interpolation functions
DESCRIPTION
Interpolation functionsTRANSCRIPT
Interpolation produces a function that matches the given data exactly. The function then can be utilized to approximate the data values at intermediate points.
Interpolation may also be used to produce a smooth graph of a function for which values are known only at discrete points, either from measurements or calculations.
Given data points
Obtain a function, P(x)
P(x) goes through the data points
Use P(x)
To estimate values at intermediate points
Given data points:
At x0 = 2, y0 = 3 and at x1 = 5, y1 = 8
Find the following:
At x = 4, y = ?
x
y
2 5
3
8
?
4
P(x)
P(x) should satisfy the following conditions:
P(x = 2) = 3 and P(x = 5) = 8
xLxLxP 10 83
P(x) can satisfy the above conditions if
at x = x0 = 2, L0(x) = 1 and L1(x) = 0 and
at x = x1= 5, L0(x) = 0 and L1(x) = 1
The conditions can be satisfied if L0(x) and L1(x) are defined in the following way.
25
2and
52
510
x
xLx
xL
At x = x0 = 2, L0(x) = 1 and L1(x) = 0 and
at x = x1= 5, L0(x) = 0 and L1(x) = 1
01
01
10
10 and
xx
xxxL
xx
xxxL
xLxLxP 10 83
1100 yxLyxLxP
1100 xfxLxfxLxP
Lagrange Interpolating Polynomial
1100 xfxLxfxLxP
825
23
52
5
xx
xP
3
15 x
xP
33363
1454 .P
The Lagrange interpolating polynomial passing through three given points; (x0, y0), (x1, y1) and (x2, y2) is:
22
1100
xfxL
xfxLxfxLxP
221100 yxLyxLyxLxP
2010
210 xxxx
xxxxxL
At x0, L0(x) becomes 1. At all other given data points L0(x) is 0.
2101
201 xxxx
xxxxxL
At x1, L1(x) becomes 1. At all other given data points L1(x) is 0.
1202
102 xxxx
xxxxxL
At x2, L2(x) becomes 1. At all other given data points L2(x) is 0.
nn yxL...........
yxLyxLyxLxP
221100
nn xfxL...........xfxL
xfxLxfxLxP
22
1100
General form of the Lagrange Interpolating Polynomial
))...()()...()((
))...()()...()(()(
nkkkkkkk
nkkk xxxxxxxxxx
xxxxxxxxxxxL
1110
1110
n
kii ik
ik xx
xxxL
0 )(
)()(
0 1 2
1 1
1
k k
n n
x x x x x x
x x x x
x x x x
Numerator of kL x
Denominator of
0 1 2
1 1
1
k k k
k k k k
k n k n
x x x x x x
x x x x
x x x x
kL x
Find the Lagrange Interpolating Polynomial using the three given points.
2504
4052
502
22
11
00
.,y,x
.,.y,x
.,y,x
2010
210 xxxx
xxxxxL
1056
42522
452
2
0
x.x
.
x.xxL
750
86
452252
42
2
1
.
xx
..
xxxL
2101
201 xxxx
xxxxxL
3
554
52424
522
2
2
x.x
.
.xxxL
1202
102 xxxx
xxxxxL
22
1100
xfxL
xfxLxfxLxP
2503
554
40750
86
501056
2
2
2
.x.x
..
xx
.x.xxP
1514250050 2 .x.x.xP
The three given points were taken from the function
x
xf1
325.0
15.13425.0305.032
P
333.03
13 f
An approximation can be obtained from the Lagrange Interpolating Polynomial as:
Newton’s Interpolating Polynomials
Newton’s equation of a function that passes through two points
00 y,x and 11 y,x is
010 xxaaxP
Set0xx
0 0 0P x y a
Set1x x
1 1 0 1 1 0P x y a a x x
010 xxaaxP
01
011 xx
yya
Newton’s equation of a function that passes through three points
00 y,x and 11 y,xis
2 2x , y
0 1 0
2 0 1
P x a a x x
a x x x x
To find
2a , set2xx
2 0 1 2 0
2 2 0 2 1
P x a a x x
a x x x x
1 02 1
2 1 1 02
2 0
y yy yx x x x
ax x
Newton’s equation of a function that passes through four points can be written by adding a fourth term .
