Projects IIST 4 Sept'2010

8/8/2019 Projects IIST 4 Sept'2010

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http://numericalmethods.eng.usf.edu1

Projects ???


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Interpolants Polynomials are the most

common choice of interpolants

because they are easy to:

EvaluateDifferentiate, andIntegrate.


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Newtons Divided

Difference Method


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f two data points are availablee as {x0,f(x0)}and {x1,f(x1)}, the points can be connected

by a straight line to obtain the simplest formof interpolation,namely,linear interpolation.(see Fig.5.4.)The formula for linearinterpolation can be derived by considering

the similar triangles ADE and ABC, which giveDE BC f1(x) f(x0) f(x1) f(x0)

= , or =

AE AC x x0 x1 x0

Newtons Divided Difference

Method


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Newtons Divided Difference

MethodLinear 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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Linear Interpolation

5 0 5 107.08

7.1

7.12

7.14

7.16

7.18

7.27.2

7.1

y s

f range( )

f x desired( )

x s1

10+x s0

10 x s range, x desired,

)()( 010 xxbbxy +=

,00.20

=x 2.7)(0

=xy

,25.41 =x 1.7)( 1 =xy

)( 00 xyb = 2.7=

01

011 )()(

xxxyxyb = 00.225.4 2.71.7 =

04444.0=


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(contd)

5 0 5 107.08

7.1

7.12

7.14

7.16

7.18

7.27.2

7.1

y s

f range( )

f x desired( )

x s1 10+x s0 10 x s range, x desired,

)()( 010 xxbbxy +=

),00.2(04444.02.7 = x 25.400.2 x

4=x)00.200.4(04444.02.7)00.4( =x

.1111.7 in=


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xamp e: n va ue o x =e . x results to estimate the value of f at

x=1

x0=0 and x1= 2f(x0)= f(0) = e0 =1.0

f(x1) = f(2) = e1.0= 2.718282

f1(x) = b0 + a1 (x-x0)a0=f(x0) = 1.0

f(x1)-f(x0) 2.718282 1.0

a1 = =

x1 x0 2.0 0.0

a1=0.859141


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. results to estimate the value of f at

x=1x

0

=0 and x1

= 2

f1(x) = 1.0 + 0.859141x

X=1.0

f1(1) = 1.0 + 0.859141(1) =1.859141

exect value :f(1) = e0.5 = 1.648721


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Quadratic Interpolation


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Given

),,( 00 yx ),,( 11 yx ),,( 22 yx

))(()()( 1020102 xxxxbxxbbxf ++=

)( 00 xfb =

01

011

)()(

xx

xfxfb

=

02

01

01

12

12

2

)()()()(

xx

xx

xfxf

xx

xfxf

b

=


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ExampleThe path of a rapid laser is given by these specifications.If the laser is traversing from x = 2 to x = 4.25 to x=5.25 using aquadratic path, find the value of y at x = 4 using the Newtons Divideddifference polynomial method.

Path of a robot

0

1

2

3

4

5

6

7

8

0 5 10 15

X

Y

x (m) y (m)

2 7.2

4.25 7.1

5.25 67.81 5

9.2 3.5

10.6 5


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(contd)

2 2.5 3 3.5 4 4.5 5 5.56

6.5

7

7.5

87.56258

6

y s

f range( )

f x desired( )

5.252 x s range, x desired,

))(()()( 102010 xxxxbxxbbxy ++=

,00.20 =x 2.7)( 0 =xy

,25.41 =x 1.7)( 1 =xy

,25.52 =x 0.6)( 2 =xy


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(contd))( 00 xyb =

2.7=

01

011 )()(

xxxyxyb

= 00.225.42.71.7

=

04444.0=

02

01

01

12

12

2

)()()()(

xxxx

xyxy

xx

xyxy

b

=00.225.5

00.225.4

2.71.7

25.425.5

1.70.6

=

25.3

04444.01.1 +=

32479.0=


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(contd)))(()()( 102010 xxxxbxxbbxy ++=

),25.4)(00.2(32479.0)00.2(04444.02.7 = xxx 25.500.2 x

,4=x)25.400.4)(00.200.4(32479.0)00.200.4(04444.02.7)00.4( =y

.2735.7 in=

a

1002735.7

1111.72735.7

=a

%2327.2=


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General Form))(()()( 1020102 xxxxbxxbbxf ++=

where

Rewriting

))(](,,[)](,[][)( 1001200102 xxxxxxxfxxxxfxfxf ++=

)(][ 000 xfxfb ==

01

01

011

)()(],[

xx

xfxfxxfb

==

02

01

01

12

12

02

0112

0122

)()()()(

],[],[],,[xx

xx

xfxf

xx

xfxf

xxxxfxxfxxxfb

===


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General Form

Given

)1( +n ( ) ( ) ( ) ( )nnnn yxyxyxyx ,,,,......,,,, 111100 ))...()((....)()( 110010 +++= nnn xxxxxxbxxbbxf

