(03/28/2015, 08:59 PM)tommy1729 Wrote: I repeat : Does this method agree with koenigs ?
What is koenigs? Are you talking about the regular iteration? If so, in the case of JmsNxn's tetration, yes, any two tetrations for bases \( 1<b<\eta \) are equal if they have the same period. See proposition 10 and corollary 8 in
this paper
I'm petty sure that this can also be proved using the techniques in JmsNxn's paper
As for the super root thing... it's not even clear if it even results in a tetration, since the super root has no nice recurrence relation we can exploit to prove it.
Quote:Im wondering if this method also simplifies towards the continuum sum.
( that might have affect on methods based on continuum sums that are in an " unfinished state " )
With this we can easily define a continuum sum:
\( \sum_{x=0}^{z-1}f(x) = \frac{d^z}{dx^z}|_{k=0} \sum_{k=1}^{\infty}\sum_{i=0}^{k-1}f(i) \frac{x^k}{k!} \)
That this works can be easily verified with the newton series identity. Using that, it can also be seen that it extends faulhaber's formula (i.e. it maps polynomials to polynomials)
Now the continuum sum formula is
\( \log_b(\frac{\mathrm{tet}_b'(z)}{\mathrm{tet}'_b(0)\ln(b)^z}) = \sum_{k=0}^{z-1}\mathrm{tet}_b(k) \)
for all positive integers z. If the LHS satisfies the bounds for Ramanujan's master theorem to apply, then this equality applies for all z in the right half-plane and is just a restatement of that fact.
I'm almost certain that the equality applies to JmsNxn's tetration, but I have no idea how to prove that.
It would also imply that the iterational form probably converges to it, too.