interpolation and polynomial approximation
TRANSCRIPT
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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
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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
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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
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Verify
ee
!
!
!
!
nji
ij
n
j
j
i
ij
nn
n
xx
xxxxx
0
1
1
0
2
)1(
10
)(
)()1(,...,,(det
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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!
!
{!
!
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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
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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
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Dr. [email protected] 8
The Divided Difference table:
nnn
nnn
n
o
wx
wx
wx
!
!
!
,1,2
..01
12
012111
01
00
....
......
.....
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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
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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
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2
67
8
377
2
35
8
15)( 23 ! xxxxP
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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.
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Dr. [email protected] 13
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
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Dr. [email protected] 14
k
kk
n
i
ii
n
i
ii
xxxRxfx
nkxfxwith
xDxNewton
xlxfxagrange
{!
!!
!
!
!
!
)()()(
..1,0)()(
)()(:
)()()(:
0
..01
0
E
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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
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Dr. [email protected] 16
Comparison between f(x) and P(x)
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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
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Dr. [email protected] 18
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
!
!
!
!
!
!
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Dr. [email protected] 19
Lagrange Formula
hsxxith
insn
isxfsP
xlxfxP
n
i
i
n
i
ii
.
)()(
)()()(
0
0
0
!
!
!
!
!
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Dr. [email protected] 20
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
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Dr. [email protected] 21
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
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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,