@GregRos how do you write your complex functions not using $u(x,y)+iv(x,y)$?

Well, say I have $\exp{(-z^{-4})}$.

That one's not written in that way.

@GregRos Do you already know the Wirtinger derivatives?

Of course, if you can just say "a composition of analytic functions ...", that is even better.

@GregRos How do you take the partial with respect to $x$?

:14625780 Well, $0$ is an essential singularity.

Sorry, it's a piecewise function. It equals $0$ at $z = 0$.

$$\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}=0$$ is another way of writing the Cauchy-Riemann equations.

Can someone told me how to compute this $(1+t)^{-1}=\sum_{q=0}^\infty(-1)^q t^q.$

@Vrouvrou that is just the taylor series

@robjohn Well, I just use the definition with $f(x + i(y + h)) - f(x + iy)$

oh so $(1+t)^{-1}$ is equal to $\sum_{q=0}^\infty(-1)^q t^q.$

by definition ?

@Vrouvrou or the generalized binomial theorem

@Vrouvrou or geometric series.

@Vrouvrou it is not by definition, but it can be gotten in any of the ways (and probably more) that I have mentioned above

i still don't understand what i have to compute

Can you show that $(1+t)(1-t+t^2-t^3+t^4)=1+t^5$

yes

Can you show that $(1+t)(1-t+t^2-t^3+t^4-t^5+t^6-t^7)=1-t^8$

yes

So as long as $|t|\lt1$, the limit says $(1+t)(1-t+t^2-t^3+t^4-t^5+\dots)=1$

I'm upset they removed our unicoin rewards.

ok so i have that $(1+t)^{-1}=\sum_{q=0}^\infty(-1)^q t^q.$

@Mike April 1 is over

@Vrouvrou that is true in the sense of formal power series and in terms of power series for $|t|<1$

you're too easily upset pal

@robjohn I know, but those gifs of unicorns upvoting or downvoting are timeless.

you want timeless? I'll give you timeless: here :D

Thanks for all your help @Mike.

Glad to have helped, if I did

For anyone, in abstract algebra we said that a map was well defined if it was independent of representative-but now that I'm just working with a simple map, what do I need to check for something to be well-defined?

The map is from ternary strings with only 0's and 2's to binary strings.

The issue with well-definedness is if you're mapping equivalence classes of something and to define your map you need to make a choice of representative

For example, if you're mapping the rationals to something, and your map requires you to say "pick a rational p/q" and you don't say "in lowest terms"

Then you need to guarantee it doesn't matter whether you pick p/q or 2p/2q etc, you'll get the same result (so your equivalence class only maps to one thing)

Your example doesn't have equivalence classes in the domain so you don't need to check well-definedness.

But he asked us to :(

Er. Can you explicitly tell me what the map is, then?

I'm taking an element of the Cantor set, considering it as a ternary string of 0's or 2's then sending that to the string of numbers made by dividing the ternary strings elements by 2.

We already showed that the Cantor set can be represented as ternary strings.

Okay, in this case well-definedness is NOT trivial

Because there are some elements of the cantor set that can be written two different ways in ternary.

@Mike the gifs are in an answer on MSO... here

hello, please an other question what is $\sum_{q=0}^\infty(-1)^q t^q \sum_{q=0}^{\infty}$ is equal to ?

@Vrouvrou that makes no sense

sup

@FernandoMartin Hello.

I am doing exercises from Lang.

@FernandoMartin Homological algebra sucks.

@Mike You suck.

Sucker.

@Mike I saw Arcade Fire live yesterday

how cool is that?

@Mike Wait till you get to spectral sequences.

@FernandoMartin Cool! =D

@Pedro you already know about spectral sequences!?

@PedroTamaroff Do you know what those are? :P

sucka

@Mike There are?

@Anthony Think $1=0.9999...$

@Mike why did you say homological alg. sucks?

@FernandoMartin It's so tedious, man.

Gravity sucks...

sorry :$\sum_{q=0}^\infty(-1)^q t^q \sum_{q=0}^{\infty}M_q t^q$ $M_q$ is a real number it depends on $q$

antigravity blows...

It doesn't depend on $q$ since $q$ is not a free variable

@FernandoMartin I'm doing some basic stuff for algebraic topology and it's getting gross fast. I can't imagine how terrible the real deal is.

I see

@Vrouvrou that depends heavily on what $M_q$ is.

