Are these solutions of $2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$ correct?

Find $x$ in $$\Large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}$$

A trick to solve this is to see that $$\large 2 = x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}} \quad\implies\quad 2 = x^{\Big(x^{x^{x^{\:\cdot^{\:\cdot^{\:\cdot}}}}}\Big)} = x^2 \quad\implies\quad x = \pm \sqrt{2}$$

Are these solutions correct? If not, why? If yes, are there other solutions?

PS: An extension of this discussion can be found in What we can say about $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$?

• How does $x^{x^{x^{x^{\ldots}}}}=2 \Rightarrow x^2=2$? – Quixotic Dec 2 '11 at 22:36
• It is usual (in work with the reals) to have $a^b$ undefined if $a$ is negative. So I would reject the solution $x=-\sqrt{2}$. As for $\sqrt{2}$, we need to define your infinite tower. A sensible definition would make it the limit of finite towers. You would need to prove that the limit exists, and show that the manipulation that got you $x^2=2$ is legitimate. If you do the details, which are not immediate, you will find that indeed $\sqrt{2}$ is correct. – André Nicolas Dec 2 '11 at 22:43
• @MaX if $x^{x^{x^{x^{\ldots}}}}=a$, then $x^{x^{x^{x^{\ldots}}}}=x^a$, because $a=x^{x^{x^{x^{\ldots}}}}=$, then if $a=2$, $x^{x^{x^{x^{\ldots}}}}=x^2$ – GarouDan Dec 2 '11 at 22:43
• Someone else will probably more directly address your specific questions, but for more than you probably want to know, see the following web pages: en.wikipedia.org/wiki/Tetration ioannis.virtualcomposer2000.com/math/exponents.html ioannis.virtualcomposer2000.com/math/IERefs.html mathdl.maa.org/images/upload_library/22/Ford/Knoebwl235-252.pdf – Dave L. Renfro Dec 2 '11 at 22:51
• Hint. This tower of powers should be treated as limit of the sequence defined by equalities $$a_1=x\quad a_{n+1}=x^{a_n}$$ This sequence have a limit iff $e^{-e}\leq x\leq e^{1/e}$. – Norbert Dec 2 '11 at 22:59

Might as well...

The power tower $x^{x^\ldots}$ is equivalent to the function $\exp(-W(-\log\,x))$, where $W(z)$ is the Lambert function, in the range $e^{-e}\leq x\leq e^{1/e}$ (as Norbert mentions in the comments; see also equation 13 in the MathWorld entry linked to). $\exp(-W(-\log\,x))$ can be inverted, like so:

\begin{align*} y&=\exp(-W(-\log\,x))\\ -\log\,y&=W(-\log\,x)\\ (\log\,y)\exp(-\log\,y)&=\log\,x\\ \frac{\log\,y}{y}&=\log\,x\\ x&=\exp\left(\frac{\log\,y}{y}\right)\\ x&=\exp\left(\log\,y^{1/y}\right)=y^{1/y} \end{align*}

If $y=2$, then $x=\sqrt2$.

Knoebel's paper establishes the interval of convergence $[\exp(-e),\exp(1/e)]$ for the power tower function, in the case of positive $z$. The paper notes that a full characterization of the region of convergence of $z^{z^\cdots}$ for complex $z$ remains to be done, but Thron, Shell (of Shellsort fame) and others have given partial results. See also this paper by Anderson for another discussion on the convergence of the power tower, this article by Cho and Park, where they discuss the inverses of the function $z^{1/z}$, and this article by Sondow and Marques.

• Alternatively: letting $h(x)$ represent the power tower function, we have the functional equation $h(x)=x^{h(x)}$. Replace $h(x)$ with $y$, and one can now solve for the inverse function $h^{(-1)}(x)$ to get \begin{align*}y&=x^y\\y^{1/y}&=x\end{align*}. Then let $y=2$... – J. M. is a poor mathematician Dec 3 '11 at 7:20
• I interpreted the question to be whether the limit $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}\dots}}}$$ actually exists. However, it appears what the argument above with the Lambert-$\mathsf{W}$ function is showing is essentially $$x^y=y\Rightarrow x=y^{1/y}$$ is it not? Heh, as soon as I saved, I saw that you commented nearly the same thing :-) – robjohn Dec 3 '11 at 7:34
• @J.M. All this tetration and lambert functions are new to me. But, before accept a answer to this question see. Using Mathematica I got: N[-(ProductLog[-Log[Sqrt]])/(Log[Sqrt]), 100]=2.00000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000 and this is ok. – GarouDan Dec 3 '11 at 18:25
• But, to $-\sqrt{2}$, out of the interval, I got: N[-(ProductLog[-Log[-Sqrt]])/(Log[-Sqrt]), 100]=0.25135029884500466069779137709679893569771463606584046004232145199569\ 62416051933646485630072430060361 + 0.3162499179777080128021379566517414370941363388551125764889642359610\ 440915347082002906655675608671575 I. A great complex value. – GarouDan Dec 3 '11 at 18:25
• Computing some values directly looks like has no convergence. Pastebin – GarouDan Dec 3 '11 at 18:28

