bo198214 Wrote:Aha and there is the answer why you considered your formula to be tetration. Because you compared it with some table values!

Now that you are catched, you have to tell where you found this table!

One of them is here: http://tetration.itgo.com/txt/table-tetbin.txt

(I think that's the only one I found for 2^^x)

bo198214 Wrote:However, to all my knowledge, there is no *the* tetration. There are different approaches which result in different tetrations that all satisfy those properties like being analytic strictly increasing and so on.

This is exactly my state of knowledge, too. And it's still hard to believe (aren't there any bounds or something?)

Ivars Wrote:May be it is worth trying ( just out of pure interpolation) without really following what has been done in this thread) something like

ln(2)-ln(2)/2*x+(ln(2)/3!)^x^2 - (ln(2)/4!)*x^3 +

or similar (may be not factorial, but just 1/2, 1/3, 1/4, etc) alternating series with ln(2) in coefficient.

I shall try that next week during the lunch breaks. I recognized I was close to ln(2) there, but didn't think that far.

bo198214 Wrote:Can you give a proof for n=0, i.e that

?

Come think of it, I could actually try that...

Recall that Newton approximated his square roots by taking the arithmetic-harmonic mean:

(2+3)/2=2.5 or 5/2 and 2/(1/2+1/3)=2.4 or 12/5

(5/2+12/5)/2=49/20 and 2/(2/5+5/12)=120/49 etc., eventually obtaining the square root of 6:

(sqr(6)+sqr(6))/2=sqr(6) and 2/(1/sqr(6)+1/sqr(6))=sqr(6)

That lies between n=1 and n=-1. Is the result really in the middle at n=0? I try that with n=1/2 and -1/2:

((sqr(2)+sqr(3))/2)^2=5/4+sqr(6)/2 and (2/(1/sqr(2)+1/sqr(3)))^2=120-48*sqr(6)

etc., I end up with the same solution. Moreover, the geometric mean of the two values at each step is exactly sqr(6), in both examples.

I hope this was somewhat understandable.

Does this count as a proof?