@FernandoMartin You should do some stuff in the dark side of topology and geometry.

@FernandoMartin Nah, Tochi knows.

Me and Karl are over here.

but it's scary @Mike

$M_q=\dim C_q(f,x)$ where $C_q(f,x)$ is the critical groups of $f$ at it's critical point $x$

@Mike What do I need to show then?

There are two elements like .000002222222222222 and .00001

I have the following problem: if $A$ is a UFD and $p$ is prime, $A_{(p)}$ is a PID.

Yes, you need to show that those both map to the same thing.

Except I can't get the second string from the Cantor set, can I?

Well, they're the same thing, so you can get both from the Cantor set. But yes, I see your point.

In which case every point has only one representative!

So your map is automatically well-defined.

@Fernando what are you taking this term?

real analysis, commutative algebra and a boring numerical calculus course I should have taken some years ago but never did

real analysis == measure theory

what do they cover in real analysis?

cms.dm.uba.ar/academico/programas/Analisis_Real

it's in Spanish but I think it's pretty readable

What is 'cambio'?

change

and yeah, I can read Spanish math as long as I have a dictionary

ah, cool

I like that you cover signed measures

</rant>

I agree^

@FernandoMartin DAWG

Are you guys in the same year?

@Mike Nope.

I am in the first quarter of my 2nd year.

@robjohn, I agree.

@JessyCat Saturn girl is here.

@Pedro How long does it take to get a degree

@Mike Six years.

Shiiiiiit.

Nah

You be so well-prepared wen you graduate

Well, CBC= 1 year, Courses = 4 years, Thesis = 1 year?

You're counting the CBC @Pedro, then you're in 3rd year

This get you a masters degree?

You can do the thesis while taking courses as well

@Mike No, it's more than that.

Measurable sets = conjuntos medibles :)

It's a Licenciate, @Mike

I didn't know there were degrees between masters and doctorate.

> In Argentina, the Licentiate degree (Spanish: Licenciatura), by which one becomes a licenciada (female) or a licenciado (male), is a four- to six-year degree. This may become six years in some cases, under the accomplishment of the "licentia doctorandi" thesis dissertation, generally equivalent to an M.Sc. or M.A. in North American universities, or Master in any country of Europe given by the Bologna Process.

> Occasionally, the achievement of the "Licentiate" degree does not require the formal writing of a thesis, although almost always, some amount of research is required. The successful defense of the "Tesis de Licenciatura" automatically habilitates the candidate to apply to a Master or Doctorate degree in a related field of science.

@Pedro, you're from Argentina?

@Mike Yeah.

@JessyCat Aha.

Ok, so you'd be roughly as trained as someone with a goo masters in the US.

@Mike I guess so? No idea.

@Pedro, it's fall there.

Yes. Today is a horrible day.

@Pedro Seems that way.

Out here in San Jose the weather is beautiful like always.

It's finally getting somewhat human here after the crazy winter. I feel your pain, @Pedro.

@Mike, :(

don't you have a bachelor's degree program?

@Mike this is a graph of the mandatory courses we have to take

to get the degree

I would bet Pedro has been a bachelor for a long time.

@Mike, he's not Saturn ;P

@skullpatrol it's annoying like mosquitoes. Each bite is merely an annoyance, but after a large number, it gets to you.

You have to take all of those @Fernando?

Yes

plus some optional courses

@robjohn Also, downvotes carry malaria.

If I survive Commutative Algebra I'll get 5 juicy points. =D

4*

@Mike that, too ;-)

4? Oh.

Still good.

@Fernando Many pf those are grad courses on the US. Functional analysis, differential geometry, probably whatever you call algebra 3....

Algebra 3 is Galois theory

Haven't taken that one yet

@robjohn We need to use a repellant of sorts :-)

That's frequently covered in undergrad algebra

Yeah, you can take that as a grad course, too. Sometimes as an undergrad course at bigger institutions.

@FernandoMartin do they explain duels, er duals?

By itself, that is,

@robjohn PUNUPNUPNUPNUPNUNUPNU. dem feels for Galois:'(

So two of your mandatory courses are grad courses.

I imagine your non-mandatory ones are too.

@Mike I've been told Topology is grad. level too.

I'm not sure about that

If most of it is point-set, it's not.

it's half point-set, half algebraic top.

OK, but it's definitely not introductory, right...?

