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