> OOF last GRE Math test of the year. I knew almost everything this time around, but I'm convinced that acing it takes a few months of just drilling speed

from twitter^

a few very torturous months :-/

the JEE stresses speed also

we have kernel symbol as \$\ker\$, but is there one for left kernel?

left kernel?

Just looked up the GRE on wikipedia, do you really have to pay 200$ to do it?!

yes it's horrible

left kernel is just a word used in one of my book, and I don't know what you meant

What is this left kernel?

it's defined as $\ker(A^T)$

@user2646 Some research I read years ago

I could try to find it, but I can't really be bothered

@AlessandroCodenotti sorry i didn't mention I'm asking about linear algebra

you should either already know or soon learn that is the same as $\text{im}(A)^\bot$

@MikeMiller I've forgotten some idea of them, but yeah, I know they're perpendicular since it can be seen as inner product when doing matrix multiplication

@AlessandroCodenotti ETS is a racket that has somehow placed themselves as essential to the application process at every level

actually academia in america is a racket

@IsanaYashiro that's a better symbol! but there's no reason not to just write $\text{ker}(A^T)$

no need to invent new notation

$test$

@IsanaYashiro Why is this not rendering

I thought my ChatJax was broken lol

@MikeMiller I'm sorry, I didn't invent new symbol, it's from my book I'm reading

Tbh the ETS probably bribes admissions committees to care about them

@IsanaYashiro I'm telling you not to invent new notation for what your book calls the "left kernel", given that it is already captured by the notation $\text{ker}(A^T)$.

@MikeMiller tl;dr america is a racket dont @ me capitalists

There is no reason to write something like $\text{lker}(A)$ - more or less the same number of keystrokes

academia is a racket in general i guess, not just US

(sorry for jumping in lol)

i dunno it elsewhere

it doesn't have to be a racket but it is and will remain until the profit motive is gone

profit? I guess you are referring to ETS, but that only applies to pre-doctoral level of academia i guess

@MikeMiller ok, maybe my english is bad but I meant: both Lker(A) and ker(A^T) are used in my book...

I had never heard of ETS before, but wiki says 580k people took the GRE in 2016, that's more 100 million in dollars, what are they doing with all that money?

they are a private for-profit company

they make money and.. pay people?

haha

@AlessandroCodenotti for party~ PARTYPARTYPARTY

even if you are in Europe there's IELTS for non english speakers

I mean sure they have costs associated with doing the exam for all that people, but it looks like a pretty big profit margin from back of the envelope estimates

Phew. That was a bit of a doozy, but I think I did pretty damned well.

certainly yeah

@IsanaYashiro What I mean is if they are the same thing, there is no reason to use the first.

That notation seems pretty isolated to your book.

@Pig yeah, luckily I avoided paying for an English exam because the university here in Bonn accepts more or less any proof of English proficiency even if it's not a standardised test

You're right, because it's not written in english, so sad. garbage book

@Pig "people" = administrators mostly

who knows if administrators are people

that's why I always download e-book written in english for clearification

@IsanaYashiro well, there are many fantastic non-english books. I just think this choice of notation on their part is misguided

lmao @MikeMiller I like how you say things

@AlessandroCodenotti you are lucky, i heard people applying to US schools, who were asked to take TOEFL even if they have studied undergrad in US and actually TA-ed in US

and was told their application won't be considered unless they have that on file

that comes back to mike miller's point i guess: who knows if admins are people

@Pig That makes no sense

the world doesn't make sense

@Mike did I tell you I applied for a PhD?

nope

Which you may or may not find a smart idea

I applied for a PhD

cool

where? doing what?

i guess you're finishing up your masters now

I did an English B2 exam as part of the requirements for my bachelor (organised by the university, not an officially recognised one) and that was enough for the people in Bonn

bonn is chill i guess

@MikeMiller I graduated in december and am doing consultancy now

But the PhD is in Amsterdam doing quantum cryptanalysis

what even is consultancy

I'm trying to solve the following problem "Show that if $B$ is infinite and $|B|\leq|A|$ then $|^AB|=|^A2|$". I don't really know what the notation $|^AB|$ means.

