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Wednesday, January 15, 2020

Fun with numbers for January 15, 2020

Infinites make for math fun



Link: https://twitter.com/3blue1brown/status/1215087264792887296

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!