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x^(1/x) = f^[oo](x) ?? - tommy1729 - 08/12/2013
Recently I started to consider an ancient idea of me again. Its about infinite recursion without an infinite amount of distinct parameters. Hence e.g. not a continued fraction form with an infinite amount of distinct integers. Also its about infinite recursion for analytic functions , not just for a number. We all know very well this function : h(x) = x^(x^(x^...)). Clearly h(x) = x^h(x) whenever h converges. Also trivial is h^[-1](x) = x^(1/x). Unlike Mandelbrot fractal these are all analytic and we did not even mention fixpoints or bifurcations. h(x) = g^[oo](x,g^[oo-1]) although that is funny notation. ( a lim form might be better notation ) As an important sidenote I mention is that the difficulty of working with functions ( such as h(x) or x^(1/x) ) mainly depends on the way THE FUNCTION IS PRESENTED. For instance if I gave h(x) or x^(1/x) as a fourrier series , an integral or a differential equation it would be harder to NOTICE and conclude these things. SO how about x^(1/x) ? Can we find an infinite recursion for this function ? Why is it usually hard to find a closed form infinite recursion for a given function AND its inverse ? ( although such things occur often in advanced math : integrals and derivatives are not equally easy to compute , checking a solution is harder than finding one etc etc ) Lets toy with this. h^[-1](x) = x^(1/x) (substitution x' = h(x) ) => x = h(x)^(1/h(x)) => x^h(x) = h(x) (substitution) => x^(x^h(x)) = h(x) =>x^(x^(... = h(x) However this does not work in general ! The main reason this worked is because we got the right amount of x's and h(x)'s in one equation. Or in other words because x^(1/x) " contains " 2 x's. Which also partially explains what I said about the form in which the functions are given and the related difficulty. TO CLARIFY : Lets toy with this. h^[-1](x) = g(x) (substitution x' = h(x) ) => x = g(h(x)) => g^[-1](x) = h(x) (substitution ???? ) => x^(x^h(x)) = h(x) ???? =>x^(x^(... = h(x) ???? Also notations such as g(x,h(x)) , Q(x) = F(g(x),h(x)) etc do not seem to help at all. Im not just talking about a nice infinite recursion form , but just an infinite recursion form at all ! ( compare to integrals , not all integrals can be done in closed form , but we have contour integrals , special cases for specific intervals , transforms , diff eq and a numerical integral Always DOES exist ) ( this might remind some of you who know me from sci.math about similar stuff where the function and its inverse did not " cooperate " easily , such as e.g. for finding the period of an analytic function ) Another way of attack might be trying to do implicit differentiation in reverse. But that seems to give similar problems. And hence also contour integration sounds like trouble ( and requiring understanding of the position of the zero's might be annoying ). Suppose the solution can only be expressed by a PDE , then one wonders how we are to find that PDE since we did nothing that involved multiple infinitesimals ? Or did we ? A table lookup seems hard since there is no big table. Hence perhaps more a dynamics approach ? I was thinking that taking the *inverse* superfunction might work : F(x) = superfunction of G(x) => G(x) = F( F^[-1](x)+1) Now F(x) = G(G(G( .... This seems , " seems " what we need. However as an example : exp(x) = super of g(x) = e x. exp(0) = 1 = e * e * e * ... exp(-oo) = lim e^n * "something". exp(-oo) = 0 and not exp(0) and besides we have to consider things as a limit and do not end up with a meaningfull iteration. SO THIS IS NOT WHAT WE WANT !? Or is it ??? Im running out of tricks ! x^(1/x) = super of G(x) G(x) = [h(x)+1]^[1/(h(x)+1)] Is this the solution we want ?? Do we need to extend h(x) beyond its Shell-Tron region ?? What is the region of convergeance for this infinite iteration ?? Also If we start at value x ( which we should ) and take G^[2](x) I get (take h) h( [h(x)+1]^[1/(h(x)+1)] ) = h(x)+1 (add 1) h(x)+2 (do x^(1/x)) (h(x)+2)^[1/(h(x)+2)] ( this makes sense since its super is x^(1/x) !! ) By induction we get (also for infinity ! ) G^[n](x) = (h(x)+n)^[1/(h(x)+n)] This seems like a nasty lim and does this really = x^(1/x) ?? I think not ; by substitution : lim (Q+n)^(1/Q+n) = n^(1/n) = 1 (or at least number on the unit circle). I might be wrong here or there , but im stuck. Should I think about fixpoints , bifurcations and fractals ? Or involve complex numbers ?? This seems harder than it looks. The only thing that seems to work is f(x) = a f(x) + x - a = x g(a) = x => iterations of f(x) + x - a Lets try a, f(a) + a - a = f(a), f(f(a)) + f(a) - a, f(f(f(a)) + f(a) - a) + f(f(a)) + f(a) - 2a, ... SO basicly this gives us the solutions : h(x) + x - a and x^(1/x) + x - a. NOTICE x^(x^(...)) is not included here !! This suggest there must be another solution for x^(1/x) too ! ( I never claimed uniqueness and It is shown that h(x) has at least 2 such forms ) Any suggestions ? regards tommy1729 RE: x^(1/x) = f^[oo](x) ?? - tommy1729 - 08/12/2013
Of course we have f(x) = a f(x) - a + t(x) = t(x) t^[-1](f(x) - a + t(x)) = x Or most(?) general f(x) = a A(B(f(x)) - B(a)) + C(x) = A(0) + C(x) C^[-1] ( A(B(f(x)) - B(a)) + C(x) - A(0) ) = x Just to be complete. regards tommy1729 RE: x^(1/x) = f^[oo](x) ?? - tommy1729 - 08/13/2013
As an additional comment I would like to say that if we wanted an iterated power tower a_n^a_(n-1)^...a_0 then this is in some sense the inverse question of my comment to gottfried(*) where we wanted the power tower a_0^a_1^a^2^... (* http://math.eretrandre.org/tetrationforum/showthread.php?tid=799 ) I know Ramanujan investigated this kind of " inversing " in particular for square root or other root iterations (and continued fraction ofcourse). http://mathworld.wolfram.com/NestedRadicalConstant.html And I have been toying with power towers too , such as my " tommyzeta function " here : http://oeis.org/A102575 A good theory for this seems not yet made ? regards tommy1729 |