Question

我在Mathematica中遇到以下问题：

values = {b -> 1, c -> 0};
Solutions := Solve[x^2 + b*x + c == 0, x]
x1 := x /. Solutions[[1]];
x2 := x /. Solutions[[2]];
"Solution 1"
x1
"Solution 2"
x2
"Choose ~preferred~ Solution, which is -1 when using values"
If[Module[{list = values}, ReplaceRepeated[x1 == -1, list]], x = x1, x = x2]
"~Preferred~ Solution"
x

当我第一次评估它时，一切正常：

Solution 1
1/2 (-b - Sqrt[b^2 - 4 c])
Solution 2
1/2 (-b + Sqrt[b^2 - 4 c])
Choose ~preferred~ Solution, which is -1 when using values
1/2 (-b - Sqrt[b^2 - 4 c])

但通过第二次评估它会发生几次错误：

Solution 1
General::ivar: 1/2 (-b-Sqrt[b^2-4 c]) is not a valid variable. >>
General::ivar: 1/2 (-b-Sqrt[b^2-4 c]) is not a valid variable. >>
ReplaceAll::reps: {1/2 b (-b-Sqrt[Plus[<<2>>]])+1/4 (-b-Power[<<2>>])^2+c==0} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >>
1/2 (-b - Sqrt[b^2 - 4 c]) /. 
1/2 b (-b - Sqrt[b^2 - 4 c]) + 1/4 (-b - Sqrt[b^2 - 4 c])^2 + c == 0
"Solution 2"
...

在我看来，if条件中的ReplaceRepeated可以全局工作，尽管它是一个Module环境。有人可以帮忙吗？我是如何解决这个问题的？

Answer 1

第一次评估你的代码

In[1]:= values = {b -> 1, c -> 0};
Solutions := Solve[x^2 + b*x + c == 0, x]
x1 := x /. Solutions[[1]];
x2 := x /. Solutions[[2]];
"Solution 1"
x1
"Solution 2"
x2
"Choose ~preferred~ Solution, which is -1 when using values"
If[Module[{list = values}, ReplaceRepeated[x1 == -1, list]], x = x1, x = x2]
"~Preferred~ Solution"
x

Out[5]= "Solution 1"
Out[6]= 1/2 (-b - Sqrt[b^2 - 4 c])
Out[7]= "Solution 2"
Out[8]= 1/2 (-b + Sqrt[b^2 - 4 c])
Out[9]= "Choose ~preferred~ Solution, which is -1 when using values"
Out[10]= 1/2 (-b - Sqrt[b^2 - 4 c])
Out[11]= "~Preferred~ Solution"
Out[12]= 1/2 (-b - Sqrt[b^2 - 4 c])

一切都很好。现在x的价值是多少？

In[13]:= x
Out[13]= 1/2 (-b - Sqrt[b^2 - 4 c])

没关系。现在开始第二次评估您的代码。

In[14]:= values = {b -> 1, c -> 0};
Solutions := Solve[x^2 + b*x + c == 0, x]

假设您要立即评估解决方案。这将评估你在x中的二次方，但是你看到x不再只是一个没有值的符号，它将在Out [13]中使用x的值，而解决方案是

In[15]:= Solutions

During evaluation of In[15]:= Solve::ivar: 1/2 (-b-Sqrt[b^2-4 c]) is not a valid variable. >>

Out[15]= Solve[1/2b(-b-Sqrt[b^2-4c])+1/4(-b-Sqrt[b^2-4c])^2+c==0, 1/2(-b-Sqrt[b^2-4c])]

这就是它失败的原因。您之前为x分配了一个值，您正在使用x，因此在您的Solve中使用该值。也许你想在评估所有这些之前清除你的x。

mathematica General :: ivar ...不是有效变量

1 个答案: