# Questions tagged [tetration]

Tetration is iterated exponentiation, just as exponentiation is iterated multiplication. It is the fourth hyperoperation and often produces power towers that evaluate to very large numbers.

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### How exactly does Knuth's Up-Arrow notation work?

I've done some research, and found this on Wikipedia. \begin{matrix}a\uparrow b=a^{b}=&\underbrace {a\times a\times \dots \times a} \\&b{\mbox{ copies of }}a\end{matrix} \begin{matrix}a\...
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### What is the maximum convergent $x$ in the power tower $x^{x^{x^{x\cdots}}}$?

In the power tower $x^{x^{x^{x\cdots}}}$ where there is an infinite stack of $x$'s, what is the maximum convergent number? I know the answer by playing with the form $x^y=y$ and using Mathematica, but ...
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### Show that the iterated $\ln^{[n]}$ of tetration(x,n) is nowhere analytic

$$f(x) = \lim_{n\to \infty} \ln^{[n]} x \uparrow\uparrow n$$ The conjecture is that $f(x)$ is monotonic and infinitely differentiable at the real axis, but nowhere analytic; because at each point on ...
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### Calculating $a^n\pmod m$ in the general case

It is well known, that $$a^{\phi(m)}\equiv1\pmod m ,$$ if $\gcd(a,m)=1.$ So, $a^n\pmod m$ can be calculated by reducing n modulo $\phi(m)$. But, for the tetration modulo $m$ a \uparrow \...
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### Smooth Elementary Function that Outgrows All Tower Functions?

This started off with me goofing off on Twitter and quickly led to this question to which I didn't know the answer. Let $T_2(x) = x^x$, $T_3(x) = x^{x{^x}}$, $T_4(x) = x^{x^{x^x}}$, and so forth. Is ...
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### How to find a function $f(x)$ such that $f(f(x))=\log_ax$?

Is there some method to do this or maybe some method to find a function such that $f(f(x))$ is at least approximately equal to $\log_{a}x$? Maybe the taylor series could be of help. So, we're looking ...
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### Is the positive root of the equation $x^{x^x}=2$, $x=1.47668433…$ a transcendental number?

I can prove using the Gelfond–Schneider theorem that the positive root of the equation $x^{x^x}=2$, $x=1.47668433...$ is an irrational number. Is it possible to prove it is transcendental?
### Seems that I just proved $2=4$.
Solving $x^{x^{x^{.^{.^.}}}}=2\Rightarrow x^2=2\Rightarrow x=\sqrt 2$. Solving $x^{x^{x^{.^{.^.}}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt 2$. Therefore, $\sqrt 2^{\sqrt 2^{\sqrt 2^{.^{.^.}}}}=2$ and ...
If $x \cdot 2 = x + x$ and $x \cdot 3 = x + x + x$ and $x^2 = x \cdot x$ and $x^3 = x \cdot x \cdot x$ Is there an operator $\oplus$ such that: $x \oplus 2 = x^x$ and $x \oplus 3 = {x^{x^x}}$? ...