Sunday, 11 March 2007

ca.analysis and odes - How to invert this series?

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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