You hire out your brain for money, on specific topics

@oskar neither do I

peterbraam.com this is krijn basically

I made kaas.gq as joke

@Krijn: Okay. @AlessandroCodenotti: Do you know?

@AlessandroCodenotti from ur writing i wouldnt even know ur not a native speaker so like ud probably have been good regardless

@OskarTegby Are you sure you have the right notation? Is it something to do with maps from one set to antohter?

and $2$ denotes the $2$-element set

Mike ninjas me all up again

@OskarTegby Functions $A\to B$

i had em backwards

Thanks, @AlessandroCodenotti. Okay. So let's see why this should be the case.

@EricSilva thanks! To be honest I wasn't worried by the exam itself, but I would have been extremely annoyed if I had to pay 200€ for it

welcome to my world dude

@OskarTegby was this a subtle way to get people to think about your problem? by first asking about notation and then keeping them there?

@MikeMiller: No.

I try to never try things. Wait! Whoah...

Some people prefer to write $^AB$ instead of $B^A$ so that when $A$ and $B$ are cardinals there are distinct notations for exponentiation and sets of functions. I don't know any situation in which this ambiguity could actually be an issue though

Okay.

@AlessandroCodenotti I always thought that was the point of the latter notation?

Isn't it a bit ambiguous to say that something is infinite? We just have a lower limit for that, right?

Hey!! It holds that $[\mathbb{Q}(\sqrt{2})(\sqrt{5}):\mathbb{Q}(\sqrt{2})]=\deg (x^2-5)=2$ and $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=\deg (x^2-2)=2$, or not?

@OskarTegby It's not at all ambiguous right?

@Krijn: Well, is it $\omega$ or $2^\omega$ or something larger?

@MaryStar Yes

Ok, thanks!! @Krijn

@Krijn Officially $\kappa^\lambda$ is some cardinal which is distinct as a set from the set of functions $\lambda\to\kappa$. But there is of course a bijection between $\kappa^\lambda$ and the set of functions $\lambda\to\kappa$

@OskarTegby It leaves options open, but it's not ambigious

@AlessandroCodenotti you mean formally, not officially

Or maybe I have a wrong understanding of ambiguous, I'm not a native speaker

I think that it means that something can be interpreted in several ways.

Do you think finite is ambiguous as well?

I guess.

@OskarTegby ok, what are the ways?

Haha. I mean, there are different finite numbers.

It's not ambiguous, it's perfectly well defined, there are just many objects that satisfy the definition

I mean "group" is not ambiguous just because there at least 2 nonisomorphic groups

Can anyone point me in the right direction for a general history of mathematics

I'm surprised that logic is so late in the game, how did people do proofs and such for so many years before that?

@MikeMiller I think about $1^A$ to be sure I get the order right

If by logic you mean stuff like "A implies B" and such, that goes back way earlier than mathematical logic

You can do a whole degree in math without ever taking a course in mathematical logic but that doesn't prevent you from writing proofs

Even within a degree of mathematics, logic is usually a second year course or so right?

Yeah

But at some point don't we have to agree on the foundations in order to accept a proof

Given e^(xy) + ye^y = 1 and asked to find dy/dx, I'm coming up with - (ye^(xy))/(xe^(xy) + ye^y), but the answer is supposed to be - (ye^(xy))/(xe^(xy) + ye^y + e^y). The formula is dy/dx = - (∂f/∂x)/(∂f/∂y). Any idea what my stupid mistake is?

Dunno, there wasn't even a proper course in mathematical logic in the undergrad curriculum at my uni

@Semiclassical how so

hot take logic sucks fight me nerds

I mean, the concept of syllogism in logic goes back to at least Aristotle

@EricSilva all of mathematics is a li'l bit circular in that eventually you need a metamathematics to talk about mathematical logic but that's ok

Is it true that $2^{|A|}=|^A2|$?

Logic is ok, model theory and set theory are the real thing

Lol logic is probably the one class I sorta regret not taking the most

Eric Silva is the Elon Musk of this chat room

@Krijn: What do you mean?

