Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was told that $\operatorname{sech} x$ has a simple pole. Could someone please explain what that means? I have looked up the definition but it involves too much jargon like holomorphic, etc. Is there a simple definition and why is this true? Thanks.

share|cite|improve this question
For an even simpler example of a simple pole: $x=0$ is a simple pole for the function $\dfrac1{x}$. – J. M. Oct 25 '11 at 15:26
Stanisław Łojasiewicz was a fantastic mathematician who proved very difficult inequalities about the growth of functions on complex analytic spaces , and yet he was very modest and always behaved like a simple Pole. – Georges Elencwajg Oct 25 '11 at 16:17
@GeorgesElencwajg: Haha :D – simpleton Oct 25 '11 at 23:45
Cross-posted to – Qmechanic Nov 19 '12 at 19:16
up vote 5 down vote accepted

Recall that $\operatorname{sech}(x) = \frac{1}{\cosh(x)}$. When $x \in \mathbb{R}$, hyperbolic cosine is non-negative, so $\operatorname{sech}(x)$ has no poles on the real axis.

Zeros of the hyperbolic cosine are all along the imaginary axis at $z_n = i \frac{\pi}{2} + i \pi n$. Consider a vicinity of such a zero: $$\begin{align} \frac{1}{\cosh(z_n + \epsilon)} &= \frac{1}{\cosh(z_n) \cosh(\epsilon) + \sinh(z_n) \sinh(\epsilon)}\\ &= \frac{1}{\sinh(z_n)} \frac{1}{\sinh(\epsilon)}\\ &\sim \frac{1}{\sinh(z_n)} \left( \frac{1}{\epsilon} + o(1) \right) \end{align} $$ The order of the pole is one, so it is called simple. But as you see, $\operatorname{sech}(x)$ has infinitely many simple poles.

Added: The series expansion for $\frac{1}{\sinh(\epsilon)}$ follows from series expansion for $\sinh(\epsilon) \sim \epsilon + \frac{\epsilon^3}{3!} + \ldots + \frac{1}{(2n+1)!}\epsilon^{2n+1} + o(\epsilon^{2n+2})$.

share|cite|improve this answer
Looks like you've covered it quite well. :) This might be of help in explaining what poles are for the OP... – J. M. Oct 25 '11 at 15:13
Thanks, Sasha and @J.M. ! – simpleton Oct 25 '11 at 15:22
@simpleton: Because of those poles, $\mathrm{sech}$ is what would be termed as a meromorphic function. – J. M. Oct 25 '11 at 15:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.