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Hi guys,

I'm working on a problem where I ended up with the following series:

$z(Q) = \exp(-Q) [ 1 + \frac{a_1}{Q} + \frac{a_2}{Q^2} + \ldots]$ valid around $Q \to \infty$

Is there a systematic way of obtaining Q as a series in z, such as:

$Q(z) = - \log(z) + \frac{b_1}{z} + \frac{b_2}{z^2} + \ldots]$ valid around $z\to 0$

How should a proceed to have a correct mathematical answer?

Thanks

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The inverse series doesn't have that form. If $z=e^{-Q} (1+Q^{-1})$ then

$$Q = - \log z + \log (1+Q^{-1}) = \log z + (\log z)^{-1} + O((\log z)^{-2}) \quad \mbox{as} \ z \to 0^{+}.$$

The general form should be

$$Q = - \log z + \sum_{i>0} b_i (\log z)^{-i}.$$

You might be able to coerce this into the Lagrange inversion form, but I don't see how right now. I would just generalize the solution above:

Write $$Q = - \log z + \log \left( 1 + \sum a_i Q^{-i} \right) = - \log z + \sum \frac{(-1)^k}{k} \left( \sum a_i Q^{-i} \right)^k.$$

Expanding this will give you a formula of the form $$Q = - \log z + \sum_{i=1}^N c_i Q^{-i} + O(Q^{-N-1}) \quad (*)$$ for any $N$ you like. If you already know that $Q = - \log z + \sum_{i=0}^{N-1} b_i (\log z)^{-i} + O((\log z)^{-N})$, then plug your known values into $(*)$ to deduce the value of $b_N$.

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  • $\begingroup$ Corrected misprint in "general form" ... it was already correct later on. $\endgroup$ – Gerald Edgar Mar 31 '10 at 14:11

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