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+--- Thread: x^(1/x) = f^[oo](x) ?? (/showthread.php?tid=808)
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