Mathematics - 2022-05-19

up to you how you attempt to define the product of all elements in a subset of R, but however you do it, unless you break what finite products mean or the usual notion of convergence that gives you infinite products, it will probably fail to give you a result unless the set is at most countable and clumps in a certain way around 1.

complex analysis books sometimes do this a little bit.

you could say "haha if 0 is in there then the product is 0" but that isn't really engaging with what the notion of product is. sets containing 0 are also a special case for countable products.

Koro

haha I didn't think of 0 ! (not factorial).
yeah, I understood that product of finitely many terms is well known; and the product of countable (infinite) terms is usually defined as limit (just as sum of countable many terms).

leslie townes

koro note that in the case of conditionally convergent series the notion of 'sum of all elements of a countable subset of R' will sometimes make sense with extra structure beyond the set itself (i.e., a linear ordering of that set) but not in general.

something like "sum S = sup {sum F: F subset S, F finite}" is only gonna give you finite sums for absolutely convergent series where the choice of ordering would not matter.

books in measure theory sometimes prove that or something slightly more general than that.

and that notion doesn't give you sums of multisets, e.g. sum {1,1,1} = sum {1} = 1 under that definition.

bumblebee

8:11 AM

Let's goooo!!!

I finished 6 exams in 3 days 💀

I'm still Ajay, I only wanted to change to bumblebee for other sites not this one. Well, gotta wait another 30 days.

Calvin Khor

8:51 AM

@Koro this reminds me of a silly riddle. Can you simplify (x-a)(x-b)(x-c)(x-d)...(x-z)=?

Prithu biswas leftmse

10:27 AM

(x-a)(x-b)(x-c)(x-d)=x^4-x^3(a+b+c+d)+x^2(ab+bc+ca+ad+bd+cd)-x(abc+abd+bcd+acd)+abcd

bumblebee

10:42 AM

No @Prithubiswasleftmse

You can't have fallen for it

letters of the alphabet!!

Koro

11:18 AM

@CalvinKhor 0

I think I'm the one who asked this riddle here long time back

😁

one potato two potato

11:49 AM

complexification $su(n)_{\Bbb C}$ of lie algebra $su(n)$ means that we first treat $su(n)$ as an $\Bbb R$ lie algebra and take a complexification?

Koro

If V is a vector space and S is the set of all subspaces of V. Then considering the binary operation as 'sum of subspaces', what elements of S have additive inverse?

Sum of subspaces A and B is defined as the set {a+b, a in A and b in B}.

I think only the zero subspace has the additive inverse.

Shaun

Hi :) I'm learning about hyperreal numbers $*\Bbb R$. Just as a preliminary question, in what sense does $\Bbb R(x)$, the set of rational functions of functions in $x$ with real coefficients, "instantiate" $*\Bbb R$?

Prithu biswas leftmse

12:09 PM

@bumblebee ???

bumblebee

what comes before y?

Prithu biswas leftmse

x

oh there is a (x-x) term there.

then the entire product is 0.

bumblebee

yuppy doo

Prithu biswas leftmse

1:36 PM

I wonder if there is a generalization of this pattern....

3 hours ago, by Prithu biswas leftmse

(x-a)(x-b)(x-c)(x-d)=x^4-x^3(a+b+c+d)+x^2(ab+bc+ca+ad+bd+cd)-x(abc+abd+bcd+acd)+abcd

There is something weird going on with the coefficients.

(1) 1
(2) a+b+c+d
(3) ab+bc+ca+ad+bd+cd
(4) abc+abd+bcd+acd
(5) abcd

Calvin Khor

3:02 PM

@Prithubiswasleftmse

Elementary symmetric polynomial

In mathematics, specifically in commutative algebra, the elementary symmetric polynomials are one type of basic building block for symmetric polynomials, in the sense that any symmetric polynomial can be expressed as a polynomial in elementary symmetric polynomials. That is, any symmetric polynomial P is given by an expression involving only additions and multiplication of constants and elementary symmetric polynomials. There is one elementary symmetric polynomial of degree d in n variables for each positive integer d ≤ n, and it is formed by adding together all distinct products of d distinct...

@Shaun i looked at hyperreals a long time ago but don't recall this. where are you learning hyperreals from?

