Mathematica 8 中函数声明的问题
这是一个奇怪的结果，在此示例中函数定义为“functionB”。有人可以解释一下吗？我想绘制 functionB[x] 和 functionB[Sqrt[x]]，它们必须不同，但此代码显示 functionB[x] = functionB [Sqrt[x]]，这是不可能的。
model = 4/Sqrt[3] - a1/(x + b1) - a2/(x + b2)^2 - a3/(x + b3)^4;
fit = {a1 -> 0.27, a2 -> 0.335, a3 -> -0.347, b1 -> 4.29, b2 -> 0.435,
b3 -> 0.712};
functionB[x_] := model /. fit
Show[
ParametricPlot[{x, functionB[x]}, {x, 0, 1}],
ParametricPlot[{x, functionB[Sqrt[x]]}, {x, 0, 1}]
]
functionB[x] 必须与 functionB[Sqrt[x]] 不同，但在这种情况下，两行是相同的（这是不正确的）。
This is a strange result with a function defined as "functionB" in this example. Can someone explain this? I want to plot functionB[x] and functionB[Sqrt[x]], they must be different, but this code shows that functionB[x] = functionB[Sqrt[x]], which is impossible.
model = 4/Sqrt[3] - a1/(x + b1) - a2/(x + b2)^2 - a3/(x + b3)^4;
fit = {a1 -> 0.27, a2 -> 0.335, a3 -> -0.347, b1 -> 4.29, b2 -> 0.435,
b3 -> 0.712};
functionB[x_] := model /. fit
Show[
ParametricPlot[{x, functionB[x]}, {x, 0, 1}],
ParametricPlot[{x, functionB[Sqrt[x]]}, {x, 0, 1}]
]
functionB[x] must different from functionB[Sqrt[x]], but in this case, the 2 lines are the same (which is incorrect).
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如果您尝试
?functionB，您会看到它存储为functionB[x_]:=model/.fit。因此，只要您现在有了functionB[y]，对于任何y，Mathematica 都会计算model/.fit，获得4/Sqrt [3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + x)^4 - 0.27/(4.29 + x)。这与使用
SetDelayed（即:=）有关。每次 Mathematica 看到模式f[_]时，functionB[x_]:=model/.fit的 rhs 都会重新计算。您将模式命名为x是无关紧要的。您想要的可以通过例如 functionC[x_] = model / 来实现。适合。即通过使用
Set(=) 而不是SetDelayed(:=) 来评估右旋。希望这足够清楚（可能不是）......
If you try
?functionB, you'll see that it is stored asfunctionB[x_]:=model/.fit. Thus, whenever you now havefunctionB[y], for anyy, Mathematica evaluatesmodel/.fit, obtaining4/Sqrt[3] - 0.335/(0.435 + x)^2 + 0.347/(0.712 + x)^4 - 0.27/(4.29 + x).This has to do with using
SetDelayed(i.e.,:=). The rhs offunctionB[x_]:=model/.fitis evaluated anew each time Mathematica sees the patternf[_]. That you have named the patternxis irrelevant.What you want could be achieved by e.g.
functionC[x_] = model /. fit. That is, by usingSet(=) rather thanSetDelayed(:=), so as to evaluate the rhs.Hope this is clear enough (it probably isn't)...
您可能想尝试在 functionB 中定义模型，因此两个地方的 x 都是相关的：
You might want to try defining the model inside functionB so x in both places are related: