f(2x) = ? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 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 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/03/2009, 06:26 PM tommy1729 Wrote: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) = ??? ... => f ( log(f(x)^2) ) = b(x) Till here I could follow you but I dont know how you derive: Quote:to get a different form : use f(log(x)) = log(f(x)) = ... Hm but let me see: $f(f(x)=\exp(x)$ $\exp(f(x))=f(f(f(x)))=f(\exp(x))$ $f(\log(x))=\log(f(x))$ ok thats also correct Quote:=> f( 2 x ) = log f ( f(f(x))^2 ) A NEW FUNCTIONAL EQUATION FOR f(x) ?!? This functional equation can be simplified to: $f(2x)=\log f( \exp(2x))$ and hence equivalent to $\exp(f(y) )=f(\exp(y))$. So it is a direct consequence from $f(f(x))=\exp(x)$. Quote: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 ) (*) 2) and 3) are consequences of 1) 1) does not follow from 2) because also a function g, with g(g(g(x)))=exp(x) satisfies 2). Surely 1) does not follow from 3). So they are not equivalent, but also no uniqueness conditions. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/03/2009, 06:41 PM bo198214 Wrote:tommy1729 Wrote: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) = ??? ... => f ( log(f(x)^2) ) = b(x) Till here I could follow you but I dont know how you derive: Quote:to get a different form : use f(log(x)) = log(f(x)) = ... Hm but let me see: $f(f(x)=\exp(x)$ $\exp(f(x))=f(f(f(x)))=f(\exp(x))$ $f(\log(x))=\log(f(x))$ ok thats also correct Quote:=> f( 2 x ) = log f ( f(f(x))^2 ) A NEW FUNCTIONAL EQUATION FOR f(x) ?!? This functional equation can be simplified to: $f(2x)=\log f( \exp(2x))$ and hence equivalent to $\exp(f(y) )=f(\exp(y))$. So it is a direct consequence from $f(f(x))=\exp(x)$. Quote: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 ) (*) 2) and 3) are consequences of 1) 1) does not follow from 2) because also a function g, with g(g(g(x)))=exp(x) satisfies 2). Surely 1) does not follow from 3). So they are not equivalent, but also no uniqueness conditions. oh well. at least i found b(x) = f ( log(f(x)^2) ) tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 05/02/2009, 09:28 PM perhaps i found more ... thinking about a potential link to change of base formulas , uniqueness criterions , fixpoint shifts , tetration limits ... potentially closed form expressions with some luck ... complicated ideas ... will explain some unspecified time later ... need testing ... ( seems in benefit of andydude / andrew robbins ideas and solutions ) regards tommy1729 « Next Oldest | Next Newest »