Elon Musk has been shitposting on his Twitter account, lately

a proof is ultimately just a logically sound argument

@Semiclassical isn't that more philosophy though

and arguments have been around since forever

@Krijn uhhhhhhh please don't be rude to eric

Valid and sound having a different meaning there compared to mathematical logic

eric's shitposting is quite different than your boy elon

pls dont compare me to a scumbag billionare thx

Isn't Elon a good guy?

no

He wants to take us all to Mars and drive electric cars and shit.

Nah, he wants to feed his own ego and bring people who can pay mad money to Mars.

he's a union-busting shitheel who is the grand wizard of iamverysmart speech

That too.

@Fargle @Eric Demonark: Did you all survive?

Ohai Ted

ya im good my man

I'm dead AMA

@Ted Without meaning to be too confident, I think I thrived.

ohai @Krijn

Rehi Ted

Rehi demonic

Demonark: What does AMA mean?

@Daminark Do u like fries?

@Krijn they are pretty great I must admit

Great @Eric @Fargle

@TedShifrin Ask Me Anything

oh

<--- one century behind the lingo

@Daminark What do you think about the Riemann Hypothesis\

Just seems like we did math for thousands of years without ever having fully defined our terms / foundations

@Daminark Would you rather fight 100 duck-sized horses or 1 horse-sized duck?

@user525966 I don't think the foundations really made 1 + 1 = 2 any more valid

@user10478 In case no one has answered your question yet, you screwed up the product rule.

It's was probably fine, I didn't really do much studying this round with classes and whatnot so I could very much see myself doing worse and relying of the first test, but eh

1 + 1 + 1 + 1 = 4 time rotations per day, you are educated stupid.

@Fargle definitely the duck-sized horses

i feel like anyone who says a horse sized duck is an idiot the duck would sit on u and ud just be gone

@Daminark I assume this is a response to "what do you think about the riemann hypothesis"

A horse-sized duck would not make very good Peking duck. Tooooo tough.

@Fargle Is this a timecube.2enp.com reference?

@Krijn Indeed.

the duck-sized horses would kick your asses too though you're gone no matter what

your only hope is to run

Have you ever fought a horse?

@MikeMiller very perceptive

horse anatomy isnt built to work on the small scale tho

@Eric: Now that you're ready to be a mathematician again, welcome back to GH.

i assume the molecules are all smaller too

@EricSilva Is duck anatomy built to work on the large scale?

@Krijn also no

@TedShifrin hell yeah

@TedShifrin Ohhh I was trying to see my error in the 1 on the RHS, thanks

that's what im gonna do as soon as i clear the gunk out of my brain

Lucky for you, GH stands for Gunk Helper.

:0

Ducks are the only species that we have observed doing necrophilic homosexytimes

Riemann HypotheSis

@Eric: To help ungunk, perhaps you can help me answer this guy who's pestering me in the comments. Surely missing the $|s|$ on the RHS of the estimate should violate differentiability.

@Krijn Surely we've observed humans doing that

Okay but the point of the story was going to be that the observer got the Ig Nobel prize for that

Speak for yourself, @Mike

And I spoke to him

And I thought that was pretty cool

lol there was a gre question like this

Oh yeah?

For a hot second I thought the GRE question was about the duck necrophilia and I was like wuuuut

not really but there was a x^2sin boi

Yeah I remember that one.

I hope I got it right

D:

idr what it was just that it happened

Oh, that boi is totally standard. This is a really interesting question, actually.

ye

hot take, $\sqrt {~\cdot} : \Bbb C \to \Bbb C$ doesn't make sense because $\operatorname{char}(\Bbb C) \ne 2$

@Fargle: You can look at that multivariable analysis question too :P

these like conditions for differentiability are way trickier than they seem

WTF @Leaky?

@Eric: I can prove that (1) and (2) imply differentiability and conversely. But I can't see what's wrong with the commenter's "proof" of Dieudonné's remark ...

@LeakyNun your argument doesn't hold for $\mathbb{R}$

the trick is to only pay attention to him when he says something interesting

@EricSilva Did you get DNE for that one?

