I suppose you can just take the tensor product of the spacial structure with the units? So energy would be R tensored with V^{N.m} (which of course is isomorphic to just V^{N.m}), and torque would be the space of axial vectors tensored with V^{N.m}.

There are plenty of academic opportunities at Harvard that are not available at state schools. Because they attract many of the top students in the world, they're able to offer highly accelerated or advanced courses that other schools can't.

I did both community college and Harvard undergrad. My experience is that while some intro classes were similarly structured, Harvard offered far more accelerated options for people who are prepared for it. You're right that the student body is a huge difference though.

That's not right either, the board and officers have fiduciary duty to act in shareholders' interest and to use reasonable business judgment. Less strict than maximizing profit, but more strict than anything-legal-goes.

You're entirely correct but I think GP was referring to personal cell: phone doesn't need to be recorded (if not trading), but you can't use your personal cell. Anything work-related must be on firm systems.

If I understand the intro correctly, the "size" they're referring to is the growth rate of a sequence, where the sequence is counting the dimensions of certain subsets of bounded denominator modular forms.

Let BDMF = bounded denominator modular form. They show congruence BDMFs grow at least N^3, but all BDMFs grow at most N^3*log(N). (The latter bound is the hard part of the proof.) To get the contradiction, they show a hypothetical noncongruence BDMF example would imply additional counterexamples that (just barely) get over the N^3*log(N) bound.

So is this a kind of result akin to the prime number theorem (https://en.wikipedia.org/wiki/Prime_number_theorem) where you have things that are asymptotically guaranteed to be/remain very close to one another?

Actually Elo does hit a maximum, because a perfect player doesn't beat an imperfect player 100% of the time: sometimes an imperfect player gets lucky and plays a perfect game.

A number of people have made conjectures on the Elo of perfect chess play, using extrapolations from data how chess programs scale. I'm not sure what the latest analysis is (things may have changed with the strength of neural network chess), but iirc they usually estimate something like 1000 Elo over current programs.

ELO does not have a maximum limit, chess does; already said this in my comment prior to your response. Also clearly stated “over significant number of matches” - not just a single match.

Perfect chess play would not set the maximum ELO, that’s not how ELO works. ELO simply ranks players that play. If a prefect chess engine existed and continued to play and there was at least one other chess engine that improved by beating its prior best version, its ELO and the prefect players ELO would continue to rise. Long, long way to go to fill the ELO ratings between current state of the art and the prefect all knowing player that’s aware of all 10^120 possible matches.