Ok, we want to compute the lower fixed point
of
for, as usual,
.
We do that by iterating:


while iterating we can not directly compute the difference
but we can compute
and we ask our self how close
must come to 1 such that

First we translate the difference into a quotient:

is decreasing. Now lets compute
^{1/y_n-1})
To make an estimate we want to know whether
,
is decreasing for
. To decide this we differentiate:
=c^{-1}\left( xc \right) ^{1/x} {\frac {1-\ln \left( xc<br />
\right) }{x^2}} )
for
)

This is true for
because then
.
So we know that
is strictly increasing for small enough
. So we get
iff ^{b^{-\epsilon}-1})
for
, equivalently:
iff
iff ^{b^{-\epsilon}})
The right side can be further simplified:
b^{-\epsilon}})
b^{-\epsilon})
The condition
is satisfied for
, i.e. on the right side is a constant
. So during the iteration we can decrease
according to
without fear that
becomes to small and the iteration does not stop.
Proposition. Let
, let
be the lower fixed point of
, let
and
then for each
:
iff
.
We do that by iterating:
while iterating we can not directly compute the difference
First we translate the difference into a quotient:
To make an estimate we want to know whether
This is true for
So we know that
for
The right side can be further simplified:
The condition
Proposition. Let