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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.

Edit

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...

Yatharth Agarwal
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    isn't this going to $\infty$? – Lost1 Apr 15 '13 at 13:15
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    If you want the limiting value, it doesn't converge. The sequence is unbounded, and increases very very quickly. – Ishan Banerjee Apr 15 '13 at 13:16
  • @Lost1 I am a dolt... – Yatharth Agarwal Apr 15 '13 at 13:19
  • @IshanBanerjee Thanks. I can see that it is a diverging series now. What confused me is that my math sir at school told me you could do it... :( – Yatharth Agarwal Apr 15 '13 at 13:19
  • Didn't you... Didn't you understand it can not converge?! – gukoff Apr 15 '13 at 13:22
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    It does converge for certain bases. See http://en.wikipedia.org/wiki/Tetration#Extension_to_infinite_heights – Ishan Banerjee Apr 15 '13 at 13:23
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    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). – Joel Cohen Apr 15 '13 at 13:36
  • Don't be so hard on yourself. Exploring areas that somehow pique your interest is a good way to learn some new mathematics. If you haven't seen things like that before, you may make some mistakes that later seem silly. You probably won't make them again. – Ross Millikan Apr 15 '13 at 15:47

6 Answers6

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

JSCB
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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)

l2=log(2)
pi2i = 2*Pi*I 
{list=matrix(20,2);
for(k=0,20-1,              \\ k contains branchno for logarithm
   x0=1+I;
   for(j=1,20,              \\ Newton-iteration
        x1=x0-(l2*x0-(log(x0)+k*pi2i))/(l2-(1/x0));
        if(abs(x1-x0)<1e-100,break(),x0=x1); );
   list[1+k,]=[k,x0];
 );}
 printp(list)
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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.

Ross Millikan
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  • 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... – Yatharth Agarwal Apr 15 '13 at 13:23
  • @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 $2^{65536}$ – Ross Millikan Apr 15 '13 at 13:25
  • Robert Munafo has a site with a "hyper calculator" where you can handle such high towers.... see http://www.mrob.com/ – Gottfried Helms Apr 15 '13 at 22:05
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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, $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^.}}}}=2.$$

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

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

where

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

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

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 :

NIntegrate[
  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:

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

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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.

Mark McClure
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