Mathematical fallacy of $x^{x^{x^{x^x...}}}$ = 2

Question

Suppose we have an equation with an infinite number of $x$'s as an exponent:

$$x^{x^{x^{x^x...}}} = 2$$

$$x^{(x^{x^{x^x...}})} = 2$$ because there are infinity $x$'s in the parentheses, which we've said equals 2:

$$x^2 = 2$$ $$x = \sqrt{2}$$

However, suppose you take a new equation:

$$y^{y^{y^{y^y...}}} = 4$$

$$y^4 = 4$$ $$y=\sqrt{2}$$ Which implies:

$$2 = x^{x^{x^{x^x...}}}= y^{y^{y^{y^y...}}}= 4$$

Which is obviously wrong.

What went wrong, and where did this mathematical fallacy come from? I was wondering if this could somehow be related to convergence in tetration, but I'm not sure how. Any ideas?

Answer 1

Observe that if the tower $x^{x^{x^{x^\ldots}}}$ has a value $u$, we have, as you note

$$ x^u = u $$

Taking the log of both sides and dividing by $u$, we have

$$ \ln x = \frac{\ln u}{u} $$

The funcion $\frac{\ln u}{u}$ is not one-to-one: For $u \geq 1$, it starts out at $(1, 0)$, rises to a peak at $(e, 1/e)$, and then falls off toward $0$ as $u \to \infty$. The "fallacy" you identified can be associated with the fact that the function attains identical values at $u = 2$ and $u = 4$.

However, as others have noted, the sequence $x, x^x, x^{x^x}, x^{x^{x^x}}, \ldots$ actually increases without bound (rather than converging) for $x > e^{1/e}$. Thus the infinite tetration, if it is to be allowed, cannot be defined for such values (at least, not in the ordinary way). For $0 < x \leq e^{1/e}$, we might permit the tetration to be defined as the limit of that sequence, and this would be the minimum value of $u$ such that $\ln x = \frac{\ln u}{u}$.

Answer 2

What went wrong (with both arguments) is that you are assuming $x^{x^{x^⋰}}$ has meaning in the first place.

Maybe it does have meaning for certain $x$, as the limit as $n\to\infty$ of $x\uparrow n$ (notation for a tower of powers).

But if you start with $y=2$, the sequence $\{y,y^y,y^{y^y},\ldots\}=\{2,4,16,\ldots\}$ does not converge. Since a real solution to $y^{y^{y^⋰}}=4$ would indeed require that $y=2$, you can conclude that $y^{y^{y^⋰}}=4$ has no solution. The condition $y=2$ is necessary, but not sufficient, to satisfy $y^{y^{y^⋰}}=4$.)

Back to $x^{x^{x^⋰}}=2$, you are right to conclude that if there is a real solution, then $x=\sqrt{2}$. It is still an issue to demonstrate that $\sqrt{2}$ actually is a solution. You need to establish that $\{\sqrt{2},\sqrt{2}^{\sqrt{2}},\sqrt{2}^{\sqrt{2}^{\sqrt{2}}},\ldots\}$ converges (which I believe you can do by showing that the sequence is increasing but bounded above by $2$ [using induction].)

Answer 3

The fallacy is that there is no such operation as this infinite exponentiation, it's ill-defined. Notice that no general definition of it is given. If it were well-defined, it should define the infinite exponentiation of a sequence of reals $(x_n)$.

Superficially, it might seem legitimate, by vague analogy with infinite sums and products. The latter can be defined as the limit of partial sums and partial products respectively, when the limits exist. For exponentiation, the grouping is to the right and not to the left. A definition of iterated exponentiation on finite sequences might be something like the following: $$\begin{align} \operatorname{ie}(\emptyset) &= 1, \\ \operatorname{ie}(\,(x_0, x_1, \dotsc, x_n)\,) &= x_0^{\operatorname{ie}((x_1, \dotsc, x_n))}. \end{align} $$ Now, for $(x_n)_{n\in \Bbb N}$ a sequence of reals, define $$ E((x_n)_{n\in \Bbb N}) = \lim_{n\to\infty} \operatorname{ie}(\,(x_0, x_1, \dotsc, x_n)\,) $$ when this limit exists.

For every real $x$ let $\vec{x}$ be the constantly-$x$ sequence. Then your infinite stack of $x$s is equal to $E(\vec{x})$: $$ $E(\vec{x}) = x^{x^{x^{x^x...}}}. $$ For some $x$ this limit does exist. But it doesn't exist for $x=2$ or $x=4$.

Answer 4

What you've actually proved is that there is no (positive real number) $y$ such that $y^{y^{y^{y^y...}}} = 4$. That is, the fallacy resides in the tacit assumption that the function

$$f(x)=x^{x^{x^{x^x...}}}$$

can be set equal to any (positive) value you like and then solved for $x$.

Somewhat more precisely, your argument, as far as it goes, shows that either $f(x)=2$ has no solution or $f(x)=4$ has no solution or neither equation has a solution, because the only possible solution, if there is one, is $x=\sqrt2$.

To show that $x=\sqrt2$ actually does solve the equation $f(x)=2$, you have to prove that the sequence $\{a_n\}$ defined by $a_{n+1}=\sqrt2^{a_n}$ with $a_1=\sqrt2$ (or, if you like, $a_0=1$) converges to $2$ as $n\to\infty$. This isn't all that hard to do, but it does need to be done.