09 numerical differentiation
DESCRIPTION
TRANSCRIPT
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Interpolation/Curve Fitting
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Measured Data
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Polynomial Fit!
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Line Fit!
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Which is better?
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
Dr. Eng. Mohammad Tawfik
Interpolation
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
Dr. Eng. Mohammad Tawfik
Acquired Data
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
But, how to get the equation of a function that passes by all the
data you have!
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Equation of a Line: Revision
xaay 21 If you have two points
1211 xaay
2212 xaay
2
1
2
1
2
1
1
1
y
y
a
a
x
x
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solving for the constants!
12
122
12
21121 &
xx
yya
xx
yxyxa
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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.
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Third-Order Polynomial
34
2321 xaxaxaay
For the four points
314
2131211 xaxaxaay
324
2232212 xaxaxaay
334
2333213 xaxaxaay
344
2434214 xaxaxaay
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
In Matrix Form
4
3
2
1
4
3
2
1
34
224
33
223
32
222
31
211
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.
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Newton's Interpolation Polynomial
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
Dr. Eng. Mohammad Tawfik
Equation of a Line: Revision
xaay 21 If you have two points
1211 xaay
2212 xaay
2
1
2
1
2
1
1
1
y
y
a
a
x
x
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
For the two points
12
12
1
1
xx
yy
xx
yy
12
12
1
1
xx
yy
xx
yxf
112
121 xx
xx
yyyxf
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
For the three points
213
121
xxxxa
xxaaxf
11 ya
12
122 xx
yya
13
12
12
23
23
3 xx
xxyy
xxyy
a
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Using a table
xiyi
x1y1
x2y2
x3y3
13
12
12
23
23
xx
xxyy
xx
yy
12
12
xx
yy
23
23
xx
yy
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
In General
• Newton’s Interpolation is performed for an nth order polynomial as follows
nn xxxxa
xxxxaxxaaxf
...... 11
213121
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Find a 3rd order polynomial to interpolate the function described by the given points
xY
-11
02
15
216
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solution: System of equations
• A third order polynomial is given by:
34
2321 xaxaxaaxf
11 4321 aaaaf
20 1 af
51 4321 aaaaf
168422 4321 aaaaf
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
322 xxxxf
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Newton’s Method
• Newton’s methods defines the polynomial in the form:
3214
213121
xxxxxxa
xxxxaxxaaxf
11
11
4
321
xxxa
xxaxaaxf
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Newton’s Method
xY
-11111
0234
1511
216
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Newton’s Method
• Finally:
11
111
xxx
xxxxf
xxxxxxf 3211
322 xxxxf
ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
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
Dr. Eng. Mohammad Tawfik
Homework #6
• Chapter 18, pp. 505-506, numbers:18.1, 18.2, 18.3, 18.5.