Prithu biswas leftmse

3:32 PM

@CalvinKhor Interesting =)

Shaun

See the following for details, @CalvinKhor.

0

Q: In what sense is $\Bbb R(x)$ an "instantiation" of the hyperreals?

I'm teaching myself about hyperreal numbers. My main motivation for doing so is that they include infinite numbers, whose existence I hear disputed & doubted often as "quantifying the unquantifiable". In this YouTube video, around twelve minutes in, it states that $$\Bbb R(x)=\left\{ \frac{f(x)}...

I have a copy of Goldblatt's, Lectures on the Hyperreals: An Introduction to Nonstandard Analysis somewhere but I can't find it.

robjohn

4:22 PM

@CalvinKhor I can, naught.

@leslietownes we can finish that with Trisquick.

Koro

4:44 PM

4

Q: Prove $\sum \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$ is convergent.

Assume $\sum\limits_{n=1}^{\infty} \dfrac{1}{a_n}$ is a convergent positive term series and $p>0$. Prove $$ \sum_{n=1}^{\infty} \frac{n^{p+1}}{a_1+2^pa_2+\cdots+n^pa_n}$$ is convergent. Since $$a_1+2^pa_2+\cdots+k^pa_k\ge \sqrt[k]{a_1\cdot2^pa_2\cdots k^pa_k}=\sqrt[k]{a_1a_2\cdots a_k}\cdot \sq...

real-analysis calculus sequences-and-series limits

Oops, you already got there @robjohn

:-)

robjohn

@Koro Is that the one that I just answered?

Koro

yeah.

I think I know the source of this problem.

robjohn

Ah, I see. I misinterpreted your comment.

Did I just answer a contest question? Hopefully, it is not ongoing, if so.

Koro

@robjohn I don't know. I meant that it looked similar to a problem in a book I have.

But now that I looked into it, it seems different.

Akiva Weinberger

5:40 PM

@Shaun Short answer is "it doesn't" lol

2

You need $\sqrt{|x|}$ to be in your set or else the transfer principle fails

and also $x^x$ and basically everything else

In fact, every function on the reals can be extended to a function on the hyperreals, no restrictions. This means we can extend, for example, the floor function $\lfloor x\rfloor$ to the hyperreals. The solutions to $\lfloor x\rfloor = x$ are called the hyperintegers.

Some infinitesimals $x$ will satisfy $\sin(1/x)=0$ and some will satisfy $\sin(1/x)=1$ etc. This (the fact that these are not infinitely close to each other) is the nonstandard analysis way of saying "$\lim_{x\to0}\sin(1/x)$ does not exist"

Shaun

5:57 PM

@AkivaWeinberger Thank you :)

Ted Shifrin

Do you know that infinitesimals satisfy this as opposed to just hyperreals in general? I.e., we have actual non-infinitesimals satisfying it, so how do we know actual infinitesimals do?

Akiva Weinberger

I'm not sure what you mean

Which part

Ted Shifrin

Your last paragraph.

Akiva Weinberger

The $\sin(1/x)$ part? Choose $n$ to be an infinite hyperinteger (these exist in the hyperreals) and the let $x=1/(2\pi n)$, now $\sin(1/x)=0$

Shaun

Please would you answer my question on the main site, @AkivaWeinberger?

Akiva Weinberger

6:00 PM

Sure, in a bit

Ted Shifrin

OK, seems right. I haven't thought about this stuff in 50 years.

Akiva Weinberger

Once you get used to it, it's basically standard analysis with new vocabulary

Like, it works for the same reason we can find a decreasing sequence of reals $x_n\to0$ with $\sin(1/x_n)=0$

Ted Shifrin

Are there limit cardinals in the infinite hyperintegers?

Akiva Weinberger

I think you can't really embed the cardinals in the hyperintegers?

Regardless, every hyperinteger $n$ has a predecessor $n-1$

I think the order type is something like continuum many copies of the integers

Ted Shifrin

If you have multiple copies of the integers, there must be ones you get not just by adding $1$ to a predecessor.