If a solution exists, you have $$x^2 = x^{(x^{x^{x^{\cdot^{{\cdot}^{\cdot}}}}})}=2$$ which means what you've got. (This part has been mentioned to be wrong for logarithmic properties misuse reasons) Not both of these are not solutions, since $$1 = \log_2(2) = \log_2(x^{x^{\dots}}) = x^{x^{\dots}} \log_2 (x) = 2 \log_2(x)$$ and in the case $x = -\sqrt 2$, $2\log_2(-\sqrt 2)$ is purely imaginary, thus cannot be $1$. (The logarithm of $\log_a(b^c) = c\log_a(b)$ part is the part that remains suspicious. As N.S. pointed out, I don't think this argument can be made right.)

One way to suggest $x=\sqrt 2$ would be to show that the sequence $$x_n = \sqrt 2^{\dots^\sqrt2}$$ where exponentiation is taken $n$ times, is strictly increasing and bounded above by $2$. Numerical evidence suggests this : up to $n = 20$ I've seen that $x_n \le 2$ and $x_n$ is increasing. Convergence is slow and very long to compute though. I wasn't quite sure we could have convergence so I computed before finding a theoretical proof. Here's one : clearly $x_n$ is increasing, and $$x_n^2 = \sqrt 2^{x_{n-1}} \times \sqrt 2^{x_{n-1}} = \sqrt 2^{2x_{n-1}} = 2^{x_{n-1}} \le 4$$ by induction, so that $x^n \le 2$ for every $n$. Since the limit exists, it must be $\sqrt 2$.

Hope that helps,

• Very interesting. But is not clear to me if the $-\sqrt{2}$ is solution or not. You did a proof, but looks like there's something more, hidden. For example, $i^i$ as did by Euler, could be infinity real values, so, maybe $(-\sqrt{2})^{(-\sqrt{2})^{(-\sqrt{2})^\ldots}}$ converges to a real number too. – GarouDan Dec 3 '11 at 15:07
• @Garou: the problem with $-\sqrt2$ is that it's not within the interval of convergence $[\exp(-e),\exp(1/e)]$... – J. M. is a poor mathematician Dec 3 '11 at 15:24
• simpler proof: $x_n = \sqrt{2}^{x_{n-1}} \leq \sqrt{2}^2 = 2$. – sdcvvc Dec 3 '11 at 19:23
• @GarouDan : There is an argument that I've written to understand why $-\sqrt 2$ doesn't work, explicitly. Taking its logarithm in base 2 explains it all. – Patrick Da Silva Dec 4 '11 at 1:01
• @PatrickDaSilva Taking logarithms of complex numbers can be tricky, I am not sure your argument is right... I think the formula $log_a(b^c)=c \log_a (b)$ is tricky when you have multiple branches... Keep in mind that your argument $1 = \log_2(2) = \log_2(x^{2}) =2 \log_2 (x) = 2 \log_2(x)$ also shows that $-\sqrt{2}$ cannot be a root of $x^2=2$.... – N. S. Dec 5 '11 at 7:32

(This should go as a comment but I doubt it would fit the box)

Also you should consider, whether you would better like to write $\small x$, $\small _bx$ , $\small _{_b}{_b}x$ ,$\small {_{...} } _{_b}{_b}x$ , because you always begin the evaluation at the top of the powertower and not at the bottom. And also then it is unambiguous to discuss $\small 2= 2$, $\small 2 = _\sqrt22$ , $\small 2= _{_\sqrt2}{_\sqrt2}2$ and $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2$ as evaluated from the top. It is then also correct to write $\small 4= 4$, $\small 4 = _\sqrt24$ , $\small 4= _{_\sqrt2}{_\sqrt2}4$ and $\small 4= {_{...} } _{_\sqrt2}{_\sqrt2}4$ as a second solution. (This is clearly no standard notation, but I really do not know why this did not become standard)

[added] Then one could also write $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}x \text{ for } -\infty \lt x \lt 4$ to note the convergence of all that initial values x, and because $\small x=\sqrt2$ is in that range we can say $\small 2= \lim {_{...} } _{_\sqrt2}{_\sqrt2}\sqrt2$

• Interesting comment. I think a interesting notation to your $\small 2= {_{...} } _{_\sqrt2}{_\sqrt2}2$ could be $2=T(2,\{\sqrt{2},\infty\})$, where $T$ would be a Tower function, and $\infty$ represents how many times the term $\sqrt2$ appears. – GarouDan Dec 3 '11 at 14:33
• Yes, in the "tetration-forum" we often use to write $\small \exp_b^{\circ h} (x)$ for a base *b* and the iteration-"height" *h* (which is also thought to be continuous, not only integer). For our example we mean $\small b=\sqrt2$ and *x* the initial value and $\small h \to \infty$. (see math.eretrandre.org/tetrationforum/index.php) – Gottfried Helms Dec 3 '11 at 16:06

This is not a answer.

