Post on 01-Jan-2016
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WHAT IS INTERPOLATION?
Given (x0,y0), (x1,y1), …, (xn,yn), finding the value of
‘y’ at a value of ‘x’ in (x0, xn) is called interpolation.
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INTERPOLANTS
Polynomials are the most common choice of interpolants because they are easy to:
Evaluate,Differentiate, and Integrate.
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LAGRANGIAN INTERPOLATION
Lagrangian interpolating polynomial is given by
n
iiin xfxLxf
0
)()()(
where ‘ n ’ in )(xf n stands for the thn order polynomial that approximates the function )(xfy
given at )1( n data points as nnnn yxyxyxyx ,,,,...,,,, 111100 , and
n
ijj ji
ji xx
xxxL
0
)(
)(xLi is a weighting function that includes a product of )1( n terms with terms of ij
omitted.
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Example
A manufacturer of thermistors makes the following observations on a thermistor. Determine the temperature corresponding to 754.8 ohms using the Lagrangian method for interpolation.
R T
Ohm C
1101.0
911.3
636.0
451.1
25.113
30.131
40.120
50.128
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Solution
We are interpolating using a cubic polynomial
3
0
)()()(i
ii RTRLRT
)()()()()()()()( 33221100 RTRLRTRLRTRLRTRL
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Solution
400 500 600 700 800 900 1000 1100 120025
30
35
40
45
50
5550.128
25.113
y s
f range( )
f x desired
1.101 103451.1 x s range x desired
32232432632832103225
30
35
40
45
50
55
y
f range( )
f xdesired
x range xdesired
113.25,0.1101 oo RTR 131.30,3.911 11 RTR
120.40,0.636 22 RTR 128.50,1.451 33 RTR
3
00 0
0 )(
jj j
j
RR
RRRL
30
3
20
2
10
1
RR
RR
RR
RR
RR
RR
3
10 1
1 )(
jj j
j
RR
RRRL
31
3
21
2
01
0
RR
RR
RR
RR
RR
RR
3
20 2
2 )(
jj j
j
RR
RRRL
32
3
12
1
02
0
RR
RR
RR
RR
RR
RR
3
30 3
3 )(
jj j
j
RR
RRRL
23
2
13
1
03
0
RR
RR
RR
RR
RR
RR
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Solution
)()(
)()()(
323
2
13
1
03
02
32
3
12
1
02
0
131
3
21
2
01
00
30
3
20
2
10
1
RTRR
RR
RR
RR
RR
RRRT
RR
RR
RR
RR
RR
RR
RTRR
RR
RR
RR
RR
RRRT
RR
RR
RR
RR
RR
RRRT
)128.50()0.6361.451)(3.9111.451)(0.11011.451(
)0.6368.754)(3.9118.754)(0.11018.754(
)120.40()1.4510.636)(3.9110.636)(0.11010.636(
)1.4518.754)(3.9118.754)(0.11018.754(
)131.30()1.4513.911)(0.6363.911)(0.11013.911(
)1.4518.754)(0.6368.754)(0.11018.754(
)113.25()1.4510.1101)(0.6360.1101)(3.9110.1101(
)1.4518.754)(0.6368.754)(3.9118.754()8.754(
T
)128.50)(11639.0()120.40)(69517.0()131.30)(51972.0()113.25)(098494.0(
35.242 C= °
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NEWTONS DIVIDED DIFFERENCE
What is divided difference?
f[x0,x1,x2] =
f[x0,x1] =
f[x1,x2] – f[x0,x1]
x2 – x1
f[x0, x1, …, xk-1, xk] = 1 2 k 0 k-1
k 0
f[x ,x - x ] - f[x ,...,x ]
x - xfor k = 3, 4, ….. n.
These Ist, IInd... and kth order differences are denoted by f, 2f, …, kf.
f[x1] – f[x0]
x1 – x0
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INTERPOLATION USING DIVIDED DIFFERENCE
The divided difference interpolation polynomial is:
P(x) = f(x0) + (x – x0) f [x0, x1] + + (x – x0) (x – xn-1) f[x0, x1, …, xn]
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Example
For the data
x: –1 0 2 5
f(x) : 7 10 22 235 Find the divided difference polynomial and estimate
f(1).
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Solution
X f f 2f 3f -1 7
3
0 10 6
1
13 2
2 22 71
5 235
P(x) = f(x0) + (x – x0) f[x0, x1] + (x – x0) ( x – x1) f [x0, x1, x2] + (x – x0) (x – x1) (x – x2) f [x0, x1, x2, x3]
= 7 + (x +1) 3 + (x +1) (x – 0) 1 + (x +1) (x –0) (x – 2) 2
= 2x3 – x2 + 10
P (1) = 11
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INTERPOLATION FOR EQUALLY SPACED POINTS
Let (X0,,Y0), (X1,,Y1), …, (Xn,,Yn) be the given points withXi+1 = Xi +h,i = 0,1,2,…, (n-1).
