The Problem

How can you evaluate (i.e., get a value for) Tetration (i.e., iterated exponentiation) to infinite heights?

For example, what would be the value of this expression?

$$ 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$

My (pathetic) Attempts

I tried equating it to $x$ and substituting it on the RHS, but no luck:

$$ x = 2^{2^{2^{2^{2^{.^{.^.}}}}}} $$ $$ x = 2^x $$ $$ x = \log_{2}{x} $$

What I think we need to do is have the RHS as a polynomial with one variable and the RHS a constant so we can solve for $x$.

I tried drawing the equation on Wolfram Alpha but the lines on the graph don't touch, so no luck there either.

Novice mathematician here. Thanks.


Sorry, I am a dolt :(

I didn't realize this was a diverging series. What confused me is my math sir told me it could be done. What he actually meant was that it could be said like this:

$$ \frac{\log{x}}{x} = \log{2} $$

but I somehow assumed there would be a numerical answer.

Alternative Questions

@Clayton's answer suggested an similar question which was a convergent series. While that wasn't what my sir meant, it practically could've been:

$$ \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}} $$

Another one I can think of would be:

$$ \sqrt{2*\sqrt{2*\sqrt{2*\sqrt{2*\sqrt{...}}}}} $$

Anyway, interesting question this has turned out to be...

  • 7
    $\begingroup$ isn't this going to $\infty$? $\endgroup$ – Lost1 Apr 15 '13 at 13:15
  • 1
    $\begingroup$ If you want the limiting value, it doesn't converge. The sequence is unbounded, and increases very very quickly. $\endgroup$ – Ishan Banerjee Apr 15 '13 at 13:16
  • $\begingroup$ @Lost1 I am a dolt... $\endgroup$ – Yatharth Agarwal Apr 15 '13 at 13:19
  • 1
    $\begingroup$ It does converge for certain bases. See en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights $\endgroup$ – Ishan Banerjee Apr 15 '13 at 13:23
  • 1
    $\begingroup$ I think, one could give a (not very interesting) meaning as a $2$-adic number. If one defines $x_0 = 1$ and $x_{n+1} = 2^{x_n}$, it diverges to infinity for the usual topology, hence $x_n = 2^{x_{n-1}}$ converges to $0$ for the $2$-adic topology. Clearly we don't have $0 = 2^0$ (but $x \mapsto 2^x$ is not continuous for the $2$-adic topology). $\endgroup$ – Joel Cohen Apr 15 '13 at 13:36

Hint: Equations $y=x$ and $y=2^x$ do not intersect, means there's no solution for $x\in\mathbb R$.

  • 1
    $\begingroup$ That's what the graph Wolfram Alpha gave me should have told me. I'll accept your answer as soon as SE lets me... $\endgroup$ – Yatharth Agarwal Apr 15 '13 at 13:20

Here are the first 20 solutions (for $\log(z)+k \cdot (2\cdot \pi\cdot i)$ such that $2^z=z$ or $z\cdot \log(2) = \log(z) + k\cdot(2 \pi i)$ : $$ \small \begin{array} {r|l} k & z : 2^z=z\\ \hline 0 & 0.82467854614+1.56743212385 \, î \\ 1 & 3.51523672192+10.8800532084 \, î \\ 2 & 4.36143141283+20.0871628060 \, î \\ 3 & 4.88885664543+29.2211855083 \, î \\ 4 & 5.27384865880+38.3277872288 \, î \\ 5 & 5.57736047492+47.4208762811 \, î \\ 6 & 5.82797084936+56.5062285608 \, î \\ 7 & 6.04142622483+65.5867042057 \, î \\ 8 & 6.22733941446+74.6638922661 \, î \\ 9 & 6.39201436790+83.7387507973 \, î \\ 10 & 6.53981045480+92.8118939379 \, î \\ 11 & 6.67386852707+101.883734634 \, î \\ 12 & 6.79652672953+110.954561406 \, î \\ 13 & 6.90957275012+120.024582285 \, î \\ 14 & 7.01440413644+129.093951274 \, î \\ 15 & 7.11213417750+138.162784952 \, î \\ 16 & 7.20366413122+147.231173287 \, î \\ 17 & 7.28973387243+156.299186877 \, î \\ 18 & 7.37095826487+165.366881942 \, î \\ 19 & 7.44785382986+174.434303834 \, î \end{array} $$

