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 f( f(x) ) = exp(x) solved ! ! ! tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/05/2009, 12:09 AM On sci.math i posted the solution to f( f(x) ) = exp(x). That is of course important towards tetration. Here is the post : http://mathforum.org/kb/thread.jspa?threadID=1891059 Its important to note that the same strategy works to compute e.g. f( f( f(x) ) ) = exp(x) And can thus be used to compute tetration on the real line numerically. Here is the post explicitly , in case sci.math is not accessible ( which happens sometimes ) : brute force tetration / greedy tetration Posted: Jan 27, 2009 9:39 AM Plain Text Reply here i introduce a brute force / greedy algoritm to compute f(f(x)) = exp(x). its works best outside the unit radius. it is based upon the approximation below , and is trivial considering the approximation. since , once we have a sequence of ever better getting approximations , taking that limit gives the desired result. the approximations are also usefull because of computational boundaries. the algoritm is the limit n -> oo note that f(f(x)) = g(x) can be computed if g(0) = 0 f(f(x)) = exp(x) - exp( -1 * 25^n * x^2 ) examples : n = 1 : f(f(x)) = exp(x) - exp( -25 * x^2 ) n = 2 : f(f(x)) = exp(x) - exp( - 625 * x^2 ) n = 3 : f(f(x)) = exp(x) - exp( - 15625 * x^2 ) ... i , tommy1729 , am the first to invent this. i will not accept shameless copies of this without mentioning me. copyright tommy1729 regards tommy1729 ( end quote ) I am the sole inventor of this. regards tommy1729 " Statisticly , i dont exist " tommy1729 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 02/05/2009, 09:17 PM First, you do a very poor job of explaining yourself, so I'm not sure if I (or anyone on the math forum) have completely understood your algorithm. I am very interested in your algorithm, but you have not shown enough of it for me to duplicate your results, or verify your solution. Until I can verify how your definition of f(x) is a solution to this equation, I'm not sure if you have a solution at all. I recommend you explain more about why you think it is a solution, rather than listing examples. Second, f(f(x)) = exp(x) has already been solved. This is what regular iteration is good at. Simply searching this forum for "regular" will give you many places to start, if you would like to know more. The problem with the solution that regular iteration gives is that it only converges for integer iterates. This means that the power series for f(x) is not analytic, so it cannot be used to find a value, but it may be good for approximations. Third, you do not need to include a license. A simple ©2009... will suffice. Also, none of what you said is part of an "invention" in the technical sense, because you do not make any claims. Basically, you need to tell people what it is before you can say you have an invention. From what you said, I have no idea what you claim to have invented. Did you invent "f" or "g"? Did you invent (25^n)? For all I know you could have just invented the word "greedy". Also, none of this would hold up in the USPTO because you can't patent math. The only way to do something similar is to publish your ideas in a well-known journal, for example, American Mathematical Monthly, or the like. Andrew Robbins tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/05/2009, 11:11 PM (This post was last modified: 02/06/2009, 03:07 AM by bo198214.) andydude Wrote:First, you do a very poor job of explaining yourself, so I'm not sure if I (or anyone on the math forum) have completely understood your algorithm. there is nothing wrong with what i said. i defined f(x) as the limit n -> oo in f(f(x)) = exp(x) - exp( -1 * 25^n * x^2 ) note that f(x) can be easily computed because f(0) = 0 like i already said in the original post. you know how to solve f(f(x)) = exp(x) - 1 then you can use the same method for f(f(x)) = exp(x) - exp( -1 * 25^n * x^2 ) Quote: I am very interested in your algorithm, but you have not shown enough of it for me to duplicate your results, or verify your solution. yes i have ! take the limit n -> oo. think in terms of taylor series ! my limit n-> oo gives a taylor series for f(f(x)) = exp(x). Quote:Until I can verify how your definition of f(x) is a solution to this equation, I'm not sure if you have a solution at all. I recommend you explain more about why you think it is a solution, rather than listing examples. Second, f(f(x)) = exp(x) has already been solved. no , it was not solved before , thats part of why this forum exists. im talking about taylor series that converge for all reals ! Quote: This is what regular iteration is good at. Simply searching this forum for "regular" will give you many places to start, if you would like to know more. The problem with the solution that regular iteration gives is that it only converges for integer iterates. This means that the power series for f(x) is not analytic, so it cannot be used to find a value, but it may be good for approximations. im aware of " regular " but my method gives analytic for all real values not just some integers ! Quote:Third, you do not need to include a license. A simple ©2009... will suffice. Also, none of what you said is part of an "invention" in the technical sense, because you do not make any claims. Basically, you need to tell people what it is before you can say you have an invention. From what you said, I have no idea what you claim to have invented. Did you invent "f" or "g"? Did you invent (25^n)? For all I know you could have just invented the word "greedy". Also, none of this would hold up in the USPTO because you can't patent math. The only way to do something similar is to publish your ideas in a well-known journal, for example, American Mathematical Monthly, or the like. i know i cannot patent math. but i can still be the inventor of an idea , an analytic solution to f(f(x)) = exp(x). compare to : you cant patent the concept or use of " integrals " but you can compute a hard integral that no one else could before you , and you can sell math software and mention that. and you can object if someone said you he knew it first , while it was yourself. now , i might sound agressive , but that is not the intention. i do feel a bit attacked though ... surely defining a limit is not " something that cannot be checked " ? Moderator's note: corrected quoting and removed the subsequent post with same content bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/06/2009, 03:22 AM (This post was last modified: 02/06/2009, 03:25 AM by bo198214.) Hey andydude, the equation $f(f(x))=\exp(x)$ was not yet really solved, though we collected several approaches on that question on the forum. What can be solved by regular iteration is $f(f(x))=\exp(x)-1$, though there are convergence issues, I think this is what you refer to. The observation of tommy was that there are methods (regular iteration!) to solve $f(f(x))=F(x)$ for $F(0)=0$. So he just modified exp by subtracting the hump $\exp(-nx^2)$ of height 1 which makes it 0 at 0. If you increase the $n$ the hump gets sharper and sharper. For sufficient large $n$ nearly everywhere except in the direct vicinity of 0 the modified function looks like $\exp$. Now he can take the regular half iterate of the modified function and hope that the modified functions will converge for increasing $n$ and he would call the result then the half iterate of $\exp$. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/06/2009, 01:28 PM bo198214 Wrote:Hey andydude, the equation $f(f(x))=\exp(x)$ was not yet really solved, though we collected several approaches on that question on the forum. What can be solved by regular iteration is $f(f(x))=\exp(x)-1$, though there are convergence issues, I think this is what you refer to. The observation of tommy was that there are methods (regular iteration!) to solve $f(f(x))=F(x)$ for $F(0)=0$. So he just modified exp by subtracting the hump $\exp(-nx^2)$ of height 1 which makes it 0 at 0. If you increase the $n$ the hump gets sharper and sharper. For sufficient large $n$ nearly everywhere except in the direct vicinity of 0 the modified function looks like $\exp$. Now he can take the regular half iterate of the modified function and hope that the modified functions will converge for increasing $n$ and he would call the result then the half iterate of $\exp$. thank you bo for the clarification. note that this can also be used to compute f(f(f(x))) = exp(x). and basicly f ^ (m/n) = exp(x) which has a huge impact on tetration. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/06/2009, 06:02 PM tommy1729 Wrote:note that this can also be used to compute f(f(f(x))) = exp(x). But slowly, slowly. Before we continue we first have to see whether your idea works. Thats not really clear. The thing is that the regular iteration at a fixed point depends strongly on the derivatives at the fixed point. However your function $\exp(-x^2)$ messes up the derivations of $\exp$ at 0. From that point of view its doubtful whether it results in something usable. However a first step to assure the substance of the idea would be to plot the approximations of $\exp^{1/2}$ and to see whether they really converge and if to what function $f$ and whether this function really satisfies $f(f(x))=\exp(x)$ (these are all open questions until now). Maybe I can do this, but not immediately. Any volunteer out there? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/06/2009, 10:33 PM bo198214 Wrote:tommy1729 Wrote:note that this can also be used to compute f(f(f(x))) = exp(x). But slowly, slowly. Before we continue we first have to see whether your idea works. Thats not really clear. The thing is that the regular iteration at a fixed point depends strongly on the derivatives at the fixed point. However your function $\exp(-x^2)$ messes up the derivations of $\exp$ at 0. From that point of view its doubtful whether it results in something usable. well , i dont take the derivates of exp(-x^2) to construct the taylor series of exp(-x^2), i take the expansion of exp(y) and substitute y = - x^2. simple and efficient. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/07/2009, 12:37 PM tommy1729 Wrote:to construct the taylor series of exp(-x^2), i take the expansion of exp(y) and substitute y = - x^2. Well thats your choice how you compute the derivatives. But I dont see the connection to the open questions. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/08/2009, 12:29 AM bo198214 Wrote:tommy1729 Wrote:to construct the taylor series of exp(-x^2), i take the expansion of exp(y) and substitute y = - x^2. Well thats your choice how you compute the derivatives. But I dont see the connection to the open questions. open questions ? surely , you must see f(f(x)) = exp(x) is connected to tetration. that cannot be denied. for instance , this lead to numerical data and computable points for tetration-related functions. regards tommy1729 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 02/08/2009, 12:15 PM 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. « Next Oldest | Next Newest »

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