Akiva Weinberger

6:06 PM

? Consider the order type of just two copies of the integers

$\dotsb<-2<-1<0<1<2<\dotsb<-2'<-1'<0'<1'<2'<\dotsb$

That's the order type of $\Bbb Z\times2$

Every element has a predecessor

To embed that in the hyperintegers, choose some infinite $n$, and then write $n-2,n-1,n,n+1,n+2$ etc for the second copy

Ted Shifrin

So what is the limit of $n_n$?

Akiva Weinberger

What's the subscript?

Ted Shifrin

which copy of the integers you're using

There are good reasons I never went very far with set theory or model theory. :)

Akiva Weinberger

There isn't a maximum copy

This is also kinda like a nonstandard model of arithmetic

Sorry, not kinda - this is a nonstandard model of arithmetic

Ted Shifrin

Somehow this feels like it should sit inside what Munkres calls $S_\Omega$ (the first uncountable ordinal).

Akiva Weinberger

6:11 PM

so we have things like $\lfloor\sqrt n\rfloor$ and $\lfloor n/2\rfloor$ and $n^2$ and $n^n$ and tetration and all that jazz

Hm, I learned that as $\omega_1$

There may be a way to relate them but I don't know it

The hyperreal are a quotient of $\Bbb R^{\Bbb N}$, anyway

Ted Shifrin

Demonic @Alessandro would laugh at my ignorance :)

Akiva Weinberger

$\Bbb R^{\Bbb N}/{\sim}$ where $\sim$ is an ultrafilter

I think that's the best way to think about it. If $n$ is the equivalence class of $(0,1,2,3,\dots)$, then $\lfloor n/2\rfloor$ is the equivalence class of $(0,0,1,1,2,\dots)$

A hyperinteger is a hyperreal that's the equivalence class of a sequence of integers

The ultrafilter properties are basically exactly what you'd want for this to work. For example, $(0,1,0,0,1,0,\dots)$ has to equal either $0$ or $1$ (by which I mean it has to be equivalent to either $(0,0,0,\dots)$ or $(1,1,1,\dots)$)

Ted Shifrin

OK, we're now exceeding my safety in the deep end of the pool.

Akiva Weinberger

so given any two subsets of the integers, the ultrafilter picks one and discards it

I should say "nonprincipal ultrafilter"

A principal ultrafilter would simply pick one of the components and say two sequences are equivalent if they agree on that component. Quotienting by this would give the usual reals

robjohn

6:54 PM

@leslietownes we had a lot of coyotes from January through March (breeding time), but the number of sightings has gone down dramatically since then. Now we have lots of bunny sightings.

Alessandro Codenotti

8:03 PM

@TedShifrin nah I would never. But Munkres's notation is nonstandard, I remember being annoyed by it before

leslie townes

robjohn: that's cute. we were still seeing bunnies here up until about a week and a half ago. but we've been seeing the coyote the whole time, so shrug

there's a large warren of thick bushes nearby that i think they live in, and if they can flee in there, a coyote wouldn't get 'em.

Semiclassical

i'm in need of a sanity check on a calculation i did yesterday

Suppose I pick a random point on the sphere. What is the distribution of z-coordinates?

i keep swinging back and forth between "it's a uniform random variable" and "that doesn't make sense"

leslie townes

how are you picking the point? e.g. uniformly?

Semiclassical

yes, uniformly on the sphere

leslie townes

(that will not result in a uniform distribution of z-coordinates)

Semiclassical

8:18 PM

hmm

7

Q: $x$-coordinate distribution on the $n$-sphere

Can we express the distribution of a coordinate of the $n$-sphere in any known distribution? In formal terms, consider $S^n = \{x\in\mathbb{R}^{n + 1}: \|x\|=1\}$ (i.e. the usual $n$-sphere). If we sample $x$ uniformly from $S^n$ what is the distribution of $x_1$? By "sampling uniformly" I mea...

probability-distributions projection

leslie townes

if it's the unit sphere, think about the area of the sphere contained in 0.9 < z < 1 vs. the area of the sphere contained in 0.0 < z < 0.1

there used to be a sci.math thread or faq or something, directed toward algorithmically generating uniform points on the sphere, that might have answered this question in as good detail or better than that

if indirectly (since its goal was not to answer that question)

Semiclassical

if i go from polar angle $\theta=\alpha$ to $\beta$, the area would be $2\pi \int_\alpha^\beta \sin\theta\,d\theta = 2\pi(\cos\beta-\cos\alpha)$

aka the area contained only depends on the vertical width, no?

aka you can write $\sin\theta\,d\theta=-d\cos\theta$

leslie townes

oh, i see.

sure.

