08/12/2013, 11:18 PM

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

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