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I saw the trick to solve $x^{x^{x^{x^{\vdots}}}}=2$, and solution is $\sqrt{2}$. I understand that $x^{x^{x}}$ is different from $(x^{x})^x$, but I still cannot see intuitively why the $x^{x^{x^{x^{\vdots}}}}$ doesn't blow out when $x$ is larger than 1.

I experimented a little, and found that all values that satisfy the equation are less than $\sqrt{2}$. What makes this range special?

$a=2:20$

$b=a^{\frac{1}{a}};b$

Thank you in advance for any help

Alex D
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Watchung
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    https://en.wikipedia.org/wiki/Tetration#Infinite_heights – user Feb 09 '21 at 17:34
  • If the tetration converges then the solution must satisfy $x^{y} = y \to y^{1/y} = x$ right? So try to plot the function $y^{1/y}$; what is the maximum value? You will find that the critical value is not $\sqrt{2}\simeq 1.414$, but slightly larger ($\sim 1.444$). – Winther Feb 09 '21 at 17:44
  • @Winther Thank you for providing the intuition! I tried that the critical value is smaller than 1.5, but had a gut feeling that it should be larger than $\sqrt{2}$. It's magical to see that the answers show it's finally connected with e. – Watchung Feb 09 '21 at 18:14

2 Answers2

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This question is analogous to asking the (equally good) question: why does the infinite series $\sum_{k=1}^\infty r^k$ blow up when $r>1$ but not when $0<r<1$? The answer, in both cases, is that the infinite expression is defined to be the limit of its finite truncations; and sometimes limits exist and sometimes they don't. In this case, the sequence $$ x,\, x^x,\, x^{x^x},\, x^{x^{x^x}}, \dots $$ happens to stabilize if $e^{-e}\le x\le e^{1/e}$ but not if $x>e^{1/e}$.

Greg Martin
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Maybe you're confusing $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ with $\left({\sqrt{2}}^{\sqrt{2}}\right)^{\sqrt{2}}. $ The latter is done by calculating ${\sqrt{2}}^{\sqrt{2}}$ and then doing this number to the power of ${\sqrt{2}}.\ $

In fact, $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ is ${\sqrt{2}}$ to the power of ${\sqrt{2}}^{\sqrt{2}},\ $ or to say it another way, you're doing ${\sqrt{2}}^{\sqrt{2}},\ $ and then doing ${\sqrt{2}}$ to the power of this number.

Expanding on this. Pretend ${\sqrt{2}} = 1.4.$

$1.4^{1.4} = 1.6.$

$1.4^{1.6} = 1.7.$

$1.4^{1.7} = 1.77.$

$1.4^{1.77} = 1.81.$

$1.4^{1.81} = 1.84.$

If the exponent is $< 2,\ $ then $1.4^{exponent} = new\_exponent$ will be less than $1.96$ which is less than $2$, so the next step in the iteration process is to do $1.4^{new\_ exponent}$, and this new exponent will again be be less than $1.96$ and so the result from the iteration process can never be greater than $1.96.$

Whereas,

$1.4^{1.4} = 1.6$

$1.6^{1.4}=1.93$

$1.93^{1.4}=2.5$

$2.5^{1.4}=3.6$

$1.93^{1.4}=6.0$

and this is growing without bound.

Adam Rubinson
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