@MikeM: I should just ignore him.

@Fargle: What was the Q, approximately?

@TedShifrin if $\operatorname{char}(F) = p$ and $F$ is algebraically complete, then $x \mapsto x^{1/p}$ is well-defined everywhere

in $\Bbb C$ you would need branch cuts

Careful that the ETS ninjas don't come swooping in

I said approximately, Demonark :P

@Leaky: You're wrong (again). You can give a well-defined square root globally.

it doesnt make sense to not have the |s| to me

uhh h,

hm

@TedShifrin but it won't be continuous, and there's more than one choice

@Fargle idr what the question was so idk

@TedShifrin It asked for whether some thing that involved an $x^2 \sin \frac{1}{x}$ term was differentiable at $0$, or maybe it was about multiply differentiable? I don't remember

@Leaky: You didn't tell me unique well-defined ....

He's just saying that the Frobenius map is an isomorphism in characteristic $p$.

Yes, I know :) perfect field and all.

That is the content of this discussion. Thanks folks have a good day

was it something about it being C^1

cuz it isnt

Whatever it was, there ended up being some constant multiple of $\sin \frac{1}{x}$ in my derivative

It's once-, but not twice-differentiable.

hi @loch

hi @LeakyNun

@MikeMiller Thanks

I just gave a generalization of this to my AoPS kids last week. (We just learned the chain rule.)

I don't think I have it exactly right in my memory unfortunately, but there was some way that it obviously failed

wasnt it one of those which are true bois

i hate functions that don't have an at least decent function space

@TedShifrin Wait why is it once-differentiable at 0? Seems like you'd get $2cx \sin \frac{1}{x} - \cos\frac{1}{x}$

u have to do

limit defn

Bad bad @Fargle.

the derivative isnt continuous

Any time you have a special definition at $x=a$ you need to use the limit definition.

frick

Typical engineers think $f'(a) = \lim\limits_{x\to a} f'(x)$ is the way you do it. I saw a UGA lecturer do that in calculus class. I threw a fit afterwards.

just kill me then tbh

2018 calculus is a travesty

there are 65 other probs my dude

Like, I know that, I'm just dumb

Yeah, but I have quite a penchant for simple errors. It's, like, my thing

@TedShifrin UGA?

i literally looked up examples of artinian rings like a couple days before incidentally and i was pumped when i saw it show up

oh gosh I hope I didn't get that one wrong too

You're kidding me — artinian rings showed up?

like II was obviously non-artinian but

finite dimensional algebras over a field my man

@LeakyNun The least known of the Grothendieck's books

They defined the term, @Ted

Oh.

@Leaky: What's your point?

@TedShifrin what is UGA?

it was one of those "definition which of the following are definition"

It's where I was a prof for 34 years, Leaky.

Artinian rings are weird. Noetherian sounds like a strong property, and Artinian looks like it should be about as strong while it's actually much stronger

@AlessandroCodenotti Ultramoderne géométrie algébrique

an artinian integral domain is a field!

i will keep this fact in my head for too long

thanks algebra qual

@MikeMiller You wikipedia'd as well?

no

oh

@MikeMiller lol

@MikeMiller Oh, that's actually surprising!

$(x^n)$ is a descending sequence that must terminate, so $x^{n+1} = ax^n$, so $(1 - ax)x^n = 0$; because $x$ was nonzero, $x^n$ is nonzero (integral domain) and so $1 = ax$, so $x$ has an inverse

@Fargle III was finite dimensional wasnt it

idr what it was

@MikeMiller neat

@EricSilva yeah it was quotient of real polys by $(X^3)$

being finite dimensional obviously kills the ability for a sequence to go on

corollary i mentioned here recently that i will also now never forget: if $K/k$ is a nontrivial finite field extension $K \otimes_k K$ is not an integral domain. for it's artinian, and it is not a field: the product on $K$ induces a homomorphism of fields $K \otimes_k K \to K$, which is nonsense because if $[K:k] = d$, this would imply $d^2 \leq d$

That was my thinking but I couldn't justify it

in retrospect it's obvious though

ya

ideals are vector subpaces!

