07 interpolationnewton

1
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Interpolation/Curve Fitting
ENEM602 Spring 2007
Objectives
• Understanding the difference between
regression and interpolation
• Knowing how to “best fit” a polynomial into
a set of data
• Knowing how to use a polynomial to
interpolate data

2
ENEM602 Spring 2007
Measured Data
ENEM602 Spring 2007
Polynomial Fit!

3
ENEM602 Spring 2007
Line Fit!
ENEM602 Spring 2007
Which is better?

4
ENEM602 Spring 2007
Curve Fitting
• If the data measured is of high accuracy
and it is required to estimate the values of
the function between the given points,
then, polynomial interpolation is the best
choice.
• If the measurements are expected to be of
low accuracy, or the number of measured
points is too large, regression would be
the best choice.
ENEM602 Spring 2007
Interpolation

5
ENEM602 Spring 2007
Why Interpolation?
• When the accuracy of your measurements
are ensured
• When you have discrete values for a
function (numerical solutions, digital
systems, etc …)
ENEM602 Spring 2007
Acquired Data

6
ENEM602 Spring 2007
But, how to get the equation of a
function that passes by all the
data you have!
ENEM602 Spring 2007
Equation of a Line: Revision
xaay 10 +=
If you have two points
1101 xaay +=
2102 xaay += 





=












2
1
1
0
2
1
1
1
y
y
a
a
x
x

7
ENEM602 Spring 2007
Solving for the constants!
12
12
1
12
2112
0 &
xx
yy
a
xx
yxyx
a
−
−
=
−
−
=
ENEM602 Spring 2007
What if I have more than two
points?
• We may fit a
polynomial of order
one less that the
number of points we
have. i.e. four points
give third order
polynomial.

8
ENEM602 Spring 2007
Third-Order Polynomial
3
3
2
210 xaxaxaay +++=
For the four points
3
13
2
121101 xaxaxaay +++=
3
23
2
3
33
2
3
43
2
ENEM602 Spring 2007
In Matrix Form














=




























4
3
2
1
3
2
1
0
3
4
2
24
3
3
2
23
3
2
2
22
3
1
2
11
1
1
1
1
y
y
y
y
a
a
a
a
xxx
xxx
xxx
xxx
Solve the above equation for the constants of the polynomial.

9
ENEM602 Spring 2007
Example
• Find a 3rd order
polynomial to
interpolate the
function described by
the given points
162
51
20
1-1
Yx
ENEM602 Spring 2007
Solution: System of equations
• A third order polynomial is given by:
( ) 3
4
2
321 xaxaxaaxf +++=
( ) 11 4321 =−+−=− aaaaf
( ) 20 1 == af
( ) 51 4321 =+++= aaaaf
( ) 168422 4321 =+++= aaaaf

10
ENEM602 Spring 2007
In matrix form














=

























 −−
16
5
2
1
8421
1111
0001
1111
4
3
2
1
a
a
a
a














=














1
1
1
2
4
3
2
1
a
a
a
a
( ) 32
2 xxxxf +++=
ENEM602 Spring 2007
Newton's Interpolation
Polynomial

11
ENEM602 Spring 2007
Newton’s Method
• In the previous procedure, we needed to solve a
system of linear equations for the unknown
constants.
• This method suggests that we may just proceed
with the values of x & y we have to get the
constants without setting a set of equations
• The method is similar to Taylor’s expansion
without differentiation!
ENEM602 Spring 2007
For the two points
( )110 xxbby −+=
( ) 0111101 byxxbby =⇒−+=
( )1
12
12
1 xx
xx
yy
yy −





−
−
+=
( )⇒−+= 12102 xxbby
( ) ( )
( )12
12
112112
xx
yy
bxxbyy
−
−
=⇒−+=

12
ENEM602 Spring 2007
Homework
• Show that the polynomial obtained by
solving a set of equations is equivalent to
that obtained by Newton method
ENEM602 Spring 2007
For the three points
( ) ( )
( )( )213
121
xxxxb
xxbbxf
−−+
−+=
10 yb =
12
12
1
xx
yy
b
−
−
=
13
12
12
23
23
2
xx
xx
yy
xx
yy
b
−
−
−
−
−
−
=

13
ENEM602 Spring 2007
Using a table
y3x3
y2x2
y1x1
yixi
13
12
12
23
23
xx
xx
yy
xx
yy
−
−
−
−
−
−
12
12
xx
yy
−
−
23
23
xx
yy
−
−
ENEM602 Spring 2007
In General
• Newton’s Interpolation is performed for an
nth order polynomial as follows
( ) ( ) ( )( )
( ) ( )nn xxxxb
xxxxbxxbbxf
−−++
−−+−+=
...... 1
212110

14
ENEM602 Spring 2007
Example
• Find a 3rd order
polynomial to
interpolate the
function described by
the given points using
Newton’s method
162
51
20
1-1
Yx
ENEM602 Spring 2007
Newton’s Method
• Newton’s methods defines the polynomial in the
form:
( ) ( ) ( )( )
( )( )( )3213
212110
xxxxxxb
xxxxbxxbbxf
−−−+
−−+−+=
( ) ( ) ( )( )
( )( )( )11
11
3
210
−++
++++=
xxxb
xxbxbbxf

15
ENEM602 Spring 2007
Newton’s Method
162
1151
4320
1111-1
Yx
ENEM602 Spring 2007
Newton’s Method
• Finally:
( ) ( ) ( )( )
( )( )( )11
111
−++
++++=
xxx
xxxxf
( ) ( ) ( ) ( )xxxxxxf −+++++= 32
11
( ) 32
2 xxxxf +++=

16
ENEM602 Spring 2007
Advantage of Newton’s Method
• The main advantage of Newton’s method
is that you do not need to invert a matrix!
ENEM602 Spring 2007
Homework #6
• Chapter 18, pp. 505-506, numbers:
18.1, 18.2, 18.3, 18.5.

07 interpolationnewton

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