06/30/2010, 03:34 PM
(This post was last modified: 06/30/2010, 04:38 PM by sheldonison.)

(06/30/2010, 02:35 PM)tommy1729 Wrote: k depends on a limit and it depends on n !By the way, I had some typos, I should've typed 2sinh where I accidentally typed sexp2. I agree that tetration can be very sensitive to input conditions, though this isn't always the case. I also agree that k has a unique value for each integer n. If one uses the exact value for the real number k, then you're ok with this equation? Or is there a different equation you would use?

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

By the way, it is very easy to get values of k accurate to 10 significant digits, which in many cases is not enough precision, but in some cases, it is. The slope of the SuperFunction at z=0 is approximately one, and the value of the Superfunction(0) is also approximately one. So generating the half-iterate of TommySexp(0) is not a problem. At the img(z)=i0.5*pi/ln(2), the SuperFunction is also pretty well behaved since there is an attracting fixed point, small slopes, small values, so the approximation is not a problem. At that contour, real(superfunction(z))=0, and img(superfunction(z)) varies between i0.8 and i2.0 (see the graph I posted), so a value of k accurate to 10 significant digits is more than enough.

There are places where the equations are much more difficult, and extremely high precision may be required. Where the SuperFunction has superexponential growth for a couple of iterations, and then the img(z)=0, so the function is going to the fixed point after that. In that case, having enough precision to do the iterated natural logarithms with branches for increasing values of n could be tricky.