tetration

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 \(1< x\le K\), \(\overbrace{x^{x^x\cdots}}^n\) is increasing in the tetration depth \(n\) and has an upper-bound (\(e\)), so \(f(x)\) is a well defined limit which is exactly \(s\).

The other half is more complicated. When \(x=0\), \(0^s=0\) for all \(s>0\) and \(0^s=1\) for \(s=0\), so there is no solution to \(s=f(x)\). This is a consequence of where \(a,b<1\), \(a^b\) is increasing in \(a\) while decreasing in \(b\). From \(x<1\), we get \(x^x>x\), \(x^{x^x} < x^x\), and so on in this alternating fashion. We also get from \(x^x < 1\) that \(x^{x^x} > x\), 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 \(L \le x < 1\), \(f(x)\) is well defined. For all \(x < L\) though, multiple solutions for \(s\) actually correspond to the curved branch of this \((s_1,s_2)\) plot, which shows solutions to \(x^{s_1}=s_2\) and \(x^{s_2}=s_1\), i.e., \(s_1^{1/s_2}=s_2^{1/s_1}\), so \(f(x)\) is not defined there. (*)

In summary, \(f(x)\) is well defined between \(1/e\) and \(e\) for \((1/e)^e\le x \le e^{1/e}\), goes to \(\infty\) over the range, and flips between two values under the range. Since \(x^{x^{x\cdots}}=4\) 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, \(f(x)\) can be formally defined in the lower range \(0 < x < L\) by the recursion \(x^{f(x)}=f(x)\) directly: in the second plot, the branch from \((1,1)\) 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 \(x\).

Comments

  1. Andrew
    June 26th, 2013 | 12:37

    This function is what I call the “exponential commutator” on the Tetration Forum. I think the fundamental problem with your proof is that you are dealing with inverse functions of functions that are not monotonic. When you try and find an inverse function of a nonmonotonic function, then you end up with a Riemann surface, which you might consider a multivalued function, and in order to think of it as a function, you must assign an index to each branch. Doing this, and going back to your proof, I would reword it as follows:

    Let c = \sqrt{2}. Let f_0(x) be the branch of the inverse function of x^{1/x} that includes the point (x=sqrt(2), f_0=2). Let f_1(x) be the branch of the inverse function of x^{1/x} that includes the point (x=sqrt(2), f_1=4). Then f_0(c) = 2. Similarly, f_1(c) = 4. But then 2 = f_0(c) != f_1(c) = 4.

    Your assumption that f_0 = f_1 is false, that is the where you went wrong.

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