Infinite Tetration!

This problem is probably incorrect.

Let $y = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}$

This simply becomes, $y=(\sqrt{2} )^y$

Now we make this into $\log y = \dfrac{y}{2} \log 2$

This is true at $y=2$ , because $(\sqrt{2} )^2 =2$

but also at $y=4$ , because $(\sqrt{2}) ^4= 2^2=4$ , and hence $4=(\sqrt{2}) ^4$

Hence this has got 2 solutions, they're $y=2$ and $y=4$ . I don't know how to decide, but I got it right when I tried both, right at the time of $\boxed{2}$ , but not satisfied.

Please say something \(\color{Purple}{\textbf{ @Sharky Kesa }}\)

This is how I thought about it:

$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...∞}}}=(2^{1/2})^{(2^{1/2})^{(2^{1/2})^{...∞}}}=2^{1^{1^{1^{...∞}}}}=2$

Let $x=^\infty \sqrt{2} \quad \Rightarrow x = \sqrt{2}^x\quad \Rightarrow x^2 = 2^x \quad \Rightarrow x = \boxed{2}$

A formal proof can be written as follows, but it may be a little boring. The only result you need to produce a rigorous mathematical proof is this fact "If a sequence $a_n$ is bounded and increasing (that means that for all $n$ , you have $a_{n+1} \geq a_n$ ) then is is convergent."

That said let's produce a formal proof :)

First of all the sequence of the exercise is best defined by recursion

$a_n = \begin{cases} \sqrt{2} & \textrm{if } n = 1 \\ \left( \sqrt{2}\right) ^{a_{n-1}} & \textrm{if } n \geq 2 \\ \end{cases}$

Now we just prove the following steps.

Step 1. For every integer $n \geq 1$ we have $\sqrt{2} \leq a_n < 2$ . This can be drawn out by induction (boring I know). For $n=1$ is is clearly true. If $\sqrt{2} \leq a_n < 2$ than we have $a_{n+1} = \left(\sqrt{2}\right)^{a_n} < \left(\sqrt{2}\right)^2 = 2$ (since the function $y = \sqrt 2 ^ x$ is strictly increasing cause the base is about 1.4 and hence greater than 1. And from the other side we have (again by the fact that the exponential function with base greater than 1 is strictly increasing) $a_{n+1} = \left(\sqrt{2}\right)^{a_n} \geq \left( \sqrt 2 \right)^{\sqrt 2} > \left( \sqrt 2 \right)^1 = \sqrt 2 .$ We got the result we wanted $\sqrt{2} \leq a_{n+1} < 2$ and by induction we proved the step 1.

Step 2. For every integer $n \geq 1$ we have $a_{n+1} > a_n$ . Again we use induction. For $n = 1$ is easy to see that is is true. Suppose now $a_{n+1} > a_n$ and let's try to show that $a_{n+2} > a_{n+1}$ . In fact $a_{n+2} = \left( \sqrt 2 \right)^{a_{n+1}} > \left( \sqrt 2 \right)^{a_{n}} = a_{n+1}$ and by induction we have shows the step 2 is true.

Now we have that $a_n$ is bounded (step 1) and is (strictly) increasing. So it converges to a finite limit we can call $L$ .

By the equality (that holds for all integers $n \geq 2$

$\left( \sqrt{2}\right) ^{a_{n-1}} = a_n$

we can derive the equality of limits

$\lim_{n \mapsto \infty} \left( \sqrt{2}\right) ^{a_{n-1}} = \lim_{n \mapsto \infty } a_n$

by the continuity of the exponential function we have

$\left( \sqrt{2}\right) ^{\lim_{n \mapsto \infty} a_{n-1}} = \lim_{n \mapsto \infty } a_n$

so we have

$\left( \sqrt{2}\right)^L = L$

Now it is algebra (notice that $L$ is positive cause is the limit of a sequence of numbers greater that 1.4)

$2^{\frac{L}{2}} = L$

squaring both the positive members

$2^L = L^2$

that has two solutions easy to find

$L=2$ and $L=4$

But $L$ cannot be 4, because it is a limit of a sequence with terms lower than 2.

So it is necessary that $L = 2$ .

Let $a^{a^{a^{.^{.^{.}}}}} = x$ where $a = \sqrt{2}$

The equation becomes $(\sqrt{2})^{x} = x$

Which gives $x = 0,2$ , but we ignore $x = 0$ because the number is more than 0. ~~~

The problem is not incorrect. It is easy to see that the limit can be evaluated from equation: $y=(\sqrt{2})^y$ .

this is true for $y=2$ and $y=4$ . However, let's give a simple proof why the solution is not $y=4$ . Let's prove by simple induction that $^{n}\left(\sqrt{2}\right)<2$ for all natural numbers $n$ .

For $n=1$ , it is obvious that $\sqrt{2}<2$ . Assume our proposition holds for all natural numbers $n$ , and let's prove that then it holds for $n+1$ .

Let $a_n =^{n}\left(\sqrt{2}\right)<2$ by inductive hypothesis. Then $a_{n+1} = (\sqrt{2})^{a_n}<(\sqrt{2})^2=2$

The inequality holds because of basic properties of exponential function, and our hypothesis. Therefore we have proven our statement for all natural numbers, so our limit can be no higher than 2, and hence it cannot be 4. Therefore $L=2$ , which is the solution to the problem.

Let A=\sqrt{2}^\sqrt{2}^\sqrt{2}^{...}

inf(sqrt(2)) = (2)inf/2 *inf/2.....................=(2)inf/2(raised to inf) = (2)inf/inf

= 2

inf = infinity