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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 1 120.00 Curve 2 Curve 3 100.00 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 0.016 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 0.025 0.009 0.002 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 å1 and å2 were able to be determined to solve the following equation: CurveX = ε1 *V 1 + ε 2 *V 2 å1 Curve 1 Curve 2 Curve 3 Curve 4 Curve 5 125.9448 -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 å2 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 å1 and å2 will be plotted in å1 and å2 space 40 5 30 20 10 0 0 50 100 150 200 -10 2 3 250 4 -20 -30 1 -40 The two inside lines were formed by connecting the origin to the #1 and #5 (å1, å2) points. The two outside lines were formed by creating a column of V1 k /V2 k . 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: ε 2 = − min ε 2 = min where min V 1k * ε1 V 2k V 1k * ε1 V 2k V 1k = 0.46 V 2k Next, we had to solve the equation: c1 * ε1 + c 2 * ε 2 = 1 where c i = ∑Vik * ∆k This created a line. Now the above graph looks like: c1 = 3.869 c 2 = 0.480 0.25 0.2 f1* 0.15 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 -0.15 -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 (å1, å2) points that we call f1 *, f1 **, f2 *, and f2 **. These intersection points are: f1 * f1 ** f2 * f2 ** å1 å2 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 å 1 and å 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