Hi everyone,

The function y_1(x)=ln(x) is an unbounded strictly increasing function going to infinity at a slow rate.

But, inside it, the function x, goes to infinity faster than ln. So, let's reduce the rate a bit more:

y_2(x)=(ln○ln)(x)=ln(ln(x))=ln(y_1(x))

You get it, we can define y_n(x)=(ln○...n-times...○ln)(x)

Thus, what is the limit if n goes to infinity?

And here's my problem:

As it is a compose of unbounded strictly increasing functions, it should be an unbounded strictly increasing function too and should go to infinity as well but, for instance, it never reaches 0. It sounds like a paradox.

Case of 0:

If x=e(0), y_1(x)=0

If x=e(e(0)), y_2(x)=0

...

If x=(e○...n-times...○e)(0), y_n(x)=0

But lim(x=(e○...n-times...○e)(0),n goes to infinity) is infinity, therefore if n tends to infinity, y_n(x) tends to 0 for this x.

And, we can proceed as well for other values of the function.

Please help me to correct my approach and to deal with this infinite compose and perhaps to try to answer the question of reaching infinity as slow as possible and by bijection as fast as possible (if there's any answer to this question).

Ciao

Nicolas

First thing is that your function f(x) =ln(.. (n times).. ln(x)) is not defined in (0, inf). To see why think why ln(ln(x)) is not defined for (0, e) and similarly ln(ln(ln(x))) is not defined in (0, e^e).

I will define f(x) =ln|ln... |ln(x)|||... (n times)... |, where |ln(x)|=abs(ln(x)) so that f(x) is defined in (0, inf).

Thanks! You are right!

The first problem is the domain. Now it appears to me the limit is defined nowhere.

I hope it is not the answer anyway...

Bye

@Nicolas CALANDRUCCIO what you wanted to ask in question?

I got diverted from answering your question as I had not understood clearly what you wanted to ask.

@Devansh Singh First, thank you for your support. In fact, I was just doing some math for fun.

I thought to ln(ln(...(ln(x))...) with its reciprocal bijection e(e(...(e(x)...)

Then, I asked myself if by this process we can create a function to go as slow as possible to infinity and conversly as fast as possible.

Obviously, I tried to evaluate a limit to this compose as n tends to infinity. But I have no idea on how to deal with it.

I tried to calculate some values, then I realize the problem to solve f(x)=0. There is also the fact that by my first construction I can see this function as a compose of increasing functions that should be increasing as well.

Now limit for f(x), I just defined can be compared with g(x) =-1/x, in (0, inf). The infinity(inf) , I am writing is infinity at which f(x) approaches zero. There are infinities bigger than this inf. f(x) is asymptotic just like -1/x to negative y and x axis in (0, inf). So for limit x tending to infinity(inf) and limit n tending to infinity(<inf) ,f(x) tends to zero.

I'm not sure to understand what you mean.

@Nicolas CALANDRUCCIO I am still working to find function which approximates f(x).

f(x) is not pointwise convergent in (0, inf). Only at x tending to inf with n tending to infinity<inf, f(x) tends to zero, it has no limit at any x >0, x<inf. Or we cannot find any m such that definition of limit holds for f(x) in (0, inf).

For f(x) =-ln(ln(ln(x))), function I found less than f(x) for (0, e^e-) is 2300/((x+1) ^5) but here 5>3, so I am not suspecting any function, g(x) which is less than f(x) such that g(x) <f(x) <epsilon(say a), here 2300/((x+1) ^5)<f(x) <a, then ln(2300/a) /ln(x+1) <5, or ln(2300/a) /ln(x+1) <m, here m=5,but m>3 . So it fails limit definition for higher n.

What work have you done on this?

I've just tried to set a recursive formula.

Take x>=e(e(...e(0)...)

Set u_0=ln(x)

Evaluate u_n+1=ln|u_n|

It works, as far as it is a finite process.

By applying a limit, it seems we are erasing the domain and the information.

So, it's like it is an impossible infinite recurrence.

The Red graph above is of f(x) =-ln(ln(ln(x))), and the graph in pink is what I am talking of, you can see that both are asymptotic to x, y axis.Curve in pink is less than f(x) for x=e^e- or approximately 14.8.can you find another function which is better than this and has degree greater than -3 with same property, if yes then it is possible that for n tending to infinity f(x) tends to zero as x tends to inf.

Now do you understand why I wanted to compare -1/x like functions with f(x) you asked in question.

Try this:

Prove that ln [ln[ln ... [ln(x)] ...] {n compositions} ---> inf as x ---> inf

1) Verify for n = 1

ln(x) ---> inf as x --> inf

Therefore true when n = 1

2) Assume ln[ln[ln[ ....[ln(x)]...] { k compositions}---> inf as x ---> inf

3) Using the assumption (2) prove that:

ln[ln[ln]...[ln(x)] ...] --> inf {k+1 compositions} as x --> inf

ln[ k+1 compositions]

= ln[ln[ k compositions]]

=ln[inf] by assumption

=inf

@Ian Fowler-ln(ln(x)) is not defined for (0, e) and similarly ln(ln(ln(x))) is not defined in (0, e^e)~(0,15.15). ln(ln(ln(ln(x)))) is not defined in (0, e^(e^e)) ~(0, 1618.17798).

ln(ln(ln(ln(ln(x))))) is not defined in (0,e^e^(e^e)).

I will define f(x) =ln|ln... |ln(x)|||... (n times)... |, n tends to inf, |ln(x)|=abs(ln(x)) so that f(x) is defined in (0, inf). Range of f(x) is (-infinity, 0), see that function can't even reach 0.It asymptotically approaches zero at inf.

limit for f(x)=ln [ln[ln ... [ln(x)] ...] {n compositions},can be compared with g(x) =-1/x, in (0, inf). The infinity(inf) , I am writing is infinity at which f(x) approaches zero. There are infinities bigger than this inf. f(x) is asymptotic just like -1/x to negative y and x axis in (0, inf). So for limit x tending to infinity(inf) and limit n tending to infinity(<inf) ,f(x) tends to zero.

You can't even reach zero in f(x) for some finite value of x, and you are talking about infinities bigger than inf, I described. Now infinities have to be compared when you are talking about limits. Can you find m for f(x) such that definition of limit is satisfied for some epsilon for f(x) tending to zero at inf or m for f(x) tending to some function g(x) from lower, upper side for x<inf for some epsilon.