f(x) = f(g(x)) tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 09/23/2010, 11:20 PM f(x) = f(g(x)) its a simple equation. and i talked about it before. as did e.g. gottfried. in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function... but that is not the end of the story. what if f(x) is given and we want to find g(x) ? well for starters g(x) might not be unique. if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i) the pattern is clear f(x) = f(g(x)) => g(x) = g(g1(x)) => f(x) = f(g1(x)) and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ... until we arrive at a moebius function.... if we ever ... i guess we could call g_n(x) the invariants of f(x). and i guess one could say : hey , this is just the construction of a riemann surface in disguise. but those are just names ! another question is the following : if we go up the chain , from g_n+1 to g_n , indefinitely , then what f(x) satisfies that ? thus : f(x) = f(g_-n(x)) for all n. => f(x) = ?? if g_(-)n is not cyclic and f(x) is coo do we have a uniqueness condition on f(x) ? is f(x) then a fractal or a constant ? ( i assume if we require f(x) to be coo and g_-n is dense ( non-repelling point ) then f(x) must be constant ) also related : is lim n-> oo g_(-)n(x) convergent or divergent ? is brouwers fixed point theorem related ? i believe lim n-> oo g_-n(x) is divergent because g(x) = g(g(x)) is the associated equation , and this has no solution. we do not accept g_(-)n(x) = x of course ... and finally of course the following f(x) = f(a(x)) = f(b(x)) which belong more too the "functional equation category" perhaps ... that equation is trickier than it seems ; at first sight it seems impossible since we have sin(super(a)/2pi) = sin(super(b)/2pi) ( each side as solution ) but then again g_2 and g_3 e.g. satisfy it. i assume lim n -> oo f(x) = f(a_g_(-)n(x)) = f(b_g_(-)n(x)) has no solution. quite a brainstorm .... tommy1729 tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 12/01/2010, 06:14 PM f(x) = f(a(x)) = f(b(x)) when a(x) and b(x) are non-linear it "seems" that a(x) and b(x) need to commute. (*) in fact P(x) = 2-periodic function ( period a and b ) S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*) and f(x) = P(S(x)) "seems" the only solution. note that we must have poles in this case because of the 2-periodic function. however , i mentioned "seems" , because i dont know many counterexamples. but if f(x) = f(exp(x)) then f(x) = f(x + 2pi i) and i excluded that by writing ' nonlinear ' above. but thats just because i know the following : f(x) = 1-periodic [(slog(x))] slog(x) = slog(exp(x)) - 1 slog(x + 2pi i) = slog(exp(x + 2pi i)) -1 = slog(x) hence f(x) = 1-periodic [(slog(x))] = 1-periodic [(slog(x + 2pi i))] thus f(x) = f(x + 2pi i) in the OP i wrote : ** f(x) = f(g(x)) its a simple equation. and i talked about it before. as did e.g. gottfried. in particular for a certain given g(x) the equation is solvable for f(x) by using the super of g(x) and a 1-periodic function... but that is not the end of the story. what if f(x) is given and we want to find g(x) ? well for starters g(x) might not be unique. if f(x) = f(exp(x)) then we must also have f(x) = f(x + 2pi i) the pattern is clear f(x) = f(g(x)) => g(x) = g(g1(x)) => f(x) = f(g1(x)) and we can now replace g(x) with g1(x) , g1(x) with g2(x) and repeat ... until we arrive at a moebius function.... if we ever ... i guess we could call g_n(x) the invariants of f(x). ** so it seems the g_n are a family of invariants. now back to the modified (!) equation f(x) = f(a_n(x)) = f(b_n(x)) where a_n and b_n are not iterations or invariants of eachother. now it seems that if a_n and b_n are not cyclic iterations , then indeed P(x) = 2-periodic function ( period a and b ) S(x) => S(a(x)) = S(x) + a , S(b(x)) = S(x) + b hence S(a(b(x))) = S(b(a(x))) = S(x) + a + b (*) and f(x) = P(S(x)) "seems" the only solution.... or not ? does this rule out ( with the similar above conditions ) f(x) = f(a_n(x)) = f(b_n(x)) = f(c_n(x)) then ? i think so. but what about algebraic functions then ? consider for instance f(x) = f(alg(x)) and the related alg^[n]alg(x). more specifly if they have closed forms , such as the following ( with thanks to "achille" ) For p, r, s with p > 0, Let Q(x) be (p^2-1)*x^2 + 2*(p+1)*r*x + s, then f(x) = p*x + sqrt(Q(x)) + r; => f^[n](x) = T_n(p)*x + U_{n-1}(p)*sqrt(Q(x)) + r*(T_n(p)-1)/(p-1); where T_n(p) and U_n(p) are the n-th Chebyshev's polynomial of first and second kind. In particular, for (p,r,s) = (3,1,20) f(x) = 3*x+ 2*sqrt(2*x^2+2*x+5) + 1 => f^(x) = 17*x + 12*sqrt(2*x^2+2*x+5) + 8 f^(x) = 99*x + 70*sqrt(2*x^2+2*x+5) + 49 ... f^[n](x) = T_n(3)*x + 2*U_{n-1}(3)*sqrt(2*x^2+2*x+5) + (T_n(3)-1)/2 makes one wonder about equations like f(x) = f(3*x+ 2*sqrt(2*x^2+2*x+5) + 1) or more complicated