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

@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.

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

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

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.

What work have you done on this?

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.

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).

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.

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).