@Jordan That does make it a bit tougher. And I’m sufficiently out of touch nowadays that I can’t even offer sensible suggestions.

Yeah I know that I am sort of screwed :P

@JM Found on the web: JORIKI is the power or strength which arises when the mind has been unified and brought to one-pointedness in Zazen concentration.

Wow. I just thought it was a foreign name.

I think the secret to good grades is aderall, I need to look into that

I think he mentioned offhand that he did Zen or something a few months ago... he does look to be terribly serene. :)

@Jordan Sounds pretty scary to me.

@JM Did you model your icon on his head? :-)

@Jordan: Adderall will help you stay awake longer (too long, in fact), but it will not give you any long-term benefit (and will in fact fux0r your sleep cycle).

and expensive, but I really need good grades. I got a C every semester I was in school so my GPA is sort of ruined unless I get all As

@MarianoSuárezAlvarez Is there a particular incident that prompted that?

that what?

@MarianoSuárezAlvarez Your comment about Arturo

I just read an old answer of him, to which I got because of a flag on another answer on the same question

nothing interesting

@BrianMScott You're bad... :P :D It's an Easter egg; I don't tease on alopecia y'know...

@MarianoSuárezAlvarez I don't think I've ever interacted with Arturo myself.

Hey, my hair’s been sliding backward for years!

@anon Yeah; amphetamines are a crapshoot with respect to ADD...

@JM :-p

@BrianMScott but you probably are not so pastel.

prescription add medications are all the rage these days as study drugs. you can typically find them passed around on campuses.

@robjohn Not so shiny, anyway.

@robjohn Eh? I did some research on these back in grad school; man those neurons post-amphetamines were screwed up in so many ways...

Is JSTOR down for anyone else? I can't access this page.

@JM I did not think that you were serious about treating ADD with uppers.

...anyway, it's the lazy treatment, way I see it. Use it to keep yourself up and wide-eyed for a few days, but ADD? Not so much...

@robjohn Adderall is an amphetamine... :)

@Antonio: I also get an error message. However the main JSTOR site is perfectly fine.

@AntonioVargas What's the title?

@AntonioVargas I can access it

I think I have test anxiety too, like my last math test I studied for an entire weekend, like 15 hours at least (although I get easily distracted) and then when I get to the test I forgot how to do everything. For an optimization problem I never took the derivative and I thought I was right, I knew it was wrong as soon as I got home. Not sure how to fix that

One success out of four attempts. Methinks a server glitch :)

Looks like robjohn wins the prize!

Works here; a bit slow though...

@Jordan Depends on what’s causing it. Do you find yourself worrying about the time and pressing hard to finish?

(Too bad there are no other AMM archives.)

@JM is that for ADD? interesting

@AntonioVargas I got in through UCLA (VPN)

@JM It's the Conrey and Ghosh article you linked, "On the Zeros of the Taylor Polynomials Associated with the Exponential Function". I can just wait and get it tomorrow.

@Brian Yeah sort of, I always feel short on time when i take tests. I think the worst part is that I just sort of get sick of taking the test so I stop double checking

@robjohn Well, Ritalin started the craze with "amphetamines for ADD", you see... then you have Concerta which is supposedly this pill that keeps you dosed all day after just one dose. Yech.

@AntonioVargas Oh, that. That's a nice one...

When I had aderall (I don't remember the mg amount) I couldn't get to sleep for two days.

@JM Since I have it, I will have to look at it :-)

@anon Sounds like me off of it ;-)

I’m going to make some pretty standard suggestions. First, always look over the whole exam right away and pick out the stuff that you know best. Concentrate on getting it done; don’t think about the rest until you’ve done that. Then go on to the next-easiest material.

@robjohn (Ack, pronouns) the pills or the paper? :)

I do sleep, just a reduced sleep time

@JM the paper :-)

It’s not true of every teacher, but I think that every good teacher would rather see you do well on $75$% of the exam than get three-quarters of the way through every problem.

@JM sorry, I rely too much on the back pointers.

Sure but I need to get As on all my tests so I just worry about getting everything

@JM My last message is now a forward pointer :-)

@anon cool

All I have to say is: yech, stimulants.

@AntonioVargas The strongest stimulant I use is tea :-)

The teas with ginseng in them I will actually crash with a few hours later :P

Caffeine doesn't work for me anymore; I had this bad habit back in the day of drinking warm coffee before sleeping...

