In here, I saw that $$x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}$$ exists as a real number (convergent) if and only if $$x\in[e^{-e}, e^\frac{1}{e}].$$ How can I prove this??

  • 6
    $\begingroup$ Have a look at en.wikipedia.org/wiki/Tetration. What you are asked has been established by Euler. $\endgroup$ Jan 3 '15 at 10:16
  • 3
    $\begingroup$ Hint: $y=x^y$. $\endgroup$
    – Lucian
    Jan 3 '15 at 10:36
  • 1
    $\begingroup$ You should define this infinite power tower as $$\lim_{n\to\infty} a_n$$ where $$a_0 = 1 \qquad \text{and} \qquad a_n=x^{a_{n-1}}$$ This lets you formulate the question precisely, unambiguously (otherwise it's not clear what exactly an infinite power tower is). $\endgroup$ Jan 3 '15 at 10:41
  • 2
    $\begingroup$ @Lucian Assuming convergence of $y$, we can say $$y=x^y$$ $\endgroup$ Jan 3 '15 at 10:42
  • 1
    $\begingroup$ These may be equivalent, but it is far from clear that one is the same question as the other, and the answers do not seem to match. $\endgroup$
    – trlkly
    Jan 6 '18 at 4:41

The equation


is not well defined, we have to find some way to define it to find its value.

There are two sensible definitions:

The first is to say that the thing in the exponent is actually the same as the entire thing, which is written as


And solving for $y$ here would give the solution.

To solve this we need a special multivalued function called the Lambert $\newcommand{\W}{\operatorname{W}}\W$ function, which satisfies $z=\W(z)\cdot\exp(\W(z))$ for all complex $z$. The solution is then

$$y=-\frac{\W(-\log x)}{-\log x}$$

If we restrict attention to real-valued $\W$, the function is defined only for $x \ge −1/e$, and is double-valued on $(−1/e, 0)$.

Since the argument is $-\log x$, which is a decreasing, and $-\log(e^{1/e})=-\frac1e$, our solution is only defined when $x\le e^{1/e}$ and when the logarithm is defined. Therefore, under this definition, the function is defined (not necessarily uniquely) and real when $x\in[0,e^{1/e}]$

There is however another definition:

Define the sequence $a_n$

$$a_0=x\qquad\qquad a_n=x^{a_{n-1}}$$

And the value were looking for is


Which can be written as $$\lim_{n\to\infty}x\uparrow\uparrow n$$

In Nicks answer, it is explained why this is the same as the definition $y=x^y$

  • 2
    $\begingroup$ Actually, it can also be informally expressed as $^n x$. $\endgroup$
    – Nick
    Jan 3 '15 at 12:28
  • $\begingroup$ @Nick sure, thats just another way to write the same thing $\endgroup$
    – Alice Ryhl
    Jan 3 '15 at 12:30

Using Knuth's up-arrow notation,

$$\newcommand{\W}{\operatorname{W}}x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}} = \lim_{n \to \infty} x\uparrow\uparrow n$$

Let $\displaystyle y = \lim_{n \to \infty} x\uparrow\uparrow n$ and let us assume that it does converge.

Then, $\displaystyle x^y = \lim_{n \to \infty} x \uparrow\uparrow(n+1)$

As $n \to \infty$, $n + 1 \to \infty$ (Roughly speaking $\infty \pm n \to \infty$)

Hence, $\displaystyle \lim_{n \to \infty} x\uparrow\uparrow n \equiv \lim_{n \to \infty} x \uparrow\uparrow(n+1) \implies \boxed{\displaystyle y = x^y}$

A trivial solution here is $(x,y) = (1,1)$

The variables $x$ and $y$ in $y = x^y$ are not separable (we cannot isolate the any of the variables here on to any one side ) in this equation by usual algebraic manipulations.

However, we can use the Lambert $\W$ function to isolate the variables, $$y = - \frac{\W(-\ln x)}{\ln x}$$

Considering the real values of $\W$ and Using the fixed point iteration method,
it can be seen that $\displaystyle x\in\left[\frac1{e^{e}}, e^\frac{1}{e}\right]$

Good Luck :)


As Lucian mentioned above, I have considered the function $$f(y)=y^{\frac{1}{y}}$$ $$\lim_{y\to 0^+}f(y)=0$$ $$\lim_{y\to \infty}f(y)=1$$ Also $$f'(y)=y^{\frac{1}{y}}\Big(\dfrac{1}{y^2}-\dfrac{\ln y}{y^2}\Big).$$ Using this, we have maximum of $f(y)$ occurs at $y=e.$ Hence $$f(y)\le e^{\frac{1}{e}},\,\,\,\,\,\, \forall y\ge 0.$$ But I did not have an idea about obtaining the lower bound $e^{-e}.$


Not the answer you're looking for? Browse other questions tagged or ask your own question.