Mittag-Leffler Expansion

Question

I am attempting to perform what is described in my notes as a "Mittag-Leffler Expansion", but first I must prove that this expansion is valid.

Given that

$$ f(z) = \frac{1}{\sin{z}} - \frac{1}{z}$$

Let $C$ be the positively oriented boundary of the rectangle $-\left(n+\frac{1}{2}\right)\pi \le x \le \left(n+\frac{1}{2}\right)\pi$, $-n\pi \le y \le n\pi$, where $z= x + iy$. Show that $|\sin(z)| = \mathcal{O}(e^{|n|\pi})$ on the top and bottom, and so $f(z) = \mathcal{O}\left(\frac{1}{n}\right)$ there. Show also that $|\sin(z)| = \cosh(y)$ on the sides, use this to bound $| f(z) |$ by a constant there, and so bound $|f(z)|$ by a constant along the entire $C$.

My work:

I was able to prove that $|\sin(z)| = \cosh(y)$ on the sides.

I got that $|\sin(z)| = \sinh(y)$ on the top and bottom by the following $$\lvert \sin(x+iy)\rvert = \lvert \frac{e^{-i(x+iy)}-e^{i(x+iy)}}{2} \rvert \le \frac{e^{|y|}-e^{-|y|}}{2} = \sinh(y) = \sinh(|n|\pi)$$

I am unsure if this is correct, as I am unsure what the script $\mathcal O$ means. I also do not see how I am supposed to "bound $| f(z) |$ by a constant", as $\cosh(y)$ is not bounded as $y$ approaches $\infty$.

I looked at the wikipedia article already. It did not really help me understand. — Snagglewhen, Apr 15 '15 at 00:06
A detailed work for expanding such functions using Mittag-Leffler expansion can be found in chapter 6 in this book. — science, Apr 15 '15 at 00:21
For example, $f(z) = {\mathcal O}(1/n)$ means that there exist $N$ and $C$ such that $|f(z)| \le C/n$ when $n > N$. — Robert Israel, Apr 15 '15 at 01:56
For the top, $y = n \pi$, what you want is not $|\sin(z)| = {\mathcal O}(e^{|n|\pi})$ but rather $1/|\sin(z)| = { \mathcal O}(e^{-|n|\pi})$. The point here is that $$|\sin(z)| \ge \dfrac{1}{2} \left( e^{n\pi} - e^{-n\pi}\right)$$ — Robert Israel, Apr 15 '15 at 02:06
Isn't $$|\sin(z)| \le \dfrac{1}{2} \left( e^{n\pi} - e^{-n\pi}\right)$$? That is what I derived above in the question anyway, and going back I don't see any mistakes. — Snagglewhen, Apr 15 '15 at 02:13

score 1 · Answer 1 · answered Jan 08 '16 at 02:15

I am unsure if this is correct, as I am unsure what the script $\mathcal O$ means.

As @RobertIsrael suggested, it helps to understand what big-$\mathcal O$ notation means. Roughly, it means a given function grows "as fast as" another function up to some constant factor.

I also do not see how I am supposed to "bound $|f(z)|$ by a constant", as $\cosh(y)$ is not bounded as $y$ approaches $\infty$.

Well … the question asks you to bound $|f(z)|$, not $\cosh(y)$ :P

Show that $|\sin(z)| = \mathcal{O}(e^{|n|\pi})$ on the top and bottom, and so $f(z) = \mathcal{O}\left(\frac{1}{n}\right)$ there.

You have already shown that

$$ |\sin(z)| \le \frac{1}{2} (e^{n \pi} - e^{-n \pi})$$

As $n \to 0$, $e^{n \pi}$ grows exponentially while $e^{-n \pi}$ decays exponentially, so for large $n$ only $e^{n \pi}$ is significant. Thus we can conclude that $|\sin(z)| = \mathcal{O}(e^{|n|\pi})$.

From this we can analyze the behavior of $f(z)$ on the top and bottom lines:

$$\begin{align} |f(z)| &= \left|\frac{1}{\sin z} - \frac{1}{z}\right| \\ &\le \frac{1}{|\sin z|} + \frac{1}{|z|} \\ &\le \frac{1}{|\sin z|} + \frac{1}{|n \pi|} \\ &= \mathcal O\left(\frac{1}{e^{|n|\pi}}\right) + \mathcal O\left(\frac{1}{|n|}\right) \\ &= \mathcal O\left(\frac{1}{n}\right) \end{align}$$

Show also that $|\sin(z)| = \cosh(y)$ on the sides, use this to bound $| f(z) |$ by a constant there, …

You already have

$$ |\sin(z)|=\cosh(y) $$

On the real axis, $\cosh$ function has a minimum of 1, so its reciprocal is at most 1. Thus

$$ \frac{1}{|\sin(z)|} \le 1 $$

We can then repeat the same procedure to analyze the left and right lines (skipping a few steps):

$$\begin{align} |f(z)| &\le 1 + \frac{1}{(n + \frac{1}{2}) \pi} \le 1 + \frac{1}{3 \pi / 2} \end{align}$$

… and so bound $|f(z)|$ by a constant along the entire $C$.

On the top and bottom lines, $|f(z)|$ decreases towards zero as $\mathcal O\left(\frac{1}{n}\right)$, while on left and right lines, $|f(z)|$ is bounded from above by a constant value $1 + \frac{2}{3 \pi}$. Hence, the value of the function is bounded on all contours by this constant.

Mittag-Leffler Expansion

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