ones ! maybe i just need a good book on invariants or alike :p regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 12/02/2010, 01:28 PM (This post was last modified: 12/02/2010, 09:35 PM by tommy1729.) maybe i sound naive. but i think that is naive. in other words , intuitively it may seem simple : f(x) = f(g(x)) we set q(g(x)) = q(x)+1 and p(x) is 1 periodic and t(x) is the inverse of p(x). then f(x) = p(q(x)) now to find g(x) from f(x) we ' believe ' that reversing the above will help : f(x) = p(q(x)) t(f(x)) = q(x) (*) t(f(x)) + 1 = q(x)+1 from (*) we know that inv q(x) = invf(p(x)). lets call that F(x). then from q(g(x)) = q(x)+1 = t(f(x)) + 1 we get g(x) = F(t(f(x)) + 1) ( assuming the correct branches are taken ) this formula looks powerfull ... lets try exp(x) = exp(g(x)) g(x) = log( sin (2pi * (1/(2pi) arcsin(exp(x)) + 1) ) ) we simplify g(x) = log( sin( arcsin(exp(x)) + 2pi) ) ) this doesnt look good ... we simplify further g(x) = log(sin(arcsin(exp(x))))) now note that we cannot work with the branches of log , since we need to know an invariant of exp to know the branches of log. if we do not know that exp(x + 2pi i) = exp(x) we do not know that log(x) + 2pi i is another branch of log(x) and visa versa ! we could find that by investigating log and exp , but that is not a general method. the riemann surfaces of a function and its inverse reveals branches and invariants , but not easily in closed form rather some numerical values. so how do we reduce g(x) = log(sin(arcsin(exp(x))))) ? sure g(x) = log(sin(arcsin(exp(x))))) implies that g(x) = x. but g(x) = x is a uselessly trivial result. my example included functions we understand well , but of course more complicated ones are not so easy to analyse. i am thus tempted to conclude that no decent theory of invariants exists without investigating the riemann surfaces. more precisely , i assume that we must find a function mapping one branch to the next , therefore requiring to know the structure of the riemann surface ( shape and cuts ) and expressing that function as a taylor series , which we compute by taking nth derivatives of the analytic continuations that take us to the next branch value of the imput. although that may work in many cases , since we need to understand the structure of the surface alot , it still might not be - or lead to - a complete theory of invariants. i conclude that , unless i am missing something important , that solving f(x) = f(g(x)) for g(x) is non-trivial. however my knowledge is limited , maybe someone can "open my eyes" with some theorems or theories. since functions like tetration and the alike are more exotic than any standard function i therefore find it justified to consider finding invariants , even if they dont have closed forms. regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,358 Threads: 330 Joined: Feb 2009 12/12/2010, 10:02 PM to make a long story short for f(z) an entire function it seems that to find at least 1 function g(z) =/= z such that f(z) = f(g(z)) we can *do* the following : consider a branch of f^[-1](z) , call it branch A. the branch below it is branch B. the domain of branch A is nonempty and connected and the domain of B is also nonempty and connected. we can map dom A or dom B uniquely and bijectively to a unit circle. RMT1(dom(A)) = unit circle RMT2(dom(B)) = unit circle. hence by the Riemann Mapping Theorem (RMT), we can map the dom A to the dom B and visa versa by : RMT2^[-1] [RMT1(dom(A))] = dom B RMT1^[-1] [RMT2(dom(B))] = dom A hence RMT2^[-1] [RMT1(z)] or RMT1^[-1] [RMT2(z)] is an invariant. these 2 solutions however can equal id(z) when dom(A) = dom(B). further , these invariants have branches too. we need to find and express the domains A and B. and we need to find the riemann mappings. and even when we have done that , we still dont have a property of the invariant ... we dont even know if its cylic under iteration , if all the other invariants are interations of it , if it is entire , if it has singularities or branches , ... but i thought id mention that once again the Riemann Mapping Theorem is involved and how it works !! note : for instance if one would try to find the period of a function by using this , it wont be that simple : if the invariant turns out to be x + C you have your period. but to find x + C you (probably*) first need to know that dom B is just dom A +/- C !! this is circular !! ( * to construct the Riemann Mappings , at least by this method ) well maybe there is a way around that , but the general case f(z) = f(g(z)) probably doesnt have a simple way around such problems ... nevertheless , we know g(z) relates to the superfunction , so we(?) are not giving up on this if i am missing something trivial about this subject , plz inform me. regards tommy1729 « Next Oldest | Next Newest »