Finding the convergent interval of $z^{{2z}^{{3z}\ldots^{{nz}\ldots}}}$

Question

I was trying to find an exact solution to $x=a^x$, and naturally derived a solution defined by infinite tetrations of $a$. I first defined a recursion equivalent to the definition of tetrations and then used invariant points to show that $$\lim_{n \to \infty} (^nx)$$ converges for $x \in [e^{-e}, e^{\frac{1}{e}}]$. Now, I am curious if I could extend the definition of tetrations and find the convergence interval of the following function. $$f(x)=x^{{2x}^{{3x}^{{...}}}}$$ I find this to be hard because I cannot, as in the case of normal tetrations, define a recursion that can be applied to define $f(x)$. Any tips on how to proceed will be appreciated. Thanks a lot!

Answer 1

I'll be using the notation you introduced $f(x) = x^{2x^{3x\dots}}$ to refer to this function.

Like I stated in my comment, I feel like trying to make $1$s appear in the exponents somewhere is a fairly straightforward way of obtaining valid solutions, and this is rather trivial for $x = 1$ or $x = \frac{1}{k}$ for some $k \in \mathbb{N}$.

More generally though :

• Consider the case $0< x < 1$. There exists some natural integer $k$ for which $kx < 1$, but $(k+1)x > 1$ (read edit), so we can infer the limit value : $$kx^{(k+1)x^{(k+2)x\dots}} = 0$$ If $k = 1$ (i.e $x < 0.5$), then $f(x) = 0$. But if $k > 1$ (i.e $x > 0.5$), then this would then imply $(k-1)x^{kx^{(k+1)x^{(k+2)x\dots}}} = 1$ and therefore $f(x) = 1$.

• For $x > 1$, it's quite clear that $f(x)$ cannot exist, and I don't think you're particularly interested in $x \leq 0$ so I won't touch on that in my answer.

• Edit : I just noticed by plotting an approximate of $f$ that there's something I forgot to adress in the case where $x = 0.5$. I believe $f(0.5)$ approches $0.5$ in this case due to the nature of my argument.

Funny, it seems like a convoluted way of defining $f(x) = \begin{cases} 0 &\text{ if } 0 < x < 0.5 \\ 0.5 &\text{ if } x = 0.5 \\ 1 &\text{ if } 0.5 < x \leq 1 \end{cases}$.

Hope this helps !

Answer 2

Lets assume $0 =< x < 1$ because otherwise things get divergent or complex.

Eventually for some $n$ we have $n x = a > 1$ and $(n-1) x = b < 1$.

Lets use $**$ for going up the power staircase.

So we get

$$x**{2x} **... ** nx **((n+1)x **...)$$

That finally reduces to

$$x**{2x} **... ** a ** ((n+1)x **...\infty)$$

and further towards

$$x**{2x} **... **b^\infty$$

Now clearly $b^\infty = 0$ and $((n-2)x)^0 = 1$.

So we get a finite power tower :

$$x**{2x} **.. **(n-3)x$$

Now if $x > 1$ we get a number larger than

$$1^{2^{3...}}$$

and if $x = 1$ we get

$$1^{2^{3...}}$$

So the sequence diverges to +infinity for $x>1$ or $x = 1$

Notice that $x<0$ gives us complex numbers.

Also $0^{0^{0^...}} = 1$ or $0$.

In conclusion, as a real function of a real variable, the only real interval that makes sense is $0= < x < 1$.

And in that case we get $f(0) = 1$ or $0$ or

$$f(x) = x**{2x} **.. **(n-3)x$$