Another proof of TPID 6

[tommy1729]

Conjecture

\( \lim_{n\to\infty} f(n) = e^{1/e} \) where \( f(n) = x \) such that \( {}^{n}x = n \)

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

• \( x = 1 \)
  
• \( x^x = 2 \)
  
• \( x^{x^x} = 3 \)
  
• \( x^{x^{x^x}} = 4 \)
  

and so on... is the sequence under discussion. The conjecture is that the limit of this sequence is \( e^{1/e} \), also known as eta (\( \eta \)). Numerical evidence indicates that this is true, as the solution for x in \( {}^{1000}x = 1000 \) is approximately 1.44.

lim n-> oo x^^n = n conj : any real x = eta

since (eta+q) ^^ n grows faster than n for any positive q , we can use the squeeze theorem

lim q -> 0 eta =< x <= eta + q

hence x = eta

see also http://en.wikipedia.org/wiki/Squeeze_theorem

QED

regards

tommy1729