arctan(z)的奇点

hbghlyj

在Mathematica中, 可以用FunctionSingularities求复变函数的奇点.
例如$\arctanh(z)$的奇点, 输出 z <= -1 || z >= 1
$\frac{d}{dz}\arctanh(z)=\frac{1}{1-z^2}$的奇点, 输出 z^2 == 1
但是$\arctan(z)$的奇点, 为什么输出 False
我觉得应该是$[i,i∞]∪[-i,-i∞]$

而且$\frac{d}{dz}\arctanh(z)=\frac{1}{1+z^2}$的奇点, 输出 z^2 == -1
况且 $i\arctanh(z)=\arctan(iz)$
见Brian文档:

For complex-valued input, arctan() is a complex analytic function that has [1j, infj] and [-1j, -infj] as branch cuts, and is continuous from the left on the former and from the right on the latter.

Czhang271828

奇点是奇点, Branch cut 是 Branch cut.

hbghlyj

Czhang271828 发表于 2023-1-10 13:12
奇点是奇点, Branch cut 是 Branch cut.

对于$\sqrt z$
FunctionSingularities[Sqrt[z], z]
输出是$z\le0$ (好像是Branch cut)

hbghlyj

hbghlyj 发表于 2023-1-10 12:55
但是$\arctan(z)$的奇点, 为什么输出 False
我觉得应该是$[i,i∞]∪[-i,-i∞]$

终于明白了, 1#代码没有加domain, 所以 False 应该是指 arctan 在 $\Bbb R$ 上没有奇点

指定了domain为complexes就没问题了

Czhang271828 发表于 2023-1-10 13:12
奇点是奇点, Branch cut 是 Branch cut.

2楼的意思应该是, FunctionDomain(函数定义域)不一定是FunctionSingularities(函数奇点)的补集, 继续以ArcTan[z]为例:
FunctionDomain[ArcTan[z], z, Complexes]
输出$(-i + z) (i + z) \ne0$
即$\Bbb C\setminus\{i,-i\}$

hbghlyj

Czhang271828 发表于 2023-1-10 13:12
奇点是奇点, Branch cut 是 Branch cut.

Wikipedia说a branch point of a multi-valued function is a point such that if the function is n-valued at that point, all of its neighborhoods contain a point that has more than $n$ values.
MathWorld说A branch point of an analytic function is a point in the complex plane whose complex argument can be mapped from a single point in the domain to multiple points in the range
例如0是$1/z$的奇点, 但不是branch point
例如0是$\sqrt z$的奇点(在0不可导), 且为branch point

1. FunctionSingularities[Sqrt[z], z]

Copy the Code

MathWorld有一个表格, 在最后一行可见$\sqrt{z}$的branch cut为$(-\infty,0)$, 是不包含0的, 所以说branch cut不一定包含branch point
它还说In addition to branch cuts, singularities known as branch points also exist. It should be noted, however, that the endpoints of branch cuts are not necessarily branch points.
即, branch cut的端点不一定是branch point.