@JM I don't think it keeps me awake at all. If it does, I'd hate to see what would happen if I stopped :-)

RIP Rob Johnson 2012. Stopped drinking caffeine.

I drink water to stay awake, it works great

@JM Caffeine has never kept me awake. If I have a very large amount of it, it may make me physically a little jittery, but it doesn’t do anything one way or the other to my sleep. Well, except increase the likelihood of interruptions due to hydraulic pressure.

@BrianMScott Agh, yeah. If your blood pressure's wonky, caffeine is a terrible idea...

On the other hand I find it amusing to find how much coffee I need to drink so that my heart beats in approximately 4/4 time... :D

@JM Not a problem that I’ve had to face, so far, anyway.

@anon $\Huge\unicode{x26B0}$

A unicode coffin? Who’da thunk it.

At least I no longer fit the description of somebody who has "blood in his caffeinestream"...

@BrianMScott Sometimes I feel the Unicode guys have way too much time on their hands...

@JM There are some curious characters there

Eh, keeps ’em off the streets.

@robjohn The glyphs, or the guys? :-)

@BrianMScott Works both ways I'd say.

@BrianMScott I was intending the glyphs, but I guess it applies to either.

@JM True, true.

@JM indeed

$\Huge\unicode{x2620}$

Now there’s an icon for you!

$\Huge\unicode{x2603}$ Frosty the Shriner... the hat is different on my machine

Does unicode in latex work on the mainsite? :O

@anon Yes it does.

@anon I believe it does

One day I'd like to see somebody use these glyphs as dummy variables in a math.SE answer...

Well then. Sometime in the future you will see an answer with snowmen and skull variables.

I was typing that as JM posted, as it happens.

We had a grad student a few years ago who worked as a (very good) tutor who had a penchant for naming variables potato and drawing them in the equations. His handwriting was bad enough that it didn’t even really look out of place.

In high school my teacher let us use hearts/spades/stars/etc as variables.

@anon Be a bit awkward if one were using one of the combinatorial principles $\diamondsuit$ or $\clubsuit$, but that’s certainly not going to come up in high school!

Back when I used to teach, for algebraic word problems I always made it a habit to use "icons" as variables. I've found kids can keep better track of what's happening when I use 'em instead of generic letters...

@Brian: I was looking at this note, and I was wondering, has the same thing been done for arbitrary rings? I'm thinking we could define a topology where the base of open sets is given by cosets r+I of ideals that are not contained in any proper ideal.

Hang on, and I’ll take a look; I’m on a slow connection, so the $1$ MB file will take a few minutes to download.

Oh, sorry. It's the topology on integers, open sets given by arithmetic sequences {an+b} with a,b coprime. (I don't see why the author restricts to positive integers though...)

If such a thing has been done, I’ve not seen it. I’ve seen that topology on the integers only as an example of a countable, connected Hausdorff space.

I think that the restriction to the positive integers is probably an artifact of his interest in the space: the applications that he mentions deal with positive integers.

Oh, that makes sense.

Hey anyone know where I can't find a good introduction to the construction of the Direct Limit? I don't have dummit and foote with me now

Direct limit of what? Groups? (I probably can’t help anyway.)

modules

@BrianMScott Apparently are two equivalences that you can quotient out by

wikipedia gives both of them

I know only one, the one that’s standard for just about any algebraic object (‘eventually equal’). What’s the other?

The number of votes in this question staggers me...

Why? This is not mathoverflow.com

@JM I’m a bit surprised, too. And the best answer (by zyx) is badly undervoted.

@DavidWallace Well, ten votes and a few hundred views is par for the course, but what you see there's quite the outlier...

And OP really crippled himself by barring the use of trigonometric and floor/ceiling/modulo constructs.

@DavidWallace Compare that question with this similar one... why would one have a crazy amount of votes+views, when both questions are essentially about oscillatory sequences?

I'm off to sleep. Night all.

@AntonioVargas See you, and thanks again!

@Brian: do you think there are better tags for this question?

@anon: that question just sounds like one of those "you have to be there" things...

lol, the RHR one?

Maybe I should just look for screencaps of the Master Hand from smash bros.

(Obviously, Lefty from Melee won't work!)