@FernandoMartin @Mike Did you see my problem?

If $A$ is a UFD and $p\in A$ is prime, $A_{(p)}$ is a PID.

The one about a prime localization of a UFD?

Yes.

@Fernando That sound like an undergrad topology course at a good US school.

hey guys I have a quick statistic question

Regardless y'all are waaaay better prepared than most US bachelors students.

Just one statistic?

I am thinking about using fischer, am i right ?

fisher*

@Dave They ate a lot between the first year and the second.

Well, I see you are very smart but I'm trying to give a mathematical explanation, @Mike

:)

I can see that some went up in weight and some went down, I am right about using fisher to solve this problem ?

@Pedro Haven't thought about it yet

@Pedro what are you taking next semester?

There are a bunch of really cool optional courses

@FernandoMartin I am not sure. I think I want to do some anlaysis. I miss it. =D

Too much algebra lately.

sad, but in some ways it is true

@FernandoMartin Oh, duh? Any ideal is of the form $(p^n)$ in $A_{(p)}$...?

@Mike

I did a good thing today.

I blocked 9gag.

block reddit while you're at it

@Dave I don't use it. =P

@DanielFischer

@PedroTamaroff Quoi?

@DanielFischer I am claiming that if A is a UFD; then the ideals of $A_{(p)}$ for $p$ a prime element are of the form $(p^n):n=0,1,2,\ldots$.

@PedroTamaroff Sounds legit.

@Pedro You need to do some geometry too.

hi .. I have a question .. if $f:\mathbb{R} \to \mathbb{R}$ is a function such that $\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists for all $x \in \mathbb{R}$ .. does it mean $f$ is differentiable in $\mathbb{R}$ ? .. this got me confused .. :(

Math isn't a dichotomy, it's a trichotomy!

Yep, that's the definition really. You have a typo though.

:-)

@Mike I should. But I don't know what geometry courses there are.

There is Diff. Geo., but I cannot take that one.

Well, you have that differential geometry course. And topology if you haven't taken it.

@pedro You should just take all the math courses available, no need to choose.

Hi @BalarkaSen

You could take complex analysis so that you could later do some complex geometry.

but the answer says no !! .. the answer really kills me ..

@Mike Yes, that is a good idea.

@Pedro FB?

@Mike I'm there.

So secretive, lol.

@r9m What exactly do the problem and the answer say?

@JasperLoy We'll talk about you.

@PedroTamaroff OK, tell me what you say OK?

@skullpatrol Hullo.

@JasperLoy "Wow, Jasper is such a douche."

That is what you say^

@Pedro Does (co)homology theory involve (complicated) topology?

@DanielFischer i'll be back in 10 mins ..

@BalarkaSen Usually.

The AMS has published three very good geometry books.

I'd say it's the other way around

@PedroTamaroff Is it? Or is it simply that we have a topological perspective towards homology/cohomology ?

@FernandoMartin Me too.

Agricola's Elementary Geometry, Hulek's Elementary Algebraic Geometry and Kuhnel's Differential Geometry.

Or I thought so.

@FernandoMartin Three very good geometry books published the AMS?

@Daniel ?

@FernandoMartin Look at the message above yours.

Ahhhh, haha, I see

I was answering Balarka's question

@FernandoMartin I guessed that.

@FernandoMartin So you studied homo/cohomo?

But, never leave a bad pun lying alone in the dirt.

No

I see.

Omnichotomy

I see a cat. Meow.

The meow killed this chat, lol.

@DanielFischer The three are all translated from German!

@JasperLoy That doesn't mean anything. I have seen good translations and terrible ones. I have seen good German books and terrible ones.

@DanielFischer Well, still, it is an amazing coincidence.

That is indeed remarkable.

@DanielFischer I'd love to read Landau's "further" Number Theory publications, but they are all in German.

I think.

@PedroTamaroff I'll let you in on a secret: To read mathematical texts, one does not need to understand the language terribly well. Learn a bit, start reading. You'll learn more on the way.

@DanielFischer Right. I will try to read Bourbaki's Topology in French. =D

German I'm a little more dubious about. =P

Though I studied it a bit when I was a kid.

@PedroTamaroff Bon courage.

@r9m I see where you are from.

@JasperLoy :)

@r9m It says that $$\lim_{h\to 0} \frac{f(x+h) - f(x-h)}{h}$$ exists. That does not imply differentiability, consider $f(x) = \lvert x\rvert$.

