Discussion of TPID 6

[JJacquelin]

Conjecture

where such that

Discussion

To evaluate f at real numbers, an extension of tetration is required, but to evaluate f at positive integers, only real-valued exponentiation is needed. Thus the sequence given by the solutions of the equations

• 
  
• 
  
• 
  
• 
  

and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is , also known as eta (). Numerical evidence indicates that this is true, as the solution for x in is approximately 1.44.

I think that the conjecture is false.
First, the numerical computation have to be carried out with much more precision.
The solution for x in is approximately 1.44467831224667 which is higher than e^(1/e)
The solution for x in is approximately 1.4446796588047 which is higher than e^(1/e)
As n increases, x increasses very slowly.
But, in any case, x is higher than e^(1/e) = 1.44466786100977

Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of is e , for n tending to infinity. So, the limit isn't = n , as expected.

[sheldonison]

....
As n increases, x increasses very slowly.
But, in any case, x is higher than e^(1/e) = 1.44466786100977

Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of is e , for n tending to infinity. So, the limit isn't = n , as expected.

I think that's a good starting point. For x=e^(1/e), the

Another limit that I think holds is that the slog(e) gets arbitrarily large as the base approaches eta from above. Note that for these bases with B>eta, sexp(z) grows super exponentially when z gets big enough.


Now lets pick 10000. Solve for base b>eta . We know there is another number n>10000, for which , because we know that super exponential growth will eventually set in, as n grows past 10000, and that . Then, for some number n>10000, . I actually have a hunch that somewhere around n=20000 or so that superexponential growth finally kicks in.

I guess what I'm trying to get at is that we can probably prove that for , solving for b as n grows arbitrarily large, b approaches eta+. For each particular base b, there is another larger number, call it "m>n", for which Andrew's equation holds. . And that might be a pretty good step in proving Andrew's lemma.
- Sheldon

[nuninho1980]

I think that the conjecture is false.
First, the numerical computation have to be carried out with much more precision.
The solution for x in is approximately 1.44467831224667 which is higher than e^(1/e)
The solution for x in is approximately 1.4446796588047 which is higher than e^(1/e)
As n increases, x increasses very slowly.
But, in any case, x is higher than e^(1/e) = 1.44466786100977

Second, on a more theoretical viewpoint, if x=e^(1/e), the limit of is e , for n tending to infinity. So, the limit isn't = n , as expected.

The solution for x in is approximately 1.44467831224667 -> is not correct! but yes - x=1.44467829141456

The solution for x in is approximately 1.4446796588047 -> is not correct but yes - x=1.4446679658595034

you mistake!?!! eheheh! lool

x=1.444667862058778534938

therefore, the conjecture is NOT false!

I calculated the numbers corrects by program "pari/gp".

[bo198214]

Hey guys,

1. please dont post discussion in the open problems survey! Its reserved for problems exclusively.

2. The conjecture is already proven:
By me here
By tommy here.

I update the stati of the problems.