首先,这看起来(来自 ContourPlot)是一个相当简单的最大化问题,为什么使用牛顿法的 FindMaximum 会出现问题?
其次,如何摆脱警告?
第三,如果我无法摆脱这些警告,我如何判断警告是否有意义,即最大化失败?
例如,在下面的代码中,使用 Newton 方法的 FindMaximum 会发出警告,而 PrimaryAxis 方法则不会
o = 1/5 Log[E^(-(h/Sqrt[3]))/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
3/10 Log[E^(h/Sqrt[3])/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/5 Log[E^(-(h/Sqrt[3]) - Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(h/Sqrt[3] - Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(-Sqrt[3] h + Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(Sqrt[3] h + Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))];
(* -1 makes more contours towards maximum *)
contourFunc[n_, p_] := Function[{min, max},
range = max - min;
Table[Exp[p (x - 1)] x range + min, {x, 0, 1, 1/n}]
];
cf = contourFunc[10, -1];
ContourPlot @@ {o, {j, -1, 1}, {h, -1, 1}, Contours -> cf}
FindMaximum @@ {o, {{j, 0}, {h, 0}}, Method -> "Newton"}
FindMaximum @@ {o, {{j, 0}, {h, 0}}, Method -> "PrincipalAxis"}
请注意,我认为可能其中一个组件的方向上的梯度为 0 是问题所在,但如果我扰乱初始点,我仍然会收到相同的警告,这是一个示例
o = 1/5 Log[E^(-(h/Sqrt[3]))/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/5 Log[E^(h/Sqrt[3])/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(-(h/Sqrt[3]) - Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
3/10 Log[E^(h/Sqrt[3] - Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(-Sqrt[3] h + Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))] +
1/10 Log[E^(Sqrt[3] h + Sqrt[2] j)/(
2 E^(-(h/Sqrt[3])) + 2 E^(h/Sqrt[3]) +
E^(-(h/Sqrt[3]) - Sqrt[2] j) + E^(h/Sqrt[3] - Sqrt[2] j) +
E^(-Sqrt[3] h + Sqrt[2] j) + E^(Sqrt[3] h + Sqrt[2] j))];
ContourPlot @@ {o, {j, -1, 1}, {h, -1, 1}}
FindMaximum @@ {o, {{j, -0.008983550852535105`}, {h,
0.06931364191023386`}}, Method -> "Newton"}