$\frac{1}{\cosh(z)}$在原点处展开式的收敛半径疑问

Hey there! Let's work through your confusion step by step—first we'll fix that incorrect expansion, then unpack the singularity and convergence radius question.

First: Your Expansion of $1/\cosh(z)$ at the Origin is Wrong #

First off, let's set the record straight: $\cosh(0) = 1$, which is non-zero. That means the origin is not a singularity of $1/\cosh(z)$—it's actually an analytic point there!

Your derived expansion ($1+\frac{2}{z^{2}+\frac{24}{z}4}+...$) has negative powers of $z$, which would be part of a Laurent expansion, but that's only valid in regions where $|z|$ is larger than the distance to the nearest singularity. For the origin neighborhood ($|z| < \pi/2$), the correct expansion is a Taylor series with no negative terms:
$$1 - \frac{z^2}{2} + \frac{5z^4}{24} - \frac{61z^6}{720} + ...$$
Your mistake likely came from applying the wrong expansion technique (or mixing up the region of validity for a Laurent series with the Taylor series at the origin).

Now: Singularities and Convergence Radius #

To answer your core question:

• The convergence radius of a Taylor series centered at a point $a$ is equal to the distance from $a$ to the nearest singularity of the function.
• If $a$ itself were a singularity (like the function $1/z$ at $z=0$), then the distance from $a$ to the nearest singularity is 0—so yes, the convergence radius would be 0. In that case, the Taylor series doesn't converge anywhere except at the point $a$ itself, and you'd need a Laurent series to describe the function around $a$.

But for $1/\cosh(z)$:
The singularities of $1/\cosh(z)$ occur where $\cosh(z) = 0$, which solves to $z = i\left(\frac{\pi}{2} + k\pi\right)$ for integer $k$. The closest of these to the origin is $z = \pm i\pi/2$, which are at a distance of $\pi/2$ from 0. So the convergence radius of the Taylor series at the origin is $\pi/2$, not 0.

To Sum Up #

Your confusion stemmed from two mix-ups:

• Thinking the origin was a singularity of $1/\cosh(z)$ (it's not—$\cosh(0)=1$ is perfectly well-defined and non-zero)
• Using a Laurent expansion (for large $|z|$) instead of the Taylor expansion valid near the origin

内容的提问来源于stack exchange，提问作者MathIsHard