NUMERICAL ANALYSIS CH4CISE301-Topic5.ppt

CISE301_Topic5 1
SE301: Numerical Methods
Topic 5:
Interpolation
Lectures 20-22:
KFUPM
Read Chapter 18, Sections 1-5

CISE301_Topic5 2
Lecture 20
Introduction to Interpolation
Introduction
Interpolation Problem
Existence and Uniqueness
Linear and Quadratic Interpolation
Newton’s Divided Difference Method
Properties of Divided Differences

CISE301_Topic5 3
Introduction
Interpolation was used for
long time to provide an
estimate of a tabulated
function at values that are
not available in the table.
What is sin (0.15)?
x sin(x)
0 0.0000
0.1 0.0998
0.2 0.1987
0.3 0.2955
0.4 0.3894
Using Linear Interpolation sin (0.15) ≈ 0.1493
True value (4 decimal digits) sin (0.15) = 0.1494

CISE301_Topic5 4
The Interpolation Problem
Given a set of n+1 points,
Find an nth
order polynomial
that passes through all points, such that:
     
)
(
,
....,
,
)
(
,
,
)
(
, 1
1
0
0 n
n x
f
x
x
f
x
x
f
x
)
(x
fn
n
i
for
x
f
x
f i
i
n ,...,
2
,
1
,
0
)
(
)
( 


CISE301_Topic5 5
Example
An experiment is used to determine
the viscosity of water as a function
of temperature. The following table
is generated:
Problem: Estimate the viscosity
when the temperature is 8 degrees.
Temperature
(degree)
Viscosity
0 1.792
5 1.519
10 1.308
15 1.140

CISE301_Topic5 6
Interpolation Problem
Find a polynomial that fits the data
points exactly.
)
V(T
V
T
a
V(T)
i
i
n
k
k
k



0
ts
coefficien
Polynomial
:
e
Temperatur
:
Viscosity
:
k
a
T
V
Linear Interpolation: V(T)= 1.73 − 0.0422 T
V(8)= 1.3924

CISE301_Topic5 7
Existence and Uniqueness
Given a set of n+1 points:
Assumption: are distinct
Theorem:
There is a unique polynomial fn(x) of order ≤ n
such that:
,...,n
,
i
for
x
f
x
f i
i
n 1
0
)
(
)
( 

n
x
x
x ,...,
, 1
0
     
)
(
,
....,
,
)
(
,
,
)
(
, 1
1
0
0 n
n x
f
x
x
f
x
x
f
x

CISE301_Topic5 8
Examples of Polynomial Interpolation
Linear Interpolation
 Given any two points,
there is one polynomial of
order ≤ 1 that passes
through the two points.
Quadratic Interpolation
Given any three points there
is one polynomial of order ≤
2 that passes through the
three points.

CISE301_Topic5 9
Linear Interpolation
Given any two points,
The line that interpolates the two points is:
Example :
Find a polynomial that interpolates (1,2) and (2,4).
   
)
(
,
,
)
(
, 1
1
0
0 x
f
x
x
f
x
 
0
0
1
0
1
0
1
)
(
)
(
)
(
)
( x
x
x
x
x
f
x
f
x
f
x
f 




  x
x
x
f 2
1
1
2
2
4
2
)
(
1 






CISE301_Topic5 10
Quadratic Interpolation
 Given any three points:
 The polynomial that interpolates the three points is:
     
)
(
,
,
)
(
,
,
)
(
, 2
2
1
1
0
0 x
f
x
and
x
f
x
x
f
x
    
0
2
0
1
0
1
1
2
1
2
2
1
0
2
0
1
0
1
1
0
1
0
0
1
0
2
0
1
0
2
)
(
)
(
)
(
)
(
]
,
,
[
)
(
)
(
]
,
[
)
(
:
)
(
x
x
x
x
x
f
x
f
x
x
x
f
x
f
x
x
x
f
b
x
x
x
f
x
f
x
x
f
b
x
f
b
where
x
x
x
x
b
x
x
b
b
x
f




















CISE301_Topic5 11
General nth
Order Interpolation
Given any n+1 points:
The polynomial that interpolates all points is:
     
