Numeriska beräkningar i Naturvetenskap och Teknik Today’s topic: Approximations Least square method Interpolations Fit of polynomials Splines

Numeriska beräkningar i Naturvetenskap och Teknik

Today’s topic:

Approximations

Least square method

Interpolations

Fit of polynomials

Splines


An exemple

80

70

60

50

40

30

20

T

83.54

48.53

76.51

71.50

60.50

19.50

56.48

)(TfL

T

L

Modell

TccTfL 10)(

20 40 60 80

48

50

52

54

Why do the measured values deviate from the mode if the measurement is correct?


How determine the ‘best’ straight line?

T

L

Model bTaTfL )(

20 40 60 80

48

50

52

54


Distance between line and measurements points...

T

L

20 40 60 80

48

50

52

54

1d

2d3d

4d5d

6d

7d


How to define the distance between the line and the measurement points?

idLargest deviation at minimum Approximation in maximum norm

2id

Sum of deviations squared as small as possible Approximation in Euclidian norm

Easier to calculate!

Norm


Matrix formulation: An example

3

2

0

1

3

x

9.0

6.0

6.0

9.0

1.2

)(

xf

x

f

xccxf 10)(

9.03

6.02

6.0

9.0

1.23

10

10

0

10

10

cc

cc

c

cc

cc

9.0

6.0

6.0

9.0

1.2

1

1

1

1

1

3

2

0

1

3

1

0

c

c

fAc with

More equations than unknowns!


Matrix formulation: An exampe

9.0

6.0

6.0

9.0

1.2

1

1

1

1

1

3

2

0

1

3

1

0

c

c

fAAcA TT

3

1

1

1

0

1

2

1

3

1

1

1

1

1

1

3

2

0

1

3

1

0

c

c

3

1

1

1

0

1

2

1

3

1

9.0

6.0

6.0

9.0

1.2



fAAcA TT

23 1

1 5

1

0

c

c

11.1

1.2

1

0

c

c

0.505

521.0

xxf 505.0521.0)(



xxf 505.0521.0)(


General Statement of the Problem:

Depending on the model, the measurement data can of course be described by other expressions than the straight line.

In general terms one seeks a function f* that approximates f’s

given values as good as possible in euclidian norm. Specifically, above we looked for a solution expressed as

xccxf 10)(

but we could as well have looked for a solution given by another function (possibly then for different data)

)cos()sin()( 10 xcxcxf

etc...


Generally one can thus write:

nncccxf ...)( 1100

f(x) is in other words a linear combination of given functions

n ,...,, 10 Where the coefficients

nccc ,...,, 10 are sought

One can in accordande with a vector space look at it so that

n ,...,, 10

Spans a function space (a space of this kind whichfulfills certain conditions is called a Hilbert space, cmp. quant.

mech)


In the case of the straight line we have

x

1

0 1

In a geometrical comparision these two functions, which can

be seen as two vectors in the function space, span a plane U:

10

x1

*f

”vector” 0

”vector” 1

Approximating function

f sought function

The smallest distance from the plane is given by a normal. The smallest

deviation between f* och f is for f*-f orthogonal to the plane U!


Normal equations

Since we are interested in fitting m measured values weleave the picture of the continuous function space and view f(x) as an m-dimensional vector with values:

)(

...

)(

)(

2

1

mxf

xf

xf

)(0

)2(0

)1(0

...m

)(1

)2(1

)1(1

...m

That should be expressed by

and0 1f

For the straight line:

1

...

1

1

0

mx

x

x

...1

1

1


The orthogonality condition now gives the equations:

0),*(

0),*(

1

0

ff

ff1100* ccf where

0),(

0),(

11100

01100

fcc

fcc

0),(),(),

0),(),(),(

111110(0

00111000

fcc

fcc

€

c0(ϕ 0,ϕ 0) + c1(1ϕ1,ϕ 0) = ( f ,ϕ 0)

c0(ϕ 0,ϕ1) + c1(ϕ1,ϕ1) = ( f ,ϕ1)

the equations for the normal:

Which gives


),(),(),(

),(),(),(

1111100

0011000

fcc

fcc

The equations for the normal :

),( ),(

),( ),(

1110

0100

1

0

c

c

),(

),(

1

0

f

f

1

0

. .

. .

. .

. .

. .

. .

10 1

0

c

c

1

0

.

.

.

.

.

.

f


Back to the exemple:

3

2

0

1

3

x

9.0

6.0

6.0

9.0

1.2

)(

xf

xccxf 10)(

1

1

1

1

10

3

2

0

1

3

)(1

x

Model:

3

2

0

1

3

3 2 0 1- 3 1 1 1 1 1

1

1

1

1

1

1

0

c

c

9.0

6.0

6.0

9.0

1.2

1 1 1 1 1 3 2 0 1- 3

TA A TAc f

Data:


Conclusion:

2||*|| ff

the minimum of

ff * is orthogonal to the basis vectors

n ...,, 3,210

Assuming the model

nncccccf ...* 33221100

Given data f

is obtained when

The coefficienterna c1, c2, c3, cn are determined from


The equations

),(),(...),(),(

...

