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I am trying to figure out for what value of $x$, is $f(x)=x^{x^{x^{...}}}=4$. $f$ is a infinite height tower of $x$.

I am able to solve for $f(x)=3$ in the following way: $$\ f(x)=x^{x^{x^{...}}}=x^{f(x)}=x^3=3 \\ \implies x= \root{3} \of{3} \approx 1.44225 $$

When I use the same technique to solve for $f(x)=4$. I get $x=\sqrt{2}\approx1.4142$. But this cannot be correct since $f$ is a monotonically increasing function. Also, $f(x)=4$ should have a solution because the function seems continuous and from Intermediate Value Theorem we can reason that there exists a $x^*$ such that $f(x^*)=4$

Any help or idea is appreciated.

Shiv Tavker
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  • I'd like to point out that $f(\sqrt[3]{3}) \ne 3$ - there's another solution to $(\sqrt[3]{3})^c = c$: https://www.wolframalpha.com/input/?i=%28%283%5E%281%2F3%29%29%5E2.478052680288302411893736516894690307868142312689099163591 – Dark Malthorp May 04 '20 at 19:21
  • If you take a sequence $\sqrt[3]{3}, (\sqrt[3]{3})^\sqrt[3]{3}, (\sqrt[3]{3})^{(\sqrt[3]{3})^\sqrt[3]{3}}, \cdots$ you will find that it converges to the other fixed point. – Dark Malthorp May 04 '20 at 19:22
  • Oh, I see thanks. https://math.stackexchange.com/questions/1855000/why-do-xxx-dots-2-and-xxx-dots-4-have-the-same-positive-ro. I checked this but still not able to figure out why is the continuous function argument wrong? Any ideas about this? – Shiv Tavker May 04 '20 at 19:24
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    The IVT only applies if you can find $c_1$ and $c_2$ such that $f(c_1) < 4$ and $f(c_2) > 4$. There is no $x$ such that $f(x) = 4$. In fact, the power tower only converges for $e^{-e} \le x \le e^{1/e}$. Indeed $f$ is monotonic and increasing on this interval, but $f(e^{1/e}) = e < 4$ – Dark Malthorp May 04 '20 at 19:27
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    https://en.wikipedia.org/wiki/Tetration#Infinite_heights – Dark Malthorp May 04 '20 at 19:28
  • @DarkMalthorp does $f(2)$ not diverge to $\infty$? Is that not a valid $c_2$? – Shiv Tavker May 04 '20 at 20:31
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    No. The IVT only applies to finite values of the function. At points of infinite value, the function is definitionally discontinuous. If $f(x)$ is finite for $x<c_2$ and $\lim\limits_{x\rightarrow c_2} f(x) = \infty$, then you can apply the IVT since this implies there exists $c_2^$ close to $c_2$ such that $f(c_2^) > 4$. But in this case you don't have a "continuous" divergence to infinity. $f(e^{1/e}) = e <\infty$, and for any $x>e^{1/e}$, you have $f(x) = \infty$ – Dark Malthorp May 04 '20 at 20:36
  • @DarkMalthorp thank you for this great help. This implies that the function is not continuous right? Precisely, at point $x=e^{1/e}$. – Shiv Tavker May 04 '20 at 21:44
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    In a sense yes. It's not defined (as a real valued function) for $x> e^{1/e}$, so if you're only thinking of $\mathbb{R}$ it is continuous on $[e^{-e},e^{1/e}]$, and is simply not defined elsewhere. If you're working with the extended reals (https://en.wikipedia.org/wiki/Extended_real_number_line) then $f$ is indeed discontinuous at $e^{1/e}$ because $f(e^{1/e} + \epsilon) = \infty$ for all $\epsilon>0$. – Dark Malthorp May 04 '20 at 21:51
  • @DarkMalthorp Great! Thanks for ur help – Shiv Tavker May 05 '20 at 13:38

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