)
(
,
...,
,
)
(
,
,
)
(
, 1
1
0
0 n
n x
f
x
x
f
x
x
f
x
        
]
,
...
,
,
[
....
]
,
[
)
(
...
...
)
(
1
0
1
0
1
0
0
1
0
1
0
2
0
1
0
n
n
n
n
n
x
x
x
f
b
x
x
f
b
x
f
b
x
x
x
x
b
x
x
x
x
b
x
x
b
b
x
f












 

CISE301_Topic5 12
Divided Differences
0
1
1
0
2
1
1
0
0
2
1
0
2
1
2
1
0
0
1
0
1
1
0
]
,...,
,
[
]
,...,
,
[
]
,...,
,
[
......
..
.
.
.
.
DD
order
Second
]
,
[
]
,
[
]
,
,
[
DD
order
First
]
[
]
[
]
,
[
DD
order
Zeroth
)
(
]
[
x
x
x
x
x
f
x
x
x
f
x
x
x
f
x
x
x
x
f
x
x
f
x
x
x
f
x
x
x
f
x
f
x
x
f
x
f
x
f
k
k
k
k
k
k












CISE301_Topic5 13
Divided Difference Table
x F[ ] F[ , ] F[ , , ] F[ , , ,]
x0 F[x0] F[x0,x1] F[x0,x1,x2] F[x0,x1,x2,x3]
x1 F[x1] F[x1,x2] F[x1,x2,x3]
x2 F[x2] F[x2,x3]
x3 F[x3]
 
 


 







n
i
i
j
j
i
n x
x
x
x
x
F
x
f
0
1
0
1
0 ]
,...,
,
[
)
(

CISE301_Topic5 14
f(xi)
0 -5
1 -3
-1 -15
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
Entries of the divided difference
table are obtained from the data
table using simple operations.

CISE301_Topic5 15
f(xi)
0 -5
1 -3
-1 -15
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
The first two column of the
table are the data columns.
Third column: First order differences.
Fourth column: Second order differences.

CISE301_Topic5 16
0 -5
1 -3
-1 -15
i
y
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
2
0
1
)
5
(
3





0
1
0
1
1
0
]
[
]
[
]
,
[
x
x
x
f
x
f
x
x
f




CISE301_Topic5 17
0 -5
1 -3
-1 -15
i
y
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
6
1
1
)
3
(
15






1
2
1
2
2
1
]
[
]
[
]
,
[
x
x
x
f
x
f
x
x
f




CISE301_Topic5 18
0 -5
1 -3
-1 -15
i
y
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
4
)
0
(
1
)
2
(
6





0
2
1
0
2
1
2
1
0
]
,
[
]
,
[
]
,
,
[
x
x
x
x
f
x
x
f
x
x
x
f




CISE301_Topic5 19
0 -5
1 -3
-1 -15
i
y
i
x
x F[ ] F[ , ] F[ , , ]
0 -5 2 -4
1 -3 6
-1 -15
)
1
)(
0
(
4
)
0
(
2
5
)
(
2 





 x
x
x
x
f
f2(x)= F[x0]+F[x0,x1] (x-x0)+F[x0,x1,x2] (x-x0)(x-x1)

CISE301_Topic5 20
Two Examples
x y
1 0
2 3
3 8
Obtain the interpolating polynomials for the two examples:
x y
2 3
1 0
3 8
What do you observe?

CISE301_Topic5 21
Two Examples
1
)
2
)(
1
(
1
)
1
(
3
0
)
(
2
2








x
x
x
x
x
P
x Y
1 0 3 1
2 3 5
3 8
x Y
2 3 3 1
1 0 4
3 8
1
)
1
)(
2
(
1
)
2
(
3
3
)
(
2
2








x
x
x
x
x
P
Ordering the points should not affect the interpolating polynomial.