@DanielFischer the limit at $0$ .. does not exist .. the left and right limit are different .. :'(

@r9m No, $$\frac{\lvert 0+h\rvert - \lvert 0-h\rvert}{h} = \frac{\lvert h\rvert - \lvert h\rvert}{h} = 0.$$

@Daniel OMG .. I was reading the question wrong all this time !!!!!!. million curses to my wretched self ..

sorry for being such a bother .. :(

@r9m Or chocolate crunches, if you prefer ;)

@PedroTamaroff I can't move my a** anymore today .. I ran more than my usual course .. and worked out like a monster :P

@r9m It gives me some solace to see that I can solve some of your question paper. Oh its a true-false quiz.

@Sawarnik awesome :D

@r9m But I can't understand more than half the questions :) Groups, matrices ..argh.

@Sawarnik the fact that you really can solve problems from that paper .. !! awesome .. :D

@Pedro

You're gonna love this

@FernandoMartin Adds Fernando to the black list.

@FernandoMartin I solved the problem.

The nonzero proper ideals are of the form $(p^n)$, $n\geqslant 1$

Why is that?

there is a mental fitness.SE

If you take some $a\in A$, and you write it as a product of irreducibles $a=up_1\cdots p_r$, then $a$ becomes a unit of none of the $p_i$ are $p$.

Else, it becomes an element of the form $u' p^n$ where $n$ is the valuation of $p$ in$ a$ and $u'$ is a unit in the localization.

Makes sense

Was that one from Lang?

Yeah.

@r9m Have you read [or watched] sherlock?

@Sawarnik I read all the stories all right .. :) seen the TV shows too the 1985 and the recent one (y)

Sigh

@r9m Good. The BBC one, right?

@Sawarnik ya .. :)

@r9m Howz it? I didn't like the last one, The Last Vow. But Reinbach Falls and The Empty Hearse are my favorites!

@Sawarnik blew my mind ... freaking cool :)

@r9m Which one! Empty Hearse?

@DanielFischer Voted to close.

Mentally voted to set on fire and watch it burn.

@Sawarnik all of em .. but my favourite is S02E03

The reichenbach fall

@r9m Mine too! Completely blew my mind! [But was a bit disappointed with S03E03]

But hopefully they'll answer the left out questions of his jump in S4.

@Sawarnik season three .. there were only a handful of moments that could be called mindblowing compared to the first two season ..

@r9m Yup, but I found Empty Hearse too be quite good as well. Not the other two, there was nothing detective in them [except for a small bit in The Sign of 3].

Hoping S4 would be better!

@Sawarnik i'm waiting for S4 as well eagerly

@Pedro That one is the worst yet since they actually wrote down limit

@Mike So sad.

Say I have two maximal independent sets, disjoint, in a connected graph, X and Y. I know that V(X)+V(Y)=n, and that the number of connections between X and Y is n-1, I can state that a+b=n,

I wonder if anybody has written to Numverphile about the problem?

And since a*b>a+b-1=n-1, I can find an item in x and an item in y so that they aren't connected?

@Mike No idea.

Assuming both have more than 1 item, i.e. 2 items or more

@Mike They made another video about that stuff.

@Sawarnik So the question is: Clueless, or reckless?

@DanielFischer youtube.com/watch?v=0Oazb7IWzbA .

Daniel is always coughing and anon is always drinking, lol.

can this be simplified further?(a/(a+4))/(a/(a-4))

if a is not 0 or 4 or -4 .. you could say its (a-4)/(a+4) .. :D

Sure, the a's- @r9m beat me to it

Well, actually can state it anyhow, isn't it so?

@Studentmath I dont follow what you said ..

Can be simplified to (a-4)/(a+4) regardless of if a is or isn't 0,4,-4

@Studentmath Wrong.

Since it would form the same issues then as with (a/(a+4))/(a(a-4))

How come?

@r9m; that was completely mistyped, sorry, its actually ((a(a+4))/(4(a-4))

Nevermind, I get it now.

@Studentmath Because if a is one of those values, the term is undefined.

Wouldn't be undefined in both cases?

No, if a=4 it isn't defined in the first case, but defined in the latter.

Got it

anyone on the "updated" version?

Seems it's the most simplified. You can play with it a bit, but gets you nothing prettier.

