NEWNON 06 December 1992 An interactive program for solving systems of nonlinear equations. Command? (H for help) Enter name of variable to set (h for help) Enter limit for number of iterations. Number of iterations set to 20 Enter name of variable to set (h for help) 0 = print out only result of iteration. 1 = print norms of x, f(x) and delta x. 2 = print x, f(x) and delta x Printout level set to 2 Enter name of variable to set (h for help) Command? (H for help) # Choose problem 1 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 2.00000 1.96 -2.00 0.301 2.00000 0.982 -2.00 -4.47 new -0.615 -2.90 1 -2.46741 -0.328 2.47 0.392 -0.900304 1.52 0.900 1.83 new 0.514 0.776 2 -0.632923 -0.811E-02 0.633 -0.199 -0.124740 0.783E-01 0.125 0.465 new 0.111E-01 0.337E-01 3 -0.168388 -0.138E-02 0.168 -0.774 -0.910302E-01 0.153E-01 0.910E-01 0.164 new 0.142E-02 0.248E-02 4 -0.454778E-02 -0.357E-04 0.455E-02 -1.05 -0.885512E-01 0.401E-03 0.886E-01 0.455E-02 new 0.358E-04 0.182E-05 5 -0.925947E-07 -0.724E-09 0.926E-07 -1.05 -0.885494E-01 0.817E-08 0.885E-01 0.931E-07 new 0.723E-09 -0.518E-03 6 0.537646E-09 0.000E+00 -0.538E-09 -1.05 -0.890676E-01 -0.477E-10 0.891E-01 0.728E-25 new 0.843E-11 0.898E-01 7 0.537646E-09 0.000E+00 -0.538E-09 -3.15 0.711277E-03 0.382E-12 -0.711E-03 Bad step!. A norm increased. -0.710E-25 new 0.544E-15 -0.711E-03 8 0.537646E-09 0.000E+00 -0.538E-09 -9.27 -0.359782E-09 0.000E+00 0.360E-09 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 9 Jacobian evaluations: 8 Matrix factorizations; 8 Linear system solves: 8 Starting point: x f(x) 2.00000 1.96337 2.00000 0.981684 Estimate of the root after 8 steps. x f(x) 0.537646E-09 0.000000E+00 -0.359782E-09 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 3.00 -3.00 0.477 3.00000 1.000 -3.00 -8.95 new -0.335 -5.97 1 -5.95360 -1.48 5.95 0.775 -2.96685 2.01 2.97 17.8 new 0.168 5.92 2 11.8443 0.735 -11.8 1.07 2.95118 4.01 -2.95 Bad step!. A norm increased. -35.4 new -0.847E-01 -5.89 3 -23.5528 -0.366 23.6 1.37 -2.93426 8.03 2.93 Bad step!. A norm increased. 70.4 new 0.426E-01 5.85 4 46.8107 0.182 -46.8 1.67 2.91589 16.1 -2.92 Bad step!. A norm increased. -140. new -0.214E-01 -5.81 5 -92.9772 -0.902E-01 93.0 1.97 -2.89583 32.1 2.90 Bad step!. A norm increased. 278. new 0.108E-01 5.77 6 184.537 0.448E-01 -185. 2.27 2.87375 64.2 -2.87 Bad step!. A norm increased. -550. new -0.544E-02 -5.72 7 -365.928 -0.222E-01 366. 2.56 -2.84926 128. 2.85 Bad step!. A norm increased. 0.109E+04 new 0.275E-02 5.67 8 724.805 0.110E-01 -725. 2.86 2.82181 257. -2.82 Bad step!. A norm increased. -0.216E+04 new -0.139E-02 -5.61 9 -1433.61 -0.543E-02 0.143E+04 3.16 -2.79067 514. 2.79 Bad step!. A norm increased. 0.426E+04 new 0.702E-03 5.55 10 2830.41 0.268E-02 -0.283E+04 3.45 2.75483 0.103E+04 -2.75 Bad step!. A norm increased. -0.840E+04 new -0.356E-03 -5.47 11 -5574.49 -0.132E-02 0.557E+04 3.75 -2.71283 0.205E+04 2.71 Bad step!. A norm increased. 0.165E+05 new 0.181E-03 5.38 12 10941.9 0.648E-03 -0.109E+05 4.04 2.66243 0.411E+04 -2.66 Bad step!. A norm increased. -0.323E+05 new -0.924E-04 -5.26 13 -21371.5 -0.316E-03 0.214E+05 4.33 -2.60011 0.821E+04 2.60 Bad step!. A norm increased. 0.628E+05 new 0.475E-04 5.12 14 41423.0 0.153E-03 -0.414E+05 4.62 2.51981 0.164E+05 -2.52 Bad step!. A norm increased. -0.121E+06 new -0.246E-04 -4.93 15 -79242.3 -0.733E-04 0.792E+05 4.90 -2.41020 0.328E+05 2.41 Bad step!. A norm increased. 0.227E+06 new 0.130E-04 4.66 16 147777. 0.342E-04 -0.148E+06 5.17 2.24737 0.653E+05 -2.25 Bad step!. A norm increased. -0.407E+06 new -0.716E-05 -4.22 17 -259417. -0.150E-04 0.259E+06 5.41 -1.97258 0.129E+06 1.97 Bad step!. A norm increased. 0.633E+06 new 0.439E-05 3.39 18 373973. 0.541E-05 -0.374E+06 5.57 1.42183 0.228E+06 -1.42 Bad step!. A norm increased. -0.551E+06 new -0.375E-05 -1.76 19 -177015. -0.640E-06 0.177E+06 5.25 -0.336502 0.563E+05 0.337 0.681E+05 new 0.175E-05 0.233 20 -108934. -0.984E-07 0.109E+06 5.04 -0.103540 0.112E+05 0.104 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 20 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 2.99963 3.00000 0.999877 Estimate of the root after 20 steps. x f(x) -108934. -0.984137E-07 -0.103540 11218.8 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 100.000 0.100E-01 -100. 2.00 1.00000 63.2 -1.00 -84.8 new -0.137E-01 -0.924 1 15.1531 0.379E-03 -15.2 1.18 0.757657E-01 1.14 -0.758E-01 -5.09 new -0.113E-02 -0.506E-01 2 10.0634 0.629E-04 -10.1 1.00 0.251585E-01 0.253 -0.252E-01 -3.36 new -0.189E-03 -0.168E-01 3 6.70610 0.105E-04 -6.71 0.826 0.838262E-02 0.562E-01 -0.838E-02 -2.24 new -0.314E-04 -0.559E-02 4 4.47052 0.175E-05 -4.47 0.650 0.279408E-02 0.125E-01 -0.279E-02 -1.49 new -0.524E-05 -0.186E-02 5 2.98033 0.291E-06 -2.98 0.474 0.931354E-03 0.278E-02 -0.931E-03 -0.994 new -0.872E-06 -0.621E-03 6 1.98607 0.477E-07 -1.99 0.298 0.310707E-03 0.617E-03 -0.311E-03 -0.698 new -0.136E-06 -0.201E-03 7 1.28779 0.750E-08 -1.29 0.110 0.109241E-03 0.141E-03 -0.109E-03 -0.580 new -0.167E-07 -0.600E-04 8 0.707640 0.135E-08 -0.708 -0.150 0.492130E-04 0.348E-04 -0.492E-04 -0.485 new -0.197E-08 -0.155E-04 9 0.222668 0.247E-09 -0.223 -0.652 0.337275E-04 0.751E-05 -0.337E-04 -0.212 new -0.259E-09 -0.158E-05 10 0.104357E-01 0.108E-10 -0.104E-01 -1.98 0.321468E-04 0.335E-06 -0.321E-04 -0.104E-01 new -0.108E-10 -0.350E-08 11 0.113635E-05 0.117E-14 -0.114E-05 -4.49 0.321433E-04 0.365E-10 -0.321E-04 -0.114E-05 new -0.117E-14 -0.201E-09 12 0.711585E-11 0.000E+00 -0.712E-11 -4.49 0.321431E-04 0.229E-15 -0.321E-04 0.000E+00 new 0.147E-19 -0.321E-04 13 0.711585E-11 0.000E+00 -0.712E-11 -11.1 -0.723022E-12 0.000E+00 0.723E-12 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 14 Jacobian evaluations: 13 Matrix factorizations; 13 Linear system solves: 13 Starting point: x f(x) 100.000 0.100000E-01 1.00000 63.2121 Estimate of the root after 13 steps. x f(x) 0.711585E-11 0.000000E+00 -0.723022E-12 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1.00000 0.632E+04 -1.00 2.00 100.000 0.100E-01 -100. -1.39 new -1.37 -38.6 1 -0.386351 -0.135E+04 0.386 1.79 61.3649 -0.630E-02 -61.4 0.406 new 1.02 -3.13 2 0.197010E-01 66.8 -0.197E-01 1.77 58.2357 0.338E-03 -58.2 -0.197E-01 new -0.591E-01 -0.753E-02 3 -0.254903E-05 -0.864E-02 0.255E-05 1.77 58.2282 -0.438E-07 -58.2 0.255E-05 new 0.765E-05 -0.249E-03 4 0.109105E-10 0.000E+00 -0.109E-10 1.77 58.2279 0.187E-12 -58.2 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 5 Jacobian evaluations: 4 Matrix factorizations; 4 Linear system solves: 4 Starting point: x f(x) 1.00000 6321.21 100.000 0.100000E-01 Estimate of the root after 4 steps. x f(x) 0.109105E-10 0.000000E+00 58.2279 0.187376E-12 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682670 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233808E-01 0.877E-02 -0.234E-01 -0.213 0.612668 0.119E-01 -0.613 -0.234E-01 new -0.989E-02 -0.248E-03 2 0.612477E-05 0.230E-05 -0.612E-05 -0.213 0.612420 0.313E-05 -0.612 -0.612E-05 new -0.259E-05 -0.105E-05 3 0.683652E-11 0.000E+00 -0.684E-11 -0.213 0.612419 0.349E-11 -0.612 0.000E+00 new 0.277E-11 -0.945 4 0.683652E-11 0.000E+00 -0.684E-11 -0.479 -0.332187 -0.215E-11 0.332 Bad step!. A norm increased. 0.000E+00 new 0.127E-11 0.373 5 0.683652E-11 0.000E+00 -0.684E-11 -1.39 0.403531E-01 0.276E-12 -0.404E-01 0.000E+00 new 0.222E-13 -0.404E-01 6 0.683652E-11 0.000E+00 -0.684E-11 -4.18 -0.657993E-04 -0.450E-15 0.658E-04 0.000E+00 new 0.592E-19 0.658E-04 7 0.683652E-11 0.000E+00 -0.684E-11 -11.2 0.649566E-12 0.000E+00 -0.650E-12 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 8 Jacobian evaluations: 7 Matrix factorizations; 7 Linear system solves: 7 Starting point: x f(x) 0.400000 0.172267 0.682670 0.218270 Estimate of the root after 7 steps. x f(x) 0.683652E-11 0.000000E+00 0.649566E-12 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682680 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233802E-01 0.877E-02 -0.234E-01 -0.213 0.612677 0.119E-01 -0.613 -0.234E-01 new -0.989E-02 -0.248E-03 2 0.612414E-05 0.230E-05 -0.612E-05 -0.213 0.612429 0.313E-05 -0.612 -0.612E-05 new -0.259E-05 -0.128E-06 3 0.830945E-12 0.000E+00 -0.831E-12 -0.213 0.612429 0.424E-12 -0.612 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 4 Jacobian evaluations: 3 Matrix factorizations; 3 Linear system solves: 3 Starting point: x f(x) 0.400000 0.172272 0.682680 0.218272 Estimate of the root after 3 steps. x f(x) 0.830945E-12 0.000000E+00 0.612429 0.424351E-12 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to problem 2 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 54.0 -3.00 0.477 3.00000 54.0 -3.00 -1.00 new 972. -1.00 1 2.00000 16.0 -2.00 0.301 2.00000 16.0 -2.00 -0.667 new 192. -0.667 2 1.33333 4.74 -1.33 0.125 1.33333 4.74 -1.33 -0.444 new 37.9 -0.444 3 0.888889 1.40 -0.889 -0.512E-01 0.888889 1.40 -0.889 -0.296 new 7.49 -0.296 4 0.592593 0.416 -0.593 -0.227 0.592593 0.416 -0.593 -0.198 new 1.48 -0.198 5 0.395062 0.123 -0.395 -0.403 0.395062 0.123 -0.395 -0.132 new 0.292 -0.132 6 0.263374 0.365E-01 -0.263 -0.579 0.263374 0.365E-01 -0.263 -0.878E-01 new 0.577E-01 -0.878E-01 7 0.175583 0.108E-01 -0.176 -0.756 0.175583 0.108E-01 -0.176 -0.585E-01 new 0.114E-01 -0.585E-01 8 0.117055 0.321E-02 -0.117 -0.932 0.117055 0.321E-02 -0.117 -0.390E-01 new 0.225E-02 -0.390E-01 9 0.780369E-01 0.950E-03 -0.780E-01 -1.11 0.780369E-01 0.950E-03 -0.780E-01 -0.260E-01 new 0.445E-03 -0.260E-01 10 0.520246E-01 0.282E-03 -0.520E-01 -1.28 0.520246E-01 0.282E-03 -0.520E-01 -0.173E-01 new 0.879E-04 -0.173E-01 11 0.346831E-01 0.834E-04 -0.347E-01 -1.46 0.346831E-01 0.834E-04 -0.347E-01 -0.116E-01 new 0.174E-04 -0.116E-01 12 0.231220E-01 0.247E-04 -0.231E-01 -1.64 0.231220E-01 0.247E-04 -0.231E-01 -0.771E-02 new 0.343E-05 -0.771E-02 13 0.154147E-01 0.733E-05 -0.154E-01 -1.81 0.154147E-01 0.733E-05 -0.154E-01 -0.514E-02 new 0.678E-06 -0.514E-02 14 0.102765E-01 0.217E-05 -0.103E-01 -1.99 0.102765E-01 0.217E-05 -0.103E-01 -0.343E-02 new 0.134E-06 -0.343E-02 15 0.685097E-02 0.643E-06 -0.685E-02 -2.16 0.685097E-02 0.643E-06 -0.685E-02 -0.228E-02 new 0.264E-07 -0.228E-02 16 0.456732E-02 0.191E-06 -0.457E-02 -2.34 0.456732E-02 0.191E-06 -0.457E-02 -0.152E-02 new 0.522E-08 -0.152E-02 17 0.304488E-02 0.565E-07 -0.304E-02 -2.52 0.304488E-02 0.565E-07 -0.304E-02 -0.101E-02 new 0.103E-08 -0.101E-02 18 0.202992E-02 0.167E-07 -0.203E-02 -2.69 0.202992E-02 0.167E-07 -0.203E-02 -0.677E-03 new 0.204E-09 -0.677E-03 19 0.135328E-02 0.496E-08 -0.135E-02 -2.87 0.135328E-02 0.496E-08 -0.135E-02 -0.451E-03 new 0.402E-10 -0.451E-03 20 0.902186E-03 0.147E-08 -0.902E-03 -3.04 0.902186E-03 0.147E-08 -0.902E-03 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 20 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 54.0000 3.00000 54.0000 Estimate of the root after 20 steps. x f(x) 0.902186E-03 0.146865E-08 0.902186E-03 0.146865E-08 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1000.00 0.200E+10 -0.100E+04 3.00 1000.00 0.200E+10 -0.100E+04 -333. new 0.120E+14 -333. 1 666.667 0.593E+09 -667. 2.82 666.667 0.593E+09 -667. -222. new 0.237E+13 -222. 2 444.444 0.176E+09 -444. 2.65 444.444 0.176E+09 -444. -148. new 0.468E+12 -148. 3 296.296 0.520E+08 -296. 2.47 296.296 0.520E+08 -296. -98.8 new 0.925E+11 -98.8 4 197.531 0.154E+08 -198. 2.30 197.531 0.154E+08 -198. -65.8 new 0.183E+11 -65.8 5 131.687 0.457E+07 -132. 2.12 131.687 0.457E+07 -132. -43.9 new 0.361E+10 -43.9 6 87.7915 0.135E+07 -87.8 1.94 87.7915 0.135E+07 -87.8 -29.3 new 0.713E+09 -29.3 7 58.5277 0.401E+06 -58.5 1.77 58.5277 0.401E+06 -58.5 -19.5 new 0.141E+09 -19.5 8 39.0184 0.119E+06 -39.0 1.59 39.0184 0.119E+06 -39.0 -13.0 new 0.278E+08 -13.0 9 26.0123 0.352E+05 -26.0 1.42 26.0123 0.352E+05 -26.0 -8.67 new 0.549E+07 -8.67 10 17.3415 0.104E+05 -17.3 1.24 17.3415 0.104E+05 -17.3 -5.78 new 0.109E+07 -5.78 11 11.5610 0.309E+04 -11.6 1.06 11.5610 0.309E+04 -11.6 -3.85 new 0.214E+06 -3.85 12 7.70735 916. -7.71 0.887 7.70735 916. -7.71 -2.57 new 0.423E+05 -2.57 13 5.13823 271. -5.14 0.711 5.13823 271. -5.14 -1.71 new 0.836E+04 -1.71 14 3.42549 80.4 -3.43 0.535 3.42549 80.4 -3.43 -1.14 new 0.165E+04 -1.14 15 2.28366 23.8 -2.28 0.359 2.28366 23.8 -2.28 -0.761 new 326. -0.761 16 1.52244 7.06 -1.52 0.183 1.52244 7.06 -1.52 -0.507 new 64.5 -0.507 17 1.01496 2.09 -1.01 0.645E-02 1.01496 2.09 -1.01 -0.338 new 12.7 -0.338 18 0.676639 0.620 -0.677 -0.170 0.676639 0.620 -0.677 -0.226 new 2.52 -0.226 19 0.451093 0.184 -0.451 -0.346 0.451093 0.184 -0.451 -0.150 new 0.497 -0.150 20 0.300729 0.544E-01 -0.301 -0.522 0.300729 0.544E-01 -0.301 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 20 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 1000.00 0.200000E+10 1000.00 0.200000E+10 Estimate of the root after 20 steps. x f(x) 0.300729 0.543944E-01 0.300729 0.543944E-01 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 -1000.00 -0.100E+10 0.100E+04 3.00 0.000000E+00 0.000E+00 0.000E+00 333. new 0.300E+13 0.000E+00 1 -666.667 -0.296E+09 667. 2.82 0.000000E+00 0.000E+00 0.000E+00 222. new 0.593E+12 0.000E+00 2 -444.444 -0.878E+08 444. 2.65 0.000000E+00 0.000E+00 0.000E+00 148. new 0.117E+12 0.000E+00 3 -296.296 -0.260E+08 296. 2.47 0.000000E+00 0.000E+00 0.000E+00 98.8 new 0.231E+11 0.000E+00 4 -197.531 -0.771E+07 198. 2.30 0.000000E+00 0.000E+00 0.000E+00 65.8 new 0.457E+10 0.000E+00 5 -131.687 -0.228E+07 132. 2.12 0.000000E+00 0.000E+00 0.000E+00 43.9 new 0.902E+09 0.000E+00 6 -87.7915 -0.677E+06 87.8 1.94 0.000000E+00 0.000E+00 0.000E+00 29.3 new 0.178E+09 0.000E+00 7 -58.5277 -0.200E+06 58.5 1.77 0.000000E+00 0.000E+00 0.000E+00 19.5 new 0.352E+08 0.000E+00 8 -39.0184 -0.594E+05 39.0 1.59 0.000000E+00 0.000E+00 0.000E+00 13.0 new 0.695E+07 0.000E+00 9 -26.0123 -0.176E+05 26.0 1.42 0.000000E+00 0.000E+00 0.000E+00 8.67 new 0.137E+07 0.000E+00 10 -17.3415 -0.522E+04 17.3 1.24 0.000000E+00 0.000E+00 0.000E+00 5.78 new 0.271E+06 0.000E+00 11 -11.5610 -0.155E+04 11.6 1.06 0.000000E+00 0.000E+00 0.000E+00 3.85 new 0.536E+05 0.000E+00 12 -7.70735 -458. 7.71 0.887 0.000000E+00 0.000E+00 0.000E+00 2.57 new 0.106E+05 0.000E+00 13 -5.13823 -136. 5.14 0.711 0.000000E+00 0.000E+00 0.000E+00 1.71 new 0.209E+04 0.000E+00 14 -3.42549 -40.2 3.43 0.535 0.000000E+00 0.000E+00 0.000E+00 1.14 new 413. 0.000E+00 15 -2.28366 -11.9 2.28 0.359 0.000000E+00 0.000E+00 0.000E+00 0.761 new 81.6 0.000E+00 16 -1.52244 -3.53 1.52 0.183 0.000000E+00 0.000E+00 0.000E+00 0.507 new 16.1 0.000E+00 17 -1.01496 -1.05 1.01 0.645E-02 0.000000E+00 0.000E+00 0.000E+00 0.338 new 3.18 0.000E+00 18 -0.676639 -0.310 0.677 -0.170 0.000000E+00 0.000E+00 0.000E+00 0.226 new 0.629 0.000E+00 19 -0.451093 -0.918E-01 0.451 -0.346 0.000000E+00 0.000E+00 0.000E+00 0.150 new 0.124 0.000E+00 20 -0.300729 -0.272E-01 0.301 -0.522 0.000000E+00 0.000E+00 0.000E+00 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 2 Newton method, analytic jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 20 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) -1000.00 -0.100000E+10 0.000000E+00 0.000000E+00 Estimate of the root after 20 steps. x f(x) -0.300729 -0.271972E-01 0.000000E+00 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to Newton method with central finite difference jacobian. # Command? (H for help) Three methods are available: B Broyden's method. F Finite difference Newton method. N Newton method. Choose a method by typing B, F, or N. Please choose the type of differencing: B Backward differencing, fixed h. C Central differencing, fixed h. F Forward differencing, fixed h. Command? (H for help) # Choose problem 1 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 2.00000 1.96 -2.00 0.301 2.00000 0.982 -2.00 -4.47 new -0.615 -2.90 1 -2.46741 -0.328 2.47 0.392 -0.900304 1.52 0.900 1.83 new 0.514 0.776 2 -0.632923 -0.811E-02 0.633 -0.199 -0.124740 0.783E-01 0.125 0.465 new 0.111E-01 0.337E-01 3 -0.168388 -0.138E-02 0.168 -0.774 -0.910302E-01 0.153E-01 0.910E-01 0.164 new 0.142E-02 0.248E-02 4 -0.454778E-02 -0.357E-04 0.455E-02 -1.05 -0.885512E-01 0.401E-03 0.886E-01 0.455E-02 new 0.358E-04 0.175E-05 5 -0.892581E-07 -0.702E-09 0.893E-07 -1.05 -0.885494E-01 0.787E-08 0.885E-01 0.890E-07 new 0.707E-09 0.292E-03 6 -0.291664E-09 0.000E+00 0.292E-09 -1.05 -0.882578E-01 0.256E-10 0.883E-01 0.368E-25 new -0.225E-11 0.889E-01 7 -0.291664E-09 0.000E+00 0.292E-09 -3.16 0.691969E-03 -0.202E-12 -0.692E-03 Bad step!. A norm increased. 0.365E-25 new -0.140E-15 -0.692E-03 8 -0.291664E-09 0.000E+00 0.292E-09 -8.55 -0.281023E-08 0.000E+00 0.281E-08 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 41 Jacobian evaluations: 0 Matrix factorizations; 8 Linear system solves: 8 Starting point: x f(x) 2.00000 1.96337 2.00000 0.981684 Estimate of the root after 8 steps. x f(x) -0.291664E-09 0.000000E+00 -0.281023E-08 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 3.00 -3.00 0.477 3.00000 1.000 -3.00 -8.95 new -0.335 -5.97 1 -5.95360 -1.48 5.95 0.775 -2.96685 2.01 2.97 17.8 new 0.168 5.92 2 11.8443 0.735 -11.8 1.07 2.95118 4.01 -2.95 Bad step!. A norm increased. -35.4 new -0.847E-01 -5.89 3 -23.5528 -0.366 23.6 1.37 -2.93426 8.03 2.93 Bad step!. A norm increased. 70.4 new 0.426E-01 5.85 4 46.8107 0.182 -46.8 1.67 2.91589 16.1 -2.92 Bad step!. A norm increased. -140. new -0.214E-01 -5.81 5 -92.9772 -0.902E-01 93.0 1.97 -2.89583 32.1 2.90 Bad step!. A norm increased. 278. new 0.108E-01 5.77 6 184.537 0.448E-01 -185. 2.27 2.87375 64.2 -2.87 Bad step!. A norm increased. -550. new -0.544E-02 -5.72 7 -365.928 -0.222E-01 366. 2.56 -2.84926 128. 2.85 Bad step!. A norm increased. 0.109E+04 new 0.275E-02 5.67 8 724.805 0.110E-01 -725. 2.86 2.82181 257. -2.82 Bad step!. A norm increased. -0.216E+04 new -0.139E-02 -5.61 9 -1433.61 -0.543E-02 0.143E+04 3.16 -2.79067 514. 2.79 Bad step!. A norm increased. 0.426E+04 new 0.702E-03 5.55 10 2830.41 0.268E-02 -0.283E+04 3.45 2.75483 0.103E+04 -2.75 Bad step!. A norm increased. -0.840E+04 new -0.356E-03 -5.47 11 -5574.49 -0.132E-02 0.557E+04 3.75 -2.71283 0.205E+04 2.71 Bad step!. A norm increased. 0.165E+05 new 0.181E-03 5.38 12 10941.9 0.648E-03 -0.109E+05 4.04 2.66243 0.411E+04 -2.66 Bad step!. A norm increased. -0.323E+05 new -0.924E-04 -5.26 13 -21371.5 -0.316E-03 0.214E+05 4.33 -2.60011 0.821E+04 2.60 Bad step!. A norm increased. 0.628E+05 new 0.475E-04 5.12 14 41423.0 0.153E-03 -0.414E+05 4.62 2.51981 0.164E+05 -2.52 Bad step!. A norm increased. -0.121E+06 new -0.246E-04 -4.93 15 -79242.3 -0.733E-04 0.792E+05 4.90 -2.41020 0.328E+05 2.41 Bad step!. A norm increased. 0.227E+06 new 0.130E-04 4.66 16 147777. 0.342E-04 -0.148E+06 5.17 2.24737 0.653E+05 -2.25 Bad step!. A norm increased. -0.407E+06 new -0.716E-05 -4.22 17 -259417. -0.150E-04 0.259E+06 5.41 -1.97258 0.129E+06 1.97 Bad step!. A norm increased. 0.633E+06 new 0.439E-05 3.39 18 373973. 