Kim Reece @kimreece@mathstodon.xyz

I'll be cleaning this up over the weeks to come, but in addition to looking for standard funding sources, I'm opening the tip jar.

https://www.ko-fi.com/feonixrift
https://www.patreon.com/feonixrift

I'm a math grad student, persistently cashflow-negative, trying to hold up the finances through my masters thesis before I can get into a phd (hence paid) position.

I plan to be more vocal with what I do, and contribute in other ways, but... Patronage of the sciences is a thing, right?

If you really want to know how my day was, I overlooked \( \sqrt{-m}\sqrt{mn}=-m\sqrt{-n} \) for several hours. And then I made a sign error.

sage: 1/8 * ((z-1/z)+sqrt(-19)*(z+1/z)) * ((z^2-1/z^2)+sqrt(-19)*(z^2+1/z^2)) *
....: ((z^3-1/z^3)+sqrt(-19)*(z^3+1/z^3))
1/8*(sqrt(-19)*(e^(6/7*I*pi) + e^(-6/7*I*pi)) + e^(6/7*I*pi) - e^(-6/7*I*pi))*(sqrt(-19)*(e^(4/7*I*pi) + e^(-4/7*I*pi)) + e^(4/7*I*pi) - e^(-4/7*I*pi))*(sqrt(-19)*(e^(2/7*I*pi) + e^(-2/7*I*pi)) + e^(2/7*I*pi) - e^(-2/7*I*pi))

...useless!

productivity theatre: me adjusting blankets as if propped up with a blanket over my feet is somehow less 'working sick from bed' than propped up with the same blanket up to my waist

My bibliography just spilled over onto the second page...

If I pull a proof out of my hat, at this late stage, ... well I'm not going to claim it's impossible. After all, I haven't proven that it can't be done.

everyone else explaining 1÷1=1: 'so you put one cat in one box,'

me explaining 1÷1=1: 'Viewing division as fractions, and fractions as quotient groups, 1/1 is a representative element of a/a which has the properties of the identity in the quotient group defined by the relation (a/b~c/d iff ad=bc), it must therefore be canonically identifiable with the identity element of the underlying group.'

.. I appear to be contaminated and should probably not be allowed near lower division students? ;)

Ok, this one is generic enough to pose...

If \( \pi \in \mathbb{Q}(\sqrt{mn}) \) a totally positive real number, letting \(\jmath : \sqrt{-m}\mapsto -\sqrt{-m}\) and \(\tau : \sqrt{-n}\mapsto -\sqrt{-n}\), it is clear that \(\pi^\jmath = \pi^\tau\). When does \((\sqrt{-\pi})^\jmath \ne (\sqrt{-\pi})^\tau\)?

Hope.

purely imaginary and totally imaginary do not mean the same thing, and you really want to be careful which one you use.

No good typos yet today. Only, I am getting a lot faster at typing in LaTeX! It can be learned! And I'm so glad that \ key on my keyboard is in a good place so I didn't have to remap it...

https://twitter.com/feonixrift/status/1305063189206441985

Very serious math threads we have over there. ;)

Strangely, no good typos today.