Here is a problem from fakalin. Find the error in the following proof:

Let \(x^{x^{x\cdots}} = 2\). Then \(x^{x^{x\cdots}} = x^{x^{x^{x\cdots}}} = 2\). Substituting, we get \(x^2=2\) and \(x=\sqrt{2}\). Now let \(x^{x^{x\cdots}} = 4\). Similarly, \(x^{x^{x^{x\cdots}}} = 4\), \(x^4=4\) and \(x=\sqrt{2}\). But then \(2=x^{x^{x\cdots}}=4\) and \(2=4\).

Obviously 2 cannot equal 4.

The problem seems to occur with solving for \(x\) by inverting the function \(f(x)=x^{x^{x\cdots}}\) via the recursion \(x^{f(x)}=f(x)\), \(x=f(x)^{1/f(x)}\). While we can set \(f(x)\) to any value \(s\) to solve \(x=s^{1/s}\), we mustn’t neglect also the condition that \(s=f(x)\) should have a solution.

Here we plot \(x=s^{1/s}\) for \(s\) from 0 to 5. (Let us only consider \(x\ge 0\), where everything is real.) Both \(s=2\) and \(s=4\) give \(x=\sqrt{2}\), but only the former satisfies \(s=f(x)\), while no \(s\) to the right of the peak has a solution. The reason is that \(f(x)\) for \(x>1\) has to be increasing by the monotonicity of exponentiation in each of its two arguments (\(a^b\) increasing in both \(a,b\) if \(a,b>1\)), and so the proper inverse (where it is defined) can only be taken from the left branch of the plot. The peak occurs at \(s=e\), \(x=K=e^{1/e}\). For every \(x>K\), \(f(x)\) diverges to \(\infty\). For \(1< x\le K[/latex], [latex]\overbrace{x^{x^x\cdots}}^n[/latex] is increasing in the tetration depth [latex]n[/latex] and has an upper-bound ([latex]e[/latex]), so [latex]f(x)[/latex] is a well defined limit which is exactly [latex]s[/latex]. The other half is more complicated. When [latex]x=0[/latex], [latex]0^s=0[/latex] for all [latex]s>0\) and \(0^s=1\) for \(s=0\), so there is no solution to \(s=f(x)\). This is a consequence of where \(a,b<1[/latex], [latex]a^b[/latex] is increasing in [latex]a[/latex] while decreasing in [latex]b[/latex]. From [latex]x<1[/latex], we get [latex]x^x>x\), \(x^{x^x} < x^x[/latex], and so on in this alternating fashion. We also get from [latex]x^x < 1[/latex] that [latex]x^{x^x} > x\), so in fact we have two non-crossing monotonic subsequences — the odd-depth tetrations increasing and the even-depth ones decreasing — that must therefore both converge. \(f(x)\) is well defined iff they converge to the same value \(s\). Suppose one subsequence has limit \(s_1\) and the other \(s_2\). Either of the cases \(s=s_1\) or \(s=s_2\) satisfies \(x^{x^s}=s\), so where there is a unique solution for a given \(x\), the limits must equal and \(f(x)\) must be well defined. From the plot (no proof) this occurs from \((1,1)\) down to the intersection at \(s=1/e\), \(x=L=(1/e)^e\). So for \(L \le x < 1[/latex], [latex]f(x)[/latex] is well defined. For all [latex]x < L[/latex] though, multiple solutions for [latex]s[/latex] actually correspond to the curved branch of this [latex](s_1,s_2)[/latex] plot, which shows solutions to [latex]x^{s_1}=s_2[/latex] and [latex]x^{s_2}=s_1[/latex], i.e., [latex]s_1^{1/s_2}=s_2^{1/s_1}[/latex], so [latex]f(x)[/latex] is not defined there. (*) In summary, [latex]f(x)[/latex] is well defined between [latex]1/e[/latex] and [latex]e[/latex] for [latex](1/e)^e\le x \le e^{1/e}[/latex], goes to [latex]\infty[/latex] over the range, and flips between two values under the range. Since [latex]x^{x^{x\cdots}}=4[/latex] can never happen, it is a vacuous assumption that can logically lead to any conclusion.

(*) Rather than looking for the limit of a sequence of tetrations, [latex]f(x)\) can be formally defined in the lower range \(\)0 < x < L[/latex] by the recursion [latex]x^{f(x)}=f(x)[/latex] directly: in the second plot, the branch from [latex](1,1)[/latex] is exactly this equilibrium (i.e., that branch is the first plot), but in this lower range the equilibrium is unstable and therefore not reachable by a finite tetration of [latex]x[/latex].