][ 00 xfb =

],[ 011 xxfb =

],,[ 0122 xxxfb =

],....,,[ 0211 xxxfb nnn =

],....,,[ 01 xxxfb nnn =


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General form

The third or

),,( 00 yx ),,( 11 yx ),,( 22 yx ),,( 33 yx

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

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


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Example The path of a rapid laser is given by thesespecifications. Find the path traversed by the laserusing the Newtons Divided Difference Polynomial

method.Path of a robot

0

1

2

3

4

5

6

7

8

0 5 10 15

X

Y

x (m) y (m)

2 7.2

4.25 7.1

5.25 67.81 5

9.2 3.5

10.6 5


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Example

The value of))()()()(())()()((

))()(())(()()(

43210532104

2103102010

xxxxxxxxxxbxxxxxxxxb

xxxxxxbxxxxbxxbbxy

++

+++=

,00.20 =x 2.7)( 0 =xy ,25.41 =x 1.7)( 1 =xy

,25.52 =x 0.6)( 2 =xy ,81.73 =x 0.5)( 3 =xy

,20.94 =x 5.3)( 4 =xy ,60.105 =x 0.5)( 5 =xy

0b 2.7= 1b 04444.0= 2b 32479.0=

3b 090198.0= 4b 023009.0= 5b 0072922.0=


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ExampleHence

))()()()(())()()((

))()(())(()()(

43210532104

2103102010

xxxxxxxxxxbxxxxxxxxb

xxxxxxbxxxxbxxbbxy

++

+++=

)2.9)(81.7)(25.5)(25.4)(2(0072922.0

)81.7)(25.5)(25.4)(2(023009.0

)25.5)(25.4)(2(090198.0

)25.4)(2(32479.0)2(04444.02.7

+

+

=

xxxxx

xxxx

xxx

xxx

6.102,0072922.023091.07862.2855.15344.41898.30)( 5432 +++= xxxxxxxy


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Example

6.102,0072922.023091.07862.2855.15344.41898.30)( 5432 +++= xxxxxxxy

Path of a robot

7.2 7.1

6

5

3.5

5

0

1

2

3

4

5

6

7

8

9

10

0 2 4 6 8 10 12

X

Y


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Newton-Gregory forwardinterpolation formula

The general Newtons interpolationformula can be simplified when the

data points are equally spacedalong x-axis. The resulting formulasare also known as Newton-Gregoryformulas. fh denotes the intervalbetween any two concecutivevalues of xi(xi+1 -xi=h) ve can write

xi

=x0

+ ih , i=1,2,.,n.


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Newton-Gregory forwardinterpolation formula

In this case, the coefficients of the interpolationpolynominal a0,a1,.,an can be expressed as,

0f0

a0=f(x0) =h0

1f0 0f0

a1 = =h1 h0


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Newton-Gregory forward

interpolation formulanf0

an =

n!hnjf0 is thejth-order forward difference(j=1,2,,n)

Using these relations we can write the nth-order

Newtons interpolation polynominal as; f0 2f0fn(x) =f(x0) + (x-x0) + (x-x0)(x-x0-h) +.

h 2!h2


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Interpolation using Splines ,Why

Splines ?2251

1)(

xxf

+=

Table : Six equidistantly spaced points in [-1, 1]

Figure : 5th order polynomial vs. exact function

x 22511x

y+

=

-1.0 0.038461

-0.6 0.1

-0.2 0.5

0.2 0.5

0.6 0.1

1.0 0.038461


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Why Splines ?

Figure : Higher order polynomial interpolation is a bad idea

Original

Function

16th Order

Polynomial

8th Order

Polynomial

4thOrder

Polynomial


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Linear InterpolationGiven ( ) ( ) ( )( )nnnn yxyxyxyx ,,,......,,,, 111100 , fit linear splines to the data. This simply involvesforming the consecutive data through s traight lines. So if the above data is given in an ascending

order, the linear splines are given by ( ))( ii xfy =

Figure : Linear splines


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Example The path of a rapid laser is given by these specifications.If the laser is traversing from x = 2 to x = 4.25 in a linear path,find the value of y at x = 4 using the linear spline interpolation

method.