Yeah, OP seems to have a real problem seeing it in his head...

@anon That would make pretzels out of his brain...

If someone asked me to explain the right-hand rule over the phone, it'd probably be messy...

now that I think of it, I could just do a vertical reflection to turn lefty into a righty..

any reflection?

@JM The number of answers is also kind of amazing

@robjohn Yeah; still, I don't see why OP had to be so restrictive...

@JM perhaps they knew a trig solution and wanted something else. They got one or two

Well, guy restricted truncation and modulo too. As Brian said, zyx's answer is underrated...

@JM You're quite right. In fact, the question that you exhibited is somewhat more interesting than the one with all the votes.

I'd say computationally it's easier to evaluate the floor function than exponentiation, but oh well...

But it's the kind of question where every person and their dog can weigh in. This is kind of "kindergarten maths" if you will.

And there's never any accounting for the behaviour of crowds.

@DavidWallace *sigh* you're right...

Morning...

Hi t.b.

Morning Theo.

Or evening, as I call it.

I call it evening too... Coffee's not ready, yet :)

@JM I love the comments on $(-1)^n$ being trig...

"why, because they're triangular?" Had to lol at that one.

@tb It's one of those "true and useless" things... :D

@anon I'd have used binomial coefficients myself just to be cute... :)

The site's unusually quiet today.

@DavidWheeler how?

@tb Good morning! It is just 30 minutes into Saturday here.

@robjohn Good morning, robjohn!

@tb well, it is $\cos(n\pi)+i\sin(n\pi)$

@JM There must be. I think that I’ll drop (functions) and add (combinatorics).

@robjohn no matter how you're going to establish that, I guess it's going to be circular :)

@tb ouch Then there is no point.

@JM: Oops; wrong question. I meant ‘add (combinatorics)’ and hope that the existing tags aren’t actually wrong.

@robjohn ...I see your angle on this...

Bah, that's it for me for Euler-Mascheroni...

@JM Judging from several recent questions, you should probably be seeing his angel.

:)

@JM what about $\gamma$?

@robjohn I wrote a second long answer to that question on numerically computing Euler-Mascheroni...

@JM oh I didn't see that. I have a page devoted to that.

@Ben: how are the proper maps doing? :P

@robjohn Yeah, I think that was the first example of your ASCII math that you showed me... :)

and your poor LaTeX-spoiled self survived?

@JM could be, but I see the question is for an efficient algorithm.

mine is not

I did compute 10000 digits, but that was slow

@tb Hey, I can read ASCII... :D I was a sci.math lurker after all.

it took most of an hour on a very slow computer

Oh, I didn't know (or didn't remember) that Mascheroni was the one who proved that every construction that you can do with straightedge and compass can be done with compass alone. Is there anything else one should know about him? The WP is a bit ... terse.

@tb ah theo I've committed heresy again

I am still struggling with direct limits

@tb Didn't know that either; I just remember him for the constant, really.

@BenjaminLim heresy?

@robjohn Ask theo he said I committed heresy by doing commutative algebra ahead of his topology problem

@JM St. Andrews has a bit more, but only a bit.

@tb It’s something that I immediately recognize as correct, but I could very rarely produce his name as the answer to the Who did it? question.

@BenjaminLim ah, I remeber that comment.

:D

So, what's the problem with direct limits? That they commute with tensor products?

(Also, I always make myself say "Euler-Mascheroni" since there are too damn many things with Euler's name on it, and having a second name helps me keep track.)

@tb Give me some time

@JM Like Newton-Raphson.

@JM I've always heard it as Euler-Mascheroni

@tb Right. :)

@robjohn Sometimes authors get lazy...

@robjohn In our analysis course they called it "Euler's gamma" (no joke). Couldn't think of a worse name :)

"The function or the constant?" ;)

But interestingly, if I Google "Euler's gamma" I get the mathworld entry

@tb Oops... I called it Euler gamma constant in my ASCII page :-(

@robjohn well, the "constant" indicates that you're not talking about $\Gamma(x)$ :)

(so it's slightly better)

I never know what to do with those suggested edits that obviously involved a lot of work but are not terribly good.

@tb If I think that an edit is a good idea, and it’s one that I can do, I usually improve the suggested edit.

@tb I close the window and pretend I never looked :-)

@robjohn Ah, the hot potato gambit... :)

@JM indeed

time to heat some water for tea.

