Infinite Power Towers

Calculus Level 2

Obviously, $1^{1^{1^{.^{.^{.}}}}}=1$

Furthermore, with a little bit of effort, we can show that $\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}}=2$

However, clearly $2^{2^{2^{.^{.^{.}}}}}=\infty$

So what is the largest real number $x$ such that $x^{x^{x^{.^{.^{.}}}}}$ is a finite number?

\frac{\pi}{2}

\frac{13}{9}

{\left(\frac{\pi}{e}\right)}^e

e^{\frac{1}{e}}

\pi^{\frac{1}{\pi}}

\sqrt{2}

\sqrt[3]{3}

\frac{3}{2}

1 solution

Andrew Ho
Jan 10, 2018

Suppose that we are given a variable x. We can use a recurrence formula to represent the infinite powers: First term = x, nth term = x to the power of the previous term. Assuming that the infinite powers does converge to a real number, as n approaches infinity, the nth term will equal the previous term. Hence, letting the n-1th term be y, $x^y=y$ . If the x value is larger than the maximum value, there will not be a solution to this equation. Putting up a graph: We indeed see that there is a maximum value that x can take. At that point, $dx/dy$ =0, and the curve will go backwards as y increases. So, all we need to do is to evaluate the derivative of x with respect of y. Solving for x, Now, we set $u$ as the exponent of e and apply the chain rule. We should find that: In all of the terms, only $1-lny$ can be exactly zero. Hence, $y=e$ and so: (Thanks to Nick Turtle)

Infinite Power Towers

1 solution

0 pending reports