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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.

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  • $\begingroup$ 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. $\endgroup$
    – TMM
    Jul 10, 2012 at 18:06
  • $\begingroup$ @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. $\endgroup$
    – Joe
    Jul 10, 2012 at 18:19

2 Answers 2

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Assuming it's $\epsilon$ that is considered as small, $$(1+\gamma/z_0 + \epsilon z_1/z_0)^{-2} = \left( 1+{\frac {\gamma}{z_{{0}}}} \right) ^{-2}-2\,z_{{1}} \left( 1 +{\frac {\gamma}{z_{{0}}}} \right) ^{-3}{z_{{0}}}^{-1}\epsilon+3\,{z_{ {1}}}^{2} \left( 1+{\frac {\gamma}{z_{{0}}}} \right) ^{-4}{z_{{0}}}^{- 2}{\epsilon}^{2}+O \left( {\epsilon}^{3} \right) $$

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  • $\begingroup$ Ah!! I see it now!! Thanks Robert Israel for this!! $\endgroup$
    – Joe
    Jul 10, 2012 at 18:23
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Note that you can use Wolfram Alpha to generate expansions of this type, getting the result

$$\frac{z_0^2}{(\gamma + z_0)^2} - \frac{2 z_0^2z_1\epsilon}{(\gamma+z_0)^3} + \frac{3z_0^2z_1^2\epsilon^2}{(\gamma+z_0)^4} - \frac{4z_0^2z_1^3\epsilon^3}{(\gamma+z_0)^5} + O(\epsilon^4)$$

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  • $\begingroup$ Thanks for the tip Chris Taylor. Wolfram Alpha wouldn't compute it for me the first time round but my commands weren't as specific as yours unfortunately! $\endgroup$
    – Joe
    Jul 10, 2012 at 18:55

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