Can get (a/4)+(8/(a-4))+2 if it's better, but again, 'uglier'.

@Studentmath, yea I think the point the book is making is to make sure ppl understand that a+4/a-4!=-1, or something

You're bringing back the Java horrors to me, @joe

Just kidding

we don't die.... we multiply!

I am bored, someone entertain me...

@JasperLoy Answer some ground fruits.

@Sawarnik I notice that high rep users tend to get more votes when they answer.

@JasperLoy Natural human tendency.

@Sawarnik No wonder I get so few votes, I have low rep, lol.

Hard to earn rep, think I will retire at 3k, lol.

Will a function $f(x_1,...,x_n)$ symmetric in $x_1,...,x_n$ that is bounded (say above) take its global maximum value always when $x_1=\cdots=x_n$?

$x_1=\cdots =x_n$ will certainly be a solution to the set of Lagrange multiplier equations, but it could be a local maximum, not global.

@Alyosha How about minimum? I don't know.

@Alyosha No, $$\min \left\{ 1, \sum_{k=1}^n \left(x_k - \frac{1}{n}\sum_{j=1}^n x_j\right)^2\right\}$$ doesn't.

@DanielFischer Okay, is it easy to classify the set of functions for which this is the case? Perhaps differentiable functions.

@Alyosha If you take a differentiable clamp instead of $\min \{1,\,\cdot\,\}$, that doesn't change much. For a symmetric function, the line $x_1 = \dotsc x_n$ is special, but it can be a minimum line or a maximum line.

@DanielFischer Oh indeed, I'm not too fussed whether it's a minimum or maximum.

I gather not much can be said about it being a global maximum/minimum?

@Alyosha Hmm, you can just use the word extremum then.

@JasperLoy Yes, though I wanted to start with a concrete example.

@Alyosha Not that I know of.

@daniel Do you have a favourite calculus book in English?

@Daniel What does it mean for a continuous function on a locally compact Hausdorff space to vanish at infinity?

@JasperLoy No, I don't know any. I don't even know what "calculus" is. I used to think it was analysis, but that seems to have been wrong. It apparently is some kind of memorisation exercise.

@Mike For every $\varepsilon > 0$, there is a compact $K$ such that $\lvert f(x)\rvert < \varepsilon$ for all $x\notin K$.

That's what I guessed, thanks.

Wow, Daniel knows everything.

@Mike Or equivalently, it can be continuously extended to the Alexandrov compactification by setting $f(\infty) = 0$.

Oh! That makes sense.

@DanielFischer I am looking for an analysis book actually with all the applications. The ones I know lack applications, like rate of change, surface area of revolution, etc.

@JasperLoy Applications? What is that? (Sorry, I don't know one.)

@GabrielR. C'est à mon tour de poser la question, RMS ? Pourquoi dit-il cela ?

@Daniel Is there much you can do with non-unital C* algebras without embedding them into a unital one?

@Mike I don't know. I've never done much with C* algebras, I'm more of a Fréchet space guy.

Jeez, here I am studying analysis and I lose access to our analysis guy :(

@robjohn have you ever tried the nice series $$\sum_{k=1}^{\infty} \arctan\left(\frac{1}{k^2}\right)$$?

It's my POTE(evening).

Ups, it's midnight ... passed ...

(time flies here)

@Chris'ssis You can solve it with the fact that $\tan(1/k^2)\sim 1/k^2$ no ?

@Julien If the problem were about proving the convergence, that would be useful.

@Chris'ssis So the problem is ?..

@Julien To find its closed form.

@Chris'ssis Okay, Have you succeed ?

@Julien I'm working now on it.

@Chris'ssis An idea : $\cot(x)-2cot(2x)=tan(x)$

@Julien Would that help?

@Julien OK, I'll look at that. Thanks.

@Mike: What are you taking this semester?

@Chris'ssis I'm working on it too :). I think it works like diminos

you're a grad student, right?

I was asking Mike

@Julien OK :-)

@PedroTamaroff

@Chris'ssis it does not work ..

@Julien don't worry about that too much.

@Chris'ssis I like series too :). I am convinced that we must use trigonometric relationships. I will try it tomorrow. If you succeed please let me know.

@Julien OK :-)

@Fernando Well, one non-math class to graduate. I hate that one. But algebraic topology and differential geometry, and C* algebras.

Nice!

Yeah, they're fun so far. The first has some tedious homological algebra though :P