@tb How do I know that the homomorphisms $\mu_i : M_i \longrightarrow \varinjlim M_i$

satisfy $\mu_i = \mu_j \circ \mu_{ij}$?

What do you mean? That's part of the definition.

No not really

the definition is that the homomorphisms $\mu_{ij}$ in the direct system satisfy

for any $i \leq j \leq k$

$\mu_{ik} = \mu_{jk} \circ \mu_{ij}$

And the definition of $\mu_i$ that you’re using?

I guess the $\mu_i$ come from the universal property..

@BrianMScott $\mu_i$ is defined above

@BenjaminLim That's the definition of a directed system only. The definition of the direct limit is a module $\varinjlim M_i$ together with maps $\mu_i: M_i \to \varinjlim M_i$ such that $\mu_i = \mu_j \mu_{ij}$ having the universal property.

@tb Ok, suppose I wish to check that what I have constructed is the direct limit of the directed system $(M_i, \mu_{ij})$

So you implement the direct limit as a quotient of $\bigoplus M_i$ modulo ...

yeah mod the submodule generated by elements

of the form $x_i - \mu_{ij}(x_i)$

@BenjaminLim No, you simply gave a property of it and asked why it satisfied that property.

@BenjaminLim So, how is $\mu_i$ defined?

@BrianMScott I was confused before, what I meant to ask was to show that the construction of taking the quotient of some submodule is actually the direct limit of the directed system $(M_i,\mu_{ij})$

@tb We have the canonical projection $\mu$ from $\bigoplus M_i \rightarrow \bigoplus M_i/D$

$\mu_i$ is just the restriction of $\mu$ to $M_i$

Hi guys

So, then pick an element $m \in M_i$ and apply $\mu_i$ and $\mu_{j}\mu_{ij}$

Hi Rajesh

@tb I tried doing that, I'l ltry it now

i was away from internet for a few days...refreshing to come back

I noticed your absence

thanks @tb

So $\mu_i(m)$ is just something in the quotient

and then $\mu_j(\mu_{ij}(m))$

well that's just $\mu_j$ applied to some element in $M_j$

Well, $m - \mu_{ij}(m) \in D$...

I almost felt guilty for not looking back at the convolution space question like I said I would. But then I remembered I'm lazy.

@tb Ok this is supposed to be $0$ right?

So that's mapped to zero when quotienting out $D$. Right.

yeah

which means that you have what you want.

huhuhuhuuhuhuhuh???????

I can't even apply $\mu_i$ or $\mu_j$ to $m - \mu_{ij}(m)$

$m \in M_i$ and $\mu_{ij}(m) \in M_j$.

yes

$\pi(m) = \mu_i(m)$ and $\pi(\mu_{ij}(m)) = \mu_j(\mu_{ij}(m))$ where $\pi$ is the quotient map

yesyesyesyesyesyesyesyesyes

$\mu_i(m) - \mu_{j}(\mu_{ij}(m)) = \pi(m) - \pi(\mu_{ij}(m)) = \pi (m - \mu_{ij}(m)) = 0$.

got it got it now

@rob: this is nice! Geogebra, I presume?

@tb I got it now. Thanks!!!

@tb I guess what was confusing was that $\mu_i$ and $\mu_{ij}$ are $\textit{different}$ types of homomorphisms

Now I will check the universal property

@JM no that was Intaglio

Bah, Mac... :D

@JM Yay, Mac!

I am so disappointed. My wife put some peppermint herbal tea bags into the jasmine green tea box. The envelopes are both green, so I didn't notice until the hot water was poured and the scent came to me... the wrong scent :-(

No caffeine... no jasmine

...you didn't notice the teabags themselves smelling different?

they were sealed in a foil envelope until I tore it open and poured water over it, I didn't smell a thing.

Ah, that's the good and bad thing about foil...

Seals in the flavors, but it might be too late to notice that you've got the wrong bag...

@JM I'll just have to examine more closely in the future. Green envelope is not enough.

@BenjaminLim done?

@tb Sorry I had to go to the kitchen to help with the chicken

Is it dead now?

...and if it's dead, is it dressed?

I thought that the point of plucking was to undress the chicken.

Yes, you undress it, then you bathe it and then you dress it again...

