NAG Library Manual, Mark 29.2
Interfaces:  FL   CL   CPP   AD 

NAG FL Interface Introduction
Example description

 D02TXF Example Program Results



 Tolerance =  0.1E-04  L =   60.000  S = 0.2400

 Used a mesh of   21 points
 Maximum error =   0.27E-07  in interval    7 for component    1

 Solution on original interval:
      x        f          g
     0.00     0.0000     0.0000
     2.00    -0.9769     0.8011
     4.00    -2.0900     1.1459
     6.00    -2.6093     1.2389
     8.00    -2.5498     1.1794
    10.00    -2.1397     1.0478
    12.00    -1.7176     0.9395
    14.00    -1.5465     0.9206
    16.00    -1.6127     0.9630
    18.00    -1.7466     1.0068
    20.00    -1.8286     1.0244
    22.00    -1.8338     1.0185
    24.00    -1.7956     1.0041
    26.00    -1.7582     0.9940
    28.00    -1.7445     0.9926
    30.00    -1.7515     0.9965
    33.00    -1.7695     1.0019
    36.00    -1.7730     1.0018
    39.00    -1.7673     0.9998
    42.00    -1.7645     0.9993
    45.00    -1.7659     0.9999
    48.00    -1.7672     1.0002
    51.00    -1.7671     1.0001
    54.00    -1.7666     0.9999
    57.00    -1.7665     0.9999
    60.00    -1.7666     1.0000


 Tolerance =  0.1E-04  L =  120.000  S = 0.1440

 Used a mesh of   21 points
 Maximum error =   0.69E-05  in interval    7 for component    2

 Solution on original interval:
      x        f          g
     0.00     0.0000     0.0000
     2.00    -1.1406     0.7317
     4.00    -2.6531     1.1315
     6.00    -3.6721     1.3250
     8.00    -4.0539     1.3707
    10.00    -3.8285     1.3003
    12.00    -3.1339     1.1407
    14.00    -2.2469     0.9424
    16.00    -1.6146     0.8201
    18.00    -1.5472     0.8549
    20.00    -1.8483     0.9623
    22.00    -2.1761     1.0471
    24.00    -2.3451     1.0778
    26.00    -2.3236     1.0600
    28.00    -2.1784     1.0165
    30.00    -2.0214     0.9775
    39.00    -2.1109     1.0155
    48.00    -2.0362     0.9931
    57.00    -2.0709     1.0023
    66.00    -2.0588     0.9995
    75.00    -2.0616     1.0000
    84.00    -2.0615     1.0001
    93.00    -2.0611     0.9999
   102.00    -2.0614     1.0000
   111.00    -2.0613     1.0000
   120.00    -2.0613     1.0000


 Tolerance =  0.1E-04  L =  240.000  S = 0.0864

 Used a mesh of   81 points
 Maximum error =   0.33E-06  in interval   19 for component    2

 Solution on original interval:
      x        f          g
     0.00     0.0000     0.0000
     2.00    -1.2756     0.6404
     4.00    -3.1604     1.0463
     6.00    -4.7459     1.3011
     8.00    -5.8265     1.4467
    10.00    -6.3412     1.5036
    12.00    -6.2862     1.4824
    14.00    -5.6976     1.3886
    16.00    -4.6568     1.2263
    18.00    -3.3226     1.0042
    20.00    -2.0328     0.7718
    22.00    -1.4035     0.6943
    24.00    -1.6603     0.8218
    26.00    -2.2975     0.9928
    28.00    -2.8661     1.1139
    30.00    -3.1641     1.1641
    51.00    -2.5307     1.0279
    72.00    -2.3520     0.9919
    93.00    -2.3674     0.9975
   114.00    -2.3799     1.0003
   135.00    -2.3800     1.0002
   156.00    -2.3792     1.0000
   177.00    -2.3791     1.0000
   198.00    -2.3792     1.0000
   219.00    -2.3792     1.0000
   240.00    -2.3792     1.0000