4

The infinitely nested power expression below has a limit of $2$:

$$x=\sqrt2^{\sqrt2^{\sqrt2^{...}}}$$

In finding this limit, we may use:

$$x=(\sqrt2)^x$$

But this expression has two solutions, $2$ and $4$.

We know that $2$ is the right answer by evaluating some finite truncations, but this $4$ is bothering me. What does $4$ mean in this expression? Is it significant in some way?

the_fox
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Flying_Banana
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  • To find the solution of $x=2$ you may use $x^2 + 8 = 6x$.This has two solutions, $2$ and $4$. Does this $4$ bother you? – Umberto P. Jan 29 '19 at 10:36
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    It's not quite a duplicate, but you might be interested in my answer to this older question and the first comment on that answer. – Mees de Vries Jan 29 '19 at 10:37
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    By writing $\sqrt2^{\sqrt2^{\sqrt2^{\cdots}}}$, do you mean $\lim_{n \to \infty} \underbrace{\sqrt2^{\sqrt2^{\sqrt2^{\cdots^{\sqrt2}}}}}_{n \text{ copies of } \sqrt2}$? – L. F. Jan 29 '19 at 10:54
  • @UmbertoP. It doesn’t, because I know in your case the 4 is introduced when you rewrote x=2 with the polynomial, and an additional root is included within the rearranged expression. Now can you tell me where did the 4 come from in my question? – Flying_Banana Jan 29 '19 at 11:46

3 Answers3

2

The problem here is that:

$$x = (\sqrt{2})^x$$

is a necessary condition but not sufficient, that is, if $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$ equals $x$ then it must hold that $x = (\sqrt{2})^x$, but you don't know if $x$ can equal $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$ in the first place.

One interesting property is that

$$x^{x^{x^{x^{\cdots}}}}$$ for positive x converges only if $x\in [e^{-e}, e^{1/e}]$ and converges to a value $y\in[e^{-1}, e]$.

So if you are working with another expression like that and you get more than one solution, keep in mind that if it does not belong to that interval then it surely is not a solution of the expression.

https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0

Here is the proof of the property aforementioned and also, a more detailed analysis of the whole problem with nested power expressions.

the_fox
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maxbp
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  • Good point, I guess now my question is that I expected to be able to rewrite x in terms of itself, but obviously I can’t since they aren’t the same thing (two answers after rewrite). Just like how we can’t rewrite $\x^2=4 \Rightarrow x=2$, without losing a solution, can you explain how rewriting a power tower expression in terms of itself introduces irrelevant solutions? – Flying_Banana Jan 30 '19 at 05:31
2

If $\sqrt{2}^{\vdots}$ is well-defined, it must be as the limit of a sequence of the form $a_{n+1}=\sqrt{2}^{a_n}$. The limit, if it exists, depends on $a_1$, but the most natural choice of $a_1$ is $\sqrt{2}$; what else do we think is "at the top of the tower", which doesn't exist?

But we can prove by induction that if $a_1\in[0,\,2]$, then, for all $n$, $a_n\le 2$, so $x:=\lim\limits_{n\to\infty}a_n$ is $x=2$, whereas $a_1=4$ implies $x=4$.

Did
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J.G.
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  • Of course that we can prove that the limit tends to 2 instead of 4, but where did the 4 come from? Why can’t we write $x = \sqrt2^x$? and expect only one answer, 2? – Flying_Banana Jan 30 '19 at 05:23
  • @Flying_Banana See edit. Incidentally, if anyone knows how to make dots go up and to the right, instead of up only viz. $\vdots$ or down and right viz. $\ddots$, I'll fix my notation. – J.G. Jan 30 '19 at 06:38
  • This post can be slightly completed to yield, in my view, the answer you are looking to: note simply that, for every $a_1$ in $[0,4)$, $\lim a_n=2$, for $a_1=4$, $\lim a_n=4$, and, for every $a_1>4$, $a_n\to\infty$. Thus, $2$ is the only "generic" limit. // Next, would be to realize that $\sqrt2$ is not specific here and that replacing it by every $c$ in $(0,e)$ would yield the same phenomenon, with "generic" limit $\ell$ the smallest of the two roots of $\ell=c^\ell$ and as "spurious" fixed point the largest of these two roots. – Did Jan 30 '19 at 07:32
  • @did - doesn't this lead to the question of "attracting fixpoints" (=2) and "repelling fixpoints" (=4). The latter is the "attracting" for the inverse of the original iterated function and should be dealt formally the same just introducing that inverse function instead. – Gottfried Helms Jan 30 '19 at 09:45
  • @GottfriedHelms No idea how the inverse function would help here. But yes, the fact that 2 is attracting and 4 repelling is what makes the results of the limit computations made in my previous comment. – Did Jan 30 '19 at 10:10
  • @did - for instance: $b=\sqrt 2$ then, say, $x_0=7$, then iterate $x_{k+1}=\log(x_k)/\log(b)$ for $k=0 \to \infty$ approaches the repelling fixpoint $4$ – Gottfried Helms Jan 30 '19 at 11:26
  • @GottfriedHelms Sorry but I got perfectly what you meant by "the inverse function", only, again, I fail to see how it is related to the present question. – Did Jan 30 '19 at 11:46
  • @did - I tried to introduce some meaningful aspect on how the occurence of the $4$ in this problem is of any (intrinsic) significance - at least that was how I interpreted the question. – Gottfried Helms Jan 30 '19 at 12:16
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It is a common misconception that an expression like

$$\sqrt2^{\sqrt2^{\sqrt2^{...}}}$$

denotes a well-defined number.

For more rigor, let us use

$$x=\lim_{n\to\infty}a_n,\text{ where }x_{n+1}=\sqrt2^{x_n}.$$

Now if $x_0=2$, $x_n=2\ \forall n$ follows. Similarly, for $x_0=4$, $x_n=4$ follows. For other initial values, the sequence may converge to $2$, but may also diverge.

So the statement "has a limit of $2$" is dubious.

  • I'd commented this sometimes earlier somewhere else... My opinion is, that the stating of the problem should firstly use a different notation, like $$x_\infty =;{;{;{ ;\cdots} b}b}x_0 \overset ?=x_0 $$ with the triple-dots where they belong due to the order of evaluation. Then set $b=\sqrt2$ and $x_0$ some chosen inital value. It is much more intuitive, that one has now two (real) choices for $x_0$ when starting the evaluation. But, ... well... times ...;) – Gottfried Helms Jan 30 '19 at 12:47