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When I studied the various known matrices of combinatorical numbers I also looked at the following idea: what if we discuss functions with a set of results, not only one number? So for instance the concept of sine and cosine gets some special charme if we look not only at f(x) = sin(x), g(x)=cos(x) but at 2x2-matrices containing cos() and sin() and the input ...


It is easier to prove that the inverse function is strictly increasing. Since the inverse function is just: $$ g(x) = \left(\frac{1}{x}\right)^{-\frac{1}{x}}$$ with a change of variable everything boils down to proving that $h(x)=x^x$ is increasing over $\left[\frac{1}{e},1\right]$. That is trivial since: $$ h'(x) = h(x)\cdot\frac{d}{dx}\log h(x) = (1+\log ...


We write $$x^{x^{x^{...}}} = \frac{W(-\ln(x))}{-\ln(x)}, x \neq 1$$ We differentiate that. This becomes $$\frac{\ln(x)W'(-\ln(x))+W(-\ln(x))}{x\ln^2(x)} , x \neq 1$$ Using the quotient rule. (W|A verification) Since $e^{\frac{1}{e}}>x>1$, the nominator is positive. Therefore we want to show that $$\ln(x)W'(-\ln(x))+W(-\ln(x))>0$$ Let ...


Here is the correct formula, where $\eta=\exp(1/e)$ and $\alpha(x)$ is the upper repelling fixed point Abel function for iterating $x \mapsto \exp(x)-1$, which is generated using Ecalle's fps solution. For details, see posts by Will Jagy on $\alpha(x)$: How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? $$\lim\limits_{n\to\infty} ...


It's not a correct proof, no. If $${x^{x^{x^{x^{x^{\dots}}}}}} = y$$ then we can say that $x$ to the power of each side is the same: $$x^{\left({x^{x^{x^{x^{x^{\dots}}}}}}\right)} = x^y$$ but then the left hand side is what exactly what we started with (provided the limit exists) so we can equate the right hand sides of each of these equations: ...


Once you prove that over the interval $\left[e^{-e},e^{\frac{1}{e}}\right]$: $$ f(x)=x^{x^{x^{x^{\ldots}}}} = \frac{W(-\log x)}{-\log x}\tag{1}$$ where $W$ is the Lambert W-function, it follows that: $$ f^{-1}(x) = x^{\frac{1}{x}}.\tag{2} $$ On the other hand, $$ x = x^{f(x)} \tag{3} $$ implies: $$ x = f^{-1}(x)^x \tag{4}$$ hence $(2)$ is quite trivial.


Really? $x^{y^z}$ equals $x^{y\cdot z}$? So you are saying that $$2=2^1=2^{1^2} = 2^{1\cdot 2} = 2^2 = 4?$$

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