Mathematica：找到复数实部为正数的条件，Reduce 的意外/冗余输出
我需要找到复数的实部为负数的条件。我认为Reduce 非常适合这个，但它提供了冗余输出（即使在简化之后）。例如：
In[543]: Reduce[{Re[-1 - Sqrt[a - b] ] < 0, a > 0, b > 0}, {a, b}, Complexes]
Out[543]: a > 0 && (0 < b < a || b >= a)
由于a和b因为出现在不等式中而被假设为实数，因此不需要对a和b之间的关系进行进一步的假设，我期望的结果是：
Out[543]: a > 0 && b > 0
是否有一个很好的理由为什么没有得到？ （在我看来）冗余结果会随着更复杂的表达式而累积，我需要减少其中相当多的结果。有什么技巧可以得到预期的结果吗？我尝试过选择 Reals 作为域并根本不选择任何域，但没有任何东西真正能给我我想要的东西。顺便说一句，我正在通过检查特征值来分析不动点的稳定性......这是一个非常常见的任务。
I need to find the conditions for the real part of a complex number to be negative. I thought Reduce would be perfect for this, but it gives redundant output (even after simplification). For example:
In[543]: Reduce[{Re[-1 - Sqrt[a - b] ] < 0, a > 0, b > 0}, {a, b}, Complexes]
Out[543]: a > 0 && (0 < b < a || b >= a)
As a and b are assumed to be real because they appear in an inequality, there needs to be no further assumption about the relation between a and b, the result I expect is:
Out[543]: a > 0 && b > 0
is there a good reason why that is not obtained?
The (in my eyes) redundant results accumulate for more complex expressions and I need to reduce quite a few of them. Is there a trick to get the expected result? I played around with choosing Reals as the domain and choosing no Domain at all, but nothing really gives me what I want. By the way I am analyzing the stability of fixed points by checking eigenvalues...a very common task.
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我不知道为什么 Mathematica 不会一步返回您期望的结果，但以下是如何分两步获取它：
一般来说，能够以通用方式处理不等式的两个函数是
Reduce和LogicalExpand。 （但我在这方面的知识非常有限！）我相信(Full)Simplify只会使用后一种。关于在
Reduce中设置域的评论：请注意，文档中说：“如果 dom 是实数，或者是整数或有理数等子集，那么所有常量和函数值也被限制为实数。”因此，如果您将域指定为
Reals（如 @belisarius 的答案），Reduce将返回00000 b <= a这也是Sqrt[ab]为实数所必需的。I don't know why Mathematica won't return the result you are expecting in one step, but here's how to obtain it in two steps:
Generally, the two functions that can deal with inequalities in a general way are
ReduceandLogicalExpand. (But my knowledge is very limited in this area!) I believe(Full)Simplifywill only use the latter one.A comment on setting domains in
Reduce:Note that the documentation says: "If dom is Reals, or a subset such as Integers or Rationals, then all constants and function values are also restricted to be real." Hence if you were to specify the domain as
Realsas in @belisarius's answer,Reducewould return0 < b <= awhich is necessary forSqrt[a-b]to be real as well.用
ComplexExpand包裹Re[...]表达式，您将得到预期结果a > 0 && b> 0 。Wrap the
Re[...]expression withComplexExpandand you'll get the expected resulta > 0 && b > 0.