Evaluating tetration to infinite heights (e.g., $2^{2^{2^{2^{.^{.^.}}}}}$)

Question

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...

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

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.

Answer 5

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)