03/03/2009, 05:28 PM
consider f(f(x)) = exp(x)
with f(x) mapping R -> R and being strictly increasing.
you may choose your favorite solution f(x).
we know that exp(2x) = a(exp(x))
where a(x) = x^2.
the logical question becomes :
f(2x) = b(f(x))
b(x) = ???
its seems logical to assume x < b(x) < a(x) ( = x^2 )
x^2 and exp(x) do not commute , so b(b(x)) =/= a(x) =/= x^2
thus b(x) cannot be x^sqrt(2).
though perhaps close ?
b(x) = ????
it growth rate seems modest compared to other functions related to tetration since x < b(x) < x^2 for most ( or all ? ) x.
we know b(f(0)) = f(0)
since f(2x) = f(x) for x = 0 since 2x = x for x = 0.
thus f(0) is a fixpoint for b(x).
f(2x) = b(f(x))
substitute x => inv f(x)
f(2 inv f(x)) = b(x)
together with the fixpoint result above :
b(f(0)) = f(0) = f( 2 inv f(f(0)) )
thus this seems consistant.
we thus can say :
f(2 inv f(x)) = b(x)
( the inv is unique since f(x) maps R -> R and is strictly increasing ( by def ) )
anything else known about b(x) ???
( YES from log(f(x)) = inv f(x) and its effects on the above equations , see below )
f(2 inv f(x)) = b(x) [1]
use log(f(x)) = inv f(x) =>
f(2 * log(f(x)) ) = b(x)
=> f ( log(f(x)^2) ) = b(x)
to get a different form :
use f(log(x)) = log(f(x)) = inv f(x) ??
=> inv f ( f(x)^2 ) = b(x) [2]
combining [1] and [2] => inv f (f(x)^2) = f ( 2 inv f(x) )
=> f( 2 x ) = log f ( f(f(x))^2 ) [ b(x) eliminated ! ]
A NEW FUNCTIONAL EQUATION FOR f(x) ?!?
if the above is all correct we have 3 functional equations for f(x) , similar to the " unique gamma function conditions "
( i once conjectured on sci.math about 3 functional equations btw )
these 3 functional equations lead to the logical questions :
are they " equivalent " ? or are they " uniqueness criterions " ?
ill list them to be clear
f(x) with R -> R and strictly increasing.
1) f(f(x)) = exp(x)
2) log(f(x)) = f(log(x)) (*)
3) f( 2 x ) = log f ( f(f(x))^2 ) (*)
( * as long as the log is taken of real numbers > e )
is all that correct ? ( headscratch )
regards
tommy1729
with f(x) mapping R -> R and being strictly increasing.
you may choose your favorite solution f(x).
we know that exp(2x) = a(exp(x))
where a(x) = x^2.
the logical question becomes :
f(2x) = b(f(x))
b(x) = ???
its seems logical to assume x < b(x) < a(x) ( = x^2 )
x^2 and exp(x) do not commute , so b(b(x)) =/= a(x) =/= x^2
thus b(x) cannot be x^sqrt(2).
though perhaps close ?
b(x) = ????
it growth rate seems modest compared to other functions related to tetration since x < b(x) < x^2 for most ( or all ? ) x.
we know b(f(0)) = f(0)
since f(2x) = f(x) for x = 0 since 2x = x for x = 0.
thus f(0) is a fixpoint for b(x).
f(2x) = b(f(x))
substitute x => inv f(x)
f(2 inv f(x)) = b(x)
together with the fixpoint result above :
b(f(0)) = f(0) = f( 2 inv f(f(0)) )
thus this seems consistant.
we thus can say :
f(2 inv f(x)) = b(x)
( the inv is unique since f(x) maps R -> R and is strictly increasing ( by def ) )
anything else known about b(x) ???
( YES from log(f(x)) = inv f(x) and its effects on the above equations , see below )
f(2 inv f(x)) = b(x) [1]
use log(f(x)) = inv f(x) =>
f(2 * log(f(x)) ) = b(x)
=> f ( log(f(x)^2) ) = b(x)
to get a different form :
use f(log(x)) = log(f(x)) = inv f(x) ??
=> inv f ( f(x)^2 ) = b(x) [2]
combining [1] and [2] => inv f (f(x)^2) = f ( 2 inv f(x) )
=> f( 2 x ) = log f ( f(f(x))^2 ) [ b(x) eliminated ! ]
A NEW FUNCTIONAL EQUATION FOR f(x) ?!?
if the above is all correct we have 3 functional equations for f(x) , similar to the " unique gamma function conditions "
( i once conjectured on sci.math about 3 functional equations btw )
these 3 functional equations lead to the logical questions :
are they " equivalent " ? or are they " uniqueness criterions " ?
ill list them to be clear
f(x) with R -> R and strictly increasing.
1) f(f(x)) = exp(x)
2) log(f(x)) = f(log(x)) (*)
3) f( 2 x ) = log f ( f(f(x))^2 ) (*)
( * as long as the log is taken of real numbers > e )
is all that correct ? ( headscratch )
regards
tommy1729