),( ),(...),(),(

),(),(...),(),(

1100

11111100

00011000

nnnnnn

nn

nn

fccc

fccc

fccc

or

fAAcA TT

Where the colomuns in A are:

n ...,, 3,210


Note 1:

The func’s Have to be linearly independent

n ...,, 3,210

(cmp vectors in a vector space)

Note 2:

993

994

996

997

999

x

9.0

6.0

6.0

9.0

1.2

)(

xf

Assume our problem would have been (x koord -996)

4958111 4979

4979- 5

1

0

c

c

2102.7

1.2

23 1

1 5

1

0

c

c

11.1

1.2

cmp to


Gauss’ elimination method:

1902568116

4464278

101694

2432

4321

4321

4321

4321

xxxx

xxxx

xxxx

xxxx

190256 81 16 1

44 64 27 8 1

10 16 9 4 1

2 4 3 2 1


Gauss’ elimination method:

190256 81 16 1

44 64 27 8 1

10 16 9 4 1

2 4 3 2 1

188 | 252 78 14 0

42 | 60 24 6 0

8 | 12 6 2 0

2 | 4 3 2 1

132 | 168 36 0 0

18 | 24 6 0 0

8 | 12 6 2 0

2 | 4 3 2 1

24 | 24 0 0 0

18 | 24 6 0 0

8 | 12 6 2 0

2 | 4 3 2 1

1

1

1

7 1

3 1

1

6 7 1

3 1

1


Error sources

1. Mätdata, Ef

0x 1x

fE

f

x


Error sources

2. Truncation error

0f

1f

ph

0x 1xx5.99 1.5

1.80

0.44 4.19 1.0

0.11 1.36

0.33 2.83 0.5

1.03

1.80 0

f f f f 32 x

These would be zero for a first degree polynomial

h


The approximation to data assumes to pass through the data points,

i.e. one assumes the errors are small.

)()( 001

010 xx

xx

fffxf

000 )()( fpfphxfxf 0f

1f

Linear interpolation

Alt for equidistant data

ph

0x 1xh

5.99 1.5

1.80

0.44 4.19 1.0

0.11 1.36

0.33 2.83 0.5

1.03

1.80 0

f f f f 32 x

Interpolation


Quadratic interpolation

Ansatz

))(()()( 1020102 xxxxcxxccxQ

00 ffxx 00 cf

11 ffxx )( 01101 xxccf h

f

xx

ffc

)( 01

011

22 ffxx ))(()( 1202202102 xxxxcxxccf

))((

)(

)())(( 1202

02

01

01

1202

022 xxxx

xx

xx

ff

xxxx

ffc

1

2

3



2

2

2012

20102

30102

1202

02

01

01

1202

022

22

2

2

22

2

2)()(

))((

)(

)())((

h

f

h

fff

h

ffff

h

hffhff

xxxx

xx

xx

ff

xxxx

ffc

3

h

fppfpfhxxp

h

ffpf

xxxxh

fxx

h

ffxQ

2)1()(

2

))((2

)()(

2

00

2

0

102

2

002



Newton’s ansatz

)(...))((

...

))((

)(

)(

110

102

01

0

mm

m

xxxxxxc

xxxxc

xxc

cxQ

mf

f

f

f

...2

1

0

Uniqueness: There is only one polynomial of order m that passes

through m+1 points.


Error interpolation

)()()( xfelxQxf

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

)(

)()()(

10

)1(

m

m

m

xxxxxxm

f

xQxfxfel

)1(2

|)(|)1(

2

|)(||))((|

2

|)(|

|)()(|)(''2''

10

''

1

ppfh

hpphf

xxxxf

xQxfxfel

Linear interpolation

81/4)1/2p01-(2p )1(

2

22 fpp

f


Exemple

Interpolation of polynomial of order 4,8,16 in equidistant points

)91/()( xexf x

Fit of polynomial of order 6 to 9 equidistant points


4th order 8th order

16th order 6th order in 9

points


Runge’s phenomenon

Interpolation in equidistant points by a polynomal of high order tends to reproduce a curve better in the central parts of the interval but gives considerable oscillations close to the end-points of the interval!

Chebychev polynomials and Chebychev abscissa

If one can select the points in which data is known (this can be

hard if the measurement values are already given… ) then the data

points should be closer close to the end-points of the interval.

An optimal choice is given by the zeros of the Chebychevpolynomial of order m which minimizes the residue above.

)21

12cos(

m

ixi


Splines

An alternative is to use a polynomial piece wise between the points. One

can e.g. set the condition that the function’s values, its derivative and

second derivative is equal in the end points of each short interval for

polynomials that meet in these points. This approach gives so-called

cubic splines. In the extreme end-points one can e.g. demand the curve to be straight.

x

f


Cubic splines

iiiiii

iii

iiiii

dcbayY

ayY

tdtctbatY

1

32

)1(

)0(

)(

iiiii

iii

iiii

dcbDY

bDY

tdtcbtY

32)1('

)0('

32)('

1

2

11

11

)(2

2)(3

iiiii

iiiii

ii

ii

DDyyd

DDyyc

Db

ya

Function Derivative

Second derivative

tdctY iii 62)(''

]1,0[:t


Insertion:

311

211

])(2[

]2)(3[)(

tDDyy

tDDyytDytY

iiii

iiiiiii

211

11

])(2[3

]2)(3[2)('

tDDyy

tDDyyDtY

iiii

iiiiii

tDDyyDDyytY iiiiiiiii ])(2[6]2)(3[2)('' 1111

Function

Derivative

Second derivative


The condtions

€

Yi(0) = y iYi−1(1) = y iY 'i−1 (1) =Y 'i (0)

Y ' 'i−1 (1) =Y ' 'i (0)

€

Y ' '0 (0) = 0

Y ' 'n−1 (1) = 0and

give


1 2

1 4 1

1 4 1

1 4 1

1 2

....

n

1-n

2

1

0

D

D

D

D

D

)(3

)(3

....

)3(y

)3(y

)3(y

1

2

03

02

01

nn

nn

yy

yy

y

y

y

the following system in matrix form:

Easy to solve! Try out MATLABs spline function on your own!

Documents

Numeriska beräkningar i Naturvetenskap och Teknik Today’s topic: Approximations Least square method Interpolations Fit of polynomials Splines