In this appendix it will be shown that for the first model discussed,
under certain assumptions is approximately 
.
The transformation for the whole cycle is 
with
s is the selection coefficient, 
. For simplicity
we assume here that 
. We will assume that n is sufficiently
big, so that 
. To compute 
we will write
Using the binomial expansion this gives
The general term in this expansion is
it can be shown that for l=2k
and for l=2k+1
looking at these terms it is apparent that will have the form
Where a, b, and c are functions of s, 
, and n. And
will then be
We wish to compute the eigenvalues of 
. This can be done
using the determinant and the trace of the matrix. We know that
(because 
), and
also
Therefore 
. So to compute the eigenvalues of it is
enough to know b. To find the eigenvalues we have solve the equation
or
The maximal eigenvalue is
The Taylor series expansion 
gives
We now turn to compute b. The bound
For l=2 k +1 gives, using equation (14), the following
expression for b:
where 
is a function of n and s, decreasing in k. This can be
written as
When we use this expression for 
,
and using our assumption that , we see that we can drop
the last term in equation (24). To find the with
the maximal eigenvalue, we will take the derivative of
equation (24).
For sufficiently large n the sign of the above expression, for a
constant 
, above 
and below 
is governed by the term 
, and is negative at , and
positive at , which shows that a root exists
between these two, or that a maximal eigenvalue is reached there. Thus
is a good approximation to .