# Stoke's Phenomena

• Oct 25th 2009, 11:06 PM
pengchao1024
Stoke's Phenomena
$\delta>0, \ \ show \ \ cosh(z) \approx \frac{1}{2}e^{z}, \ \ |arg(z)|<\frac{\pi}{2}-\delta$

and

$cosh(z)\approx \frac{1}{2}e^{-z}, \ \ \frac{\pi}{2}+\delta

not need to go along rays.
• Oct 26th 2009, 10:37 AM
Laurent
For instance, for the first one: $\cosh z=\frac{e^z}{2}(1+e^{-2z})$ and $|e^{-2z}|=e^{-2\Re(z)}=e^{-2|z|\cos({\rm arg}(z))}$ $\leq e^{-2|z|\cos(\frac{\pi}{2}-\delta)}=e^{-2|z|\sin\delta}$ for all $z$ such that $|{\rm arg}(z)|<\frac{\pi}{2}-\delta$.

This proves that $\cosh z = \frac{e^z}{2}(1+o(1))$ as $|z|\to\infty$ and $|{\rm arg}(z)|<\frac{\pi}{2}-\delta$, uniformly for $z$ in this domain.
• Oct 26th 2009, 11:08 AM
pengchao1024
wat
what about the case $\delta=0$
• Oct 26th 2009, 11:31 AM
Laurent
Quote:

Originally Posted by pengchao1024
what about the case $\delta=0$

Then this is false. If you choose $|z|\to\infty$ that gets very close to the imaginary axis, $\cosh z$ gets close to cosine of a real number, hence it can be bounded ( $\cosh(ix)=\cos(x)$).