%This is a previous version.
%The version which is going to appear is currently not available.










\magnification=\magstep1
\input amstex
\documentstyle{amsppt}
\headline{\hfil --- \ \folio\ \  --- \hfil}
\TagsOnRight
\font\sc=cmcsc10

\topmatter
\title
asymptotics of the luria-Delbr\"uck distribution
\endtitle
\author
Helmut Prodinger
\endauthor
\affil
Technical University of Vienna\\
Department of Algebra and Discrete Mathematics
\endaffil
\address
\flushpar
Helmut Prodinger\newline
Technical University of Vienna\newline
Department of Algebra and Discrete Mathematics\newline
Wiedner Hauptstrasse 8--10\newline
A-1040 Vienna \newline
Austria
\endaddress
\email proding\@rsmb.tuwien.ac.at \endemail










\endtopmatter

\def\adrienne{{2}}

\def\flaodl{{1}}


\def\({\left(}
\def\){\right)}



\document

\define\lu{Luria--Delbr\"uck\ }
\define\ff{{\frac1{1-z}}}

We consider the \lu distribution, as described in \cite\adrienne\ and previous
writings, cited therein. It is given by the probability generating function
$$
G(z)=(1-z)^{m(1-z)/z} =\sum_{n\ge0}p_nz^n,
$$
where $m$ is a parameter.

We show that with {\sl singularity analysis of generating functions}
\cite\flaodl\ (in principle) a full asymptotic expansion of
$p_n$ and $P_n:=\sum_{k>n}p_k$ can be achieved. In this way we generalize
results given in   \cite\adrienne.

%\font\sc{cmcsm10}

The basic idea of this powerful method is to realize that $G(z)$ has
radius of convergence 1 and 1 as the only singularity on the radius of
convergence. Then there is a {\sl local expansion} about $z=1$
(produced by {\sc Maple}), 
$$\aligned
G(z)&=1-m(1-z)\log\ff-m(1-z)^2\log\ff\\&+\frac{m^2}2(1-z)^2\log^2\ff+
\Cal O\((1-z)^3\log^3\ff\),     
\endaligned\tag1
$$
which can be translated {\sl term--by--term} into an asymptotic expansion
of the coefficients under some mild technical conditions which are
satisfied in out instance; we refer again to \cite\flaodl\ for a full
treatment.

Therefore we only have to know about the expansion of the standard 
functions appearing in \thetag1.

\define\poch#1#2{#1^{\underline{#2}}}

We use the following notions: $[z^n]f(z)$ means the coefficient
of $z^n$ in (the Taylor expansion of) $f(z)$,  the symbol
$\poch nk$ 
means $n(n-1)\dots(n-k+1)$ and $H_n=1+\frac12+\frac13+\dots+
\frac1n$ means the $n$th harmonic number. Then we have for $n$ large enough
$$
\aligned
[z^n](1-z)^i\log\ff&=\frac{(-1)^ii!}{\poch n{i+1}}, \\
[z^n]\log^2\ff&=
\frac2nH_{n-1}, \\
[z^n](1-z)^2\log^2\ff&=
\frac2nH_{n-1}-
\frac4{n-1}H_{n-2}+
\frac2{n-2}H_{n-3}                                  .
\endaligned
$$

Hence
$$
p_n=\frac m{\poch n2}+\frac{2m^2}{\poch n3}
+\frac {m^2}2\(
\frac2nH_{n-1}-
\frac4{n-1}H_{n-2}+
\frac2{n-2}H_{n-3}                                  
               \)
+\Cal O\(\frac{\log^2n}{n^4}\)
.$$
{\sc Maple} also translates this expansion into more traditional
terms,
$$
p_n=                 \frac m{n^2}
+2m^2\frac{\log n}{n^3}+\frac{2m^2\gamma-3m^2-m}{n^3}+
\Cal O\(\frac{\log^2n}{n^4}\)
.$$

To get the asymptotic expansions for the tails $P_n$, we just note
that its generating function is
$$
\aligned
\ff-\ff\, G(z)&=
m\log\ff+m(1-z)\log\ff\\ & -\frac{m^2}2(1-z)\log^2\ff+
\Cal O\((1-z)^2\log^3\ff\),     
\endaligned
$$
from which we conclude
$$ \aligned
P_n&=\frac mn-\frac m{\poch n2}-\frac{m^2}2\(\frac2nH_{n-1}-
\frac2{n-1}H_{n-2}\)
+\Cal O\(\frac{\log^2n}{n^3}\)\\
&=\frac mn+m^2\frac{\log n}{n^2}+\frac{m^2\gamma-m^2-m}{n^2}
+\Cal O\(\frac{\log^2n}{n^3}\)
.\endaligned
$$



\Refs
\ref \no \flaodl\paper Singularity
Analysis of Generating Functions
\by P.~Flajolet and A.~Odlyzko \jour SIAM J.Discrete
Mathematics\vol3\yr1990\pages216--240
\endref

\ref\no\adrienne\paper Comments on the Luria--Delbr\"uck 
Distribution\jour J.~Appl.~Prob.\vol31\yr1994\pages822-828
\by A.W.~Kemp
\endref

\endRefs

\enddocument