Semiclassical

kk. i'm trying to compare the very simple answer i got (for a larger question) with another person's solution which comes out very complicated

but i think this reassures me that this part of it is at least not too strange

leslie townes

no, it's not.

Semiclassical

8:23 PM

(the bigger problem is not entirely clear on what it's asking for so it might just be a matter of different interpretations)

leslie townes

i can't visualize for shit, so i'm terrible at these things. sanity checks always needed.

Semiclassical

same. the post there says "oh it's just Archimedes" but i don't trust myself on that

leslie townes

one time in grad school we had a problem set with a slightly unclear problem, where the ambiguity was very difficult to notice. i think everyone in the class got slightly different solutions.

Semiclassical

oof

leslie townes

it was supposedly from physics (i have no way of knowing if this is true, but it feels right)

Semiclassical

8:25 PM

a slightly weird problem i ran into a week ago was along these ilnes

a common problem is to find the ground state of a hydrogen-like atom. the equation for that, modulo constants, is $-u''(r)+(1/r) u(r)=\epsilon u(r)$

small adjustment: replace $1/r$ with $2/r$. then the solution with smallest $\epsilon$ is given by $u(r)=re^{-r}$ with $\epsilon=-1$

the problem of interest here was to perturb that ODE with a term $f(r)$ which vanishes for $r<R$

the weird part here is: what exactly is the perturbation parameter?

the only one that really makes sense here is $R$ itself, i.e., an expansion in powers of $R$

but usually you can write the preturbation as $\lambda \cdot V(r)$ where $V(r)$ is order-1 and $\lambda$ is small. That description doesn't seem to work here

so the usual description of perturbation theory doesn't really make sense here

bleh, i have a typo above

leslie townes

8:54 PM

well, that's why the physics people make the big bucks

robjohn

@leslietownes That wouldn't be an unclear nuclear problem, would it?

leslie townes

haha, it might have been

robjohn

So, it's uncertain.

leslie townes

the ambiguity was something like, it was asking for coefficients of something in a basis, but not posed that way. and depending on how you read the problem there were at least two bases worth choosing, or you could come up with basis-free formulations of the problem that were much harder

it probably had something to do with nuclei

everything kind of does, after all

Ted Shifrin

9:12 PM

@Semiclassical Your very own version of the Bertrand paradox.

Semiclassical

lol

Ted Shifrin

@leslietownes You have this wrong. The areas are equal. How does that address the issue?

The area of a zone of a sphere depends only on its height, not on its location.

leslie townes

ted: you'll see that i realized that a moment later

Ted Shifrin

You can't expect me to read everything!

leslie townes

you have a knack for locating all of the wrong stuff i say

Semiclassical

9:18 PM

i repeat, lol

Ted Shifrin

@Semiclassic So are we sampling on a circle with random $x$ value or with random $\theta$ value?

leslie townes

i mean, there's enough of it

Semiclassical

theta

Ted Shifrin

There's so much, @leslie, that I have a huge probability of stumbling into it.

Semiclassical

wait

it's on the sphere, so i'm not sure talking about a circle is going to give the right resutl

Ted Shifrin

9:19 PM

I'm asking by analogy.

You mean uniform distribution with respect to the area measure of the sphere.

Semiclassical

yeah

and that seems to be equivalent to random $x$ value

Ted Shifrin

Careful. The area of a zone thing I'm talking about works only on the $2$-sphere, not in higher dimensions.

Semiclassical

yeah, i'm only worrying about the 2-sphere

Ted Shifrin

OK. Then, yeah.

leslie townes

semi: famous last words.

Semiclassical

9:21 PM

which makes me wonder about the source i'm comparing with. they only worry about the angle $\theta$, and while that's not exactly wrong, it does make me worry they've managed to get the wrong distribution

Ted Shifrin

Yeah, you're looking at $\cos\theta$.