@tb No you don't need to kill the chicken it is already dead I bought it from ALDI.

@BrianMScott ALDI = (kinda like cosco)

You got ALDI in Australia?

I know Aldi. We have it here. Well, I know of it; I’ve not actually been inside one.

There was also a long article on the Aldi brothers in Der Spiegel a while back.

@tb Yes ALDI is like super super super super super super cheap

It's cheaper than Coles and Woolworths, both of which have like 70% of the market

@BrianMScott You have ALDI in the US and A?

Yes. Also Trader Joe’s, which is a branch of Aldi.

@BrianMScott I am not familiar with that. Have you heard of Woolworths/Coles?

@BenjaminLim Limited but good selection at excellent prices.

If you are british you will know of Tesco

@BrianMScott The produce at ALDI is kinda limit as you said but it is super cheap

For me Aldi always was the mother of all supermarkets I like to avoid and a German conception. It was the first time I saw red wine sold in TetraPak. I'm still in therapy after that shock.

for students like me that is perfect as we are usually poor

Not British, but I’ve shopped in a Tesco or two.

@tb Wine sold in tetrapack is called goon in Australia

@tb "It was the first time I saw red wine sold in TetraPak." - eww...

@JM People like to get drunk on goon

Grape juice, sure, but wine?

@JM I am in the process of learning cebuano

@BenjaminLim The accents matter, I'm told.

@BenjaminLim this euphemism really makes my day :)

@JM Someone was trying to explain to me that kayu = "all of you" in tagalog, in a different tone it means fire in cebuano

@tb What euphemism??

@BenjaminLim Yes, something like that. "False friends" abound...

Goon. Not so much euphemism as slang.

@BrianMScott @tb Never heard of the term before?

Nope.

christmas in australia : youtube.com/watch?v=BQ9J_o7hI94

@BenjaminLim isn't a goon a bully, someone who terrorizes? That's a mild way of describing this monstrosity.

oh ok!!!

@tb Not if it’s from the The Goon Show!

@tb People here who go surfing like to say: "Yeah we were rippin' it bra!!"

Now let me get back to checking if we have any other $A$ - module $C$ and maps $\tau_i : M_i \rightarrow C$ such that $\tau_i = \tau_i \mu_{ij}$

then $\bigoplus M_i/D$ is actually contained in the kernel of $\tau_i$

this doesn't make sense, does it?

@tb Why not I want to show that the $\tau_i$ factor through $\bigoplus M_i/D = L$

So I want to show that $L$ is contained in the kernel of $\tau_i$

oh crap

that won't work

the kernel of $\tau_i$ lives in $M_i$

but $L$ does not even live in here!

That's why I said it didn't make sense. What must be in the kernel of what?

hmmmm

Well, first you get a map $\bigoplus M_i \to C$, right?

@tb Not half as much of a shock as drinking red wine from a tetrapak.

yeah they have to be in $\bigoplus M_i$

so I want that the kernel of....

wait

@tb I get it

first we show that $\bigoplus M_i$ is contained in the kernel of $\tau_i$

we get an induced map (say $f$ )from $ \bigoplus M_i$ to $C$

You apply the universal property of the direct sum to get a map from there. Now you want to check that the kernel of that map contains $D$.

Then we show that $D$ is contained in the kernel of $f$

to get finally an induced map (say $g$ ) from $\bigoplus M_i/D$ to $C$

You induce twice

crap now I will have to relearn all these direct sums, direct products etc in terms of universal properties

@tb What I said above about inducing twice is correct no?

@JM I can't resist: tonight I'm gonna have Lasagne :)

Is this normally something to be resisted?

@Ben: yes. I would say factor twice, but that doesn't matter much.

@tb Eat one in my stead. :)

@BrianMScott I can't resist making him jealous...

Ah, now I understand.

@BrianMScott He can't exactly see me turning green, y'know... :)

Well, you're pale, as far as I'm concerned...

@JM Shouldn’t eat it, or can’t get it?

@tb You know I just realised something

@BrianMScott I can't afford the ingredients these days... oh well.

All the time I have used universal properties to invoke existence of this map, that map, etc

But then when I one to compute something concrete with the maps

I am like crap.............

@JM That is unfortunate. I’m just too lazy to fix it.

