1
$\begingroup$

I was trying to solve and ODE and while doing some asymptotics, I bumped into something like this $\left(1+\frac{\gamma}{z_{0}}+\epsilon \frac{z_{1}}{z_{0}}\right)^{-2}$

where $\gamma$ $\,$ is of $\mathcal{O}(1)$. To my understanding, the expansion of the latter would be $\left(1+\frac{\gamma}{z_{0}}\right)-2\epsilon\frac{z_{1}}{z_{0}}+\ldots$

The above doesn't feel right to me though! How would one go about something like this?

$\gamma$,$z_{0}$ and $z_{1}$ are all $\mathcal{O}(1)$ apart from $\epsilon$ where we take the latter as small.

  • 0
    Usually it's a matter of Tayloring, and being careful which quantity is small. For example, $1/(1 + x) \approx 1 - x$ for small $x$. But unless we know more about $\gamma, z_0, z_1$ it's hard to say which are small.2012-07-10
  • 0
    @TMM I added some extra information about $\gamma$, $z_{0}$ and $z_{1}$ and to be honest, it is that extra $\frac{\gamma}{z_{0}}$ that confused me.2012-07-10

2 Answers 2