@Mike yeah that's how I ruled out one of the GRE things because it came up on an algebra pset

@Fargle the q you never completed :P

@Eric @Fargle: Can you guys give me a continuous function $f(x,y)$ with $\partial f/\partial y$ continuous near the origin and $\partial f/\partial x(0,0)$ non-existent?

i can't because i won't spend 15 seconds thinking about that for the rest of my life B)

hi @Piggy

Notice I did not ask you, Mike.

@TedShifrin :(

hi @TedShifrin

hmmm

Oh, wait. That's too easy.

Just have a function only of x

:P

I want the estimate $|f(h,k)-f(h,0)-f(0,k)+f(0,0)|\le\epsilon(|h|+|k|)$ to fail.

I remember there was a problem in Rudin chapter 9 that involved this trick somehow

In addition.

Demonark: Any function of the form $g(x)+h(y)$ will do it, with $g$ suitably bad.

But for that function, the estimate I just wrote down is trivial.

@TedShifrin i assumed this is what you meant

I thought it should be easy, but so far I don't have one.

Products $g(x)h(y)$ seem not to work, either. So it needs to be something intermeshing the variables.

OK, I have one. And a product does work.

wut is it

So what goes wrong with that guy's "proof"?

$f(x,y) = \sqrt{|x|}\cos y$

$x$ non-Lipschitz was my idea

im a lil confused about the control of |D_y (a + s, y_1) - D_y(a, b)|

Hi all.

i dont see why the y partial should have nice controls

to get the kind of lipschitz-y kind of estimate he wants

He's just using the MVT and using uniform continuity to get a bound on the difference of the $y$-partials. But I don't think you get $x$ control. The $y$ pulls out with the $\Delta y$ by MVT.

heya @CaptainAmerica

I did finally see your ping. I believe the answer is not 1/2.

@TedShifrin I knew it :(

Back to the drawing board. I refuse to give up!

Did you actually do the summation explicitly (or turn it into an integral)?

ya i meant to say i dont see why the y partials should have any nice control wrt x

@Ted Here's a nice one. I gave it to Balarka a while back, he got the same solution I did. I only know of one approach.

@Eric: I just posted my function for him to work on :)

word

all right im gonna take a nap before doing any more math today

see y'all

Night :)

@TedShifrin I was trying to do it explicitly. I messed up somewhere.

cya

Let $\lambda_n$ enumerate the positive solutions to $\tan x = x$. Show $$\sum \frac 1{\lambda_n^2} = \frac 1{10}.$$

? @Mike

Show sum + 1/10

@CaptainAmerica: So with $A=(1,0)$ and $B=(\cos\theta,\sin\theta)$, you had the probability that $C$ works is $\theta/(2\pi)$.

Ok Mike.

Ah.

Oops!

Oh, that's interesting.

what are you covering in your AoPS class these days? @TedShifrin

@TedShifrin Yes.

Particularly interesting since we don't know precisely what the $\lambda_n$ are.

Implicit differentiation tomorrow @user2236.

:O

@TedShifrin My calc class just reached differentiation.

tough topic for the young

@CaptainAmerica: So now take $N$ equally spaced $\theta$ around the circle for $B$ and compute.

Nah, there's no proof of the implicit function theorem, of course. I'll mention it, equally of course.

And refer them to my book/lectures :P

:P

heya @Oskar

are you using Leibniz notation?

for implicit differentiation

I'll use both. I've already lectured them on how the $f$'s on the two sides of the chain rule are different functions.

how did they handle the delta-epsilon proofs?

Not so well. And I'm not being successful at getting them to write up one or two of my problems for me to red-mark. Oh well. I miss college teaching.

In fairness, it takes a week of $\delta$-$\epsilon$ to get the hang of it. But it'll come back later and I wanted some groundwork laid before we get to integrals and sequences/series.

have you touched on hyperbolic functions?

Nope. I don't usually do them in calculus -- not enough time. I usually taught them in differential geometry, actually.

right

@CaptainAmerica16: I bet you did the right calculation but overcounted.