Semiclassical

right

they work in terms of $d\Omega$ so I think they're safe

Ted Shifrin

Solid angle?

Semiclassical

(though the physics habit of writing expressions like $d \sigma/d\Omega$ always makes me paranoid)

yeah

differential cross-sections in scattering always make my eyebrow twitch

roughly speaking, i have a function $F=A+B\cos\theta$, where $\theta$ is the scattering angle of some process, and we're told that every scattering direction is equally ilkely

so I think that means that $F$ should be uniformly distributed over $[A-B,A+B]$

the solution i'm looking at gets a much worse answer by a more circuitous route

Ted Shifrin

Not enough people know the zonal property I enunciated. I first had that proved to me in high school geometry and I never forgot it.

Semiclassical

9:27 PM

nice

it's clear enough from the solid angle formula

leslie townes

we barely got to the 1-sphere in high school geometry

Ted Shifrin

Well, I think almost all our course was about polygons and circles.

But our teacher did talk about the Archimedes limiting stuff for area/volume formulas.

leslie townes

our teacher mostly talked about the last week's volleyball game (that she coached)

she also really, really wanted to be friends with the girls in class that she regarded as popular

we did get triangles and parallelograms and i think a bit of lines and circles

Ted Shifrin

My teacher had a Ph.D. (in mathematical logic) for what it was worth. She was not a good teacher — she punished me for my curiosity and natural questions. She also chose to stop teaching the calculus course because she wanted to teach students who needed her (laudable ... but ...) and then told one of those "regular" classes that she didn't expect them to understand X because they weren't honors.

leslie townes

we call this "ph.d. syndrome"

Ted Shifrin

9:32 PM

That made me livid. It's OK to tell students stuff is hard and it's ok to say that you don't have to understand it completely. It's not OK to say "you're too stupid for this."

Semiclassical

ugh

Ted Shifrin

Even as a high school senior, I was enough of a natural teacher to know she was full of s**t.

leslie townes

i had a number of unpleasant teachers in hs, but none who were academically pompous. they'd make you hate them for reasons unrelated to your abilities or performance in the subject matter

Ted Shifrin

Well, most of our math teachers were dumb as rocks and coaches. The guy that took over the calculus class — whom I'd had junior year for math analysis — was a nice guy, but he literally had never had calculus in college many years earlier. So that was a disaster. He dismissed me and two other people and we did it on our own. Ultimately, I came back and taught the class for the last month or so.

This was 1969-70, before they had AP classes, but I took the AP BC test. It even had a $\delta$-$\epsilon$ proof on it.

I don't remember doing any practice tests or anything. Those were the days before "civilization."

Colourful Spacetime

my math teacher in hs told me that I shouldn't go to uni cause I wasn't good enough apparently

Ted Shifrin

9:43 PM

Did that motivate you to prove him/her wrong ?

leslie townes

sounds like a delightful person

Ted Shifrin

I did suggest to a couple of students in my 36+ yrs of teaching that perhaps math wasn’t for them .

Colourful Spacetime

It did and it still does motivate me, although this is clearly not the reason why I keep going

Nah that's fine, my gf for example is quite incompetent in math

leslie townes

it also seems different in college, where students have more time and room to plan a course of study. there are many options and figuring out what works is part of it.

Ted Shifrin

Probably because of years of bad teachers.

Colourful Spacetime

9:45 PM

He just didn't like me because I wasn't paying attention to his class, never did homework and still aced the exams ;)

I had some wonderful teachers, he was an exception

Ted Shifrin

Shame on him for writing exams like that!

Colourful Spacetime

😂

Ted Shifrin

I did get annoyed with a very bright student who skipped almost all my Calc w Theory classes and did well. I didn’t ban him from office hours, but I should have. When he got to diff geo a few years later, I warned him and it didn’t go well.

leslie townes

i think hs teachers obviously should help students meet/exceed graduation requirements with a view to what they might do next, but it isn't their place to decide what those next steps are or aren't going to be. if someone is underperforming that is an advising issue for planning next hs steps, not a "the door to X future is closed" issue.

and obviously fine for hs teachers to tell kids that what they're getting away with in hs might not work in college. we heard a lot of that.

although it does raise the question of how they're able to 'get away with it'.

i guess now a lot of formulaic curricula can take the blame and not the teacher.