Path of a robot

0

1

2

3

4

5

6

7

8

0 5 10 15

X

Y

x (m) y (m)

2 7.2

4.25 7.1

5.25 67.81 5

9.2 3.5

10.6 5


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5 0 5 107.08

7.1

7.12

7.14

7.16

7.18

7.27.2

7.1

y s

f range( )

f x desired( )

x s1

10+x s0

10 x s range, x desired,

,00.20 =x

2.7)( 0 =xy

,25.41 =x 1.7)( 1 =xy

)()()()()( 001

010 xx

xx

xyxyxyxy +=

)00.2(00.225.4

2.71.72.7

+= x

)00.2(04444.02.7)( = xxy

,4=x

)00.200.4(04444.02.7)00.4( =y

.1111.7 in=


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(contd)

The linear sp),00.2(04444.02.7)( = xxy 25.400.2 x

),25.4(1.11.7)( = xxy 25.525.4 x

),25.5(390625.00.6)( = xxy 81.725.5 x

),81.7(07914.10.5)( = xxy 20.981.7 x

),20.9(07143.15.3)( += xxy 60.1020.9 x

Find the path of the robot if it follows linear splines.


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(contd)Find the length of the path traversed by therobot following linear splines.

The length of the robots path can be found by simply adding the length

of the line segments together. The length of a straight line from one point

),( 00 yx to another point ),( 11 yx is given by2

01

2

01 )()( yyxx + .

Hence, the length of the linear splines from x = 2.00 to x = 10.60 is

22

2222

2222

)5.30.5()20.960.10(

)0.55.3()81.720.9()0.60.5()25.581.7(

)1.70.6()25.425.5()2.71.7()00.225.4(

++

++++

+++=L

L=10.584 m


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Exemple: find a linear spline to fit

the following data :i

0

12

3

4

xi

2.0

3.0

6.5

8.0

12.0

f(xi)

14.0

20.017.0

16.0

23.0

Use the results toestimate the value of fat x= 7.0.


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solutionThe data points can be joined by straightlines to linear spline.since x=7.0 lies inthe interval (6.5 to 8.0) we can findthe equation of the straight linebetween x=6.5 and x=8.0 as

f(x3) f(x2)

f3(x) = f(x2) + (x-x2)x3 x2


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Solution(con)f3(x) = 17.0 +((16.0 17.0) / (8.0 -6.5)) (x-6.5)

= 21.3333 0.6667xThe value of f at x=7.0 can be estimated as

f(7.0)=21.3333-0.6667(7.0) = 16.6664


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Q d i I l i


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Quadratic Interpolation(contd)

Each quadratic spline goes through two consecutive data points

)( 01012

01xfcxbxa =++

)(1111

2

11xfcxbxa =++ .

.

.

)(11

2

1 =++ iiiiii xfcxbxa

)(2

iiiiii xfcxbxa =++ .

.

.

)(11

2

1 =++ nnnnnn xfcxbxa

)(2

nnnnnn xfcxbxa =++

This condition gives 2n equations


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Quadratic Splines (contd)The first derivatives of two quadratic splines are continuous at the interior points.

For example, the derivative of the first spline

11

2

1

cxbxa

++is

11

2 bxa

+

The derivative of the second spline

22

2

2 cxbxa ++ is 222 bxa +

and the two are equal at1

xx = giving

21211122 bxabxa +=+

022212111 =+ bxabxa


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Quadratic Splines (contd)Similarly at the other interior points,022 323222 =+ bxabxa

.

.

.

022 11 =+ ++ iiiiii bxabxa

.

.

.

022 1111 =+ nnnnnn bxabxa

We have (n-1) such equations. The total number of equations is )13()1()2( =+ nnn .

We can assume that the first spline is linear, that is 01 =a


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Quadratic Splines (contd)This gives us 3n equations and 3n unknowns. Once we find the 3n constants,we can find the function at any value of x using the splines,

,)( 112

1 cxbxaxf ++= 10 xxx

,222

2 cxbxa ++= 21 xxx

.

.

.

