06/30/2010, 02:35 PM

(06/30/2010, 01:12 AM)sheldonison Wrote:(06/29/2010, 10:45 PM)tommy1729 Wrote: it is clear that iterations of exp and 2sinh are not exactly equal and thus n and z are not interchangeable ...

regards

tommy1729

I'm confident in my equations, so we'll need to be more precise to figure out where we agree, and where we disagree. I assume the first equation, for the limit of the SuperFunction of sexp2 is agreed upon, but one never knows. Using this equation, the value of SuperFunction(0)=1.058049043.... I think the two equations after that are also fairly obvious, and I assume you also agree with them.

Then substitute n with the inverse superfunction of n. Here the intent is to normalize so that SuperFunction(0)=n. And finally substitute n with the iterated exponentiation of n. This is equation is the equation I used to generate TommySexp; the only remaining step is iterating the ln function, and the equation I used for k. Do you still disagree with this equation? If so, why is this equation invalid, and is there another equation would you use instead? And does that alternative equation give identical values? I don't read minds, and I would be extremely interested in an alternative equation if you have one, because I would like to compare the two equations! They're probably equal!

The other thing I posted was the shorthand substitution equation for k, which quickly converges to k+n, where k converges to k=0.067838366... for values of n>=2. At that point, after applying the iterating ln function, we have the equation you disagreed with, which still seems completely correct to me.

-Sheldon

k depends on a limit and it depends on n !

iterations of exp cannot simply be replaced by iterations of 2 sinh , your formula is almost equal to mine but only if k starts approaching its limit , that is for large z and large n.

the convergence of 'k' is merely due to that fact that 2 sinh and exp are close , but that is only for large values , not around the fixpoints of 2 sinh or exp.

so you cannot write k for small n and small z.

since tetration grows so fast , carefully selecting parameters can lead to numerically undetectible differences.

( and btw , you havent even shown that k converges for large n and large z , although it makes sense that it does ... but what if it grows like e.g. slog(slog(slog(n))) ... though unlikely yes )

regards

tommy1729