Colourful Spacetime

I wonder, why would a professor care if a student attends his/her classes? I never quite understood that.

leslie townes

9:52 PM

in courses without a lot of check-in mid semester (or where students can copy assignments freely with one another), not attending classes can make it difficult to spot disasters before they happen. which can waste a lot of the student's own later time and money, even if they're at fault.

this may be US-specific. in other jurisdictions it might not waste much of the student's money.

Colourful Spacetime

This is probably US-specific indeed, since in UK and Netherlands no one really cares what you are doing.

It's different of course for laboratory classes

leslie townes

i would also distinguish between expressly penalizing students on the basis of attendance (which i think is counterproductive, although is not uncommon in the US) vs. attempting to intervene in nonattendance to avoid potential student disasters (which ought to be done).

some students just drop off the face of the earth, get an F on the final, and you get an email "can you regrade that? i need at least a D to graduate." and taking another unexpected term means another X months of unplanned rent + enrollment expenses, or interfering with some job. well, whoops.

best to catch that stuff early, but you can't always catch it if they aren't there.

i wonder how all of this worked out with remote learning during the pandemic. my wife has horror stories about this.

Colourful Spacetime

Ok this has to be US-specific... XD

Let me give you my point of view

Ted Shifrin

@ColourfulSpacetime I taught classes with a lot of student interaction. Not at all the European-style lecture.

leslie townes

i can guess. europeans who come to the US often say, why is there homework. why is there attendance. if you can't do the exams, then go away. there are a million competing explanations for why that hasn't been the system in the US.

Colourful Spacetime

10:00 PM

Whenever I attend a class, either the lecture was "easy" as hell and I wasted a lot of time, or there were crucial points which I didn't understand and the professor just skipped it, the result was to get lost and not understand anything from that point on

The most productive thing for me was to just self-study, so I never set foot on lectures ever again

leslie townes

similar with grade inflation. one of the competing explanations, as hinted above, is the relative expense of college in the US. there might be others.

in the 60s being in college could mean not getting sent to vietnam. some people locate the beginning of a lot of what we might think of as an intensified paternalism to that era

Colourful Spacetime

do you europeans say "if you can't do the exams, then go away"? I got confused if that's your opinion or that's the "europe's" opinion

@TedShifrin I see I see

leslie townes

colorful: i'm exaggerating. the idea i get from some europeans is "why does it matter if the student doesn't understand the material, they're the only one affected by it."

if people in my classes were doing fine on midterms and such i didn't care where they were, but if people would not show up at lecture and get middling grades, i'd usually schedule at least one meeting and say, hey, you're on track to get X grade, is that consistent with your expectations, and would a lower grade throw them off,

and if not, attending lecture was always part of the recommended solution

Colourful Spacetime

Imo exam is the most important factor, and indeed if you can't solve the exam then there is something wrong.

Ted Shifrin

I also spent 7-10 office hours a week with a largely full office, helping students. Also not the European model (or the US model at most research universities).

Exams are just to check you’ve learned the basics, not to push the challenging boundaries. So they were just a part of my courses.

Colourful Spacetime

10:06 PM

@TedShifrin Very well explained, I will agree with that

But I would say that the exam should also consist of a "hard problem" so that you can differentiate the excellent students from the average

Ted Shifrin

That’s why weekly homework is there, with varying levels of problems.

On a 3-hour final there might be time for one problem for the A students, but you have to put enough for the FDCB students to do!

Colourful Spacetime

Hmm I see, idk how much the homework is worth, but in my universities it's about 20%. We have the choice of "exam counts for 100%" or exam counts for "80%" (+ the 20% homework)

leslie townes

interesting game theory. do they have to make that choice before the exams?

Colourful Spacetime

That's true, usually this "hard problem" is worth like 10-20%

leslie townes

:)

Colourful Spacetime

10:11 PM

no, after the exam

You can choose which one favors you more

leslie townes

or perhaps a formula in a spreadsheet can make it for them

Ted Shifrin

In real math courses, I counted homework 35 percent or so.

leslie townes

i sometimes used formulas where a lowest midterm exam score could sometimes be adjusted up if there was improvement on the final, but HW was always a pretty big chunk.

for people who don't find HW annoying, they really like it, because it's a piece of the grade they have the most control over

Colourful Spacetime

I really like homework because it provides me with a lot of problems. But I consider it unfair to have it to count towards your final grade without having the choice not to. My argument is that, for some students (like me), it's not always possible to submit the homework on time, and we have better lack to show our skills long term, i.e. in the final exam, not weekly.

