Project 13
An Examination of the Self-Modeling Curve Resolution Paper we been working with indicates that the
authors have worked out a method and provided an example.
Work out that example using a program in any language you wish. Clearly demonstrate that your program
will generate fundamental vectors and that your fundamentals match the papers.
Solution
First of all, the paper gave five curves to model
Curve 1
0.92
4.41
5.49
6.53
4.98
4.90
3.88
3.60
3.50
4.88
9.99
16.74
27.34
40.15
52.74
54.80
51.26
46.78
39.83
30.27
22.46
15.80
11.35
7.95
4.76
2.81
2.07
1.59
0.96
0.67
Here are the five curves plotted
Curve 2
2.48
8.01
9.01
11.90
9.42
8.42
6.43
6.16
4.78
5.43
9.95
16.60
27.72
44.04
62.57
73.00
80.16
87.51
74.96
64.95
51.50
34.88
25.73
17.90
11.27
7.32
4.49
2.81
1.44
0.47
Curve 3
1.24
6.85
10.11
11.59
10.31
8.24
6.85
6.26
5.27
5.72
9.68
17.76
28.52
42.57
58.09
66.86
72.41
73.45
68.35
57.61
45.97
34.44
25.08
17.70
11.68
7.14
4.54
2.66
1.32
0.08
Curve 4
0.41
5.08
6.39
9.01
7.48
6.45
4.84
4.49
3.80
4.92
8.95
15.82
26.65
42.02
55.16
62.94
66.70
66.94
60.62
52.68
39.97
30.15
21.30
15.09
9.80
5.84
3.70
2.42
1.26
0.14
Curve 5
2.77
11.92
18.39
20.97
16.68
13.91
10.88
10.03
8.44
7.53
9.07
12.85
20.08
33.32
48.75
70.26
88.63
98.35
97.80
88.24
72.80
55.35
41.19
30.14
20.10
13.63
8.07
5.13
2.83
0.55
Curve 1
1 20. 00
Curve 2
Curve 3
1 00. 00
Curve 4
Curve 5
80. 00
60. 00
40. 00
20. 00
0. 00
0
5
10
15
20
25
30
35
As shown in a previous Project, two representative eigenvectors can be found for the 30 x 5 matrix, V1 and
V2
V1
0.009
0.041
0.056
0.068
0.055
0.047
0.037
0.034
0.029
0.031
0.050
0.082
0.134
0.210
0.290
0.349
0.389
0.402
0.377
0.327
0.261
0.191
0.140
0.100
0.065
0.042
0.026
V2
0.015
0.048
0.103
0.103
0.080
0.058
0.045
0.041
0.030
-0.004
-0.078
-0.169
-0.291
-0.400
-0.477
-0.302
-0.074
0.094
0.218
0.284
0.285
0.254
0.192
0.150
0.109
0.085
0.044
0.016
0.009
0.002
0.025
0.014
-0.002
All five of the curves can be represented by the two representative eigenvectors, V1 and
V2.