It's just a helper to discuss some things about the question, because is too large for the comments.

Looks like $-\sqrt{2}$ isn't a solution for the equation, but I'm not sure. Looks like too, the power tower of a number should converge only on a specific interval ($[e^{−e},e^{1/e}]$).

But using Mathematica and the ProductLog function (wich the Lambert $W(z)$ function) we find some strange things:

Using $h(z)=z^{z^{z^{\ldots}}}=-\frac{W(-\log (z))}{\log (z)}$ (h[z_]:=(-ProductLog[-Log[z]])/Log[z])

Calculating the power tower to $\sqrt{2}$ we have N[h[Sqrt], 10]=2.000000000

And the power tower to $-\sqrt{2}$ we have N[h[-Sqrt], 10]=0.2513502988 + 0.3162499180 I

Calculating explicity, by iteration

${-\sqrt{2}},{(-\sqrt{2})}^{({-\sqrt{2}})},{(-\sqrt{2})}^{({-\sqrt{2})}^{\ldots}}$ we have

Table[N[Re[PowerTower[-Sqrt, i]], 30] + I*N[Im[PowerTower[-Sqrt, i]], 5], {i, 1, 15}] // TableForm

-1.41421356237309504880168872421
-0.163093997943414854921937604558+0.59044 I
0.140921295793052749536215801866-0.044791 I
1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
0.894980750563013739735614892750-1.1090 I
-33.5835630157562847787187418023+29.118 I
6.49187847255812829134661655850*10^-46-1.5181*10^-45 I
1.00000000000000000000000000000+1.5134*10^-45 I
-1.41421356237309504880168872421-2.2930*10^-44 I
-0.163093997943414854921937604558+0.59044 I
0.140921295793052749536215801866-0.044791 I
1.10008630700672531426983704055+0.50079 I
-0.268168781568546776692908102136-0.14235 I
0.894980750563013739735614892750-1.1090 I

Ploting the real and imaginary part of the function $h$, we have:

To the real part:

Plot[Re[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt, N[Re[h[-Sqrt]]]}]}] and to the imaginary part:

Plot[Im[N[h[x]], {x, -2, 0}, Epilog -> {PointSize[0.01], Point[{-Sqrt, N[Im[h[-Sqrt]]]}]}] So looks like the function converges, but, unfortunally not to $2$.

I will post this for now, but, maybe I will create a new question just to treat this convergence and I will embrace a answer from here.

Please if someone can clarify this a bit, left a comment.

Thx.

• Hmm, assume you have found some z for base $\small b= -\sqrt2$ such that $\small b^z=z$ then obviously also $\small \log(z)/\log(b) = z$. With this you can iterate from a point z0 near the true value z and iterate $\small z_0=\log(z_0)/\log(b)$ until convergence. It means, that z is a "repelling" fixpoint for iterated exponentiation with the given base b - and thus an "attracting" fixpoint for the inverse of exponentiation (aka logarithm) and its iteration. "repelling" means: iteration from nearby z doesn't converge, "attracting" means: iteration from nearby z converges. – Gottfried Helms Dec 3 '11 at 21:47
• I looked whether z=2 is an attracting fixpoint for base b; it seems not to be true. A key is, that by iterations from near z you've to take fractional powers of a negative base, so you deal with branches of the logarithm. There is a discussion in the tetration-forum with some generalization of the Kneser-solution for the fractional exp-iteration. It might be possible, that some procedures allow to use negatives bases as well, because they deal already with the branches of the log. Perhaps you ask Sheldonison or Mike3 for details. See math.eretrandre.org/tetrationforum/index.php – Gottfried Helms Dec 3 '11 at 22:00
• It is possible that the sequence $a^{a^{a^{\dots}}}$ converges for some $a \neq \sqrt{2}$; however, if it converges to 2, then $a=\sqrt{2}$. – sdcvvc Dec 4 '11 at 1:15
• @GottfriedHelms , what's the admin mail in math.eretrandre.org? I'd like register but the site tells me to send a e-mail to bo198214(at)eretrandre.org, is it correct? with the brackets? – GarouDan Dec 4 '11 at 14:05
• @Garou: you are supposed to replace the "(at)" with the symbol that is standard for e-mail addresses... – J. M. is a poor mathematician Dec 4 '11 at 14:26