I have a serious case of OCD and sometimes I can't study for like 2 weeks

And then for 1 week I will study 12 hours per day to cover everything

By the time I've caught up, I've missed like 3 homeworks

And that happens multiple times per semester

leslie townes

yeah. one thing i noticed a bit when teaching was that substantial homework could be very tough on people who had a lot of other stuff going on, but easy for people who didn't. it wasn't always measuring what you'd want it to measure.

someone might be doing well on the homework just because that class is the only 'real' class they are taking that semester, even if they are spending a worryingly long time to do it. and busy people could miss assignments or not just get stuff done, irrespective of understanding.

it's hard.

similar thing with exams, too, limited time vs. take home. can be really unfair depending on someone's final exam calendar, which at most schools is random.

Colourful Spacetime

10:20 PM

Yep..

leslie townes

the way to solve this at the upper levels is just to give everyone As and have grades not matter. but you gotta do something to filter and provide feedback at the lower levels. at least i think so.

some of my wife's friends think that colleges shouldn't give grades. it might be easier to do that in a field that isn't math or physics or something that needs a lot of prerequisites to jump into.

i don't have any conception of what understanding makes a C in a history class vs. an A in a history class, for example. but with any of the sciences or engineering, i can form a pretty good guess.

Colourful Spacetime

Somewhat unrelated, I wanted to ask experienced mathematicians if the following is considered "normal": I don't know basic linear algebra (doing calculations) but I know "advanced" algebra (proving theorems etc). For example i just started learning Lie Algebras and I'm doing fine, but I'm not familiar with properties of matrices, eigenvalues etc

And I think it's important to note that I'm a theoretical physicist

leslie townes

that seems pretty normal to me. if the 'calculations' you don't know are things like matrix diagonalization, eigenvalues, whatever. you can do a ton of stuff without the calculational side coming into play.

and indeed in physics sometimes it doesn't even help

it was pretty common in my math grad program for people never to have gotten the 'calculational' version of an elementary subject because they never had to take it in college.

Colourful Spacetime

Ahh ok!

That makes me feel a bit better

I gotta say though, I really didn't like the undergraduate courses in UK

For physicists

The math we did were a joke

Compared to what they do e.g. in Greece

Or netherlands

leslie townes

10:37 PM

one time my officemate asked me something, and it was basically, what is the rank of some 4x4 matrix. and i said "can't you just compute it?" and he didn't know how to do it from the matrix. he was trying to realize it as the dimension of something that he understood better. i'm not kidding. he knew what the rank was conceptually, but had never learned the algorithm, row operations, anything.

he picked it up very quickly, of course, and would have had to teach it if he had been a TA in the right class. but he hadn't done that.

colorful: i wonder if that depends at least somewhat on the school? my impression of physics undergrad in the US is that it varies widely in mathematical sophistication, sometimes depending on the interests of the faculty or size of the department. even at research schools.

Colourful Spacetime

Hmm.. maybe, I just have experience from one uni. But it's a joke in Greece: "Do a master's in UK for the fun"

leslie townes

i wonder if it's some vestige of an earlier stereotypic aristocratic tradition of how it's ungentlemanly to be too good at something.

i read somewhere that the PhD degree was regarded in the UK as a continental affectation for a really long time

Colourful Spacetime

I hadn't thought about that, quite interesting!

On the other hand, netherlands is considered very hard for math masters (compared to greece)

Idk if all of that is true, that's just what I've heard from people

Anyways, I gotta go to sleep. It was a pleasure to meet you both! @leslietownes @TedShifrin

leslie townes

this would have been a while ago. like the hardy-littlewood era. i think both of them only had masters degrees, and this was not unusual for that time.

cheers!

Ted Shifrin

10:55 PM

Cheers

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$Mathematics$ Mathematics