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=x(xc)^{1/x-1}=c^{-1}(xc)^{1/x})
, 
is decreasing for 
. To decide this we differentiate:
=c^{-1}\left( xc \right) ^{1/x} {\frac {1-\ln \left( xc<br />
\right) }{x^2}} )
>0)
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 ^{b^{-\epsilon}-1})
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 b^{-\epsilon} )
.
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