0.541E-05 -0.374E+06 5.57 1.42183 0.228E+06 -1.42 Bad step!. A norm increased. -0.551E+06 new -0.375E-05 -1.76 19 -177015. -0.640E-06 0.177E+06 5.25 -0.336502 0.563E+05 0.337 0.681E+05 new 0.175E-05 0.233 20 -108934. -0.984E-07 0.109E+06 5.04 -0.103540 0.112E+05 0.104 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 2.99963 3.00000 0.999877 Estimate of the root after 20 steps. x f(x) -108934. -0.984136E-07 -0.103540 11218.8 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 100.000 0.100E-01 -100. 2.00 1.00000 63.2 -1.00 -84.8 new -0.137E-01 -0.924 1 15.1531 0.379E-03 -15.2 1.18 0.757657E-01 1.14 -0.758E-01 -5.09 new -0.113E-02 -0.506E-01 2 10.0634 0.629E-04 -10.1 1.00 0.251585E-01 0.253 -0.252E-01 -3.36 new -0.189E-03 -0.168E-01 3 6.70610 0.105E-04 -6.71 0.826 0.838262E-02 0.562E-01 -0.838E-02 -2.24 new -0.314E-04 -0.559E-02 4 4.47052 0.175E-05 -4.47 0.650 0.279408E-02 0.125E-01 -0.279E-02 -1.49 new -0.524E-05 -0.186E-02 5 2.98033 0.291E-06 -2.98 0.474 0.931354E-03 0.278E-02 -0.931E-03 -0.994 new -0.872E-06 -0.621E-03 6 1.98607 0.477E-07 -1.99 0.298 0.310706E-03 0.617E-03 -0.311E-03 -0.698 new -0.136E-06 -0.201E-03 7 1.28779 0.750E-08 -1.29 0.110 0.109240E-03 0.141E-03 -0.109E-03 -0.580 new -0.167E-07 -0.600E-04 8 0.707630 0.135E-08 -0.708 -0.150 0.492130E-04 0.348E-04 -0.492E-04 -0.485 new -0.197E-08 -0.155E-04 9 0.222653 0.247E-09 -0.223 -0.652 0.337279E-04 0.751E-05 -0.337E-04 -0.212 new -0.259E-09 -0.158E-05 10 0.104357E-01 0.108E-10 -0.104E-01 -1.98 0.321472E-04 0.335E-06 -0.321E-04 -0.104E-01 new -0.108E-10 -0.350E-08 11 0.113736E-05 0.118E-14 -0.114E-05 -4.49 0.321437E-04 0.366E-10 -0.321E-04 -0.114E-05 new -0.117E-14 0.396E-07 12 -0.140113E-08 0.000E+00 0.140E-08 -4.49 0.321833E-04 -0.451E-13 -0.322E-04 0.000E+00 new -0.146E-17 -0.322E-04 13 -0.140113E-08 0.000E+00 0.140E-08 -8.85 0.571775E-09 0.000E+00 -0.572E-09 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 66 Jacobian evaluations: 0 Matrix factorizations; 13 Linear system solves: 13 Starting point: x f(x) 100.000 0.100000E-01 1.00000 63.2121 Estimate of the root after 13 steps. x f(x) -0.140113E-08 0.000000E+00 0.571775E-09 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1.00000 0.632E+04 -1.00 2.00 100.000 0.100E-01 -100. -1.39 new -1.37 -38.6 1 -0.386351 -0.135E+04 0.386 1.79 61.3649 -0.630E-02 -61.4 0.406 new 1.02 -3.13 2 0.197010E-01 66.8 -0.197E-01 1.77 58.2357 0.338E-03 -58.2 -0.197E-01 new -0.591E-01 -0.753E-02 3 -0.254824E-05 -0.864E-02 0.255E-05 1.77 58.2282 -0.438E-07 -58.2 0.255E-05 new 0.764E-05 -0.196E-02 4 0.856541E-10 0.000E+00 -0.857E-10 1.77 58.2262 0.147E-11 -58.2 0.000E+00 new -0.857E-10 58.2 5 0.856541E-10 0.000E+00 -0.857E-10 2.07 116.452 0.736E-12 -116. Bad step!. A norm increased. 0.000E+00 new -0.857E-10 116. 6 0.856541E-10 0.000E+00 -0.857E-10 2.37 232.905 0.368E-12 -233. Bad step!. A norm increased. 0.000E+00 new -0.857E-10 233. 7 0.856541E-10 0.000E+00 -0.857E-10 2.67 465.810 0.184E-12 -466. Bad step!. A norm increased. 0.000E+00 new -0.857E-10 466. 8 0.856541E-10 0.000E+00 -0.857E-10 2.97 931.619 0.919E-13 -932. Bad step!. A norm increased. 0.000E+00 new -0.857E-10 932. 9 0.856541E-10 0.000E+00 -0.857E-10 3.27 1863.24 0.460E-13 -0.186E+04 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.186E+04 10 0.856541E-10 0.000E+00 -0.857E-10 3.57 3726.48 0.230E-13 -0.373E+04 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.373E+04 11 0.856541E-10 0.000E+00 -0.857E-10 3.87 7452.96 0.115E-13 -0.745E+04 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.745E+04 12 0.856541E-10 0.000E+00 -0.857E-10 4.17 14905.9 0.575E-14 -0.149E+05 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.149E+05 13 0.856541E-10 0.000E+00 -0.857E-10 4.47 29811.8 0.287E-14 -0.298E+05 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.298E+05 14 0.856541E-10 0.000E+00 -0.857E-10 4.78 59623.6 0.144E-14 -0.596E+05 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.596E+05 15 0.856541E-10 0.000E+00 -0.857E-10 5.08 119247. 0.718E-15 -0.119E+06 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.119E+06 16 0.856541E-10 0.000E+00 -0.857E-10 5.38 238495. 0.359E-15 -0.238E+06 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.238E+06 17 0.856541E-10 0.000E+00 -0.857E-10 5.68 476989. 0.180E-15 -0.477E+06 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.477E+06 18 0.856541E-10 0.000E+00 -0.857E-10 5.98 953978. 0.898E-16 -0.954E+06 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.954E+06 19 0.856541E-10 0.000E+00 -0.857E-10 6.28 0.190796E+07 0.449E-16 -0.191E+07 Bad step!. A norm increased. 0.000E+00 new -0.857E-10 0.191E+07 20 0.856541E-10 0.000E+00 -0.857E-10 6.58 0.381591E+07 0.224E-16 -0.382E+07 Bad step!. A norm increased. Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 1.00000 6321.21 100.000 0.100000E-01 Estimate of the root after 20 steps. x f(x) 0.856541E-10 0.000000E+00 0.381591E+07 0.224465E-16 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682670 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233808E-01 0.877E-02 -0.234E-01 -0.213 0.612668 0.119E-01 -0.613 -0.234E-01 new -0.989E-02 -0.248E-03 2 0.612479E-05 0.230E-05 -0.612E-05 -0.213 0.612420 0.313E-05 -0.612 -0.612E-05 new -0.259E-05 -0.658E-04 3 0.426573E-09 0.000E+00 -0.427E-09 -0.213 0.612354 0.218E-09 -0.612 -0.506E-25 new 0.866E-10 -0.944 4 0.426573E-09 0.000E+00 -0.427E-09 -0.479 -0.332050 -0.134E-09 0.332 Bad step!. A norm increased. 0.000E+00 new 0.397E-10 0.372 5 0.426573E-09 0.000E+00 -0.427E-09 -1.39 0.403002E-01 0.172E-10 -0.403E-01 -0.802E-25 new 0.692E-12 -0.404E-01 6 0.426573E-09 0.000E+00 -0.427E-09 -4.18 -0.655372E-04 -0.280E-13 0.655E-04 -0.481E-25 new 0.183E-17 0.655E-04 7 0.426573E-09 0.000E+00 -0.427E-09 -8.94 0.114673E-08 0.000E+00 -0.115E-08 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 36 Jacobian evaluations: 0 Matrix factorizations; 7 Linear system solves: 7 Starting point: x f(x) 0.400000 0.172267 0.682670 0.218270 Estimate of the root after 7 steps. x f(x) 0.426573E-09 0.000000E+00 0.114673E-08 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682680 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233802E-01 0.877E-02 -0.234E-01 -0.213 0.612677 0.119E-01 -0.613 -0.234E-01 new -0.989E-02 -0.248E-03 2 0.612458E-05 0.230E-05 -0.612E-05 -0.213 0.612429 0.313E-05 -0.612 -0.612E-05 new -0.259E-05 -0.348E-04 3 0.225832E-09 0.000E+00 -0.226E-09 -0.213 0.612395 0.115E-09 -0.612 0.000E+00 new 0.458E-10 -0.945 4 0.225832E-09 0.000E+00 -0.226E-09 -0.479 -0.332135 -0.710E-10 0.332 Bad step!. A norm increased. 0.000E+00 new 0.210E-10 0.372 5 0.225832E-09 0.000E+00 -0.226E-09 -1.39 0.403332E-01 0.910E-11 -0.403E-01 0.000E+00 new 0.367E-12 -0.404E-01 6 0.225832E-09 0.000E+00 -0.226E-09 -4.18 -0.657035E-04 -0.148E-13 0.657E-04 0.000E+00 new 0.975E-18 0.657E-04 7 0.225832E-09 0.000E+00 -0.226E-09 -8.83 -0.147582E-08 0.000E+00 0.148E-08 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 36 Jacobian evaluations: 0 Matrix factorizations; 7 Linear system solves: 7 Starting point: x f(x) 0.400000 0.172272 0.682680 0.218272 Estimate of the root after 7 steps. x f(x) 0.225832E-09 0.000000E+00 -0.147582E-08 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to problem 2 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 54.0 -3.00 0.477 3.00000 54.0 -3.00 -1.000 new 972. -1.000 1 2.00000 16.0 -2.00 0.301 2.00000 16.0 -2.00 -0.667 new 192. -0.667 2 1.33333 4.74 -1.33 0.125 1.33333 4.74 -1.33 -0.444 new 37.9 -0.444 3 0.888889 1.40 -0.889 -0.512E-01 0.888889 1.40 -0.889 -0.296 new 7.49 -0.296 4 0.592593 0.416 -0.593 -0.227 0.592593 0.416 -0.593 -0.198 new 1.48 -0.198 5 0.395062 0.123 -0.395 -0.403 0.395062 0.123 -0.395 -0.132 new 0.292 -0.132 6 0.263374 0.365E-01 -0.263 -0.579 0.263374 0.365E-01 -0.263 -0.878E-01 new 0.577E-01 -0.878E-01 7 0.175583 0.108E-01 -0.176 -0.756 0.175583 0.108E-01 -0.176 -0.585E-01 new 0.114E-01 -0.585E-01 8 0.117055 0.321E-02 -0.117 -0.932 0.117055 0.321E-02 -0.117 -0.390E-01 new 0.225E-02 -0.390E-01 9 0.780369E-01 0.950E-03 -0.780E-01 -1.11 0.780369E-01 0.950E-03 -0.780E-01 -0.260E-01 new 0.445E-03 -0.260E-01 10 0.520246E-01 0.282E-03 -0.520E-01 -1.28 0.520246E-01 0.282E-03 -0.520E-01 -0.173E-01 new 0.879E-04 -0.173E-01 11 0.346831E-01 0.834E-04 -0.347E-01 -1.46 0.346831E-01 0.834E-04 -0.347E-01 -0.116E-01 new 0.174E-04 -0.116E-01 12 0.231220E-01 0.247E-04 -0.231E-01 -1.64 0.231220E-01 0.247E-04 -0.231E-01 -0.771E-02 new 0.343E-05 -0.771E-02 13 0.154147E-01 0.733E-05 -0.154E-01 -1.81 0.154147E-01 0.733E-05 -0.154E-01 -0.514E-02 new 0.678E-06 -0.514E-02 14 0.102765E-01 0.217E-05 -0.103E-01 -1.99 0.102765E-01 0.217E-05 -0.103E-01 -0.343E-02 new 0.134E-06 -0.343E-02 15 0.685097E-02 0.643E-06 -0.685E-02 -2.16 0.685097E-02 0.643E-06 -0.685E-02 -0.228E-02 new 0.264E-07 -0.228E-02 16 0.456732E-02 0.191E-06 -0.457E-02 -2.34 0.456732E-02 0.191E-06 -0.457E-02 -0.152E-02 new 0.522E-08 -0.152E-02 17 0.304488E-02 0.565E-07 -0.304E-02 -2.52 0.304488E-02 0.565E-07 -0.304E-02 -0.101E-02 new 0.103E-08 -0.101E-02 18 0.202992E-02 0.167E-07 -0.203E-02 -2.69 0.202992E-02 0.167E-07 -0.203E-02 -0.677E-03 new 0.204E-09 -0.677E-03 19 0.135328E-02 0.496E-08 -0.135E-02 -2.87 0.135328E-02 0.496E-08 -0.135E-02 -0.451E-03 new 0.402E-10 -0.451E-03 20 0.902186E-03 0.147E-08 -0.902E-03 -3.04 0.902186E-03 0.147E-08 -0.902E-03 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 54.0000 3.00000 54.0000 Estimate of the root after 20 steps. x f(x) 0.902186E-03 0.146865E-08 0.902186E-03 0.146865E-08 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1000.00 0.200E+10 -0.100E+04 3.00 1000.00 0.200E+10 -0.100E+04 -333. new 0.120E+14 -333. 