interpolation and polynomial approximation

8/8/2019 Interpolation and Polynomial Approximation

1/22

Dr. [email protected]

1

Interpolation and Polynomial Approximation

Problem: Let x0, x1,xn be (n+1) distinct points

on the real axis. We wish to construct apolynomial P(x) of degree


2/22


2

Existence and uniqueness of solution

Let P(x) = a0.xn + a1.x

n-1 ++an-1.x + an

Linear system (n+1) equations in (n+1) unknown:

nnnn

n

n

n

n

nn

nn

nn

nn

waxaxaxa

waxaxaxa

waxaxaxa

!

!!

......

..............

......

......

)1(

1

10

11)1(

1

1110

00)1(

1

0100


3/22


3

The linear system has a solution if :

The determinant is called determinant Van der

Monde.

0

1.

....

.....

1.

1.

),..,,(det

1

1

1

11

0

1

00

10

{

-

!

n

n

n

n

n

nn

nn

n

xxx

xxx

xxx

xxxV


4/22


4

Verify

ee

!

!

!

!

nji

ij

n

j

j

i

ij

nn

n

xx

xxxxx

0

1

1

0

2

)1(

10

)(

)()1(,...,,(det


5/22


5

Lagrange Formula:

The polynomial P(x) of degree at most n that passes

through the (n+1) points (x0,w0), (x1,w1),(xn,wn) andhas the form:

))...()()...((

))...()()...(()(

)(.)(

110

110

0

0

niiiiii

nii

ji

jn

jij

i

n

iii

xxxxxxxx

xxxxxxxx

xx

xxxL

ithxLxP

!

4!

!

{!

!


6/22


6

Example: P(-1) = 3 ; P(1) =1 ; P(2) = 3 et P(3) = 7

1)(

)(.)(.)(.)(.)(

)1)(2)(4()2)(1)(1(

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

)1)(1)(3(

)3)(1)(1(

))()((

))()((

)2)(1)(2(

)3)(2)(1(

))()((

))()((

)4).(3).(2(

)3).(2).(1(

))()((

))()((

2

33221100

231303

2103

321202

3102

312101

320

1

302010

3210

!

!

!

!

!

!

!

!

!

!

xxxP

xlxlxlxlxP

xxxxxxxxxxxxxxxl

xxx

xxxxxx

xxxxxxl

xxx

xxxxxx

xxxxxx

l

xxx

xxxxxx

xxxxxxl


7/22

Dr. [email protected] 7

Newton Interpolation Polynomial

!

!

!

!!

!!

!

!

!

!

in

njinjk

nijk

ij

ij

ij

ii

j

i

jij

j

n

j

j

xx

DDD

xx

DDD

xx

DDD

WDD

and

xxxxxxxxx

x

ith

xDxp

)1...(...

...

01

0101

00

1

0

110

0

0

...01

,

,

))...()(()()(

1)(

)(.)(

E

E

E


8/22


The Divided Difference table:

nnn

nnn

n

o

wx

wx

wx

!

!

!

,1,2

..01

12

012111

01

00

....

......

.....


9/22


Example: p(1) = 2, P(3)=-1, P(4)=5, P(5) =1

15

4554

8

156

2

5

13

2

321

0123

012

01

0

!

!

!

!

D

D

D

D


10/22


2

67

8

377

2

35

8

15)(

)4)(3)(1())()((

)3)(1()).((1)(

1

.)(

23

2103

102

01

0

3

0

30123201210100...01

!

!!

!!

!!

!

!!!

xxxxP

xxxxxxxxx

xxxxxxxxx

DDDDDxPj

jj

E

E

E

E

EEEEE


11/22


2

67

8

377

2

35

8

15)( 23 ! xxxxP


12/22


Algorithm for interpolation with a Lagrangian polynomial

To interpolate for f(x), given x and a set ofN+1 data

pairs, (xi, fi), I = 0,,N:

Set SUM = 0.

DO For I = 0 to N:

SET P = 1.

DO FOR J = 0 to N:

IF J I:

SET P = P*(x - x(J) )/ (x(I) - x(J)).

ENDO (J).

SET SUM = SUM + P*fi

ENDO (I).

SUM is the interpolated value.


13/22


The Error of the interpolating polynomial

Let x0, x1,.. ,xn be (n+1) distinct points on the real axis

and let f(x) be a real-valued function defined on someinterval [a, b] containing these points, we wish toconstruct P(x) of degree


14/22


k

kk

n

i

ii

n

i

ii

xxxRxfx

nkxfxwith

xDxNewton

xlxfxagrange

{!

!!

!

!

!

!

)()()(

..1,0)()(

)()(:

)()()(:

0

..01

0

E


15/22


Interpolation Error

f(x) = 1/x, x0 = 2, x1 = 2.5, x2 = 4

f(2) = 0.5, f(2.5) = 0.4, f(4) = 0.25

P(x) = 0.05x2 - 0.425x + 1.15

P(3) = 0.325 f(3) = 0.333 |R(3)|= 0.008

031.02

6.

12

1)(

12

1)3(

)(

)!3(

)(

4

)3(

)3(210

!

e

!

!

tf

tfxxxxxx

x


16/22


Comparison between f(x) and P(x)


17/22


The Error of the interpolating polynomial Cauchy Th. : Let f(x) be a real-valued function

defined on [a,b] and (n+1) times differentiable on[a,b]. If P(x) is the polynomial of degree


18/22


Interpolation in a function based on equally

spaced points

Note : xk = x0 + k.h K= 0,1,,n h: step

and x = x0 +s.h s : variable

)(1

.)()!1(

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

2

)1.(

2;

1

!

)1)..(1.(;1

0

)1()1()1(tf

n

shtf

n

hnshshssR

ssss

s

j

jsss

j

ssNote

nnn

!

!

!

!

!

!


19/22


Lagrange Formula

hsxxith

insn

isxfsP

xlxfxP

n

i

i

n

i

ii

.

)()(

)()()(

0

0

0

!

!

!

!

!


20/22


Newton Formula

!!

(

!!

"((!((

!!(

!(

!(

n

i

in

i

ii

j

i

j

i

j

i

j

j

i

xfi

sxDxP

finde

iforxfxfxf

ixfxf

Define

xfxfxf

xfhxfxfote

0

0

0

..01

1

1

11

010

)(.)()(

0)()()(

0)()(

:

)()()(

)()()(

E


21/22


Let f(x) = x.e-x

1) Find the interpolation polynomial p(x)

which interpolates f(x) at x0 = 3, x1=4

and x2

= 5

2) Find a bound for the interpolating

error at point x = 3.5


22/22


Let f(x) = ex

1) Find the interpolation polynomial p(x)which interpolates f(x) at x0 = 1, x1=2,x2= 3 and x3=4.

2) Find a bound for the interpolating

error at point x = 2.5Where:

e1 = 2.7183,

e2

= 7.389,e3 = 20.085,

e4 = 54.598,

interpolation and polynomial approximation

Documents