(11/27/2012, 05:13 PM)Nasser Wrote: solve this limit ....I don't know if there is a software support tetration!!
I posted a pari-gip routine that generates sexp(z) for real bases greater than
)
here,
http://math.eretrandre.org/tetrationforu...hp?tid=486.
By definition, T(a,0) = 1, since sexp(0) is defined to be 1. If T is analytic, then for each value of a, T has a Taylor series expansion around 0, corresponding to the Taylor series for sexp(z) around 0. Define
)
as the first derivitive of that Taylor series.
^{1/x} \approx (1 + k_a x) ^ {1/x} )
^{1/x}) \approx \frac{1}{x} \log (1 + k_a x) \approx \frac{k_a x}{x} = k_a )
^{1/x} = \exp(k_a) )
There is an unproven conjecture that
)
is analytic in the base=a for complex values of a, with a singularity at base
. For real values of a, if
, then sexp(z) goes to infinity at the real axis as z increases. If
, then iterating
)
converges towards the attracting fixed point as n goes to infinity, but this is a different function than tetration. Then for base>
, we can have a taylor series for the any of the derivatives of
, with the radius of convergence =
.
I posted such a the taylor series for the first derivative of the base. For base=e, the first derivative ~= 1.0917673512583209918013845500272. The post includes pari-gp code to calculate sexp(z) for complex bases; the code for complex bases isn't as stable as the code for real bases, and doesn't always converge. If you're interested in a Taylor series for
for your limit, search for "the Taylor series of the first derivative of sexp_b(z), developed around b=2" in this post:
http://math.eretrandre.org/tetrationforu...e=threaded.
- Sheldon
![[Image: gif.latex?lim_{x\rightarrow 0}(T(a,x))^\frac{1}{x}]](http://latex.codecogs.com/gif.latex?lim_{x\rightarrow 0}(T(a,x))^\frac{1}{x})
![[Image: gif.latex?f(x)=(T(a,x))^\frac{1}{x}]](http://latex.codecogs.com/gif.latex?f(x)=(T(a,x))^\frac{1}{x})