CISE301_Topic5 22
Properties of Divided Difference
]
,
,
[
]
,
,
[
]
,
,
[ 0
1
2
0
2
1
2
1
0 x
x
x
f
x
x
x
f
x
x
x
f 

Ordering the points should not affect the divided difference:

CISE301_Topic5 23
Example
 Find a polynomial to
interpolate the data.
x f(x)
2 3
4 5
5 1
6 6
7 9

CISE301_Topic5 24
Example
x f(x) f[ , ] f[ , , ] f[ , , , ] f[ , , , , ]
2 3 1 -1.6667 1.5417 -0.6750
4 5 -4 4.5 -1.8333
5 1 5 -1
6 6 3
7 9
)
6
)(
5
)(
4
)(
2
(
6750
.
0
)
5
)(
4
)(
2
(
5417
.
1
)
4
)(
2
(
6667
.
1
)
2
(
1
3
4















x
x
x
x
x
x
x
x
x
x
f

CISE301_Topic5 25
Summary
.
polynomial
ing
interpolat
affect the
not
should
points
the
Ordering
methods
Other
-
2]
18.
[Section
ion
Interpolat
Lagrange
-
]
18.1
[Section
Difference
Divided
Newton
-
it
obtain
to
used
be
can
methods
Different
*
unique.
is
Polynomial
ing
interpolat
The
*
...,
,
2
,
1
,
0
)
(
)
(
:
Condition
ing
Interpolat n
i
for
x
f
x
f i
n
i 


CISE301_Topic5 26
Lecture 21
Lagrange Interpolation

CISE301_Topic5 27
The Interpolation Problem
Given a set of n+1 points:
Find an nth
order polynomial:
that passes through all points, such that:
     
)
(
,
....,
,
)
(
,
,
)
(
, 1
1
0
0 n
n x
f
x
x
f
x
x
f
x
)
(x
fn
n
i
for
x
f
x
f i
i
n ,...,
2
,
1
,
0
)
(
)
( 


CISE301_Topic5 28
Problem:
Given
Find the polynomial of least order such that:
Lagrange Interpolation Formula:
n
i
for
x
f
x
f i
i
n ,...,
1
,
0
)
(
)
( 

)
(x
fn
….
….
1
x n
x
0
y 1
y n
y
i
x
i
y
 
 
 









n
i
j
j j
i
j
i
n
i
i
i
n
x
x
x
x
x
x
x
f
x
f
,
0
0
)
(
)
(
)
(


0
x

CISE301_Topic5 29






j
i
j
i
x
x
j
i
th
i
1
0
)
(
:
s
polynomial
order
n
are
cardinals
The
cardinals.
the
called
are
)
(



CISE301_Topic5 30
Lagrange Interpolation Example
x 1/3 1/4 1
y 2 -1 7
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
)
4
/
1
)(
3
/
1
(
2
7
)
1
)(
3
/
1
(
16
1
)
1
)(
4
/
1
(
18
2
)
(
4
/
1
1
4
/
1
3
/
1
1
3
/
1
)
(
1
4
/
1
1
3
/
1
4
/
1
3
/
1
)
(
1
3
/
1
1
4
/
1
3
/
1
4
/
1
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
2
1
2
1
0
2
0
2
2
1
2
0
1
0
1
2
0
2
1
0
1
0
2
2
1
1
0
0
2











































x
x
x
x
x
x
x
P
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
f
x
x
f
x
x
f
x
P







CISE301_Topic5 31
Example
Find a polynomial to interpolate:
Both Newton’s interpolation
method and Lagrange
interpolation method must
give the same answer.
x y
0 1
1 3
2 2
3 5
4 4

CISE301_Topic5 32
Newton’s Interpolation Method
0 1 2 -3/2 7/6 -5/8
1 3 -1 2 -4/3
2 2 3 -2
3 5 -1
4 4

CISE301_Topic5 33
Interpolating Polynomial
4
3
2
4
4
8
5
12
59
8
95
12
115
1
)
(
)
3
)(
2
)(
1
(
8
5
)
2
)(
1
(
6
7
)
1
(
2
3
)
(
2
1
)
(
x
x
x
x
x
f
x
x
x
x
x
x
x
x
x
x
x
f

