Finite Difference Operators
● Forward difference operatorf(xi) = f (xi + h) – f (xi)
● Backward difference operatorf(xi) = f (xi) – f (xi – h)
● Central difference operator
f(xi) = f - f
2
hx i
2
hx i
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INTERPOLATION FOR EQUALLY SPACED POINTS
● Shift operators
● Averaging operator
E f(xi) = f(xi +h) Er f(xi) = f (xi +rh)
f(xi) = 2
1 [ f
2
hx i + f
2
hx i ]
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NEWTON GREGORY FORWARD INTERPOLATION
- -D D D 2 3
0 o 0 0 0
p(p – 1) p(p 1)(p 2)P(x + ph) = y + p y + y + y + +
2! 3!
0x - x
hFor convenience we put p = and f0 = y0. Then we have
- - -Dn
0
p(p 1)(p 2)L (p n + 1)y
n!
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Example
Estimate f (3.17)from the data using Newton Forward Interpolation.
x: 3.1 3.2 3.3 3.4 3.5
f(x): 0 0.6 1.0 1.2 1.3
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Solution
First let us form the difference table
x y y 2y 3y 4y 3.1 0
0.6 3.2 0.6 - 0.2
0.4 0 3.3 1.0 - 0.2 0.1
0.2 0.1 3.4 1.2 -0.1
0.1 3.5 1.3
Here x0 = 3.1, x = 3.17, h = 0.1.
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Solution
p = h
xx 0 = 1.0
07.0 = 0.7
Newton forward formula is:
P(x) = y0 + py0+ !2
)1( pp 2y0+ !3
)2p)(1p(p 3y0+ !4
)3p)(2p)(1p(p 4y0
P(3.17)=0+0.70.6+ 2
)17.0(7.0 (-0.2)+ 6
)27.0)(17.0(7.0 0+ 24
)37.0)(27.0)(17.0(7.0 0.1
= 0.4384
Thus f(3.17) = 0.4384.
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NEWTON GREGORY BACKWARD INTERPOLATION FORMULA
Taking p = h
xx n , we get the interpolation formula as:
P(xn+ph) = y0 + pyn + !2
)1p(p 2yn + !3
)2p)(1p(p 3yn + +
!n
1)n (p 2) (p 1)(p p nyn
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Estimate f(42) from the following data using newton backward interpolation.
x: 20 25 30 35 40 45
f(x): 354 332 291 260 231 204
Example
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SolutionThe difference table is:
x f f 2f 3f 4f 5f 20 354
- 22 25 332 - 19
- 41 29 30 291 10 -37
- 31 - 8 45 35 260 2 8
- 29 0 40 231 2
- 27 45 204
Here xn = 45, h = 5, x = 42
and p = - 0.6
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Solution
Newton backward formula is:
P(x) = yn+pyn+p(p +1)
2!2yn+
p(p +1)(p + 2)
3!3yn+
p(p +1)(p + 2)(p + 3)
4!4yn +
p(p +1)(p + 2)(p + 3)(p + 4)
5!5yn
P(42)=204+(-0.6)(-27)+ (-0.6)(0.4)
22+ (-0.6)(0.40(1.4)
60+ (-0.6)(0.4)(1.4)(2.4)
24 8 +
(-0.6)(0.4)(1.4)(2.4)(3.4)
120 45 = 219.1430
Thus, f(42) = 219.143
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INTERPOLATION USING CENTRAL DIFFERENCES
Suppose the values of the function f (x) are known at the points a -3h, a – 2h, a – h, a, a +h, a + 2h, a + 3h, ... etc. Let these values be y-3, y-2, y-1, y0, y1, y2, y3 ..., and so on. Then we can form the central difference table as:
x f(x) f 2f 3f 4f 5f 6f
a-3h y-3
y-3
a-2h y-2 2y-3
y-2 3y-3
a-h y-1 2y-2 4y-3
y-1 3y-2 5y-3
a y0 2y-1 4y-2 6y-3
y0 3y-1 5y-2
a+h y1 2y0 4y-1
y1 3y0
a+2h y2 2y1 y2
a+3h y3
We can relate the central difference operator with and E using the operator relation = E½.
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GAUSS FORWARD INTERPOLATION FORMULA
P(x) = y0+
1
py0+
2
p2y-1+
3
1p3y-1+
4
1p4y-2 +
5
2p5y-2 + where
p = 0x-xh
and ( )pr = p( p-1) ( p-2) ( p-r+1)
r!
