tetration and cardinality ?
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From: amy666 on 20 Dec 2008 23:32 i wrote : > matt271829 wrote : > > > On Dec 21, 1:06 pm, amy666 <tommy1...(a)hotmail.com> > > wrote: > > > while thinking about matrices and tetration i > came > > to the following equation : > > > > > > f(x) = exp( f(log(x)) ) > > > > > > of course this has the trivial solution f(x) = x > > > > > > as the title hints , i wonder if there are other > > solutions > > > > > > f(x) =/= x ?? > > > > > > maybe using lambert-W and tetration and pentation > ? > > > > erm... > > > > f(x) = exp(x) > > > > (also f(x) = exp(exp(x)), f(x) = exp(exp(exp(x))) > > etc.) > > hmm right. > > so i guess > > f(x) = exp(f(log(x))) > > has as solutions > > any y'th iterate of exp(x). > > in fact this might work as a way to define FRACTIONAL > ITERATIONS !!?? > > ( generalized ) > > f(x) = g( f(g^-1(x)) ) > > then f(x) is a fractional iteration of g(x). > > and if the fractional iterations can be ' connected ' > we have a REAL amount of iterations ... > > > this might be intresting ; > > f(f(x)) = exp(x) > > has to satisfy : f(x) = exp( f(log(x)) ) > > and also f(x) = log( f(exp(x)) ) > > does this match the current data for tetration ? > > > ( also intresting for A^B and other ideas ) > > thank you matt271829 > > (although the result is trivial and i should have > seen it myself ...) > > > regards > > tommy1729 perhaps it is intresting to think of cardinality here ! ( finally an application of set theory ??? :) ) f(x) = g( f(g^-1(x)) ) show that for any non-real-periodic g(x) that has no finite solution for g(x) = 0 the cardinality of the solutions for f(x) is either finite or uncountable. ( not even sure about that , a lemma conjecture ) regards tommy1729
From: amy666 on 21 Dec 2008 07:49 i wrote : > i wrote : > > > matt271829 wrote : > > > > > On Dec 21, 1:06 pm, amy666 > <tommy1...(a)hotmail.com> > > > wrote: > > > > while thinking about matrices and tetration i > > came > > > to the following equation : > > > > > > > > f(x) = exp( f(log(x)) ) > > > > > > > > of course this has the trivial solution f(x) = > x > > > > > > > > as the title hints , i wonder if there are > other > > > solutions > > > > > > > > f(x) =/= x ?? > > > > > > > > maybe using lambert-W and tetration and > pentation > > ? > > > > > > erm... > > > > > > f(x) = exp(x) > > > > > > (also f(x) = exp(exp(x)), f(x) = > exp(exp(exp(x))) > > > etc.) > > > > hmm right. > > > > so i guess > > > > f(x) = exp(f(log(x))) > > > > has as solutions > > > > any y'th iterate of exp(x). > > > > in fact this might work as a way to define > FRACTIONAL > > ITERATIONS !!?? > > > > ( generalized ) > > > > f(x) = g( f(g^-1(x)) ) > > > > then f(x) is a fractional iteration of g(x). > > > > and if the fractional iterations can be ' connected > ' > > we have a REAL amount of iterations ... > > > > > > this might be intresting ; > > > > f(f(x)) = exp(x) > > > > has to satisfy : f(x) = exp( f(log(x)) ) > > > > and also f(x) = log( f(exp(x)) ) > > > > does this match the current data for tetration ? > > > > > > ( also intresting for A^B and other ideas ) > > > > thank you matt271829 > > > > (although the result is trivial and i should have > > seen it myself ...) > > > > > > regards > > > > tommy1729 > > perhaps it is intresting to think of cardinality here > ! > > ( finally an application of set theory ??? :) ) > > f(x) = g( f(g^-1(x)) ) > > show that for any non-real-periodic g(x) that has no > finite solution for g(x) = 0 > > the cardinality of the solutions for f(x) is either > finite or uncountable. > > > ( not even sure about that , a lemma conjecture ) > > > regards > > tommy1729 since you guys are silent , i guess you dont believe it ? a counterexample in mind ?
From: Denis Feldmann on 21 Dec 2008 18:22
amy666 a écrit : > i wrote : > >> i wrote : >> >>> matt271829 wrote : >>> >>>> On Dec 21, 1:06 pm, amy666 >> <tommy1...(a)hotmail.com> >>>> wrote: >>>>> while thinking about matrices and tetration i >>> came >>>> to the following equation : >>>>> f(x) = exp( f(log(x)) ) >>>>> >>>>> of course this has the trivial solution f(x) = >> x >>>>> as the title hints , i wonder if there are >> other >>>> solutions >>>>> f(x) =/= x ?? >>>>> >>>>> maybe using lambert-W and tetration and >> pentation >>> ? >>>> erm... >>>> >>>> f(x) = exp(x) >>>> >>>> (also f(x) = exp(exp(x)), f(x) = >> exp(exp(exp(x))) >>>> etc.) >>> hmm right. >>> >>> so i guess >>> >>> f(x) = exp(f(log(x))) >>> >>> has as solutions >>> >>> any y'th iterate of exp(x). >>> >>> in fact this might work as a way to define >> FRACTIONAL >>> ITERATIONS !!?? >>> >>> ( generalized ) >>> >>> f(x) = g( f(g^-1(x)) ) >>> >>> then f(x) is a fractional iteration of g(x). >>> >>> and if the fractional iterations can be ' connected >> ' >>> we have a REAL amount of iterations ... >>> >>> >>> this might be intresting ; >>> >>> f(f(x)) = exp(x) >>> >>> has to satisfy : f(x) = exp( f(log(x)) ) >>> >>> and also f(x) = log( f(exp(x)) ) >>> >>> does this match the current data for tetration ? >>> >>> >>> ( also intresting for A^B and other ideas ) >>> >>> thank you matt271829 >>> >>> (although the result is trivial and i should have >>> seen it myself ...) >>> >>> >>> regards >>> >>> tommy1729 >> perhaps it is intresting to think of cardinality here >> ! >> >> ( finally an application of set theory ??? :) ) >> >> f(x) = g( f(g^-1(x)) ) >> >> show that for any non-real-periodic g(x) that has no >> finite solution for g(x) = 0 >> >> the cardinality of the solutions for f(x) is either >> finite or uncountable. >> >> >> ( not even sure about that , a lemma conjecture ) >> >> >> regards >> >> tommy1729 > > since you guys are silent , i guess you dont believe it ? > > a counterexample in mind ? What are the "solutions for f(x)"? Besides this, here is a construction of f such that f(x)=exp(f(log(x))) which gives an infinity of f (on, say, positive x), so it seems hard to believe those f are all of the shape above... First, find a bijection h such that exp(h(x))=h(x+1) fos say, all positive x. This is easy if you dont ask for much smoothness for h ; for instance, take any monotonic h defined on [0,1] such that h(1)=exp(h(0)), and define h by induction on [n,n+1]. Now, take any function k such that k(x+1)=k(x)+1, like k(x)=x+sin (2pi*x). Then f(x)= h(k(h^-1(x))) will satisfy f(x)=exp(f(log(x))) Oh, and one last HAHAHA : how comes you are not able to find this kind of things by yourself? |