1 666.667 0.593E+09 -667. 2.82 666.667 0.593E+09 -667. -222. new 0.237E+13 -222. 2 444.444 0.176E+09 -444. 2.65 444.444 0.176E+09 -444. -148. new 0.468E+12 -148. 3 296.296 0.520E+08 -296. 2.47 296.296 0.520E+08 -296. -98.8 new 0.925E+11 -98.8 4 197.531 0.154E+08 -198. 2.30 197.531 0.154E+08 -198. -65.8 new 0.183E+11 -65.8 5 131.687 0.457E+07 -132. 2.12 131.687 0.457E+07 -132. -43.9 new 0.361E+10 -43.9 6 87.7915 0.135E+07 -87.8 1.94 87.7915 0.135E+07 -87.8 -29.3 new 0.713E+09 -29.3 7 58.5277 0.401E+06 -58.5 1.77 58.5277 0.401E+06 -58.5 -19.5 new 0.141E+09 -19.5 8 39.0184 0.119E+06 -39.0 1.59 39.0184 0.119E+06 -39.0 -13.0 new 0.278E+08 -13.0 9 26.0123 0.352E+05 -26.0 1.42 26.0123 0.352E+05 -26.0 -8.67 new 0.549E+07 -8.67 10 17.3415 0.104E+05 -17.3 1.24 17.3415 0.104E+05 -17.3 -5.78 new 0.109E+07 -5.78 11 11.5610 0.309E+04 -11.6 1.06 11.5610 0.309E+04 -11.6 -3.85 new 0.214E+06 -3.85 12 7.70735 916. -7.71 0.887 7.70735 916. -7.71 -2.57 new 0.423E+05 -2.57 13 5.13823 271. -5.14 0.711 5.13823 271. -5.14 -1.71 new 0.836E+04 -1.71 14 3.42549 80.4 -3.43 0.535 3.42549 80.4 -3.43 -1.14 new 0.165E+04 -1.14 15 2.28366 23.8 -2.28 0.359 2.28366 23.8 -2.28 -0.761 new 326. -0.761 16 1.52244 7.06 -1.52 0.183 1.52244 7.06 -1.52 -0.507 new 64.5 -0.507 17 1.01496 2.09 -1.01 0.645E-02 1.01496 2.09 -1.01 -0.338 new 12.7 -0.338 18 0.676639 0.620 -0.677 -0.170 0.676639 0.620 -0.677 -0.226 new 2.52 -0.226 19 0.451093 0.184 -0.451 -0.346 0.451093 0.184 -0.451 -0.150 new 0.497 -0.150 20 0.300729 0.544E-01 -0.301 -0.522 0.300729 0.544E-01 -0.301 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 1000.00 0.200000E+10 1000.00 0.200000E+10 Estimate of the root after 20 steps. x f(x) 0.300729 0.543944E-01 0.300729 0.543944E-01 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 -1000.00 -0.100E+10 0.100E+04 3.00 0.000000E+00 0.000E+00 0.000E+00 333. new 0.300E+13 0.000E+00 1 -666.667 -0.296E+09 667. 2.82 0.000000E+00 0.000E+00 0.000E+00 222. new 0.593E+12 0.000E+00 2 -444.444 -0.878E+08 444. 2.65 0.000000E+00 0.000E+00 0.000E+00 148. new 0.117E+12 0.000E+00 3 -296.296 -0.260E+08 296. 2.47 0.000000E+00 0.000E+00 0.000E+00 98.8 new 0.231E+11 0.000E+00 4 -197.531 -0.771E+07 198. 2.30 0.000000E+00 0.000E+00 0.000E+00 65.8 new 0.457E+10 0.000E+00 5 -131.687 -0.228E+07 132. 2.12 0.000000E+00 0.000E+00 0.000E+00 43.9 new 0.902E+09 0.000E+00 6 -87.7915 -0.677E+06 87.8 1.94 0.000000E+00 0.000E+00 0.000E+00 29.3 new 0.178E+09 0.000E+00 7 -58.5277 -0.200E+06 58.5 1.77 0.000000E+00 0.000E+00 0.000E+00 19.5 new 0.352E+08 0.000E+00 8 -39.0184 -0.594E+05 39.0 1.59 0.000000E+00 0.000E+00 0.000E+00 13.0 new 0.695E+07 0.000E+00 9 -26.0123 -0.176E+05 26.0 1.42 0.000000E+00 0.000E+00 0.000E+00 8.67 new 0.137E+07 0.000E+00 10 -17.3415 -0.522E+04 17.3 1.24 0.000000E+00 0.000E+00 0.000E+00 5.78 new 0.271E+06 0.000E+00 11 -11.5610 -0.155E+04 11.6 1.06 0.000000E+00 0.000E+00 0.000E+00 3.85 new 0.536E+05 0.000E+00 12 -7.70735 -458. 7.71 0.887 0.000000E+00 0.000E+00 0.000E+00 2.57 new 0.106E+05 0.000E+00 13 -5.13823 -136. 5.14 0.711 0.000000E+00 0.000E+00 0.000E+00 1.71 new 0.209E+04 0.000E+00 14 -3.42549 -40.2 3.43 0.535 0.000000E+00 0.000E+00 0.000E+00 1.14 new 413. 0.000E+00 15 -2.28366 -11.9 2.28 0.359 0.000000E+00 0.000E+00 0.000E+00 0.761 new 81.6 0.000E+00 16 -1.52244 -3.53 1.52 0.183 0.000000E+00 0.000E+00 0.000E+00 0.507 new 16.1 0.000E+00 17 -1.01496 -1.05 1.01 0.645E-02 0.000000E+00 0.000E+00 0.000E+00 0.338 new 3.18 0.000E+00 18 -0.676639 -0.310 0.677 -0.170 0.000000E+00 0.000E+00 0.000E+00 0.226 new 0.629 0.000E+00 19 -0.451093 -0.918E-01 0.451 -0.346 0.000000E+00 0.000E+00 0.000E+00 0.150 new 0.124 0.000E+00 20 -0.300729 -0.272E-01 0.301 -0.522 0.000000E+00 0.000E+00 0.000E+00 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 1.0000000000000000E-05 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) -1000.00 -0.100000E+10 0.000000E+00 0.000000E+00 Estimate of the root after 20 steps. x f(x) -0.300729 -0.271972E-01 0.000000E+00 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Now compare the performance of the finite difference jacobians for a tiny h, and a large h=0.01. The first run uses the default h, which should be roughly 1.0e-16 for double precision # Command? (H for help) Enter name of variable to set (h for help) Enter absolute error tolerance. Absolute error tolerance set to 0.0000000000000000E+00 Enter name of variable to set (h for help) Command? (H for help) Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.100000E-01 0.200E-05 -0.100E-01 -2.00 0.100000E-01 0.200E-05 -0.100E-01 -0.333E-02 new 0.120E-06 -0.333E-02 1 0.666667E-02 0.593E-06 -0.667E-02 -2.18 0.666667E-02 0.593E-06 -0.667E-02 -0.222E-02 new 0.237E-07 -0.222E-02 2 0.444444E-02 0.176E-06 -0.444E-02 -2.35 0.444444E-02 0.176E-06 -0.444E-02 -0.148E-02 new 0.468E-08 -0.148E-02 3 0.296296E-02 0.520E-07 -0.296E-02 -2.53 0.296296E-02 0.520E-07 -0.296E-02 -0.988E-03 new 0.925E-09 -0.988E-03 4 0.197531E-02 0.154E-07 -0.198E-02 -2.70 0.197531E-02 0.154E-07 -0.198E-02 -0.658E-03 new 0.183E-09 -0.658E-03 5 0.131687E-02 0.457E-08 -0.132E-02 -2.88 0.131687E-02 0.457E-08 -0.132E-02 -0.439E-03 new 0.361E-10 -0.439E-03 6 0.877915E-03 0.135E-08 -0.878E-03 -3.06 0.877915E-03 0.135E-08 -0.878E-03 -0.293E-03 new 0.713E-11 -0.293E-03 7 0.585277E-03 0.401E-09 -0.585E-03 -3.23 0.585277E-03 0.401E-09 -0.585E-03 -0.195E-03 new 0.141E-11 -0.195E-03 8 0.390184E-03 0.119E-09 -0.390E-03 -3.41 0.390184E-03 0.119E-09 -0.390E-03 -0.130E-03 new 0.278E-12 -0.130E-03 9 0.260123E-03 0.352E-10 -0.260E-03 -3.58 0.260123E-03 0.352E-10 -0.260E-03 -0.867E-04 new 0.549E-13 -0.867E-04 10 0.173415E-03 0.104E-10 -0.173E-03 -3.76 0.173415E-03 0.104E-10 -0.173E-03 -0.578E-04 new 0.109E-13 -0.578E-04 11 0.115610E-03 0.309E-11 -0.116E-03 -3.94 0.115610E-03 0.309E-11 -0.116E-03 -0.385E-04 new 0.214E-14 -0.385E-04 12 0.770735E-04 0.916E-12 -0.771E-04 -4.11 0.770735E-04 0.916E-12 -0.771E-04 -0.257E-04 new 0.423E-15 -0.257E-04 13 0.513823E-04 0.271E-12 -0.514E-04 -4.29 0.513823E-04 0.271E-12 -0.514E-04 -0.171E-04 new 0.836E-16 -0.171E-04 14 0.342549E-04 0.804E-13 -0.343E-04 -4.47 0.342549E-04 0.804E-13 -0.343E-04 -0.114E-04 new 0.165E-16 -0.114E-04 15 0.228366E-04 0.238E-13 -0.228E-04 -4.64 0.228366E-04 0.238E-13 -0.228E-04 -0.761E-05 new 0.326E-17 -0.761E-05 16 0.152244E-04 0.706E-14 -0.152E-04 -4.82 0.152244E-04 0.706E-14 -0.152E-04 -0.507E-05 new 0.645E-18 -0.507E-05 17 0.101496E-04 0.209E-14 -0.101E-04 -4.99 0.101496E-04 0.209E-14 -0.101E-04 -0.338E-05 new 0.127E-18 -0.338E-05 18 0.676640E-05 0.620E-15 -0.677E-05 -5.17 0.676640E-05 0.620E-15 -0.677E-05 -0.226E-05 new 0.252E-19 -0.226E-05 19 0.451093E-05 0.184E-15 -0.451E-05 -5.35 0.451093E-05 0.184E-15 -0.451E-05 -0.150E-05 new 0.497E-20 -0.150E-05 20 0.300729E-05 0.544E-16 -0.301E-05 -5.52 0.300729E-05 0.544E-16 -0.301E-05 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.4901161193847656E-08 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 0.100000E-01 0.200000E-05 0.100000E-01 0.200000E-05 Estimate of the root after 20 steps. x f(x) 0.300729E-05 0.543947E-16 0.300729E-05 0.543947E-16 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) Command? (H for help) # Now we reset h to 0.01 and try again. Since the difference between the iterates and the true solution itself is of order 0.01, we expect poor performance. # Command? (H for help) Command? (H for help) Enter name of variable to set (h for help) Enter a small, positive number for use for approximating jacobians by finite differences. Difference parameter for jacobians= 1.0000000000000000E-02 Enter name of variable to set (h for help) Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.100000E-01 0.200E-05 -0.100E-01 -2.00 0.100000E-01 0.200E-05 -0.100E-01 -0.285E-02 new 0.212E-06 -0.285E-02 1 0.715104E-02 0.731E-06 -0.715E-02 -2.15 0.715104E-02 0.731E-06 -0.715E-02 -0.179E-02 new 0.832E-07 -0.179E-02 2 0.535960E-02 0.308E-06 -0.536E-02 -2.27 0.535960E-02 0.308E-06 -0.536E-02 -0.113E-02 new 0.433E-07 -0.113E-02 3 0.423348E-02 0.152E-06 -0.423E-02 -2.37 0.423348E-02 0.152E-06 -0.423E-02 -0.728E-03 new 0.285E-07 -0.728E-03 4 0.350526E-02 0.861E-07 -0.351E-02 -2.46 0.350526E-02 0.861E-07 -0.351E-02 -0.494E-03 new 0.219E-07 -0.494E-03 5 0.301142E-02 0.546E-07 -0.301E-02 -2.52 0.301142E-02 0.546E-07 -0.301E-02 -0.352E-03 new 0.184E-07 -0.352E-03 6 0.265907E-02 0.376E-07 -0.266E-02 -2.58 0.265907E-02 0.376E-07 -0.266E-02 -0.263E-03 new 0.164E-07 -0.263E-03 7 0.239604E-02 0.275E-07 -0.240E-02 -2.62 0.239604E-02 0.275E-07 -0.240E-02 -0.204E-03 new 0.151E-07 -0.204E-03 8 0.219214E-02 0.211E-07 -0.219E-02 -2.66 0.219214E-02 0.211E-07 -0.219E-02 -0.163E-03 new 0.142E-07 -0.163E-03 9 0.202916E-02 0.167E-07 -0.203E-02 -2.69 0.202916E-02 0.167E-07 -0.203E-02 -0.134E-03 new 0.136E-07 -0.134E-03 10 0.189560E-02 0.136E-07 -0.190E-02 -2.72 0.189560E-02 0.136E-07 -0.190E-02 -0.112E-03 new 0.131E-07 -0.112E-03 11 0.178388E-02 0.114E-07 -0.178E-02 -2.75 0.178388E-02 0.114E-07 -0.178E-02 -0.950E-04 new 0.127E-07 -0.950E-04 12 0.168883E-02 0.963E-08 -0.169E-02 -2.77 0.168883E-02 0.963E-08 -0.169E-02 -0.820E-04 new 0.125E-07 -0.820E-04 13 0.160681E-02 0.830E-08 -0.161E-02 -2.79 0.160681E-02 0.830E-08 -0.161E-02 -0.716E-04 new 0.122E-07 -0.716E-04 14 0.153517E-02 0.724E-08 -0.154E-02 -2.81 0.153517E-02 0.724E-08 -0.154E-02 -0.632E-04 new 