0 1 0
2 0 1
P x a a x x
a x x x x
3 0 1 2 a x x x x x x
0 1 0
2 0 1
3 0 1 2
P x a a x x
a x x x x
a x x x x x x
The fourth term will vanish at all three previous points and, therefore, leaving all three previous coefficients intact.
Divided differences and the coefficientsf
ix if xThe divided difference of a function,
with respect to is denoted as
It is called as zeroth divided difference and is simply the value of the function, fat ix
ii xfxf
1i if x , x
fThe divided difference of a function,
called as the first divided difference, is denoted ixwith respect to and 1ix
11
1
i ii i
i i
f x f xf x , x
x x
fThe divided difference of a function,
called as the second divided difference, is denoted as
ixwith respect to and1ix , 2ix
1 2i i if x , x , x
1 2 11 2
2
i i i ii i i
i i
f x , x f x , xf x , x , x
x x
1 2 3
1 2 3 1 2
3
i i i i
i i i i i i
i i
f x , x , x , x
f x , x , x f x , x , x
x x
The third divided difference with respect to
ix 1ix 2ix 3ix , and ,
The coefficients of Newton’s interpolating polynomial are:
00 xfa
101 x,xfa
2102 x,x,xfa
32103 x,x,x,xfa
432104 x,x,x,x,xfa and so on.
x xf First
divided differences
Second
divided differences
Third
divided differences
0x 0xf
01
0110 xx
xfxfx,xf
1x 1xf 02
1021210 xx
x,xfx,xfx,x,xf
12
1221 xx
xfxfx,xf
03
2103213210 xx
x,x,xfx,x,xfx,x,x,xf
2x 2xf 13
2132321 xx
x,xfx,xfx,x,xf
23
2332 xx
xfxfx,xf
14
3214324321 xx
x,x,xfx,x,xfx,x,x,xf
3x 3xf 24
3243432 xx
x,xfx,xfx,x,xf
34
3443 xx
xfxfx,xf
25
4325435432 xx
x,x,xfx,x,xfx,x,x,xf
4x 4xf 35
4354543 xx
x,xfx,xfx,x,xf
45
4554 xx
xfxfx,xf
5x 5xf
ExampleFind Newton’s interpolating polynomial to approximate a function whose 5 data points are given below.
f x
2.0 0.85467
2.3 0.75682
2.6 0.43126
2.9 0.22364
3.2 0.08567
x
i ix ixf ii x,xf 1 iii x,x,xf 12 ii x,,xf 3 ii x,,xf 4
0 2.0 0.85467
-0.32617
1 2.3 0.75682 -1.26505
-1.08520 2.13363
2 2.6 0.43126 0.65522 -2.02642
-0.69207 -0.29808
3 2.9 0.22364 0.38695
-0.45990
4 3.2 0.08567
The 5 coefficients of the Newton’s interpolating polynomial are:
0 0 0 85467a f x .
1 0 1 0 32617a f x , x .
2 0 1 2 1 26505a f x , x , x .
3 0 1 2 3 2 13363a f x , x , x , x .
4 0 1 2 3 4 2 02642a f x , x , x , x , x .
0 1 0
2 0 1
3 0 1 2
4 0 1 2 3
P x a a x x
a x x x x
a x x x x x x
a x x x x x x x x
0 85467 0 32617 2 0
-1.26505 2 0 2 3
2 13363 2 0 2 3 2 6
2 02642 2 0 2 3 2 6 2 9
P x . . x .
x . x .
. x . x . x .
. x . x . x . x .
P(x) can now be used to estimate the value of the function f(x) say at x = 2.8.
2 8 0 85467 0 32617 2 8 2 0
-1.26505 2 8 2 0 2 8 2 3
2 13363 2 8 2 0 2 8 2 3 2 8 2 6
2 02642 2 8 2 0 2 8 2 3 2 8 2 6 2 8 2 9
P . . . . .
. . . .
. . . . . . .
. . . . . . . . .
2 8 2 8 0 275f . P . .