+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: Solve this limit (/showthread.php?tid=757)
Solve this limit - Nasser - 11/27/2012
Hi
solve this limit
I tried to sovle it but no way until now and I searched for solving formula but no results found.
this limit
a>0
the result must be f(a)
if there is no soving formula,
then try to draw this function,
I don't know if there is a software support tetration!!
with various values of "a" and check the curve at x =0
thank you
RE: Solve this limit - sheldonison - 11/27/2012
(11/27/2012, 05:13 PM)Nasser Wrote: solve this limitWhat is T(a,x)? Perhaps super exponentiation base a of x? We usually say sexp(0)=1, which works for all bases. Also, in my quote, I modified your comment to use the tex tag.
....
\( \lim_{x\rightarrow 0}(T(a,x))^{\frac{1}{x}} \)
a>0 ...
- Sheldon
RE: Solve this limit - Nasser - 11/28/2012
(11/27/2012, 06:28 PM)sheldonison Wrote: What is T(a,x)? Perhaps super exponentiation base a of x?You are right
thank you
RE: Solve this limit - sheldonison - 11/28/2012
(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 \( \eta=\exp(1/e) \) here, http://math.eretrandre.org/tetrationforum/showthread.php?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 \( k_a=T'(a,0) \) as the first derivitive of that Taylor series.
\( \lim_{x \to 0} T(a,x)^{1/x} \approx (1 + k_a x) ^ {1/x} \)
\( \log(\lim_{x \to 0} T(a,x)^{1/x}) \approx \frac{1}{x} \log (1 + k_a x) \approx \frac{k_a x}{x} = k_a \)
\( \lim_{x \to 0} T(a,x)^{1/x} = \exp(k_a) \)
There is an unproven conjecture that \( \text{sexp}_a(z) \) is analytic in the base=a for complex values of a, with a singularity at base \( \eta=\exp(1/e) \). For real values of a, if \( a>\eta \), then sexp(z) goes to infinity at the real axis as z increases. If \( a<=\eta \), then iterating \( \exp^{[on]}_a(0) \) converges towards the attracting fixed point as n goes to infinity, but this is a different function than tetration. Then for base>\( \eta \), we can have a taylor series for the any of the derivatives of \( \text{sexp}_a(z) \), with the radius of convergence = \( a-\eta \).
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 \( k_a \) 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/tetrationforum/showthread.php?tid=729&pid=6325&mode=threaded.
- Sheldon
RE: Solve this limit - Nasser - 12/03/2012
You found an approximated solution.
It is ok, but this will not help me, because I tried to find the first derive of b^^x and x^^x and other related functions like for example b^^(x^2) by using differentiation fundamentals concepts, and I am just facing this problem to finish my work.
I may post my work here for discussion.
thank you Sheldonison.