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Discussion of TPID 6 - Printable Version

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Discussion of TPID 6 - JJacquelin - 10/22/2010

(10/07/2009, 12:03 AM)andydude Wrote: 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.


RE: Limit of self-super-roots is e^1/e. TPID 6 - sheldonison - 10/22/2010

(10/22/2010, 11:27 AM)JJacquelin Wrote: ....
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



RE: Limit of self-super-roots is e^1/e. TPID 6 - nuninho1980 - 10/23/2010

(10/22/2010, 11:27 AM)JJacquelin Wrote: 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". Smile


RE: Discussion of TPID 6 - bo198214 - 10/24/2010

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.