,2 nnn cxbxa ++= nn xxx 1


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ExampleThe path of a rapid laser is given by thesespecifications in Cartesian co-ordinates. Find thelength of the path traversed by the robot using

quadratic splines.Path of a robot

0

1

2

3

4

5

6

7

8

0 5 10 15

X

Y

x (m) y (m)

2 7.2

4.25 7.1

5.25 67.81 5

9.2 3.5

10.6 5


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Solution

2 4 6 8 10 123

4

5

6

7

87.2

3.5

y

10.62 x

Since ther

fi d

,)( 1121 cxbxaxy ++= 25.400.2 x

,222

2 cxbxa ++= 25.525.4 x

,332

3 cxbxa ++= 81.725.5 x

,4424 cxbxa ++= 20.981.7 x

,552

5 cxbxa ++= 60.1020.9 x


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Solution (contd)Setting up the eq

Each quadratic s

11

2

1cxbxa ++

2.7)00.2()00.2( 112

1 =++ cba

1.7)25.4()25.4( 112

1 =++ cba

Similarly,

1.7)25.4()25.4( 222

2 =++ cba

0.6)25.5()25.5( 222

2 =++ cba

0.6)25.5()25.5( 332

3 =++ cba

0.5)81.7()81.7( 332

3 =++ cba

0.5)81.7()81.7( 442

4 =++ cba

5.3)20.9()20.9( 442

4 =++ cba

5.3)20.9()20.9( 552

5 =++ cba

0.5)60.10()60.10( 552

5 =++ cba


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Solution (contd)

Quadratic spline

At x = 4.25

0)25.4(2)25.4(2 2211 =+ baba

0)25.5(2)25.5(2 3322 =+ baba

0)81.7(2)81.7(2 4433 =+ baba

0)20.9(2)20.9(2 5544 =+ baba

11

2

1 cxbxa ++

01 =a

2 2.5 3 3.5 4 4.5 5 5.56

6.5

7

7.5

87.56258

6

y s

f range( )

f x desired( )

5.252 x s range, x desired,


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Solution (contd)

=

0

0

0

0

0

0.5

5.3

5.3

0.5

0.5

0.6

0.6

1.7

1.7

2.7

000000000000001

014.18014.18000000000

0000162.150162.15000000

000000015.10015.10000

000000000015.8015.8

16.1036.112000000000000

12.964.84000000000000

00012.964.84000000000

000181.7996.60000000000

000000181.7996.60000000

000000125.5563.27000000

000000000125.5563.27000

000000000125.4063.18000

000000000000125.4063.18

000000000000124

5

5

5

4

4

4

3

3

3

2

2

2

1

1

1

c

b

a

c

b

a

c

b

a

c

b

a

c

b

a


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Solution (contd)

Solving theia ib ic


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Solution (contd),2889.704444.0)( += xxy 25.400.2 x

,777.119278.80556.1 2 += xx 25.525.4 x

,319.363945.968943.0 2 += xx 81.725.5 x

,40.113945.287651.1 2 += xx 20.981.7 x

,34.314042.642886.3 2 += xx 60.1020.9 x


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Solution (contd)

2 4 6 8 100

2

4

6

88

0

y

fquadratic range( )

111.5 x range,

The length of a cu

)(xfy =

+=b

a

dxdx

dyL

2

1

( ),2889.704444.0 += xdx

d

dx

dy25.400.2 x

( ),777.119278.80556.1 2 += xxdx

d25.525.4 x

( ),319.363945.968943.0 2 += xxdxd 81.725.5 x

( ),40.113945.287651.1 2 += xxdx

d20.981.7 x

( ),34.314042.642886.3 2 += xxdx

d60.1020.9 x


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Solution (contd)

++

++

++

++

+=

60.10

20.9

220.9

81.7

281.7

25.5

225.5

25.4

225.4

00.2

2

11111dx

dy

dx

dy

dx

dy

dx

dydx

dx

dyL

( ) ( ) ( )

( ) ( )

+++++

++++++=

60.10

20.9

220.9

81.7

2

81.7

25.5

225.5

25.4

225.4

00.2

2

042.645772.61945.285302.31

3945.93788.119278.81111.2104444.01

dxxdxx

dxxdxxdx

8080.36067.26596.35499.12522.2 ++++=

876.13=


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ComparisonCompare the answer from part (a) to linearspline result and fifth order polynomial result.

We can find the

6.102,0072923.023091.07862.2855.15344.41898.30)(5432 +++= xxxxxxxy

+=b

a

dxdxdyL

2

1

( ) +++=60.10

00.2

432036461.092364.03586.8710.31344.411 dxxxxx

133.13=


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C programming#include/* interpolation using splines */int main(void){

int n,i,i1,i2;float x[30],fx[30],xx,ffx,a,b;printf("Enter number of datapoints\n");scanf("%d",&n);printf("enter data points (values of x) \n");

for( i=0;i


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Documents

Projects IIST 4 Sept'2010