0.120E-07 -0.632E-04 15 0.147194E-02 0.638E-08 -0.147E-02 -2.83 0.147194E-02 0.638E-08 -0.147E-02 -0.563E-04 new 0.119E-07 -0.563E-04 16 0.141564E-02 0.567E-08 -0.142E-02 -2.85 0.141564E-02 0.567E-08 -0.142E-02 -0.505E-04 new 0.117E-07 -0.505E-04 17 0.136512E-02 0.509E-08 -0.137E-02 -2.86 0.136512E-02 0.509E-08 -0.137E-02 -0.457E-04 new 0.116E-07 -0.457E-04 18 0.131947E-02 0.459E-08 -0.132E-02 -2.88 0.131947E-02 0.459E-08 -0.132E-02 -0.415E-04 new 0.115E-07 -0.415E-04 19 0.127797E-02 0.417E-08 -0.128E-02 -2.89 0.127797E-02 0.417E-08 -0.128E-02 -0.379E-04 new 0.114E-07 -0.379E-04 20 0.124004E-02 0.381E-08 -0.124E-02 -2.91 0.124004E-02 0.381E-08 -0.124E-02 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 0 Newton method, central f-d jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 101 Jacobian evaluations: 0 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 0.100000E-01 0.200000E-05 0.100000E-01 0.200000E-05 Estimate of the root after 20 steps. x f(x) 0.124004E-02 0.381363E-08 0.124004E-02 0.381363E-08 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to Broyden method, with initial matrix the identity # Command? (H for help) Three methods are available: B Broyden's method. F Finite difference Newton method. N Newton method. Choose a method by typing B, F, or N. Please choose how Broyden's method will start: I Initial iteration matrix is identity. J Initial iteration matrix is jacobian. Command? (H for help) # Choose problem 1 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 2.00000 1.96 -2.00 0.301 2.00000 0.982 -2.00 -1.96 new 1.00 -0.982 1 0.366313E-01 0.380E-01 -0.366E-01 0.788E-02 1.01832 0.232E-01 -1.02 -0.387E-01 new 0.980 -0.237E-01 2 -0.211118E-02 -0.209E-02 0.211E-02 -0.234E-02 0.994618 -0.133E-02 -0.995 0.202E-02 new 1.03 0.129E-02 3 -0.916078E-04 -0.909E-04 0.916E-04 -0.178E-02 0.995906 -0.579E-04 -0.996 0.919E-04 new 0.989 0.584E-04 4 0.247863E-06 0.246E-06 -0.248E-06 -0.176E-02 0.995964 0.157E-06 -0.996 -0.248E-06 new 0.992 -0.158E-06 5 0.134567E-09 0.000E+00 -0.135E-09 -0.176E-02 0.995964 0.850E-10 -0.996 -0.251E-12 new 0.992 -0.852E-10 6 0.134317E-09 0.000E+00 -0.134E-09 -0.176E-02 0.995964 0.848E-10 -0.996 -0.134E-09 new 0.185E-02 -0.456E-07 7 0.104486E-17 0.000E+00 -0.104E-17 -0.176E-02 0.995964 0.660E-18 -0.996 Bad step!. A norm increased. -0.104E-17 new 0.185E-02 -0.355E-15 8 0.811281E-26 0.000E+00 -0.811E-26 -0.176E-02 0.995964 0.512E-26 -0.996 -0.811E-26 new 0.185E-02 -0.276E-23 9 0.000000E+00 0.000E+00 0.000E+00 -0.176E-02 0.995964 0.000E+00 -0.996 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 10 Jacobian evaluations: 1 Matrix factorizations; 9 Linear system solves: 9 Starting point: x f(x) 2.00000 1.96337 2.00000 0.981684 Estimate of the root after 9 steps. x f(x) 0.000000E+00 0.000000E+00 0.995964 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 3.00 -3.00 0.477 3.00000 1.000 -3.00 -3.00 new 1.00 -1.000 1 0.370229E-03 0.148E-02 -0.370E-03 0.301 2.00012 0.182E-03 -2.00 -0.148E-02 new 1.000 -0.182E-03 2 -0.111156E-02 -0.445E-02 0.111E-02 0.301 1.99994 -0.546E-03 -2.00 Bad step!. A norm increased. 0.111E-02 new 4.00 0.136E-03 3 -0.538525E-07 -0.214E-06 0.539E-07 0.301 2.00008 -0.264E-07 -2.00 0.536E-07 new 4.00 0.670E-08 4 -0.254330E-09 0.000E+00 0.254E-09 0.301 2.00008 -0.125E-09 -2.00 -0.113E-10 new 4.00 0.123E-09 5 -0.265644E-09 0.000E+00 0.266E-09 0.301 2.00008 -0.130E-09 -2.00 Bad step!. A norm increased. 0.266E-09 new -0.178 -0.290E-08 6 0.313585E-18 0.000E+00 -0.314E-18 0.301 2.00008 0.154E-18 -2.00 Bad step!. A norm increased. -0.314E-18 new -0.178 0.342E-17 7 -0.386642E-27 0.000E+00 0.387E-27 0.301 2.00008 -0.190E-27 -2.00 0.387E-27 new -0.178 -0.422E-26 8 0.000000E+00 0.000E+00 0.000E+00 0.301 2.00008 0.000E+00 -2.00 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 9 Jacobian evaluations: 1 Matrix factorizations; 8 Linear system solves: 8 Starting point: x f(x) 3.00000 2.99963 3.00000 0.999877 Estimate of the root after 8 steps. x f(x) 0.000000E+00 0.000000E+00 2.00008 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 100.000 0.100E-01 -100. 2.00 1.00000 63.2 -1.00 -0.100E-01 new 1.00 -63.2 1 99.9900 38.7 -100.0 2.00 -62.2121 -1.61 62.2 -37.8 new 1.03 1.57 2 62.2389 59.1 -62.2 1.79 -60.6445 -1.03 60.6 Bad step!. A norm increased. 107. new -0.561 -1.83 3 169.034 23.1 -169. 2.23 -62.4725 -2.71 62.5 Bad step!. A norm increased. 62.6 new -0.354 2.53 4 231.637 15.5 -232. 2.36 -59.9430 -3.86 59.9 103. new -0.134 9.80 5 334.568 7.51 -335. 2.52 -50.1402 -6.67 50.1 Bad step!. A norm increased. 43.5 new -0.935E-01 11.9 6 378.089 3.87 -378. 2.58 -38.2602 -9.88 38.3 Bad step!. A norm increased. -11.4 new -0.118 6.32 7 366.681 2.78 -367. 2.56 -31.9414 -11.5 31.9 Bad step!. A norm increased. 29.7 new 0.108 -2.69 8 396.383 3.03 -396. 2.60 -34.6324 -11.4 34.6 Bad step!. A norm increased. 598. new 0.513E-02 -35.2 9 993.956 4.90 -994. 3.00 -69.7895 -14.2 69.8 Bad step!. A norm increased. -0.181E+05 new -0.175E-03 943. 10 -17153.0 -44.5 0.172E+05 4.23 873.382 -19.6 -873. Bad step!. A norm increased. 0.200E+05 new 0.167E-02 -0.102E+04 11 2893.78 7.21 -0.289E+04 3.46 -144.446 -20.0 144. Bad step!. A norm increased. 0.294E+04 new 0.146E-02 -114. 12 5834.09 11.4 -0.583E+04 3.77 -258.356 -22.6 258. Bad step!. A norm increased. 0.647E+04 new 0.455E-03 -201. 13 12300.0 17.1 -0.123E+05 4.09 -459.246 -26.8 459. Bad step!. A norm increased. 0.635E+04 new 0.230E-03 -143. 14 18646.5 19.4 -0.186E+05 4.27 -602.046 -31.0 602. Bad step!. A norm increased. -0.269E+05 new -0.709E-04 629. 15 -8269.61 -0.869E-01 0.827E+04 3.92 26.8148 -308. -26.8 Bad step!. A norm increased. 0.273E+05 new 0.472E-02 -591. 16 19051.4 16.7 -0.191E+05 4.28 -564.342 -33.8 564. Bad step!. A norm increased. 0.102E+04 new 0.456E-02 19.5 17 20066.8 14.8 -0.201E+05 4.30 -544.864 -36.8 545. Bad step!. A norm increased. -0.299E+04 new -0.234E-02 -8.49 18 17072.9 17.9 -0.171E+05 4.23 -553.354 -30.9 553. Bad step!. A norm increased. -0.190E+04 new -0.143E-02 42.3 19 15169.4 17.2 -0.152E+05 4.18 -511.082 -29.7 511. -0.598E+05 new -0.441E-04 0.131E+04 20 -44601.2 -14.5 0.446E+05 4.65 803.738 -55.5 -804. Bad step!. A norm increased. Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 100.000 0.100000E-01 1.00000 63.2121 Estimate of the root after 20 steps. x f(x) -44601.2 -14.4838 803.738 -55.4922 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1.00000 0.632E+04 -1.00 2.00 100.000 0.100E-01 -100. -0.632E+04 new 1.00 -0.100E-01 1 -6320.21 -1.58 0.632E+04 3.80 99.9900 -63.2 -100.0 1.58 new 1.00 63.2 2 -6318.62 -4.21 0.632E+04 3.80 163.183 -38.7 -163. 10.9 new 0.387 100. 3 -6307.73 -11.0 0.631E+04 3.80 263.223 -24.0 -263. Bad step!. A norm increased. 40.8 new 0.143 167. 4 -6266.91 -29.6 0.627E+04 3.80 430.434 -14.6 -430. Bad step!. A norm increased. 148. new 0.475E-01 322. 5 -6119.36 -92.4 0.612E+04 3.79 752.129 -8.14 -752. Bad step!. A norm increased. 0.189E+04 new 0.484E-02 0.257E+04 6 -4231.75 -0.260E+04 0.423E+04 3.63 3319.68 -1.27 -0.332E+04 Bad step!. A norm increased. -0.213E+04 new -0.587E-01 -0.271E+04 7 -6365.66 -57.9 0.637E+04 3.80 607.192 -10.5 -607. Bad step!. A norm increased. -290. new -0.543E-01 -133. 8 -6655.56 -33.7 0.666E+04 3.82 473.860 -14.0 -474. 0.214E+04 new 0.881E-02 806. 9 -4511.34 -363. 0.451E+04 3.65 1279.96 -3.52 -0.128E+04 Bad step!. A norm increased. -0.667E+04 new -0.434E-02 -0.197E+04 10 -11180.5 -42.9 0.112E+05 4.05 -692.225 16.2 692. Bad step!. A norm increased. 0.249E+04 new -0.686E-02 604. 11 -8694.62 -0.902 0.869E+04 3.94 -88.5346 98.2 88.5 Bad step!. A norm increased. -0.288E+04 new 0.420E-01 -740. 12 -11577.2 -59.3 0.116E+05 4.06 -828.705 14.0 829. Bad step!. A norm increased. -340. new 0.375E-01 -141. 13 -11917.1 -78.9 0.119E+05 4.08 -969.827 12.3 970. Bad step!. A norm increased. -992. new 0.927E-02 -531. 14 -12909.1 -174. 0.129E+05 4.11 -1500.33 8.60 0.150E+04 Bad step!. A norm increased. 101. new 0.963E-02 -99.4 15 -12808.3 -200. 0.128E+05 4.11 -1599.71 8.01 0.160E+04 Bad step!. A norm increased. -591. new -0.525E-02 -27.8 16 -13398.8 -198. 0.134E+05 4.13 -1627.55 8.23 0.163E+04 Bad step!. A norm increased. -0.336E+05 new -0.905E-04 -0.274E+04 17 -47011.1 -406. 0.470E+05 4.67 -4370.86 10.8 0.437E+04 Bad step!. A norm increased. 0.618E+05 new 0.108E-03 0.471E+04 18 14835.7 7.82 -0.148E+05 4.17 340.623 43.6 -341. Bad step!. A norm increased. -0.193E+05 new 0.157E-03 -0.330E+04 19 -4500.52 -0.194E+04 0.450E+04 3.65 -2958.55 1.52 0.296E+04 Bad step!. A norm increased. 0.181E+05 new 0.195E-02 848. 