CISE301_Topic5 34
Interpolating Polynomial Using
Lagrange Interpolation Method
24
)
3
)(
2
)(
1
(
)
3
4
(
)
3
(
)
2
4
(
)
2
(
)
1
4
(
)
1
(
)
0
4
(
)
0
(
6
)
4
)(
2
)(
1
(
)
4
3
(
)
4
(
)
2
3
(
)
2
(
)
1
3
(
)
1
(
)
0
3
(
)
0
(
4
)
4
)(
3
)(
1
(
)
4
2
(
)
4
(
)
3
2
(
)
3
(
)
1
2
(
)
1
(
)
0
2
(
)
0
(
6
)
4
)(
3
)(
2
(
)
4
1
(
)
4
(
)
3
1
(
)
3
(
)
2
1
(
)
2
(
)
0
1
(
)
0
(
24
)
4
)(
3
)(
2
)(
1
(
)
4
0
(
)
4
(
)
3
0
(
)
3
(
)
2
0
(
)
2
(
)
1
0
(
)
1
(
4
5
2
3
)
(
)
(
4
3
2
1
0
4
3
2
1
0
4
0
4












































































x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
f
x
f
i
i
i












CISE301_Topic5 35
Lecture 22
Inverse Interpolation
Error in Polynomial Interpolation

CISE301_Topic5 36
given
is
where
,
)
(
:
that
such
Find
values
of
a table
Given
:
Problem
k
k y
y
x
f
x 
….
….
1
x n
x
0
y 1
y n
y
i
x
i
y
One approach:
Use polynomial interpolation to obtain fn(x) to interpolate the
data then use Newton’s method to find a solution to x
k
n y
x
f 
)
(
0
x

CISE301_Topic5 37
….
….
1
x n
x
0
y 1
y n
y
i
x
i
y
Inverse interpolation:
1. Exchange the roles
of x and y.
2. Perform polynomial
Interpolation on the
new table.
3. Evaluate
)
( k
n y
f
x 
….
….
i
y 0
y 1
y n
y
i
x 0
x 1
x n
x
0
x

CISE301_Topic5 38
x
y
y
x

CISE301_Topic5 39
Question:
What is the limitation of inverse interpolation?
• The original function has an inverse.
• y1, y2, …, yn must be distinct.

CISE301_Topic5 40
Example
5
.
2
)
(
that
such
Find
table.
the
Given
:
Problem

x
f
x
x 1 2 3
y 3.2 2.0 1.6
3.2 1 -.8333 1.0417
2.0 2 -2.5
1.6 3
2187
.
1
)
5
.
0
)(
7
.
0
(
0417
.
1
)
7
.
0
(
8333
.
0
1
)
5
.
2
(
)
2
)(
2
.
3
(
0417
.
1
)
2
.
3
(
8333
.
0
1
)
(
2
2














f
x
y
y
y
y
f
x

CISE301_Topic5 41
Errors in polynomial Interpolation
 Polynomial interpolation may lead to large
errors (especially for high order polynomials).
BE CAREFUL
 When an nth
order interpolating polynomial is
used, the error is related to the (n+1)th
order
derivative.

CISE301_Topic5 42
-5 -4 -3 -2 -1 0 1 2 3 4 5
-0.5
0
0.5
1
1.5
2
true function
10 th order interpolating polynomial
10th
Order Polynomial Interpolation

CISE301_Topic5 43
1
)
1
(
)
1
(
)
1
(
4
)
(
)
(
:
Then
points).
end
the
(including
b]
[a,
in
points
spaced
equally
1
at
es
interpolat
that
n
degree
of
polynomial
any
be
Let
.
)
(
and
b],
[a,
on
continuous
is
)
(
:
that
such
function
a
be
)
(
Let








 





n
n
n
n
a
b
n
M
x
-P
x
f
n
f
P(x)
M
x
f
x
f
x
f
Errors in polynomial Interpolation
Theorem

CISE301_Topic5 44
Example
9
10
1
)
1
(
th
10
34
.
1
9
6875
.
1
)
10
(
4
1
)
1
(
4
9
,
1
0
1
.
[0,1.6875]
interval
in the
points)
spaced
equally
0
1
(using
f(x)
e
interpolat
to
polynomial
order
9
use
want to
We
sin

















 







f(x)-P(x)
n
a
b
n
M
f(x)-P(x)
n
M
n
for
f
(x)
f(x)
n
n

CISE301_Topic5 45
Summary
 The interpolating polynomial is unique.
 Different methods can be used to obtain it.
 Newton’s divided difference
 Lagrange interpolation
 Others
 Polynomial interpolation can be sensitive to
data.
 BE CAREFUL when high order polynomials
are used.

NUMERICAL ANALYSIS CH4CISE301-Topic5.ppt

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