@BrianMScott Like I said, "when life gives you lemons..." :)

@BenjaminLim welcome to the club! We all meet once a year in the coliseum.

@tb To be thrown to the lions?

@BrianMScott No, I prefer to be stabbed to death by the guy with the pitchfork and the net.

The retiarius, if I remember correctly.

The bad thing is that the pitchfork is unsterilized...

Yes, retiarius was the name I was looking for.

@BenjaminLim Got it now? And have you checked uniqueness of the factorization?

@ tb sorry.....i fell asleep

Oh, no! My bad influence again!

How can that be? I thought that you were the one who stayed up till odd hours.

it was that "perfect map" problem you posed the other day.....i finally figured it out, but it was...convoluted

Its proof is pretty much the same as the one of what you and Ben called the "tube lemma", I think.

@BrianMScott David was complaining about my having a bad influence on his sleeping habits.

@Gigili: Ha! I guessed even before I hovered over the new icon. Good morning.

'Ello Gigili

well, what i did was: take an open cover of $f^{-1}(C)$, which is an open cover of $\{f^{-1}(y) : y \in C\}$

'Ello Tee-Bee, @BrianMScott.

(I can’t believe that I actually combined the words good and morning.)

so we have an C-indexed subcover of $f^{-1}(C)$

@BrianMScott Haha, very true. You can't combine them like that since mornings are awful.

@Gigili They’re not too bad if they come at the end of one’s day.

for each y, since f is perfect, we have a finite number of open sets in our open cover covering $f^{-1}(y)$

@Gigili 'pends. Sometimes sunrises are nice to see... assuming that when you're seeing 'em, you've just woken up and not about to sleep.

@BrianMScott They're bad, no matter when they come.

@tb Why, do you constantly jab him whenever his eyes droop?

@JM Nonono, you're wrong.

@BrianMScott BTW: is daylight savings still in place there?

if i call these sets$\{(U_y)_i\}$, i take the complements of the image of their complements

@JM Started a few weeks ago. I wish that we’d stay on it year-round.

@JM apparently. Must be my mean subconscious...

(I'm still following, David. You're on the right track)

@BrianMScott Silly me, I had your seasons in reverse... I had thought you switch from daylight to standard around this time.

No, we’re finally starting to get a reasonable amount of daylight again.

which gives me a open cover of C. now C is compact, so we take a finite subcover of the cover we obtained, and pull that back via the pre-image, which gives a finite subcover of our original cover, showing $f^{-1}(C)$ is compact.

This is absolutely the right idea, however you should say a word what exactly you mean by "pull that back via the pre-image" and how exactly you get your finite subcover from that.

well if i call the open sets i get in Y, $(V_y)_i$, i get a finite subcollection (which i can "index by the y's", say $V_{y_1},\dots,V_{y_k}$

so then i take the U's indexed by these points of C.

hey

sorry I had to go man

chicken!

yes

that is, $f^{-1}(C) \subset \{f^{-1}(V_{i_j})\} \subset \{(U_{y_j})_i\}$ (i hope i wrote that right)

There was some emergency well my uncle put the chicken on the barbecue and then the oil dripped into the fire underneath and then the whole thing flared up

@DavidWheeler Now it’s good.

@BenjaminLim I hope the chicken was still edible...

@tb No it wasn't thank god the fire alarms did not go off

Oh, no! But nothing worse happened, no injuries?

@DavidWheeler I see where you're going. Just one more thing: for each $y \in C$ you got a finite collection of sets from your original cover which cover $f^{-1}\{y\}$, right? You didn't quite say this explicitly.

it seems to me this ought to be able to be used as the definition of a perfect map as well, since we need all those hypotheses

What do you mean? You can of course require that pre-images of compact sets are compact, but that's a priori substantially stronger than having compact fibres.

@tb, yes the U's were part of our original cover, each U covers a pre-image of a V

But how do you ensure that you cover everything in the end?

writing out the indexing sets explicitly is tedious

I would build a new cover as an intermediate step: For each finite subset $F = \{i_1,\ldots,i_n\} \subset I$ write $W_{F} = U_{i_1} \cup \cdots \cup U_{i_n}$.

what do you mean by "fibre"?

The pre-image of a point.

Inverse image of a point.

Now for each $y \in C$ you have $F(y) \subset I$ finite such that $f^{-1}\{y\} \subset W_{F(y)}$