Counting Is Hard

@b0rk@social.jvns.ca you can do it without maths at all. Say I want to give you my bicycle, but our schedules just will never work out for me to give you it directly. Well, we pick a place both of us can access, say outside the train station. I lock my bike up at a designated spot and go about my day. You come along with your own lock and use it to fasten the bike to the same spot without interfering with my lock, then go about your day. At some point I remove my lock, after which you can remove yours and gain a bicycle. As I understand it, such a protocol was the inspiration for DH key exchange. Picking a spot is picking a generator. Locking is exponentiating. Locks not interfering is the commutativity property. It's the explanation I always use when TAing cryptography

@edwinb@types.pl Maybe it could be amended to, "Enjoy Yourself (It's Later than You Think)" 😃 https://www.youtube.com/watch?v=6liD8TdlJFA

@flippac Being able to quote that entire list was one of my useless talents

@mc@mathstodon.xyz Not that it doesn't go both ways, but it's funny how a set theory person trying to give a foundation for mathematics basically ignores computing? No mention of work on types, explicit mathematics, computability at all (there is a cite that doesn't appear to be referenced?). 2nd order arithmetic gets a nod at least (so transitively reverse maths). Maybe it's good work, I can't judge it on its own merits, but I'd be sceptical of anyone who wants to give "an antidote to godel"

@hallasurvivor the laplace transform link has always been there, in e.g. feller, but I always felt it was treated as a kind of magic trick by most undergrad probability classes (when they bothered to include it)

@hallasurvivor Pick a, b, c such that \( a^2 + b^2 + c^2 = 1 \), then \( ai + bj + ck \) is a root of -1. Clearly uncountably many solutions. I'm guessing this is the solution the majority of people come up with. Edit. Reinstating my previous claim which was fine actually One property that is no longer true of polynomials is that if they agree on infinitely many places then they are equal. Take \( (x^2 + 1) p(x) \) for a counterexample for two distinct p. You could also observe that, x^4 and x^8 agree infinitely often. I'm not sure what to make of that, although it is basic to how we treat polynomials.

@danielmclaury I get your point, they definitely should have called the class a little early and picked it up next time

@ZachWeinersmith I've often found it slightly amusing that I run into so many people who are the brown-haired guy in panel 1, and yet who really love puzzle games of all descriptions. My dude, if you can finish "Baba is You" you can definitely handle Calc 1

@flippac I have a different pun in mind for that, but I'm maybe hoping I can figure out how to tweak sagan's "a mote of dust suspended in a sunbeam"

@liamoc@types.pl I keep a diary of everything I do phd-wise, but I don't publish it. It's currently a 5421 line org mode file