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 f( f(x) ) = exp(x) solved ! ! ! tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/08/2009, 03:59 PM bo198214 Wrote:Pretending that there are no open questions does not help here. Especially for you tommy I list them again: Do the approximations of $\exp^{1/2}$ converge? I.e. we have approximations of $\exp$ these are the functions $g_n(x)=\exp(x)-\exp(-n x^2)$. These converge pointwise to $\exp$ (except at 0). Then we take the regular iteration of these functions $f_n={g_n}^{1/2}$. The question is whether they converge to anything. If they converge to $f$, is $f$ a differentiable or even analytic function? If they converge to $f$, does $f$ really satisfy $f(f(x))=\exp(x)$? As I already told some pictures would make a good start if proofs could not provided yet. question # n will be answered by "#n )" 1) yes they do converge. because , clearly g_n does. f_n(f_n) = g_n its clear that f_n cannot be to far away from f_(n-1) and f_(n+1). e.g. because both must be strictly rising for x > 1 and both are larger than the line y = x for x > 1. in fact that distance ( not near 0 at least ) is bounded by O ( exp(x) - g_n(x) ) 2) since they converge , f is analytic. because both sequences f_n and g_n are continu and analytic. think of it like this : ( disregarding the neigbourhood at 0 for convenience ) suppose the opposite : f_n is not analytic for some n. since g_n is analytic and regular iterations of strictly rising functions preserve analytic properties(*). ( * if we also have f(0) = 0 which we do , we can construct taylor series , which are always analytic were they converge ) thus for all finite n , f_n is analytic. thus only candidate for not being analytic is f_oo. but that is absurd. since lim n-> oo f_oo(x) - f_n(x) = 0. if a function is not analytic it cannot have a non-positive real distance from an an analytic function everywhere ! 3) yes f(f(x)) = exp(x) is really satisfied , at least for x = / = 0. this is because lim n -> oo of g_n = for x =/= 0 => exp(x) for x = 0 => 0. ---------- there are still other questions than those 3. for instance , is my method equivalent to another method ? since my solution is analytic it can only be equivalent to other ( potentially ) analytic methods of course ... another remark is , that my method applies to base e tetration. it relates to other bases too , but how is not exactly clear yet. a 'change of base formula' is not yet found. also intresting are series and integrals of tetration. etc etc etc however im working on those too. ( it seems i will probably agree with most that has been written about it here ) one of the problems encountered in f(f(x)) = a^x is that i might have an annoying zero : a^x = x. for a real x. ( so that my 'bumpy method' does not apply , at least not without a modification ) regards tommy1729 " Statisticly , i dont exist " tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/08/2009, 10:08 PM (This post was last modified: 02/08/2009, 10:09 PM by bo198214.) Tommy these are far apart from proofs. I looked up your profile to see your math education. You wrote "professor" ... What do you think this question is for? To discriminate people with not much math education? Its the opposite; this point is for being prepared at which level you can expect and give math explanations to that person. And I expect that you correct your wrong declaration. The level we are currently talking is high school maths. Nothing wrong about it, but it lacks the punch of real proofs, the certainty that the findings are solid, etc. To only give an example how shallow reasoning may fail: tommy1729 Wrote:2) since they converge , f is analytic. because both sequences f_n and g_n are continu and analytic. ... thus for all finite n , f_n is analytic. thus only candidate for not being analytic is f_oo. but that is absurd. since lim n-> oo f_oo(x) - f_n(x) = 0. Wrong. You can have sequences $f_n$ of analytic functions that converge to a function that is not analytic, not even continuous. And curiously your example $g_n(x)=\exp(x)-\exp(-nx^2)$ is such an example. It converges to the function that is $\exp(x)$ for $x\neq 0$ and that is 0 at 0. This function is not continuous at 0. Or consider the sequence $f_n(x)=\sqrt{\frac{1}{n}+x^2}$ it converges to $f(x)=|x|$. Which is surely not differentiable at 0, while each $f_n$ is analytic. And your "proofs" of 1) and 3) do not contain any usuable conclusion that would answer these questions. Quote:one of the problems encountered in f(f(x)) = a^x is that i might have an annoying zero : a^x = x. for a real x. For you its a problem for others its a luck. You know you can apply regular iteration whenever a function has fixed point... tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/08/2009, 10:56 PM tommy1729 Wrote:bo198214 Wrote:Tommy these are far apart from proofs. I looked up your profile to see your math education. You wrote "professor" ... What do you think this question is for? To discriminate people with not much math education? Its the opposite; this point is for being prepared at which level you can expect and give math explanations to that person. And I expect that you correct your wrong declaration. The level we are currently talking is high school maths. Nothing wrong about it, but it lacks the punch of real proofs, the certainty that the findings are solid, etc. To only give an example how shallow reasoning may fail: tommy1729 Wrote:2) since they converge , f is analytic. because both sequences f_n and g_n are continu and analytic. ... thus for all finite n , f_n is analytic. thus only candidate for not being analytic is f_oo. but that is absurd. since lim n-> oo f_oo(x) - f_n(x) = 0. Wrong. You can have sequences $f_n$ of analytic functions that converge to a function that is not analytic, not even continuous. And curiously your example $g_n(x)=\exp(x)-\exp(-nx^2)$ is such an example. It converges to the function that is $\exp(x)$ for $x\neq 0$ and that is 0 at 0. This function is not continuous at 0. Or consider the sequence $f_n(x)=\sqrt{\frac{1}{n}+x^2}$ it converges to $f(x)=|x|$. Which is surely not differentiable at 0, while each $f_n$ is analytic. And your "proofs" of 1) and 3) do not contain any usuable conclusion that would answer these questions. Quote:one of the problems encountered in f(f(x)) = a^x is that i might have an annoying zero : a^x = x. for a real x. For you its a problem for others its a luck. You know you can apply regular iteration whenever a function has fixed point... you removed my reply to this ?? you do realise sqrt is not an entire function !? thus not a counterexample. