Infinites make for math fun
I did as GS suggested and figured out the solution myself. (That's the point of following recreational math accounts, after all.)
It wasn't difficult, since the behavior of a function $ f_n(x) = x^{x^{x\ldots}}$ where there are $n-1$ exponentiations, when $n \rightarrow \infty$ is likely to be divergent for any $x>1$. This was my intuition, and based on that intuition, I assumed there couldn't exist $x$ and $y$ such that $f_{\infty}(x) = 2$ and $f_{\infty}(y) = 4$. Therefore the premise is false and the reasoning fails because of that.
But I didn't prove it. I did play around with a spreadsheet to check the behavior of the function around $x=1$, since $f_n(1) =1$ for all $n$ and the functions are continuous in $x$. A simple spreadsheet shows the behaviors for $x$ between 0 and 1 and for $x$ above 1. The rows are increasing $n$, the relevant part is the diff-in-diff column, a discrete version of a second derivative w.r.t. $n$:
What these results (and several others, the point of a spreadsheet model being to play around with the numbers, or as a responsible adult would say it "do sensitivity analysis") show is that for $0 < x < 1$ the function is increasing and "concave," therefore most likely converging to a number in the [0,1] interval; for numbers above 1, the function is increasing and "convex," therefore most likely diverging to infinity.
Still not a proof, but confidence is high. (And confirmed later by the rest of the thread.)
True to my origins as a Prolog (and Lisp, occasionally) programmer, I feel compelled to write a formal definition of this function, thusly:
$\qquad f_1(x) = x;$
$\qquad f_n(x) = x^{f_{n-1}(x)} \quad\text{for $n>1$} $
As usual, recursions FTW!
Go beyond one-step thinking to understand executive pay
Scott Manley, whose space videos are among the most informative on YouTube, expressed a common complaint about executive severance pay, on the occasion of Boeing's change of the guard:
My response is the précis of why you have to pay outgoing executives significant severance: it's a signal to the incoming executives, not a reward to the outgoing executives.
(This was particularly obvious in the case of PG&E in California, which paid a lot to its executives in what most people thought scandalous and well-informed people recognized as a move to retain talent under circumstances when most top managers were considering outside options.)
The responses to SM's tweet contained a lot of misconceptions about executive-level, also called C-suite, management, and this one, which was a response to my response illustrates the three most important ones:
(As I try to be more positive, I'll be anonymizing tweets that I'm critiquing.)
Yet another Rotten Tomatoes calculation (Doctor Who)
Given these numbers, it's
5,872,182,639,638,860,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
times more likely that critics and audience use opposite criteria than the same criteria.
Never give up; never surrender!
