简化 Mathematica 中的绝对值
形式的术语
Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]
我目前有一个大型表达式，其中有许多我从问题的几何形状中知道的
-2 b + 2 d1 m + l Tan[\[Theta]] > 0
，但是，当我尝试简化我的表达式时，
Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + l Tan[\[Theta]] > 0]
我只是回到
Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]
如何使 Mathematica 简化不必要的绝对值？
编辑1
我试图简化的完整表达式是
-(1/(2 (m - Tan[\[Theta]])))
Sqrt[1 + m^2] (B2 Sqrt[(-2 b + 2 d1 m + l Tan[\[Theta]])^2] +
B4 Sqrt[(-2 b + 2 d2 m + l Tan[\[Theta]])^2] +
B5 Sqrt[(2 b + 2 d3 m + l Tan[\[Theta]])^2] +
B7 Sqrt[(2 b + 2 d4 m + l Tan[\[Theta]])^2] +
B1 Sqrt[(2 b - 2 (d1 + l) m + l Tan[\[Theta]])^2] +
B3 Sqrt[(2 b - 2 (d2 + l) m + l Tan[\[Theta]])^2] +
B6 Sqrt[(-2 (b + (d3 + l) m) + l Tan[\[Theta]])^2] +
B8 Sqrt[(-2 (b + (d4 + l) m) + l Tan[\[Theta]])^2])
每个根号下的平方项已知是正实数。
I currently have a large expression with many terms of the form
Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]
I know, from the geometry of my problem, that
-2 b + 2 d1 m + l Tan[\[Theta]] > 0
However, when I try to simplify my expression,
Simplify[Abs[-2 b + 2 d1 m + l Tan[\[Theta]]], -2 b + 2 d1 m + l Tan[\[Theta]] > 0]
I just get back
Abs[-2 b + 2 d1 m + l Tan[\[Theta]]]
How can I make Mathematica simplify out the unnecessary absolute value?
EDIT 1
The full expression which I'm trying to simplify is
-(1/(2 (m - Tan[\[Theta]])))
Sqrt[1 + m^2] (B2 Sqrt[(-2 b + 2 d1 m + l Tan[\[Theta]])^2] +
B4 Sqrt[(-2 b + 2 d2 m + l Tan[\[Theta]])^2] +
B5 Sqrt[(2 b + 2 d3 m + l Tan[\[Theta]])^2] +
B7 Sqrt[(2 b + 2 d4 m + l Tan[\[Theta]])^2] +
B1 Sqrt[(2 b - 2 (d1 + l) m + l Tan[\[Theta]])^2] +
B3 Sqrt[(2 b - 2 (d2 + l) m + l Tan[\[Theta]])^2] +
B6 Sqrt[(-2 (b + (d3 + l) m) + l Tan[\[Theta]])^2] +
B8 Sqrt[(-2 (b + (d4 + l) m) + l Tan[\[Theta]])^2])
The terms being squared under each of the radicals is known to be a positive real number.
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评论（4）
由于已知这些项都是实数和正数，因此平方和取平方根只会得到相同的数字。因此，您可以执行类似的操作
，其中
expr是上面的巨型表达式。Since the terms are all known to be real and positive, squaring and taking the square-root will only give you the same number. Hence, you could do something like
where
expris your giant expression above.这里有两个想法：
1)
2)
复杂度函数 f 使得 Abs 比 Times 更昂贵。请参阅文档以了解简化。这有帮助吗？
Here are two ideas:
1)
2)
Th complexity function f makes Abs more expensive than Times. See docu for Simplify. Does that help?
如果您只想删除绝对值的特定实例，您可以按照以下方式执行操作：
这样它只会从您指定的任何表达式中删除绝对值：
我将看看是否可以找到比这更好的解决方案。
If you only wish to remove specific instances of absolute value, you could do something along these lines:
That way it only removes the absolute value from any expressions you specify:
I'll see if I can find a nicer looking solution than this.
我经常被诸如
Abs[a]^2之类的东西所刺激，而诸如使用Assuming与a\[Element]Reals之类的东西则不会。没有帮助。我在这里找到了一些帮助WolframMathWorld - 绝对平方与
ComplexExpand[Abs[a ]^2，目标函数-> {Conjugate}]，但有时它仍然返回诸如Conjugate[Sqrt[a^2 + b^2]]之类的内容，并且我发现包装了第二个ComplexExpand （不带参数）周围有帮助。I'm constantly stimied by things like
Abs[a]^2, and stuff like usingAssumingwitha\[Element]Realsdoesn't help.I found some help here WolframMathWorld - Absolute Square with
ComplexExpand[Abs[a]^2, TargetFunctions -> {Conjugate}], but sometimes it still returns stuff likeConjugate[Sqrt[a^2 + b^2]]and I've found wrapping a secondComplexExpand(without parameters) around that helps.