(Using Pari/GP , more than 100 digits precision)

pi2i = 2*Pi*I 
for(k=0,20-1,              \\ k contains branchno for logarithm
   for(j=1,20,              \\ Newton-iteration
        if(abs(x1-x0)<1e-100,break(),x0=x1); );

Any finite height tower can (in theory) be evaluated. It will equate to some natural number. If the tower is only moderately tall, it will be an enormous number.Wolfram Alpha shows that a tower only five layers high has $19729$ digits. If the tower height is infinite, the value diverges (quickly) to infinity and the value cannot be evaluated.

Your trick of equating to $x$ and substituting will find the limit if it exists. In this case, it does not.

  • $\begingroup$ Sorry, see my edit. Could you please explain a bit more about the 'height' of 'towers'? I know what a diverging series is and can see that this was one, but I don't know what those terms mean... $\endgroup$ – Yatharth Agarwal Apr 15 '13 at 13:23
  • $\begingroup$ @YatharthROCK: A tower four layers high is $2^{2^{2^2}}$ which evaluates to $65536$. Five layers is then $2^{2^{2^{2^2}}}$ which partially evaluates to $2^{65536}$ $\endgroup$ – Ross Millikan Apr 15 '13 at 13:25
  • $\begingroup$ Robert Munafo has a site with a "hyper calculator" where you can handle such high towers.... see mrob.com $\endgroup$ – Gottfried Helms Apr 15 '13 at 22:05

Perhaps your 'math sir' at school meant to tell you $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}$$can be evaluated. In fact, it can be evaluated in the following sense; for $x>0$, $$x^{x^{x^{.^{.^.}}}}=2\Longrightarrow x^2=2\Longrightarrow x=\sqrt{2}.$$ Hence, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$

  • 2
    $\begingroup$ This shows the equation holds if it converges. $\endgroup$ – GEdgar Apr 15 '13 at 14:01
  • $\begingroup$ Even that is not showed (the implication is reversed). $\endgroup$ – Did Apr 15 '13 at 14:05
  • $\begingroup$ While that is not what he meant, +1 for the answer :) $\endgroup$ – Yatharth Agarwal Apr 15 '13 at 15:45
  • 3
    $\begingroup$ The issue with this is that could also say: $x^{x^{x^{\cdots}}}=4\Rightarrow x^4=4\Rightarrow x=\sqrt{2}\Rightarrow \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}=4$ $\endgroup$ – L. F. Apr 15 '13 at 21:38
  • $\begingroup$ @L.F. Wha??? How can $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}$ equal both 2 and 4? For stuff like $0^0$ indecision is fine but here? $\endgroup$ – Yatharth Agarwal Apr 16 '13 at 14:28

In fairness, WolframAlpha does give you the answer when you type in "x=2^x". Here's the output I see:

enter image description here

Now, the "Solution" might look a little strange but, if I hit the "Approximate form" I see that it's approximately $0.824679+1.56743 i$ - a complex number.

  • 1
    $\begingroup$ ... and there are infinitely many more, depending on the multivaluedness of the (complex) log-function and the branch, which you select. $\endgroup$ – Gottfried Helms Apr 15 '13 at 21:20
  • $\begingroup$ @GottfriedHelms Which we would see in the WolframAlpha result, if my laptop had a larger screen. :) $\endgroup$ – Mark McClure Apr 15 '13 at 21:21
  • $\begingroup$ I've added 20 solutions. I've put it in a new answer, because the margin of this comment is too small... :-) $\endgroup$ – Gottfried Helms Apr 15 '13 at 21:46
  • 2
    $\begingroup$ @GottfriedHelms If only Fermat used Math.SE... $\endgroup$ – Yatharth Agarwal Jul 21 '14 at 15:09

The limit as $n$ approaches infinity of $\, ^nz$ (the $n$ times iterated exponential or power tower function) converges for the bases