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/09/2009, 12:27 AM tommy1729 Wrote:you do realise sqrt is not an entire function !? thus not a counterexample. So what, then consider $f_n(x)=\exp(-x^{2n})$. Do you guess what it converges to? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/10/2009, 11:17 PM bo198214 Wrote:tommy1729 Wrote:you do realise sqrt is not an entire function !? thus not a counterexample. So what, then consider $f_n(x)=\exp(-x^{2n})$. Do you guess what it converges to? ok but , does the limit f_n ( f_n (x) ) converge to a strictly rising entire function that maps R to a subset of R that has no real fixpoints ? ( disregarding removable singularities ) my f oo does : exp(x) ( again disregarding removable singularities ( x = 0 ) ) high regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/10/2009, 11:48 PM tommy1729 Wrote:bo198214 Wrote:tommy1729 Wrote:you do realise sqrt is not an entire function !? thus not a counterexample. So what, then consider $f_n(x)=\exp(-x^{2n})$. Do you guess what it converges to? ok but , does the limit f_n ( f_n (x) ) converge to a strictly rising entire function that maps R to a subset of R that has no real fixpoints ? ( disregarding removable singularities ) my f oo does : exp(x) ( again disregarding removable singularities ( x = 0 ) ) high regards tommy1729 keep in mind that we know : if g(x) is a monotonic function defined on an interval I, then g(x) is differentiable almost everywhere on I , i.e. the set of numbers x in I such that g(x) is not differentiable in x has at most Lebesgue measure zero. if f(f(x)) = g(x) then f(x) is also a monotonic function. ( we may replace g and f with 'my' f_n and g_n ) high regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/10/2009, 11:52 PM tommy1729 Wrote:tommy1729 Wrote:bo198214 Wrote:tommy1729 Wrote:you do realise sqrt is not an entire function !? thus not a counterexample. So what, then consider $f_n(x)=\exp(-x^{2n})$. Do you guess what it converges to? ok but , does the limit f_n ( f_n (x) ) converge to a strictly rising entire function that maps R to a subset of R that has no real fixpoints ? ( disregarding removable singularities ) my f oo does : exp(x) ( again disregarding removable singularities ( x = 0 ) ) high regards tommy1729 keep in mind that we know : if g(x) is a monotonic function defined on an interval I, then g(x) is differentiable almost everywhere on I , i.e. the set of numbers x in I such that g(x) is not differentiable in x has at most Lebesgue measure zero. if f(f(x)) = g(x) then f(x) is also a monotonic function. ( we may replace g and f with 'my' f_n and g_n ) high regards tommy1729 also note if g(x) is one-to-one so is f(x) in f(f(x)) = g(x) this to show no jump points can occur. right ? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/11/2009, 09:08 AM I am not here to find more and more refined counterexamples. Perhaps there is a misunderstanding: The counterexample was to show that your "proof" is deficient, not to show that your method is wrong. Actually I dont know whether it works, and I dont even have a hint, because there are no pictures provided. So if you can not prove it, why not convince us with some graphs of your approximation functions that show them converging? PS: And please dont quote everything, but only the part you are replying to. There are Code:[quote] and Code:[/quote] that enclose each quote. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/11/2009, 02:35 PM bo198214 Wrote:Actually I dont know whether it works, and I dont even have a hint, because there are no pictures provided. So if you can not prove it, why not convince us with some graphs of your approximation functions that show them converging? 1) i dont have math software and cannot make those graphs. 2) you probably do have the software to make those plots and since you get the idea, you could make those plots yourself. that way your also certain that the data is not manipulated or carefully selected. 3) the truth of math does not depend upon someones capability of making graphs. call me oldfashion. 4) i admit more research is neccessary and also proofs of uniqueness and existance. however i believe most here assume analytic tetration exists and is unique ? 5) i believe a lot is clarified if we can prove f_n(x) does not intersect with f_m(x) for any integer n and m and x > 1. a graph might show if it does or not. 6) yes , im stubborn , dont use capitals at the beginning of a senteance and i make spelling mistakes regards tommy1729 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 02/11/2009, 03:28 PM tommy1729 Wrote:1) i dont have math software and cannot make those graphs.Sage is a free enormous powerful math software. Quote:2) you probably do have the software to make those plots and since you get the idea, you could make those plots yourself. Yes but its some effort, no math software has implemented half iterates via regular iteration. Quote:that way your also certain that the data is not manipulated or carefully selected. I think this is more an issue in other sciences not in mathematics. Quote:3) the truth of math does not depend upon someones capability of making graphs. call me oldfashion. The truth of math depends on nothing. However if you want to show that truth to someone, then you have to convince him. Usually this is done by giving a proof. If that can not be done, then other ways should be sought. Quote:however i believe most here assume analytic tetration exists and is unique ? Existence was shown by Kneser. Uniqueness is a mostly unresearched yet. Sure is that it is not unique by analyticity alone. « Next Oldest | Next Newest »

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