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GAUSS FORWARD INTERPOLATION FORMULA
• The value p is measured forwardly from the origin and 0<p<1.
• The above formula involves odd differences below the central horizontal line and even differences on the line. This is explained in the following figure.
y0 2y-1 4y-2 6y-3
y0 3y-1 5y-2
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Example
Find f(30) from the following table values using Gauss forward difference formula:
x: 21 25 29 33 37
F(x): 18.4708 17.8144 17.1070 16.3432 15.5154
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Solution
The difference table is
x f f 2f 3f 4f
21 18.4708
-0.6564
25 17.8144 - 0.0510
-0.7074 - 0.0054
29 17.1070 0.0564 - 0.0022
-0.7638 - 0.0076
33 16.3432 - 0.0640
-0.8278
37 15.5154
P(x) = y0 +
1
py0 +
2
p2y-1 +
3
1p3y-1 +
4
1p4y-2 +
f (30) = 16.9217
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GAUSS BACKWARD INTERPOLATION FORMULA
This formula is used to interpolate the value of the function for a negative value of p and – 1<p<0. By substituting for y0, 2y0, 3y0 , ....... in terms of central difference in the Newton Forward Difference formula, we get
P(x) = y0 +
1
p y-1 +
2
1p 2y-1 +
3
1p 3y-2 +
4
2p 4y-2 +…
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GAUSS BACKWARD INTERPOLATION FORMULA
This formula involves odd differences above the central
horizontal line and even differences on the central line.
y-1 3y-2 5y-3
y0 2y-1 4y-2 6y-3
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Example
Estimate cos 51 42' by Gauss backward interpolation from the following data:
x: 50 51 52 53 54
cos x: 0.6428 0.6293 0.6157 0.6018 0.5878
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Solution
x0 = 52, x = 51 42' = 51.7 , h = 1
P = h
xx 0 = 1
527.51 = -0.3
The difference table is
X y y 2y 3y 4y
50 0.6428
- 0.0135
51 0.6293 - 0.0001
- 0.0136 - 0.0002
x0 = 52 0.6157 - 0.0003 0.0004
- 0.0139 - 0.0002
53 0.6018 - 0.0001
- 0.0140
54 0.5878
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Solution
The Gauss backward formula is:
P(x) = y0 +
1
p y-1 +
2
1p 2y-1 +
3
1p 3y-2 +
3
2p 4y-2
P(51.7) = 0.6198
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STIRLING’S FORMULA
This formula gives the average of the values obtained by Gauss forward and backward interpolation formulae. For using this formula we should have – ½ < p< ½.
We can get very good estimates if - ¼ < p < ¼.The formula is:
P(x) = y0+p
2
yy 10 +!2
p 2
2y-1+ !3
)1p(p 2
2
yy 23
13
+!4
)1p(p 2 4y-2 + ...
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Example
Using Sterling Formula estimate f (1.63) from the following table:
x: 1.50 1.60 1.70 1.80 1.90
f(x):17.609 20.412 23.045 25.527 27.875
Solution
1.63 – 1.60p = = 0.3
0.1
X0 = 1.60, x = 1.63, h = 0.1
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Solution
Difference table
x y y 2y 3y 2y 1.50 17.609 2.803
x0= 1.60 20.412 -0.170 2.633 0.019 1.70 23.045 -0.151 -0.02 2.482 0.017 1.80 25.527 -0.134 2.348 1.90 27.875
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BESSELS’ INTERPOLATION FORMULA
This formula involves means of even difference on and below the central line and odd difference below the line.
The formula is
P(x) =
2 20 1 1 0
0
y y 1 p (p – 1)p – y
2 2 2! 2
y y-æ ö+ D + Dæ ö+ D +ç ÷ ç ÷è ø è ø
+
1 p (p – 1)2
3!
pæ ö-ç ÷è ø
3y-1 +
4 4–2 –1(p 1) (p – 1) (p – 2) y y
4! 2
æ ö+ D + Dç ÷è ø
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Note When p = 1/2 , the terms containing odd differences vanish. Then we get the formula in a more simple form:
2 20 1 1 0y y p (p – 1)
P(x)2 2! 2
y y-æ ö+ D +D= + ç ÷è ø
BESSELS’ INTERPOLATION FORMULA
4 4–2 –1(p 1) p(p – 1)(p – 2) y y
4! 2
æ ö+ D +D+ +ç ÷è ø
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ExampleUse Bessels’ Formula to find (46.24)1/3 from the following table of x1/3.
X: 41 45 49 53 X1/3: 3.4482 3.5569 3.6593 3.7563