Using a non-linear regression program, POLYMATH 5.0, the two functions ε1 and ε2 were able to be
determined to solve the following equation:
CurveX  1*V1   2 *V 2
Curve 1
Curve 2
Curve 3
Curve 4
Curve 5
ε1
ε2
125.9448
-34.325112
204.47594
185.79134
169.36686
235.93268
-6.7631867
-8.5203729
-13.552098
40.768568
The following is the exact POLYMATH results:
POLYMATH 5.0 Results
No title 12-05-2001
Nonlinear regression (mrqmin)
Model: Y1 = E1*V1+E2*V2
Variable
E1
E2
Ini guess
0.5
0.5
Value
125.9448
-34.325112
Conf-inter
0.158128
0.1576398
Value
204.47594
-6.7631867
Conf-inter
0.3180027
0.3170207
Nonlinear regression settings
Max # iterations = 64
Tolerance = 0.0001
Precision
R^2
R^2adj
Rmsd
Variance
Chi-Sq
Alamda
=
=
=
=
=
=
0.9981953
0.9981308
0.1361498
0.5958247
1668.309
1.0E-05
POLYMATH 5.0 Results
No title 12-05-2001
Nonlinear regression (mrqmin)
Model: Y2 = E1*V1+E2*V2
Variable
E1
E2
Ini guess
0.5
0.5
Nonlinear regression settings
Max # iterations = 64
Tolerance = 0.0001
Precision
R^2
R^2adj
Rmsd
Variance
Chi-Sq
Alamda
=
=
=
=
=
=
0.996942
0.9968327
0.2738034
2.4096958
6747.1484
1.0E-07
POLYMATH 5.0 Results
No title 12-05-2001
Nonlinear regression (mrqmin)
Model: Y3 = E1*V1+E2*V2
Variable
E1
E2
Ini guess
0.5
0.5
Value
185.79134
-8.5203729
Conf-inter
0.1073739
0.1070424
Value
169.36686
-13.552098
Conf-inter
0.1471911
0.1467366
Nonlinear regression settings
Max # iterations = 64
Tolerance = 0.0001
Precision
R^2
R^2adj
Rmsd
Variance
Chi-Sq
Alamda
=
=
=
=
=
=
0.9995572
0.9995414
0.09245
0.274725
769.22998
1.0E-05
POLYMATH 5.0 Results
No title 12-05-2001
Nonlinear regression (mrqmin)
Model: Y4 = E1*V1+E2*V2
Variable
E1
E2
Ini guess
0.5
0.5
Nonlinear regression settings
Max # iterations = 64
Tolerance = 0.0001
Precision
R^2
R^2adj
Rmsd
Variance
Chi-Sq
Alamda
=
=
=
=
=
=
0.9990465
0.9990125
0.126733
0.5162545
1445.5127
1.0E-05
POLYMATH 5.0 Results
No title 12-05-2001
Nonlinear regression (mrqmin)
Model: Y5 = E1*V1+E2*V2
Variable
E1
E2
Ini guess
0.5
0.5
Value
235.93268
40.768568
Conf-inter
0.079529
0.0792834
Nonlinear regression settings
Max # iterations = 64
Tolerance = 0.0001
Precision
R^2
R^2adj
Rmsd
Variance
Chi-Sq
Alamda
=
=
=
=
=
=
0.9998492
0.9998438
0.0684752
0.1507132
421.99689
1.0E-07
Graphical Representation of the POLYMATH results are:
Next, the five values of ε1 and ε2 will be plotted in ε1 and ε2 space
40
5
30
20
10
0
0
50
1 00
1 50
200
250
4
-1 0
2
3
-20
-30
1
-40
The two inside lines were formed by connecting the origin to the #1 and #5 (ε1, ε2)
points.
The two outside lines were formed by creating a column of V1k/V2k. Where k is the
elements of the V1 and V2 vectors. Then, the minimum value was found.
The two outer lines were determined to be:
 2   min
 2  min
where min
V 1k
* 1
V 2k
V 1k
* 1
V 2k
V 1k
 0.46
V 2k
Next, we had to solve the equation:
c1 * 1  c2 *  2  1
where ci  Vik * k
c1  3.869
c 2  0.480
This created a line.
Now the above graph looks like:
0. 25
0. 2
f1*
0. 1 5
0. 1
0. 05
f1**
0
-0. 1
0. 1
0. 3
0. 5
0. 7
0. 9
1 .1
1 .3
1 .5
f2**
-0. 05
-0. 1
-0. 1 5
-0. 2
f2*
-0. 25
From this line, one can find f1*, f1**, f2*, and f2** which is the intersection points of
the lines.
This gives us four (ε1, ε2) points that we call f1*, f1**, f2*, and f2**.
These intersection points are:
f 1*
f1**
f 2*
f2**
ε1
ε2
0.244484
0.253098
0.267051
0.274079
0.112462
0.043027
-0.06943
-0.12608
It was shown that the true solution of the problem is between these two extremes.
So plotting these ε 1 and ε 2 values gives the following graphs.
0.160
0.140
f1*
0.120
f2**
0.100
f2*
0.080
0.060
f1**
0.040
0.020
0.000
0
5
10
15
20
25
30
However, one last assumption given in equation 30, 31, and 32 eliminates the f1** and
f2** lines to give one solution to the problem
0.160
0.140
f1*
0.120
0.100
f2*
0.080
0.060
0.040
0.020
0.000
0
5
10
15
20
25
30