# Asymptotics of Power Series With Branch Singularities

I am wondering if there are analytic tools to find asymptotic formulae for the coefficients of a complex power series of a function with branch singularities. For example, it is possible to show using elementary means that, for $q>1$, the coefficients of $\frac{1}{1-z} log(\frac{1}{1-qz})$ are asymptotic to $\frac{1}{1-q^{-1}} \frac{q^n}{n}$, but I would like to know if it is possible to see this from analytic properties of the function itself.

Motivation: There are nice asymptotic formulae for the coefficients of power series of meromorphic functions. As a simple example, the coefficients of an entire function must be $O(\epsilon^n)$ for any $\epsilon$. In general, the sum of the principal parts of the poles of smallest modulus will provide a very good first approximation, and contour integration can be applied to get more delicate bounds, see e.g. Wilf's Generatingfunctionology.

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Much of the second half of Flajolet and Sedgewick is dedicated to answering such questions: algo.inria.fr/flajolet/Publications/book.pdf – Qiaochu Yuan Nov 13 '09 at 21:29
Ah, you're a fan of Shining Force I see? – Vandermonde Apr 5 at 21:09

The answer is "Quite often yes, but the error terms are seldom as good as in the meromorphic case". The reason the asymptotics for the meromorphic functions works is that we know the exact coefficients for $c_k(z-z_0)^{-k}$ and can always remove the singularity by subtracting a few terms of this kind thus increasing the radius of convergence and boosting the decay of the coefficients, so that the error is exponentially small compared to the main terms.
When you have some branching singularity $S(z)$ (like $\log(z-z_ 0)$ or $\sqrt[]{z-z_ 0}$) for which you know the coefficients when it stands alone, you can still play the trick to get the asymptotics for a function $F(z)S(z)$ where $F$ is alnalytic in some larger disk by writing the sum as $F(z_ 0)S(z)+(F(z)-F(z_ 0))S(z)$ and noting that the second term is smoother than the first one on the boundary circle, so its Fourier coefficients decay a bit faster. If you take more terms in the Taylor series for $F$ at $z_ 0$, you can get any degree of smoothness in the remainder you wish, so the error terms can be made of size $n^{-k}$ times the leading terms with any given $k$, but that's about as far as you can go unless you want integral expressions instead of algebraic quantities.