${\displaystyle \textstyle (e^{-1})^{e}\leq z\leq e^{e^{-1}}}, \tag*{}$


$\displaystyle e^{-e}=\frac{1}{e^e} \approx 0.0659880358453125 \tag*{}$

$\large e^{e^{-1}}= e^{ \frac{1}{e}} \approx 1.44466786100977 \tag*{}$

It can be verified that when the value of $z$ is between the two numerical values above, the infinite iterated exponential $\, ^{\infty }z$ or $z^{z^{z^{\cdots\infty}}}$ (determined by the relation involving the product log shown below) has a real valued numerical solution.

The infinite power tower or infinite tetration ${\displaystyle z^{z^{z^{\cdot ^{\cdot ^{\cdot }}}}}\!}$ can be extended to complex numbers or to the complex plane. This infinite power tower has the general value

${\displaystyle h(z)={\frac {W(-\ln(z))}{-\ln(z)}}} \quad (1), \tag*{}$

where $\ln(z)$ is the principal branch of the complex logarithmic function, and $W(u)$ is the Lambert W function or the product log function, defined as:

${\displaystyle u = y e^y \Leftrightarrow y = W(u).} \tag*{}$

For additional clarification, here is the proof of the relation $(1)$ above:

Let ${\displaystyle z^{h}=h} \tag*{}$ Then ${\displaystyle z=h^{1/h}} \tag*{}$ ${\displaystyle z^{-1}=h^{-1/h}} \tag*{}$ ${\displaystyle {\frac {1}{z}}=\left({\frac {1}{h}}\right)^{1/h}} \tag*{}$ ${\displaystyle -\ln(z)=\left({\frac {1}{h}}\right)\ln \left({\frac {1}{h}}\right)} \tag*{}$ ${\displaystyle -\ln(z)=e^{\ln \left({\frac {1}{h}}\right)}\ln \left({\frac {1}{h}}\right)} \tag*{}$ ${\displaystyle \ln \left({\frac {1}{h}}\right)=W(-\ln(z))} \tag*{}$ ${\displaystyle {\frac {1}{h}}=e^{W(-\ln(z))}} \tag*{}$ ${\displaystyle {\frac {1}{h}}={\frac {-\ln(z)}{W(-\ln(z))}}} \tag*{}$

${\displaystyle h={\frac {W(-\ln(z))}{-\ln(z)}}.} \tag*{}$

Below is a plot of $(1)$ (from wolfram Alpha):

enter image description here

For $z = 2$ we have the following result:

$\boxed{\displaystyle {2^{2^{2^{2 ...}}}= h(2) = -\frac{W(-\ln (2))}{\ln (2)}}=\frac{W_0(-\ln (2))}{\ln (2)}}, \tag*{}$

where $W_0$ is the main branch of the Lambert W function.

The numerical value of $h(2)$ is (verified with Mathematica):

$\displaystyle \begin{align} 2^{2^{2^{2 ...}}} &= h(2) \\ &\approx 0.8246785461420742223140645943816032399746074201816 \\ &\quad -1.5674321238496478610585743911929869275333075742042 i \end{align} \tag*{}$

Wolfram Alpha providess the following symbolic result :

$\displaystyle -\frac{W(-\ln (2))}{\ln (2)}= \frac{1}{\pi \ln(2)} \int_{-\infty }^{-\frac{1}{e}} \ln \left(\frac{x+\ln (2)}{x}\right) \Im\left(\frac{\partial W(x)}{\partial x}\right) \, dx ,\tag*{}$

where $\Im(z)$ is the imaginary part of $z$ .

The last result above can be numerically verified and approximated with Mathematica by typing :

  Im[D[ProductLog[x], {x}]]*
   Log[(x + Log[2])/x], {x, -Infinity, -(1/E)}]/(Pi*Log[2])

For another value such as $z = \sqrt{2},$ we get:

$\displaystyle \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}} = h(\sqrt{2})= -\frac{W(-\ln (\sqrt{2}))}{\ln (\sqrt{2})}=2 \tag*{}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.