20 13646.1 326. -0.136E+05 4.14 -2110.63 -6.47 0.211E+04 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 1.00000 6321.21 100.000 0.100000E-01 Estimate of the root after 20 steps. x f(x) 13646.1 326.449 -2110.63 -6.46543 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682670 0.218 -0.683 -0.172 new 1.00 -0.218 1 0.227733 0.479E-01 -0.228 -0.333 0.464400 0.951E-01 -0.464 -0.766E-01 new 0.625 -0.152 2 0.151128 0.146E-01 -0.151 -0.506 0.312139 0.449E-01 -0.312 -0.472E-01 new 0.354 -0.122 3 0.103904 0.373E-02 -0.104 -0.721 0.190020 0.194E-01 -0.190 -0.282E-01 new 0.211 -0.845E-01 4 0.756790E-01 0.840E-03 -0.757E-01 -0.977 0.105522 0.794E-02 -0.106 -0.171E-01 new 0.128 -0.549E-01 5 0.585860E-01 0.150E-03 -0.586E-01 -1.23 0.506634E-01 0.296E-02 -0.507E-01 -0.947E-02 new 0.817E-01 -0.313E-01 6 0.491210E-01 0.184E-04 -0.491E-01 -1.31 0.193556E-01 0.951E-03 -0.194E-01 -0.431E-02 new 0.559E-01 -0.144E-01 7 0.448067E-01 0.108E-05 -0.448E-01 -1.35 0.491849E-02 0.220E-03 -0.492E-02 -0.128E-02 new 0.431E-01 -0.431E-02 8 0.435221E-01 0.158E-07 -0.435E-01 -1.36 0.603637E-03 0.263E-04 -0.604E-03 -0.173E-03 new 0.379E-01 -0.582E-03 9 0.433489E-01 0.193E-10 -0.433E-01 -1.36 0.211372E-04 0.916E-06 -0.211E-04 -0.626E-05 new 0.366E-01 -0.210E-04 10 0.433426E-01 0.381E-15 -0.433E-01 -1.36 0.938588E-07 0.405E-08 -0.939E-07 -0.278E-07 new 0.364E-01 -0.934E-07 11 0.433426E-01 0.796E-20 -0.433E-01 -1.36 0.428798E-09 0.000E+00 -0.429E-09 -0.382E-19 new 0.364E-01 -0.966E-19 12 0.433426E-01 0.796E-20 -0.433E-01 -1.36 0.428798E-09 0.000E+00 -0.429E-09 -0.848E-10 new 0.164E-10 -0.214E-09 13 0.433426E-01 0.199E-20 -0.433E-01 -1.36 0.214399E-09 0.000E+00 -0.214E-09 Bad step!. A norm increased. -0.283E-10 new 0.123E-10 -0.715E-10 14 0.433426E-01 0.885E-21 -0.433E-01 -1.36 0.142933E-09 0.000E+00 -0.143E-09 -0.226E-10 new 0.684E-11 -0.572E-10 15 0.433426E-01 0.318E-21 -0.433E-01 -1.36 0.857597E-10 0.000E+00 -0.858E-10 -0.127E-10 new 0.438E-11 -0.322E-10 16 0.433426E-01 0.124E-21 -0.433E-01 -1.36 0.535999E-10 0.000E+00 -0.536E-10 -0.815E-11 new 0.267E-11 -0.206E-10 17 0.433426E-01 0.471E-22 -0.433E-01 -1.36 0.329845E-10 0.000E+00 -0.330E-10 -0.497E-11 new 0.166E-11 -0.126E-10 18 0.433426E-01 0.181E-22 -0.433E-01 -1.36 0.204189E-10 0.000E+00 -0.204E-10 -0.309E-11 new 0.102E-11 -0.781E-11 19 0.433426E-01 0.689E-23 -0.433E-01 -1.36 0.126115E-10 0.000E+00 -0.126E-10 -0.190E-11 new 0.633E-12 -0.482E-11 20 0.433426E-01 0.263E-23 -0.433E-01 -1.36 0.779648E-11 0.000E+00 -0.780E-11 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 0.400000 0.172267 0.682670 0.218270 Estimate of the root after 20 steps. x f(x) 0.433426E-01 0.263211E-23 0.779648E-11 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682680 0.218 -0.683 -0.172 new 1.00 -0.218 1 0.227728 0.479E-01 -0.228 -0.333 0.464408 0.951E-01 -0.464 -0.766E-01 new 0.625 -0.152 2 0.151123 0.146E-01 -0.151 -0.506 0.312150 0.449E-01 -0.312 -0.472E-01 new 0.354 -0.122 3 0.103898 0.373E-02 -0.104 -0.721 0.190032 0.194E-01 -0.190 -0.282E-01 new 0.211 -0.845E-01 4 0.756719E-01 0.840E-03 -0.757E-01 -0.977 0.105534 0.794E-02 -0.106 -0.171E-01 new 0.128 -0.549E-01 5 0.585779E-01 0.150E-03 -0.586E-01 -1.23 0.506741E-01 0.296E-02 -0.507E-01 -0.947E-02 new 0.817E-01 -0.313E-01 6 0.491115E-01 0.184E-04 -0.491E-01 -1.31 0.193630E-01 0.951E-03 -0.194E-01 -0.432E-02 new 0.559E-01 -0.144E-01 7 0.447959E-01 0.108E-05 -0.448E-01 -1.35 0.492192E-02 0.220E-03 -0.492E-02 -0.129E-02 new 0.430E-01 -0.432E-02 8 0.435105E-01 0.159E-07 -0.435E-01 -1.36 0.604398E-03 0.263E-04 -0.604E-03 -0.173E-03 new 0.379E-01 -0.583E-03 9 0.433370E-01 0.194E-10 -0.433E-01 -1.36 0.211840E-04 0.918E-06 -0.212E-04 -0.627E-05 new 0.366E-01 -0.211E-04 10 0.433307E-01 0.384E-15 -0.433E-01 -1.36 0.942117E-07 0.408E-08 -0.942E-07 -0.280E-07 new 0.364E-01 -0.943E-07 11 0.433307E-01 0.102E-21 -0.433E-01 -1.36 -0.486228E-10 0.000E+00 0.486E-10 -0.491E-21 new 0.364E-01 -0.124E-20 12 0.433307E-01 0.102E-21 -0.433E-01 -1.36 -0.486228E-10 0.000E+00 0.486E-10 Bad step!. A norm increased. 0.961E-11 new -0.186E-11 0.243E-10 13 0.433307E-01 0.256E-22 -0.433E-01 -1.36 -0.243119E-10 0.000E+00 0.243E-10 Bad step!. A norm increased. 0.320E-11 new -0.140E-11 0.810E-11 14 0.433307E-01 0.114E-22 -0.433E-01 -1.36 -0.162077E-10 0.000E+00 0.162E-10 0.256E-11 new -0.776E-12 0.648E-11 15 0.433307E-01 0.409E-23 -0.433E-01 -1.36 -0.972475E-11 0.000E+00 0.972E-11 0.144E-11 new -0.496E-12 0.365E-11 16 0.433307E-01 0.160E-23 -0.433E-01 -1.36 -0.607775E-11 0.000E+00 0.608E-11 0.924E-12 new -0.303E-12 0.234E-11 17 0.433307E-01 0.606E-24 -0.433E-01 -1.36 -0.374035E-11 0.000E+00 0.374E-11 0.564E-12 new -0.188E-12 0.143E-11 18 0.433307E-01 0.232E-24 -0.433E-01 -1.36 -0.231520E-11 0.000E+00 0.232E-11 0.350E-12 new -0.116E-12 0.885E-12 19 0.433307E-01 0.886E-25 -0.433E-01 -1.36 -0.143032E-11 0.000E+00 0.143E-11 0.216E-12 new -0.717E-13 0.546E-12 20 0.433307E-01 0.338E-25 -0.433E-01 -1.36 -0.884214E-12 0.000E+00 0.884E-12 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 0.400000 0.172272 0.682680 0.218272 Estimate of the root after 20 steps. x f(x) 0.433307E-01 0.338456E-25 -0.884214E-12 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to problem 2 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 54.0 -3.00 0.477 3.00000 54.0 -3.00 -54.0 new 1.00 -54.0 1 -51.0000 -0.265E+06 51.0 1.71 -51.0000 -0.265E+06 51.0 Bad step!. A norm increased. 54.0 new 0.491E+04 54.0 2 2.98901 53.4 -2.99 0.476 2.98901 53.4 -2.99 -0.109E-01 new 0.491E+04 -0.109E-01 3 2.97814 52.8 -2.98 0.474 2.97814 52.8 -2.98 -0.989 new 53.4 -0.989 4 1.98904 15.7 -1.99 0.299 1.98904 15.7 -1.99 Bad step!. A norm increased. -0.420 new 37.5 -0.420 5 1.56933 7.73 -1.57 0.196 1.56934 7.73 -1.57 -0.405 new 19.1 -0.405 6 1.16423 3.16 -1.16 0.660E-01 1.16422 3.16 -1.16 -0.280 new 11.3 -0.280 7 0.884688 1.38 -0.885 -0.532E-01 0.884708 1.38 -0.885 -0.219 new 6.34 -0.219 8 0.666144 0.591 -0.666 -0.176 0.666116 0.591 -0.666 -0.163 new 3.63 -0.163 9 0.503341 0.255 -0.503 -0.298 0.503343 0.255 -0.503 -0.124 new 2.06 -0.124 10 0.379815 0.110 -0.380 -0.420 0.379820 0.110 -0.380 -0.931E-01 new 1.18 -0.931E-01 11 0.286755 0.472E-01 -0.287 -0.542 0.286760 0.472E-01 -0.287 -0.703E-01 new 0.671 -0.703E-01 12 0.216452 0.203E-01 -0.216 -0.665 0.216457 0.203E-01 -0.216 -0.531E-01 new 0.382 -0.531E-01 13 0.163398 0.873E-02 -0.163 -0.787 0.163403 0.873E-02 -0.163 -0.401E-01 new 0.218 -0.401E-01 14 0.123344 0.375E-02 -0.123 -0.909 0.123348 0.375E-02 -0.123 -0.302E-01 new 0.124 -0.302E-01 15 0.931098E-01 0.161E-02 -0.931E-01 -1.03 0.931135E-01 0.161E-02 -0.931E-01 -0.228E-01 new 0.707E-01 -0.228E-01 16 0.702861E-01 0.694E-03 -0.703E-01 -1.15 0.702896E-01 0.695E-03 -0.703E-01 -0.172E-01 new 0.403E-01 -0.172E-01 17 0.530571E-01 0.299E-03 -0.531E-01 -1.28 0.530604E-01 0.299E-03 -0.531E-01 -0.130E-01 new 0.230E-01 -0.130E-01 18 0.400513E-01 0.129E-03 -0.401E-01 -1.40 0.400544E-01 0.129E-03 -0.401E-01 -0.982E-02 new 0.131E-01 -0.982E-02 19 0.302335E-01 0.553E-04 -0.302E-01 -1.52 0.302365E-01 0.553E-04 -0.302E-01 -0.741E-02 new 0.746E-02 -0.741E-02 20 0.228223E-01 0.238E-04 -0.228E-01 -1.64 0.228252E-01 0.238E-04 -0.228E-01 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 54.0000 3.00000 54.0000 Estimate of the root after 20 steps. x f(x) 0.228223E-01 0.237773E-04 0.228252E-01 0.237803E-04 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1000.00 0.200E+10 -0.100E+04 3.00 1000.00 0.200E+10 -0.100E+04 -0.200E+10 new 1.00 -0.200E+10 1 -0.200000E+10 -0.160E+29 0.200E+10 9.30 -0.200000E+10 -0.160E+29 0.200E+10 Bad step!. A norm increased. LUFACT - Fatal error! The iteration matrix is singular! NEWCON - Warning! Error occurred in linear system routine. The iteration must stop! Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 2 Jacobian evaluations: 1 Matrix factorizations; 2 Linear system solves: 1 Starting point: x f(x) 1000.00 0.200000E+10 1000.00 0.200000E+10 Estimate of the root after 2 steps. x f(x) -0.200000E+10 -0.160000E+29 -0.200000E+10 -0.160000E+29 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 -1000.00 -0.100E+10 0.100E+04 3.00 0.000000E+00 0.000E+00 0.000E+00 0.100E+10 new 1.00 0.000E+00 1 0.999999E+09 0.100E+28 -0.100E+10 9.00 0.000000E+00 0.000E+00 0.000E+00 Bad step!. A norm increased. -0.100E+10 new 0.100E+19 0.000E+00 2 -1000.00 -0.100E+10 0.100E+04 3.00 0.000000E+00 0.000E+00 0.000E+00 0.100E-08 new 0.100E+19 0.000E+00 3 -1000.000 -0.100E+10 1000. 3.00 0.000000E+00 0.000E+00 0.000E+00 333. new 0.300E+07 0.000E+00 4 -666.688 -0.296E+09 667. 2.82 0.000000E+00 0.000E+00 0.000E+00 Bad step!. A norm increased. 140. new 0.211E+07 0.000E+00 5 -526.327 -0.146E+09 526. 2.72 0.000000E+00 0.000E+00 0.000E+00 136. new 0.107E+07 0.000E+00 6 -390.366 -0.595E+08 390. 2.59 0.000000E+00 0.000E+00 0.000E+00 93.7 new 0.635E+06 0.000E+00 7 -296.667 -0.261E+08 297. 2.47 0.000000E+00 0.000E+00 0.000E+00 73.3 new 0.356E+06 0.000E+00 8 -223.367 -0.111E+08 223. 2.35 0.000000E+00 0.000E+00 0.000E+00 54.6 new 0.204E+06 0.000E+00 9 -168.783 -0.481E+07 169. 2.23 0.000000E+00 0.000E+00 0.000E+00 41.4 new 0.116E+06 0.000E+00 10 -127.361 -0.207E+07 127. 2.11 0.000000E+00 0.000E+00 0.000E+00 31.2 new 0.662E+05 0.000E+00 11 -96.1565 -0.889E+06 96.2 1.98 0.000000E+00 0.000E+00 0.000E+00 23.6 new 0.377E+05 0.000E+00 12 -72.5823 -0.382E+06 72.6 1.86 0.000000E+00 0.000E+00 0.000E+00 17.8 new 0.215E+05 0.000E+00 13 -54.7919 -0.164E+06 54.8 1.74 0.000000E+00 0.000E+00 0.000E+00 13.4 new 0.122E+05 0.000E+00 14 -41.3609 -0.708E+05 41.4 1.62 0.000000E+00 0.000E+00 0.000E+00 10.1 new 0.698E+04 0.000E+00 15 -31.2225 -0.304E+05 31.2 1.49 0.000000E+00 0.000E+00 0.000E+00 7.65 new 0.398E+04 0.000E+00 16 -23.5691 -0.131E+05 23.6 1.37 0.000000E+00 0.000E+00 0.000E+00 5.78 new 0.227E+04 0.000E+00 17 -17.7918 -0.563E+04 17.8 1.25 0.000000E+00 0.000E+00 0.000E+00 4.36 new 0.129E+04 0.000E+00 18 -13.4306 -0.242E+04 13.4 1.13 0.000000E+00 0.000E+00 0.000E+00 3.29 new 736. 0.000E+00 19 -10.1385 -0.104E+04 10.1 1.01 0.000000E+00 0.000E+00 0.000E+00 2.49 new 419. 0.000E+00 20 -7.65332 -448. 7.65 0.884 0.000000E+00 0.000E+00 0.000E+00 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 3 Broyden method, B(0)=Identity. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) -1000.00 -0.100000E+10 0.000000E+00 0.000000E+00 Estimate of the root after 20 steps. x f(x) -7.65332 -448.281 0.000000E+00 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to Broyden method, with initial matrix the jacobian # Command? (H for help) Three methods are available: B Broyden's method. F Finite difference Newton method. N Newton method. Choose a method by typing B, F, or N. Please choose how Broyden's method will start: I Initial iteration matrix is identity. J Initial iteration matrix is jacobian. Command? (H for help) # Choose problem 1 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 2.00000 1.96 -2.00 0.301 2.00000 0.982 -2.00 -4.47 new -0.615 -2.90 1 -2.46741 -0.328 2.47 0.392 -0.900304 1.52 0.900 -54.4 new -0.524E-01 -21.2 2 -56.8887 -8.58 56.9 1.76 -22.0903 2.58 22.1 Bad step!. A norm increased. -4.55 new -0.496E-01 2.75 3 -61.4364 -6.09 61.4 1.79 -19.3408 3.18 19.3 8.46 new 0.346 2.24 4 -52.9801 -5.52 53.0 1.72 -17.0997 3.10 17.1 Bad step!. A norm increased. -0.391E+04 new -0.760E-03 -828. 5 -3960.78 -180. 0.396E+04 3.60 -845.339 4.69 845. Bad step!. A norm increased. 0.353E+04 new -0.795E-02 765. 6 -430.163 -14.8 430. 2.63 -79.8784 5.39 79.9 -462. new -0.913E-02 -72.2 7 -892.274 -25.9 892. 2.95 -152.077 5.87 152. Bad step!. A norm increased. -856. new -0.321E-02 -98.3 8 -1747.83 -35.9 0.175E+04 3.24 -250.368 6.98 250. Bad step!. A norm increased. -0.448E+04 new -0.516E-03 -448. 9 -6227.43 -78.3 0.623E+04 3.79 -698.271 8.92 698. Bad step!. A norm increased. 0.263E+04 new -0.125E-02 303. 10 -3602.04 -43.4 0.360E+04 3.56 -395.268 9.11 395. -0.139E+04 new -0.265E-02 -114. 11 -4993.91 -51.9 0.499E+04 3.70 -508.930 9.81 509. Bad step!. A norm increased. -0.585E+04 new -0.510E-03 -411. 12 -10840.8 -78.1 0.108E+05 4.04 -920.101 11.8 920. Bad step!. A norm increased. -0.128E+04 new -0.418E-03 -26.5 13 -12120.4 -73.9 0.121E+05 4.08 -946.649 12.8 947. 0.160E+04 new 0.168E-02 93.0 14 -10523.6 -69.3 0.105E+05 4.02 -853.696 12.3 854. Bad step!. A norm increased. -0.174E+05 new -0.170E-03 -922. 15 -27923.8 -113. 0.279E+05 4.45 -1775.44 15.7 0.178E+04 Bad step!. A norm increased. 0.420E+04 new -0.224E-03 300. 16 -23721.0 -91.8 0.237E+05 4.38 -1475.29 16.1 0.148E+04 -0.333E+04 new -0.108E-02 -156. 17 -27053.9 -98.3 0.271E+05 4.43 -1631.03 16.6 0.163E+04 Bad step!. A norm increased. -0.168E+05 new -0.178E-03 -685. 18 -43837.2 -122. 0.438E+05 4.64 -2315.73 18.9 0.232E+04 Bad step!. A norm increased. -0.162E+05 new -0.906E-04 -552. 19 -60074.9 -137. 0.601E+05 4.78 -2867.95 20.9 0.287E+04 Bad step!. A norm increased. -0.152E+06 new -0.872E-05 -0.498E+04 20 -212513. -290. 0.213E+06 5.33 -7844.79 27.1 0.784E+04 Bad step!. A norm increased. Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 2.00000 1.96337 2.00000 0.981684 Estimate of the root after 20 steps. x f(x) -212513. -289.586 -7844.79 27.0897 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 3.00 -3.00 0.477 3.00000 1.000 -3.00 -8.95 new -0.335 -5.97 1 -5.95360 -1.48 5.95 0.775 -2.96685 2.01 2.97 230. new 0.153E-01 98.5 2 223.771 40.8 -224. 2.35 95.4925 2.34 -95.5 Bad step!. A norm increased. -274. new -0.826E-01 -113. 3 -50.2023 -6.06 50.2 1.70 -17.4403 2.88 17.4 Bad step!. A norm increased. -50.4 new -0.703E-01 -13.9 4 -100.586 -9.79 101. 2.00 -31.3804 3.21 31.4 Bad step!. A norm increased. -136. new -0.192E-01 -28.6 5 -236.619 -15.2 237. 2.37 -59.9802 3.94 60.0 Bad step!. A norm increased. -291. new -0.616E-02 -49.1 6 -527.585 -22.6 528. 2.72 -109.090 4.84 109. Bad step!. A norm increased. -84.0 new -0.481E-02 -3.13 7 -611.609 -20.6 612. 2.79 -112.220 5.45 112. 98.5 new 0.272E-01 13.9 8 -513.131 -18.8 513. 2.71 -98.3408 5.22 98.3 Bad step!. A norm increased. -0.104E+04 new -0.285E-02 -130. 9 -1554.99 -33.5 0.155E+04 3.19 -228.347 6.81 228. Bad step!. A norm increased. 65.9 new -0.305E-02 22.9 10 -1489.06 -28.3 0.149E+04 3.17 -205.433 7.25 205. -66.7 new -0.472E-01 -7.07 11 -1555.73 -29.0 0.156E+04 3.19 -212.503 7.32 213. Bad step!. A norm increased. -0.140E+04 new -0.214E-02 -129. 12 -2959.57 -39.3 0.296E+04 3.47 -341.014 8.68 341. Bad step!. A norm increased. -0.129E+04 new -0.111E-02 -95.6 13 -4253.76 -44.8 0.425E+04 3.63 -436.646 9.74 437. Bad step!. A norm increased. -0.126E+05 new -0.104E-03 -885. 14 -16849.6 -104. 0.168E+05 4.23 -1321.25 12.8 0.132E+04 Bad step!. A norm increased. 0.117E+05 new -0.152E-02 847. 15 -5115.85 -44.0 0.512E+04 3.71 -474.342 10.8 474. -0.211E+04 new -0.185E-02 -127. 16 -7223.79 -50.0 0.722E+04 3.86 -600.900 12.0 601. Bad step!. A norm increased. 0.286E+06 new 0.138E-04 0.168E+05 17 278737. 943. -0.279E+06 5.45 16210.4 17.2 -0.162E+05 Bad step!. A norm increased. -0.289E+06 new -0.155E-02 -0.169E+05 18 -9763.32 -53.5 0.976E+04 3.99 -722.708 13.5 723. Bad step!. A norm increased. -0.365E+04 new -0.153E-02 -183. 19 -13413.8 -61.2 0.134E+05 4.13 -906.066 14.8 906. Bad step!. A norm increased. -0.311E+05 new -0.161E-03 -0.150E+04 20 -44532.1 -130. 0.445E+05 4.65 -2409.79 18.5 0.241E+04 Bad step!. A norm increased. Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 2.99963 3.00000 0.999877 Estimate of the root after 20 steps. x f(x) -44532.1 -130.403 -2409.79 18.4796 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 100.000 0.100E-01 -100. 2.00 1.00000 63.2 -1.00 -84.8 new -0.137E-01 -0.924 1 15.1531 0.379E-03 -15.2 1.18 0.757657E-01 1.14 -0.758E-01 -1.41 new -0.135E-01 -0.263E-01 2 13.7433 0.178E-03 -13.7 1.14 0.494602E-01 0.679 -0.495E-01 -2.29 new -0.513E-02 -0.353E-01 3 11.4575 0.176E-04 -11.5 1.06 0.141919E-01 0.163 -0.142E-01 Bad step!. A norm increased. -0.810 new -0.379E-02 -0.105E-01 4 10.6477 0.126E-05 -10.6 1.03 0.366091E-02 0.390E-01 -0.366E-02 -0.270 new -0.284E-02 -0.330E-02 5 10.3782 0.128E-07 -10.4 1.02 0.364519E-03 0.378E-02 -0.365E-03 -0.295E-01 new -0.256E-02 -0.354E-03 6 10.3488 0.105E-10 -10.3 1.01 0.104032E-04 0.108E-03 -0.104E-04 -0.864E-03 new -0.249E-02 -0.104E-04 7 10.3479 0.895E-16 -10.3 1.01 0.304340E-07 0.302E-06 -0.304E-07 -0.243E-05 new -0.248E-02 -0.292E-07 8 10.3479 0.153E-18 -10.3 1.01 0.125630E-08 0.000E+00 -0.126E-08 0.636E-15 new -0.248E-02 0.203E-20 9 10.3479 0.153E-18 -10.3 1.01 0.125630E-08 0.000E+00 -0.126E-08 Bad step!. A norm increased. -0.197E-03 new 0.803E-14 -0.628E-09 10 10.3477 0.381E-19 -10.3 1.01 0.628122E-09 0.000E+00 -0.628E-09 Bad step!. A norm increased. -0.655E-04 new 0.602E-14 -0.209E-09 11 10.3476 0.169E-19 -10.3 1.01 0.418749E-09 0.000E+00 -0.419E-09 -0.524E-04 new 0.335E-14 -0.168E-09 12 10.3476 0.610E-20 -10.3 1.01 0.251247E-09 0.000E+00 -0.251E-09 -0.295E-04 new 0.214E-14 -0.942E-10 13 10.3475 0.238E-20 -10.3 1.01 0.157029E-09 0.000E+00 -0.157E-09 -0.189E-04 new 0.131E-14 -0.604E-10 14 10.3475 0.902E-21 -10.3 1.01 0.966332E-10 0.000E+00 -0.966E-10 -0.115E-04 new 0.811E-15 -0.368E-10 15 10.3475 0.346E-21 -10.3 1.01 0.598205E-10 0.000E+00 -0.598E-10 -0.716E-05 new 0.500E-15 -0.229E-10 16 10.3475 0.132E-21 -10.3 1.01 0.369479E-10 0.000E+00 -0.369E-10 -0.442E-05 new 0.309E-15 -0.141E-10 17 10.3475 0.504E-22 -10.3 1.01 0.228405E-10 0.000E+00 -0.228E-10 -0.273E-05 new 0.191E-15 -0.873E-11 18 10.3475 0.193E-22 -10.3 1.01 0.141149E-10 0.000E+00 -0.141E-10 -0.169E-05 new 0.118E-15 -0.539E-11 19 10.3475 0.735E-23 -10.3 1.01 0.872382E-11 0.000E+00 -0.872E-11 -0.104E-05 new 0.730E-16 -0.333E-11 20 10.3475 0.281E-23 -10.3 1.01 0.539154E-11 0.000E+00 -0.539E-11 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 100.000 0.100000E-01 1.00000 63.2121 Estimate of the root after 20 steps. x f(x) 10.3475 0.280925E-23 0.539154E-11 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1.00000 0.632E+04 -1.00 2.00 100.000 0.100E-01 -100. -1.39 new -1.37 -38.6 1 -0.386351 -0.135E+04 0.386 1.79 61.3649 -0.630E-02 -61.4 0.601 new -1.55 4.51 2 0.214587 910. -0.215 1.82 65.8761 0.326E-02 -65.9 -0.167 new -2.81 -2.04 3 0.478738E-01 195. -0.479E-01 1.81 63.8384 0.750E-03 -63.8 -0.533E-01 new -2.23 -0.524 4 -0.544736E-02 -21.8 0.545E-02 1.80 63.3147 -0.860E-04 -63.3 0.560E-02 new -2.48 0.520E-01 5 0.147642E-03 0.593 -0.148E-03 1.80 63.3667 0.233E-05 -63.4 -0.147E-03 new -2.55 -0.138E-02 6 0.461747E-06 0.185E-02 -0.462E-06 1.80 63.3653 0.729E-08 -63.4 -0.462E-06 new -2.54 -0.432E-05 7 -0.137866E-09 0.000E+00 0.138E-09 1.80 63.3653 -0.218E-11 -63.4 0.270E-09 new -2.54 -0.909E-09 8 0.132589E-09 0.000E+00 -0.133E-09 1.80 63.3653 0.209E-11 -63.4 -0.133E-09 new -4.98 0.446E-09 9 -0.969688E-21 0.000E+00 0.970E-21 1.80 63.3653 -0.153E-22 -63.4 0.970E-21 new -4.98 -0.326E-20 10 -0.682031E-32 0.000E+00 0.682E-32 1.80 63.3653 -0.108E-33 -63.4 0.682E-32 new -4.98 -0.229E-31 11 0.000000E+00 0.000E+00 0.000E+00 1.80 63.3653 0.000E+00 -63.4 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 12 Jacobian evaluations: 1 Matrix factorizations; 11 Linear system solves: 11 Starting point: x f(x) 1.00000 6321.21 100.000 0.100000E-01 Estimate of the root after 11 steps. x f(x) 0.000000E+00 0.000000E+00 63.3653 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682670 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233808E-01 0.877E-02 -0.234E-01 -0.213 0.612668 0.119E-01 -0.613 -0.224E-01 new -0.198 -0.225E-02 2 0.963611E-03 0.359E-03 -0.964E-03 -0.214 0.610414 0.491E-03 -0.610 -0.963E-03 new -0.189 -0.920E-04 3 0.775916E-06 0.289E-06 -0.776E-06 -0.214 0.610322 0.395E-06 -0.610 -0.776E-06 new -0.189 -0.740E-07 4 0.343003E-10 0.000E+00 -0.343E-10 -0.214 0.610322 0.175E-10 -0.610 -0.461E-10 new -0.189 0.300E-10 5 -0.117707E-10 0.000E+00 0.118E-10 -0.214 0.610322 -0.600E-11 -0.610 0.118E-10 new -0.254 -0.767E-11 6 0.280306E-21 0.000E+00 -0.280E-21 -0.214 0.610322 0.143E-21 -0.610 -0.280E-21 new -0.254 0.183E-21 7 0.229071E-32 0.000E+00 -0.229E-32 -0.214 0.610322 0.117E-32 -0.610 -0.229E-32 new -0.254 0.149E-32 8 0.000000E+00 0.000E+00 0.000E+00 -0.214 0.610322 0.000E+00 -0.610 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 9 Jacobian evaluations: 1 Matrix factorizations; 8 Linear system solves: 8 Starting point: x f(x) 0.400000 0.172267 0.682670 0.218270 Estimate of the root after 8 steps. x f(x) 0.000000E+00 0.000000E+00 0.610322 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 0.400000 0.172 -0.400 -0.166 0.682680 0.218 -0.683 -0.377 new -0.209 -0.700E-01 1 0.233802E-01 0.877E-02 -0.234E-01 -0.213 0.612677 0.119E-01 -0.613 -0.224E-01 new -0.198 -0.225E-02 2 0.963562E-03 0.359E-03 -0.964E-03 -0.214 0.610424 0.491E-03 -0.610 -0.963E-03 new -0.189 -0.920E-04 3 0.775843E-06 0.289E-06 -0.776E-06 -0.214 0.610332 0.395E-06 -0.610 -0.776E-06 new -0.189 -0.740E-07 4 0.356892E-10 0.000E+00 -0.357E-10 -0.214 0.610332 0.182E-10 -0.610 -0.479E-10 new -0.189 0.312E-10 5 -0.122470E-10 0.000E+00 0.122E-10 -0.214 0.610332 -0.624E-11 -0.610 0.122E-10 new -0.254 -0.798E-11 6 0.303454E-21 0.000E+00 -0.303E-21 -0.214 0.610332 0.155E-21 -0.610 -0.303E-21 new -0.254 0.198E-21 7 0.258021E-32 0.000E+00 -0.258E-32 -0.214 0.610332 0.131E-32 -0.610 -0.258E-32 new -0.254 0.168E-32 8 -0.342114E-48 0.000E+00 0.342E-48 -0.214 0.610332 -0.174E-48 -0.610 0.342E-48 new -0.254 -0.223E-48 9 0.000000E+00 0.000E+00 0.000E+00 -0.214 0.610332 0.000E+00 -0.610 Iteration halted, point accepted. Command? (H for help) Problem being solved: Homework problem 1. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 10 Jacobian evaluations: 1 Matrix factorizations; 9 Linear system solves: 9 Starting point: x f(x) 0.400000 0.172272 0.682680 0.218272 Estimate of the root after 9 steps. x f(x) 0.000000E+00 0.000000E+00 0.610332 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # Switch to problem 2 # Command? (H for help) There are 3 problems available: 1: Homework problem 1. 2: Homework problem 2. 3: Dennis and Schnabel example Which one would you like to work on? Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 3.00000 54.0 -3.00 0.477 3.00000 54.0 -3.00 -1.00 new 972. -1.00 1 2.00000 16.0 -2.00 0.301 2.00000 16.0 -2.00 -0.421 new 684. -0.421 2 1.57895 7.87 -1.58 0.198 1.57895 7.87 -1.58 -0.408 new 347. -0.408 3 1.17107 3.21 -1.17 0.686E-01 1.17107 3.21 -1.17 -0.281 new 206. -0.281 4 0.889980 1.41 -0.890 -0.506E-01 0.889980 1.41 -0.890 -0.220 new 115. -0.220 5 0.670083 0.602 -0.670 -0.174 0.670083 0.602 -0.670 -0.164 new 66.1 -0.164 6 0.506336 0.260 -0.506 -0.296 0.506336 0.260 -0.506 -0.124 new 37.6 -0.124 7 0.382075 0.112 -0.382 -0.418 0.382075 0.112 -0.382 -0.936E-01 new 21.4 -0.936E-01 8 0.288462 0.480E-01 -0.288 -0.540 0.288462 0.480E-01 -0.288 -0.707E-01 new 12.2 -0.707E-01 9 0.217741 0.206E-01 -0.218 -0.662 0.217741 0.206E-01 -0.218 -0.534E-01 new 6.96 -0.534E-01 10 0.164372 0.888E-02 -0.164 -0.784 0.164372 0.888E-02 -0.164 -0.403E-01 new 3.97 -0.403E-01 11 0.124080 0.382E-02 -0.124 -0.906 0.124080 0.382E-02 -0.124 -0.304E-01 new 2.26 -0.304E-01 12 0.936652E-01 0.164E-02 -0.937E-01 -1.03 0.936652E-01 0.164E-02 -0.937E-01 -0.230E-01 new 1.29 -0.230E-01 13 0.707057E-01 0.707E-03 -0.707E-01 -1.15 0.707057E-01 0.707E-03 -0.707E-01 -0.173E-01 new 0.734 -0.173E-01 14 0.533742E-01 0.304E-03 -0.534E-01 -1.27 0.533742E-01 0.304E-03 -0.534E-01 -0.131E-01 new 0.418 -0.131E-01 15 0.402909E-01 0.131E-03 -0.403E-01 -1.39 0.402909E-01 0.131E-03 -0.403E-01 -0.988E-02 new 0.238 -0.988E-02 16 0.304147E-01 0.563E-04 -0.304E-01 -1.52 0.304147E-01 0.563E-04 -0.304E-01 -0.746E-02 new 0.136 -0.746E-02 17 0.229594E-01 0.242E-04 -0.230E-01 -1.64 0.229594E-01 0.242E-04 -0.230E-01 -0.563E-02 new 0.774E-01 -0.563E-02 18 0.173315E-01 0.104E-04 -0.173E-01 -1.76 0.173315E-01 0.104E-04 -0.173E-01 -0.425E-02 new 0.441E-01 -0.425E-02 19 0.130832E-01 0.448E-05 -0.131E-01 -1.88 0.130832E-01 0.448E-05 -0.131E-01 -0.321E-02 new 0.251E-01 -0.321E-02 20 0.987621E-02 0.193E-05 -0.988E-02 -2.01 0.987621E-02 0.193E-05 -0.988E-02 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 3.00000 54.0000 3.00000 54.0000 Estimate of the root after 20 steps. x f(x) 0.987621E-02 0.192664E-05 0.987621E-02 0.192664E-05 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 1000.00 0.200E+10 -0.100E+04 3.00 1000.00 0.200E+10 -0.100E+04 -333. new 0.120E+14 -333. 1 666.667 0.593E+09 -667. 2.82 666.667 0.593E+09 -667. -140. new 0.844E+13 -140. 2 526.316 0.292E+09 -526. 2.72 526.316 0.292E+09 -526. -136. new 0.429E+13 -136. 3 390.356 0.119E+09 -390. 2.59 390.356 0.119E+09 -390. -93.7 new 0.254E+13 -93.7 4 296.660 0.522E+08 -297. 2.47 296.660 0.522E+08 -297. -73.3 new 0.142E+13 -73.3 5 223.361 0.223E+08 -223. 2.35 223.361 0.223E+08 -223. -54.6 new 0.817E+12 -54.6 6 168.779 0.962E+07 -169. 2.23 168.779 0.962E+07 -169. -41.4 new 0.464E+12 -41.4 7 127.358 0.413E+07 -127. 2.11 127.358 0.413E+07 -127. -31.2 new 0.265E+12 -31.2 8 96.1542 0.178E+07 -96.2 1.98 96.1542 0.178E+07 -96.2 -23.6 new 0.151E+12 -23.6 9 72.5805 0.765E+06 -72.6 1.86 72.5805 0.765E+06 -72.6 -17.8 new 0.860E+11 -17.8 10 54.7906 0.329E+06 -54.8 1.74 54.7906 0.329E+06 -54.8 -13.4 new 0.490E+11 -13.4 11 41.3598 0.142E+06 -41.4 1.62 41.3598 0.142E+06 -41.4 -10.1 new 0.279E+11 -10.1 12 31.2217 0.609E+05 -31.2 1.49 31.2217 0.609E+05 -31.2 -7.65 new 0.159E+11 -7.65 13 23.5686 0.262E+05 -23.6 1.37 23.5686 0.262E+05 -23.6 -5.78 new 0.906E+10 -5.78 14 17.7914 0.113E+05 -17.8 1.25 17.7914 0.113E+05 -17.8 -4.36 new 0.517E+10 -4.36 15 13.4303 0.484E+04 -13.4 1.13 13.4303 0.484E+04 -13.4 -3.29 new 0.294E+10 -3.29 16 10.1382 0.208E+04 -10.1 1.01 10.1382 0.208E+04 -10.1 -2.49 new 0.168E+10 -2.49 17 7.65314 896. -7.65 0.884 7.65314 896. -7.65 -1.88 new 0.956E+09 -1.88 18 5.77718 386. -5.78 0.762 5.77718 386. -5.78 -1.42 new 0.545E+09 -1.42 19 4.36106 166. -4.36 0.640 4.36106 166. -4.36 -1.07 new 0.310E+09 -1.07 20 3.29207 71.4 -3.29 0.517 3.29207 71.4 -3.29 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) 1000.00 0.200000E+10 1000.00 0.200000E+10 Estimate of the root after 20 steps. x f(x) 3.29207 71.3571 3.29207 71.3571 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) The previous iteration is now cancelled. Enter new x Command? (H for help) step x fx step It-matrix Det error log(e-norm) 0 -1000.00 -0.100E+10 0.100E+04 3.00 0.000000E+00 0.000E+00 0.000E+00 333. new 0.300E+13 0.000E+00 1 -666.667 -0.296E+09 667. 2.82 0.000000E+00 0.000E+00 0.000E+00 140. new 0.211E+13 0.000E+00 2 -526.316 -0.146E+09 526. 2.72 0.000000E+00 0.000E+00 0.000E+00 136. new 0.107E+13 0.000E+00 3 -390.356 -0.595E+08 390. 2.59 0.000000E+00 0.000E+00 0.000E+00 93.7 new 0.635E+12 0.000E+00 4 -296.660 -0.261E+08 297. 2.47 0.000000E+00 0.000E+00 0.000E+00 73.3 new 0.356E+12 0.000E+00 5 -223.361 -0.111E+08 223. 2.35 0.000000E+00 0.000E+00 0.000E+00 54.6 new 0.204E+12 0.000E+00 6 -168.779 -0.481E+07 169. 2.23 0.000000E+00 0.000E+00 0.000E+00 41.4 new 0.116E+12 0.000E+00 7 -127.358 -0.207E+07 127. 2.11 0.000000E+00 0.000E+00 0.000E+00 31.2 new 0.662E+11 0.000E+00 8 -96.1542 -0.889E+06 96.2 1.98 0.000000E+00 0.000E+00 0.000E+00 23.6 new 0.377E+11 0.000E+00 9 -72.5805 -0.382E+06 72.6 1.86 0.000000E+00 0.000E+00 0.000E+00 17.8 new 0.215E+11 0.000E+00 10 -54.7906 -0.164E+06 54.8 1.74 0.000000E+00 0.000E+00 0.000E+00 13.4 new 0.122E+11 0.000E+00 11 -41.3598 -0.708E+05 41.4 1.62 0.000000E+00 0.000E+00 0.000E+00 10.1 new 0.698E+10 0.000E+00 12 -31.2217 -0.304E+05 31.2 1.49 0.000000E+00 0.000E+00 0.000E+00 7.65 new 0.398E+10 0.000E+00 13 -23.5686 -0.131E+05 23.6 1.37 0.000000E+00 0.000E+00 0.000E+00 5.78 new 0.227E+10 0.000E+00 14 -17.7914 -0.563E+04 17.8 1.25 0.000000E+00 0.000E+00 0.000E+00 4.36 new 0.129E+10 0.000E+00 15 -13.4303 -0.242E+04 13.4 1.13 0.000000E+00 0.000E+00 0.000E+00 3.29 new 0.736E+09 0.000E+00 16 -10.1382 -0.104E+04 10.1 1.01 0.000000E+00 0.000E+00 0.000E+00 2.49 new 0.419E+09 0.000E+00 17 -7.65314 -448. 7.65 0.884 0.000000E+00 0.000E+00 0.000E+00 1.88 new 0.239E+09 0.000E+00 18 -5.77718 -193. 5.78 0.762 0.000000E+00 0.000E+00 0.000E+00 1.42 new 0.136E+09 0.000E+00 19 -4.36106 -82.9 4.36 0.640 0.000000E+00 0.000E+00 0.000E+00 1.07 new 0.776E+08 0.000E+00 20 -3.29207 -35.7 3.29 0.517 0.000000E+00 0.000E+00 0.000E+00 Maximum number of steps taken. Use the I command to continue iteration. Command? (H for help) Problem being solved: Homework problem 2. Number of equations = 2 Absolute error tolerance= 0.0000000000000000E+00 Relative error tolerance= 1.0000000000000000E-05 Machine precision= 2.2204460492503130E-16 Difference parameter for jacobians= 1.0000000000000000E-02 Solution method number: 4 Broyden method, B(0)=Jacobian. Output level for iterations= 2 Maximum number of iterations before pause= 20 Number of steps before the iteration matrix is updated= 1 No damping is used. The vector norm used is the L-infinity norm. Work so far for this problem: Function evaluations: 21 Jacobian evaluations: 1 Matrix factorizations; 20 Linear system solves: 20 Starting point: x f(x) -1000.00 -0.100000E+10 0.000000E+00 0.000000E+00 Estimate of the root after 20 steps. x f(x) -3.29207 -35.6786 0.000000E+00 0.000000E+00 Exact solution: x f(x) 0.000000E+00 0.000000E+00 0.000000E+00 0.000000E+00 Command? (H for help) # That's all! # Command? (H for help) NEWNON has been requested to stop.