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22:00
I luckily quit teaching before COVID. My teaching style and COVID would have been disastrous.
Covid makes it difficult to get to know teachers and classmates well
Agreed.
So there won't be many letters of recommendation for students this year
I haven't written any in 4 years ... I'm done.
I died of covid
22:12
is this usage of "faculty" to mean "staff member" a feature of american english?
Leaky, no. "staff member" includes all administration, support staff, not just instructional faculty.
i'm referring to how i've always known it to mean something like a sub-department
like it refers to an organization instead of to people
I have no idea what you're talking about.
use of faculty to mean sub department is definitely not US english. "faculty of applied mathematics" that kind of thing.
yeah, faculty of mathematics may be under the department of natural sciences, for example
22:15
but "faculty" with reference to people usually means instructional/research human beings and excludes other roles.
We say department, not faculty, in that context.
I think that may be more European or something.
My +250 bounty expires in one day and none answered :( math.stackexchange.com/questions/4265472/…
We do refer to teaching faculty.
similarly "staff", at least as i've normally seen it, generally excludes research/instructional roles
@TedShifrin is it more difficult to do a PhD in Europe, in the US or in Canada?
22:16
I can't answer that.
why?
I would say US for the past year at least
because of covid
too many variables.
I mean in average
Because I don't know enough and because it's so variable. PhD quality varies hugely.
22:17
there is no 'average' phd in the us. too many variables.
I think the average PhD in math in the US never publishes past the thesis, if that.
I haven't followed this in many years.
there's also stuff like, people going into phds in the us usually have less training than people going into similar programs in europe. so it's not a direct comparison even if there were computable averages.
what about Canada vs US?
is it the same?
i don't know of any meaningful differences. i don't see geography as affecting this stuff as much as specific people and situations.
when i was in grad school, just within my department there was a lot of variation. it generally couldn't be traced to some professors being 'easier' or 'harder' than others, or students being more or less talented.
it was a complicated mix of a lot of things.
I heard of a teacher in Canada named Marlene Frigon whose exams are very tough and she looks like a Harry Potter character
22:23
Europe is just different. College in Europe is like grad school here for most students.
Hi guys, are you fine?

I need to solve a linear algebra question:

Given a pseudo vectorial space, I must prove that is in fact a vectorial space.
$$V = \left\{ (a,b) \in \mathbb{R}^2 |\; a,b > 0 \right\}$$
* $(a,b) \oplus (c,d) = (ac, bd)\;\; \forall (a,b),(c,d) \in V$
* $ \alpha (a,b) = (a^{\alpha}, b^{\alpha})\;\;\forall \alpha \in \mathbb{R} \text{ and } \forall (a,b) \in V$

My solution:

The first pseudo vectorial space limits the properties of adition operation:
$$(ac, bd) = (a + c, b + d) \\ c = \frac{a}{a-1}\;\; d = \frac{b}{b-1}$$
Where are you getting these formulas for $c$ and $d$, @Matheus?
$\oplus$ has nothing to do with usual addition.
You have to check all the axioms. What is the zero vector? What is the additive inverse? Is addition commutative, associative, etc.
Why is 1+1=2?
It's not.
It is, because of Peano axioms
22:29
do the given recipes even define binary operations on $V$. maybe this is assumed. perhaps worth dropping a line about it even if only to observe that they do.
Yes, one will definitely need to use $a,b,c,d>0$ a few places.
@leslie Does Olivia have any words of wisdom for Screech? I have to get her shortly.
today the munchkin announced that there was a spider in her closet. she hadn't been in her closet so this was make believe. i asked what she wanted to do about it. the answer was "wipe it off." i said, do you know what happens when we wipe the spider off. she said "i dunno. it's gone!" we'll save the rest of that for when she's older.
olivia says, if you play your cards right you can get treats, on account of the recent procedure.
Ah, yes. A few might be in order.
@TedShifrin Ah, now I understood. Thanks man. I was trying to redefine the sum and product operations
I only have to prove these properties
OK, cool, @Matheus.
22:33
the shelter we adopted olivia from just liked two of her photos on social media. they are strong supporters of olivia, i think because they don't want us to bring her back.
LOL ... they know how much trouble you are, @leslie, and you've only passed it on to your progeny.
she's taking a very late nap, but she's horrible without her nap, so we're allowing it.
she's also horrible with her nap.
23:00
Anywhere I can read up more on modular arithmetic and montgomery reduction for computing $2^n \bmod x$?
I've read this, and I can't say I'm making great progress. en.wikipedia.org/wiki/Montgomery_modular_multiplication
I can't say that the variable naming is consistent between the section "Montgomery form" and "The REDC algorithm".
Nor do I understand if there is a way to properly interpret these in what apparently has been called "standard form".
As usual, here's a (messy) workspace showing where I'm at. desmos.com/calculator/4ucs2rhrhe
23:22
Correction: I mean the phrasing is odd to me and not that consistent for me to follow between the two sections. I also don't know what a residue is.
See I learned that you can actually compute individual bits of the reciprocal of $x$ in parallel by computing $2^n \bmod x$ to arbitrary precision. Then for each base two digit $d_n$ standing in for the numerator of $(0.d_n)_x$ as $\frac{d_n}{x}$, the value of the bit associated with $d_n$ is $[2 d_n \geq 1]$ using Iverson bracket notation.
Using addition-chain exponentiation, you can get a relatively fast computation using arithmetic natively in base $x$ with the value stored in a binary encoding for the digits, however, the number of multiplies unfortunately still makes it rather slow, even though it is still a big improvement over the usual cost for HW divide.
The lead, then, is to look at $1\bmod x$ and finding a way to compute this. Current ideas are either by a bit of symbolic computation via some comparisons, or something else. This function is actually very easy to compute if you know which integer "reciprocal group" that $x$ belongs to.
And, for the record, $2^n\bmod x = 2^n(1\bmod (2^{-n} x))$

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