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 Solve this limit Nasser Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2012 11/27/2012, 05:13 PM 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 $lim_{x\rightarrow 0}(T(a,x))^\frac{1}{x}$ 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!! $f(x)=(T(a,x))^\frac{1}{x}$ with various values of "a" and check the curve at x =0 thank you sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 11/27/2012, 06:28 PM (This post was last modified: 11/27/2012, 06:30 PM by sheldonison.) (11/27/2012, 05:13 PM)Nasser Wrote: solve this limit .... $\lim_{x\rightarrow 0}(T(a,x))^{\frac{1}{x}}$ a>0 ...What 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. - Sheldon Nasser Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2012 11/28/2012, 07:14 AM (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 sheldonison Long Time Fellow Posts: 663 Threads: 23 Joined: Oct 2008 11/28/2012, 02:22 PM (This post was last modified: 11/28/2012, 05:05 PM by sheldonison.) (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/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 $k_a=T'(a,0)$ 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/tetrationforu...e=threaded. - Sheldon Nasser Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2012 12/03/2012, 07:46 AM 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. « Next Oldest | Next Newest »

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