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 tommy equation tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 02/16/2015, 02:59 AM g is an entire function. " tommy equation " g(f(a x^b)) = f(c x^d) a,b,c,d integers >0. prototype solution Taylor(ln(x),x) ... maybe more later. with thanks to Kouznetsov. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 03/16/2015, 11:30 PM Ok I tried some things with the tommy equation. First a usefull link with thanks to my friend mick : http://math.stackexchange.com/questions/...lor-series So this explains what type of series expansions I am talking about ; the Tommy-Kouznetsov expansion of some order. So back to the tommy equation and considering parabolic fixpoints. Without going into detail we can recenter the parabolic fixpoint at 0. THE CORRECT SERIES EXPANSIONS FOR PARABOLIC FIXPOINTS ARE NOT KNOWN ?? Anyway Kouznetsov conjectured that for large x , Tommy-Kouznetsov expansions of order 1 are asymptotically correct. However I commented that by taylor's theorem and/or fake function theory ( and A few others ) this is also true for the simpler Taylor series. So I considered these expansions as EXACT solutions for real x > 0. That lead to the tommy equation ( as explained/introduced in post 1 ) , with g now having a parabolic fixpoint at 0. However after playing around I seemed to be getting into trouble. There were no Tommy-Kouznetsov expansions that solved the equation in most cases. In a kind of pseudocode this is what I wrote ( in simple Notepad ). ( if unclear let me know ) *** fix at 0 ( fix a = f(a) => f* (0) = 0 ) larger then x asymp x x + x^2 = p(x) p(f(x)) = f(ex) f(x) + f(x)^2 = f(ex) (b + c ln(x)) x +o(xx) + ((b + c ln(x)) x + o(xx))^2 = (eb + ec + ec ln(x)) x + o(xx) => b = eb + ec c = ec => error restart (b + c ln(x) + d ln(x)^2) x = (b + c + c ln(x) + d (ln^2 + 2 ln + 1) ) ex => = (eb + ec + ed + (ec + 2ed) ln + ed ln^2 ) x ed = d => error etc Same problem higher order Same problem p(f(A x^B)) = f(C x^D) for nontrivial cases of A,B,C,D. *** Notice o(xx) means o(x^2) and this goes to 0 very fast for small x , faster then the other terms hence they are ignored. Then it should be clear to you that no solution exists in general. SO I REPEAT : THE CORRECT SERIES EXPANSIONS FOR PARABOLIC FIXPOINTS ARE NOT KNOWN ?? Or am I wrong here ? Maybe we need a new equation to replace the tommy equation. I call it the tommy-abc equation for now. And if it works I call it the shadow equation. Here it is : p( f(x) ) = f(ax) + b f(cx) Hopefully this works better. I think the tommy-abc equation or shadow equation is new. regards tommy1729 " Truth is whatever does not go away when we stop believing in it " tommy1729 tommy1729 Ultimate Fellow Posts: 1,640 Threads: 366 Joined: Feb 2009 03/17/2015, 11:45 PM The problem might be p( f(x) ) = f(ax) + b f(cx) is nice but does it mean p^[z] ( f(x) ) = f(a^z x) + b^z f(c^z x) or does p^[z] ( f(x) ) = f(a^z x) + b f(c^z x) or something else ? I noted that since p(0) = 0 : Let f(A) = 0 p(f(A)) = f(aA) + b f(cA) = 0 Not sure what to make of that ... regards tommy1729 sheldonison Long Time Fellow Posts: 684 Threads: 24 Joined: Oct 2008 03/18/2015, 08:52 AM (03/16/2015, 11:30 PM)tommy1729 Wrote: ...considering parabolic fixpoints. Without going into detail we can recenter the parabolic fixpoint at 0. THE CORRECT SERIES EXPANSIONS FOR PARABOLIC FIXPOINTS ARE NOT KNOWN Is this related to Jean Ecalle's formal solution for the Abel function for the parabolic case? See for example: Jean Ecalle's solution - Sheldon « Next Oldest | Next Newest »

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