Limit of a hyperpower function

Question

i have a question regarding this class of equations:

Let $\gamma(x)=x^x$

Let $\Psi_n(x)=\underbrace{\gamma(x)\circ\gamma(x)\circ\gamma(x)}_n$, such that $\Psi_1(x)=\gamma(x)$ and $\Psi_2(x)=(\gamma\circ\gamma)(x)$

Let a final function, $\Sigma(x)$ be the infinite composition like so: $$\Sigma(x)=\lim_{n\to\infty}{\Psi_n(x)},\phantom{.}\{n=2\sigma|\phantom{.}\sigma\in \mathbb Z\}$$ This indicates that there can be no odd values of n.

In this case, $\Sigma(x)$ will have an extrema (a minimum). Can someone identify what is the exact point $P(x,\Sigma)$ where the minima occurs $\left(\frac{d\Sigma}{dx}=0\right)$ or explain why $P$ may not exist? Thanks!

Answer 1

You can see that $\forall x \in[0,1],0\le \gamma(x) \le 1$, $\gamma(0)=1$ and $\gamma(1)=1$.

From that, you can deduce that $\forall n\in\mathbb N,\forall x \in[0,1], 0\le\Psi_{n}(x)\le 1$, $\Psi_{n}(0)=1$ and $\Psi_{n}(1)=1$. So assuming $\Sigma$ is well defined on $]0,1[$, $\Sigma(0)=1$, $\Sigma(1)=1$ and $\forall x \in[0,1],0\le \Sigma(x) \le 1$, $\Sigma(0)=1$.

So $\Sigma$ has a minimum on $[0,1]$, and since $\Sigma(x)$ must be greater than 1 for $x\ge 1$ (if defined at all), it has a global minimum.

Answer 2

It seems to me that you are looking for the "infinite tower function"

$$\displaystyle{}^{\infty}x:=\lim_{n\to +\infty}x^n=x^{x^{x^{\dots}}}$$

I say this because looking in your picture I can clearly see the left-bracketing scheme. In fact those function plotted are tower functions and are related to tetration: $${}^{0}x:=1$$ $${}^{n+1}x:=x^{{}^{n}x}$$

Some examples:

${}^{1}x:=x$

${}^{2}x:=x^x$

${}^{3}x:=x^{(x^x)}$

${}^{4}x:=x^{(x^{(x^x)})}$

In your plot we can see the following functions ${}^{2n}x$ with $n\in\mathbb N-\{0\}$ or in other words the power towers with $n$ even.

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Note Notice however that those functions are different from the functions $\Psi_n$ you have defined!

In fact we have:

$\Psi_1(x)=\gamma(x)={}^{2}x=x^x$

$\Psi_2(x)=(x^x)^{(x^x)}=x^{x\cdot x^x}=x^{x^{x+1}}\neq x^{(x^{(x^x)})}={}^{4}x$

$\displaystyle\Psi_3(x)=(x^x)^{{(x^x)}^{{(x^x)}+1}}=x^{x\cdot {x}^{x\cdot({{}^{2}x}+1)} }=x^{x\cdot {x}^{({x^{x+1}}+x)} }=x^{ {x}^{({x^{x+1}}+x+1)} }$

As you can see, that kind of bracketing of the "exponential tower" $\Psi_n$ is totally different from the ${}^n x$ one.

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If you are interested in the functions $\Psi_n$ I do not know the answer to your question but if you are talking about the ${}^n x$ functions then we have that the sequences odd and the even towers both converge on a single function evne if only over a restricted domain. That function is the infinite tower

$$\displaystyle{}^{\infty}x:=\lim_{n\to +\infty}x^n={W(-\ln(x))\over -\ln (x)}$$

where $W$ is the Lambert W function (or product logarithm).

As user2345215 said in the comments, the sequence of functions converges on the interval $[e^{-e},e^{1/e}]$ and it is oscillating (not convergent) in the interval $(0,e^{-e})$ (it is decreasing if $n$ is even and increasing if $n$ is odd). It gets infinite if $x$ is bigger than $e^{1/e}$.

Since it is strictly increasing over the interval its lowest value is at $e^{-e}$ $${}^{\infty}(e^{-e})=1/e$$

You can read more about this in the Wikipedia page about Tetration.