0 0 0

2

( 1) 0.1 ( 1)( ) ( ) ( ) ( ) 18.5 0.3

2! 2!

0.1(0.5)( 0.5)( 3.5) 18.5 0.3(0.5) 18.5 0.15 0.0125 18.66252

s s s sT z T z s T z T z s

T

The absolute relative approximate error obtained the resultfrom the first order and the second order polynomial is

a

18.6625 18.65

100 0.06%18.6625a x

8/14/2019 Lec 6BUE

10/21

Given (x0,y0), (x1,y1), (x2,y2), (x3,y3)

2 2

3 0 0 0 0

( 1) ( 1)( 2)( ) ( ) ( ) ( ) ( )

2! 3!

s s s s sf x f x s f x f x f x

Cubic Interpolation (contd)For the third order polynomial interpolation

Using the given data, we see the largest change in temperatureis between z =-4 m, z =-3 m, z=-2m and z=-1m. Determine thevalue of the temperature at z =-3.5 m using Newton's forward

divided method of interpolation of a cubic order polynomial.

8/14/2019 Lec 6BUE

11/21

z T(z)

-4 18.5

0.3

-3 18.8 -0.10.2 -0.3

-2 19 0.2

0.4

-1 19.4

T 2T 3T

0z

1z

2z

3z

3

0.1(0.5)( 0.5) 0.3(0.5)( 0.5)( 1.5)( 3.5) 18.5 0.3(0.5) 18.64375

2 6T

The absolute relative approximate error obtained the resultfrom the second order and the third order polynomial is

a

18.64375 18.6625100 0.1%

18.64375a x

8/14/2019 Lec 6BUE

12/21

Comparison Table

Order of

Polynomial toNewtons forward

1 2 3

T(z=-3.5) 18.65 18.6625 18.64375

Absolute RelativeApproximate Error

---------- 0.0669 % 0.1%

8/14/2019 Lec 6BUE

13/21

0 0

1

2

0 0 0 0

( 1)( 2)....( 1)( ) ( ) ( )!

( 1) ( 1)( 2)....( 1)( ) ( ) ( )..... ( )

2! !

nk

n

k

n

s s s s kx f x f xk

s s s s s s nf x s f x f x f x

n

General Form

For the n+1 order polynomial Newton's forward interpolationGiven (x0,y0), (x1,y1), (x2,y2),. (xn,yn)

8/14/2019 Lec 6BUE

14/21

Newton's backward Divided

Difference

8/14/2019 Lec 6BUE

15/21

: Given pass a linearLinear interpolation

interpolant to backward divided difference is

),,( 00 yx ),,( 11 yx

1( ) ( ) ( )n nx f x s f x

Such that , the inverted delta symbol which

is called nabla.

nx x

s h

8/14/2019 Lec 6BUE

16/21

in the same example the linear polynomial Newton'sbackward divided difference is

z T(z)

-4 18.5

0.3

-3 18.8

0z

( )T z

1( 3.5) 18.8 0.3( 0.5) 18.8 0.15 18.65T

1z

1( ) ( ) ( )n nf x f x s f x

3.5 ( 3)

0.51s

8/14/2019 Lec 6BUE

17/21

z T(z)

-4 18.5

0.3

-3 18.8 -0.1

0.2

-2 19

in the same example the second order polynomialNewton's backward divided difference is

0z

1

z

2z

( )T z 2 ( )T z

2

2

( 1)( ) ( ) ( ) ( )

2!n n n

s sx f x s f x f x

2

( 0.5)(0.5)( 0.1)( 3.5) 19 0.2( 0.5)

2

19 0.1 0.0125 18.8875

T

8/14/2019 Lec 6BUE

18/21

The absolute relative approximate error obtained the result

from the first order and the second order polynomial is

a

18.8875 18.65100 1.25%

18.8875a x

8/14/2019 Lec 6BUE

19/21

z T(z)

-4 18.5

0.3

-3 18.8 -0.1

0.2 -0.3

-2 19 0.2

0.4

-1 19.4

in the same example the third order polynomialNewton's backward divided difference is

0z

1z

2z

3z

( )T z 2 ( )T z 3 ( )T z

3

( 0.5)(0.5)(0.2) ( 0.5)(0.5)(1.5)( 0.3)( 3.5) 19 0.4( 0.5)

2 6

19 0.2 0.025 0.01875 18.79375

T

8/14/2019 Lec 6BUE

20/21

Comparison Table

Order of

Polynomial

1 2 3

T(z=-3.5) 18.65 18.8875 18.79375

Absolute RelativeApproximate Error

---------- 1.25 % 0.4%

Since the value z=-3.5 is near the first of the table, to get thecorresponding value of z we must to use Newton's forward

interpolation.

8/14/2019 Lec 6BUE

21/21

General Form

For the n+1 order polynomial Newton's backward interpolationGiven (x0,y0), (x1,y1), (x2,y2),. (xn,yn)

1

2

( 1)( 2)....( 1)( ) ( ) ( )!

( 1) ( 1)( 2)....( 1)( ) ( ) ( )..... ( )

2! !

nk

n n n

k

n

n n n n

s s s s kx f x f xk

s s s s s s nf x s f x f x f x

n