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The solution to the first equation is $x = 2018^{ \frac{ 1} { 2018} }$ .
The solution to the second equation is not $x = 2018^{ \frac{ 1} { 2018} }$ . In fact, there is no (positive real) solution to this equation.

Find the error in this approach

First, prove that $x = 2018^{ \frac{1}{2018} }$ is a solution to the first problem. This can be done by showing that

• On $x > 1$ , $x^{ x^{ 2018} }$ is an increasing function on $x > 1$ , hence there is at best a unique answer on $x > 1$ . Manually verify that $x = 2018^{ \frac{1}{2018} }$ works.
• On $0 < x < 1$ , $x^{ x^{2018} } < 1$ , hence there are no solutions.

Let $y = x^{x ^ \ldots } = 2018$ .
By substituting into the power, we get $y = x^ y$ .
Thus, $x = y^{\frac{1}{y} } = 2018^{ \frac{1}{2018} }$ .
Hence, the 2 solutions are the same.
What is the error made?

Explanation of the error

The error is made in saying "Let $y = x^{x^\ldots } = 2018$ ." without justifying that it is a valid step. Just because you write down a string of symbols, doesn't mean that it must make sense. Specifically, this value does not converge for some values of $x$ (e.g. $x = 2$ ), hence we do not yet know if there is a value of $x$ which will make this statement valid. This results in a circular argument of "Let's assume a solution exists. Hence a solution exists".

Instead, the proper conclusion is "If there is a solution, then the solution can only be $x = 2018^{\frac{1}{2018} }$ ". We now have to verify if it is indeed a solution. By looking at the first few terms, for $x = 2018^{\frac{1}{2018} } \approx 1.00378$ , we have $x^{x^x} \approx 1.00 379, x^{x^{x^x}} \approx 1.00 379$ , and so this sequence converges to $1.00 379$ . Hence, this value of $x$ is not a solution to the equation that we're looking for.

More generally, how can we know what $x^{ x^\ldots }$ converge to? The previous discussion shows that we will require $x = n ^ \frac{1}{n}$ . If there is no solution, then clearly it will not converge. But, if a solution exists, must it converge? And what happens if multiple solutions exist?
To fully answer this requires a bit more theory which I will not go into as yet. Here's an outline of the result:

• For $x > e^ \frac{1}{e}$ , there is no solution to $x = n^\frac{1}{ n }$ , and (expectedly) the tower does not converge.
• For $x = e^ \frac{1}{e}$ , the tower converges to $e$ .
• For $e^ \frac{1}{e} > x > 1$ , there are 2 solutions to the equation $x = n^{ \frac{1}{n} }$ , and the tower converges to the smaller of these values. In particular, it will be smaller than $e$ . As an example, $x = 2018^{ \frac{1}{2018}} = 1.00379^{ \frac{1}{1.00379 }}$ . Hence, we get $x^{x^\ldots }$ converting to $1.00379$ .
• For $1 \geq x \geq \frac{1}{ e^e }$ , there is 1 solution to $x = n^\frac{1}{ n }$ , which the tower converges to.
• For $\frac{1}{e^e } > x > 0$ , there is 1 solution to $x = n^\frac{1}{ n }$ , the tower does not converge,

Notes from the discussions.

1. This error is very similar to the instance where we squared both sides of an equation and potentially introduced an extraneous solution, and thus have to check whether or not it is a valid solution. Due to the one-way implication, we have a necessary condition that restricts what the solutions could be, and must then verify if the solutions are indeed valid.

2. Infinite powers can exist/converge even though $x > 1$ . As an example, a well-known fact (which I will not prove) is that if $x = \sqrt{2}$ then $x ^ { x ^ \ldots } = 2$ . Indeed, in this scenario $x ^ 2 = 2$ .

3. However, if $x$ is "too large", then infinite towers might not exist. As an example, $2^{2^\ldots }$ will not converge.

4. If we take $x = 2018^{ \frac{1}{2018} } \approx 1.00378$ then the sequence $x, x^x, x^{x^x}$ does converge. However, it does not converge to 2018. This shouldn't be too surprising since $1.00 378 < \sqrt{2}$ , so the limit should be bounded above by $\sqrt{2}^{ \sqrt{2} ^ \ldots } = 2$ .

5. What is the range of $x$ for which $x^{ x^\ldots }$ converges?

Regarding the second equation:

Euler showed that $\displaystyle \lim_{n \ \to \ \infty} \underbrace{x^{x^{\cdot ^{\cdot ^{x}}}}}_{\text{n times}}$ converges for $\displaystyle \large \frac{1}{e^e} \leq x \leq e^{\frac1e}$ , which bounds $\displaystyle \large y = \underbrace{x^{x^{\cdot ^{\cdot ^{x}}}}}_{\text{n times}}$ to $\displaystyle \large \frac 1e \leq y \leq e$ . The limit, should it exist, is a positive real solution of the equation $\large y = x^y$ . Thus, $\large x = y^{\frac 1y}$ .

The limit defining the infinite tetration of $\large x$ fails to converge for $\large x > e^{\frac1e}$ because the maximum of $\large y$ is $\large e$ . Thus, there exists no (positive real) solution for the equation $x^{x^{\cdot ^{\cdot ^{x}}}} = 2018$ .

The second infinite power sequence is either 1 or diverges, therefore there is no solution.