equation-brent 実行例

help出力 equation-brent.py --help

Solution of EF by Brent's method

  EFmin =   0.00000000  dQmin =   -1.201e+19
  EFmax =   1.00000000  dQmax =      2.1e+18
  Iter 0: EFa,b =   1.00000000 -   0.85118392  dQa,b =      2.1e+18 - -2.1449726e+15
     EFs =   0.00000000  dQs =       0.8512
     Ne=6.6409e+15  Nh=6.0254e+04  NA-=1.0000e+15  ND+=9.7859e+15  dQ=-2.1450e+15
  Iter 1: EFa,b =   1.00000000 -   0.85133578  dQa,b =      2.1e+18 - -2.1046118e+15
     EFs =   1.00000000  dQs =       0.8513
     Ne=6.6801e+15  Nh=5.9901e+04  NA-=1.0000e+15  ND+=9.7847e+15  dQ=-2.1046e+15
  Iter 2: EFa,b =   0.85133578 -   0.85924693  dQa,b =   -2.105e+15 - 3.6172485e+14
     EFs =   2.00000000  dQs =       0.8592
     Ne=9.0716e+15  Nh=4.4109e+04  NA-=1.0000e+15  ND+=9.7098e+15  dQ=3.6172e+14
  Iter 3: EFa,b =   0.85529136 -   0.85924693  dQa,b =   -9.655e+14 - 3.6172485e+14
     EFs =   3.00000000  dQs =       0.8553
     Ne=7.7845e+15  Nh=5.1402e+04  NA-=1.0000e+15  ND+=9.7500e+15  dQ=-9.6546e+14
  Iter 4: EFa,b =   0.85924693 -   0.85726915  dQa,b =    3.617e+14 - -3.2719712e+14
     EFs =   4.00000000  dQs =       0.8573
     Ne=8.4034e+15  Nh=4.7616e+04  NA-=1.0000e+15  ND+=9.7306e+15  dQ=-3.2720e+14
  Iter 5: EFa,b =   0.85924693 -   0.85820848  dQa,b =    3.617e+14 - -6.5555368e+12
     EFs =   5.00000000  dQs =       0.8582
     Ne=8.7144e+15  Nh=4.5917e+04  NA-=1.0000e+15  ND+=9.7209e+15  dQ=-6.5555e+12
  Iter 6: EFa,b =   0.85820848 -   0.85822734  dQa,b =   -6.556e+12 - 3.1855438e+09
     EFs =   6.00000000  dQs =       0.8582
     Ne=8.7208e+15  Nh=4.5884e+04  NA-=1.0000e+15  ND+=9.7207e+15  dQ=3.1855e+09
  Iter 7: EFa,b =   0.85822734 -   0.85822733  dQa,b =    3.186e+09 -     -1159776
     EFs =   7.00000000  dQs =       0.8582
     Ne=8.7207e+15  Nh=4.5884e+04  NA-=1.0000e+15  ND+=9.7207e+15  dQ=-1.1598e+06
  Success: Convergence reached at EF = 0.8582273368600042

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