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So far when I looked at tetration I noticed it had a recursive relation. It's $t_2=2^{(t_1)}.$

For example if we start at point $(0,1)$, we can take the x-value of $0$, and $2^0=1$, then we take $1$ and get $2^1=2$, and so on, for $^{x} 2$.

$$\begin{align} & (0,1)\\ & (1,2) \\ & (2,4) \\ & (4,16) \\ & (16,65536) \\ \end{align} $$

If you take this relation you basically get all the integers for $^{x} 2$. If it is all raised by $2$. We can find $1/2$ between each integer by using $x^{x^{(t_1)}}=(t_2) $, which will give the intermediate values.

However, unlike exponents, and addition, there is no similarity to the "x's" between each integer.

For example, if we take the function $2^x$, and take $1.5$, you would get $2\sqrt{2}$, and if you multiply by $\sqrt{2}$ you get $4$. For each $1/2$ between each integer the factor was by $\sqrt{2}$.

However the (x's) for tetrations are different. Also if you divide by $1/4^{th}$ you would $x^{x^{x^{x^{(t_1)}}}}=2^{(t_1)} $. However if you take $x^{x^{(2^{(t_1)})}} $, it's not equal to the half intervals.

Also if you take the inverse of $2^{(t_1)}$ you would get $\log_2(t_1)$ , so lets take $\log_2(t_1)=1 $, we get $0$, but if we take $\log_2(t_1)= 0$ , it would be undefined, (some would argue it's $-\infty$).

So what are all the requirements for a real continuous tetration function? What are the problems with Kneser's method, if any? Do you personally think there is a solution to a "tetration function"?

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  • $\begingroup$ Arbuja, here is a MathJax tutorial : meta.math.stackexchange.com/questions/5020/… . [Please check the edit, I replaced 2^^x with 2^x ,I'm not sure whether 2^^x is anything.] $\endgroup$ – Kushashwa Ravi Shrimali Jun 30 '14 at 1:10
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    $\begingroup$ 2^^x refers to the tetration function. It should look in standard form a subscript on the upper left, and the base on the normal level right. $\endgroup$ – Arbuja Jun 30 '14 at 2:01
  • $\begingroup$ I already made the edits, thank you so much Kushashwa. $\endgroup$ – Arbuja Jun 30 '14 at 2:20
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    $\begingroup$ Oh, fine! I didn't about the tetration function, sorry! And you're most welcome :) $\endgroup$ – Kushashwa Ravi Shrimali Jun 30 '14 at 5:24
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    $\begingroup$ Kneser's method is the best, and other analytic methods give the same results (to the limit of numerical accuracy); for example, Kouznetsov's Cauchy method, and Peter Walker's matrix approach, which calculates the slog inverse of Tetration with systems of numerical equations. There are some interesting methods that turn out to be $C^\infty$, but not analytic. There are trivial piecemeal approaches that are continuous, and differentiable but have discontinuous 2nd derivatives. Kneser's method is really the best. Check out math.eretrandre.org/tetrationforum/index.php for details. $\endgroup$ – Sheldon L Jul 1 '14 at 14:51

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