No... this was only the first day of an intro complex variables course

I agree with the spirit of that remark, though it’s not quite right: f(z)=1/z blows up at zero and so isn’t analytic there

I guess it would be better to say analytic over its domain, rather than over $\mathbb{C}$?

But the Cauchy-Riemann equations give conditions on the real/imaginary parts for an analytic function f(z)

Ah, cool.

Eh, you can still have trouble with something like sqrt(z)

Then the first derivative blows up at the origin but not the function itself

But you’ll see that stuff eventualy

I see, so it's not so straightforward

Okay, if I have to find all complex numbers that satisfy a certain equality, is it better to write it in terms of $x+iy$ or leave it as $z$?

Like, $\frac{z}{1-z}=1-5i$

If I write it in terms of $x+iy$, it gets ugly real fast.

Definitely solve for z

You can usually rearrange at the end to get x,y

But you really don’t want to juggle them early on if you don’t have to

Ah, end up with a linear system of equations for x and y.

Yeah. It’s not terrible just tedious

What do you get?

Doing it right now, 1 second.

$y = x -5$

$2x = 1 - 5y$

$x = \frac{26}{7}$

$y = \frac{26}{7}-5$.

Brb

Oops, that was wrong.

Back

Sorry, I made a mistake with complex multiplication, I'm redoing it

Kk, no rush

Hey @semiclassical, think you can answer a quick question about 4yr universities?

I'm also at a four year university!

Ok @orbit I'll shoot this at you too:

I'm currently at a tiny little community college. Next semester I'll finish the calculus series, differential equations, and linear algebra. I'll be pretty much ready to transfer, except that most of universities I'd really like to go to only accept transfers during the fall.

darn, I keep messing up.

@CookieToast okay, so what's the question?

There is a local state school nearby, but it's math department doesn't have the greatest reputation. I could begin my upper division coursework there, and then transfer to a major university next year, but I worry that by taking the most important classes like Real Analysis at a less rigorous school, I'll be setting myself up for a more difficult education down the road

So would it be a better idea to just take the next year off and self study?

Do you know for sure that school is less rigorous? If they're using Rudin for their analysis course, I'd imagine it'd be okay.

If you could look at their course offerings and books/assignments/exams, that would be helpful

I just looked, and Real Analysis isn't even offered in the spring

Give me a moment and I can check their course materials for the fall

Got it! $y = \frac{-5}{29}$ and $x = \frac{27}{29}$.

Gah, this takes a little time to get used to

@CookieToast you could always go and supplement your learning with harder material

It looks like they use a combination of Rudin/Bartle/Goldberg. Not sure which is the primary text. Would it hurt my chances of graduate school at all to have a "less prestigious" state school on my resume?

I have no idea about the answer to that question. I'm still in undergrad.

Haha well either way, I appreciate the advice.

The math department at my school caters mainly to engineering/compsci majors, so its difficult to get information and advice regarding a career in math

Hey @ted!

Heya @Cookie.

It's not just a question of prestige. It's a question of the caliber of the students/faculty and what's expected of you in coursework. When I was Associate Department Head, I dealt with tons of transfer student issues. Typically, students who transferred serious math courses from lesser 4-year schools hadn't covered — or been pushed — comparably to courses at UGA. So it's a difficult question. But, generally, transfer students struggled in our upper-division curriculum.

yo @Ted

yo Eric

so that low dimensional topology class was wayyyyyy too hard

Grad class?

yeah

possibly assuming some other geometric topology coursework, not just 1 quarter of Alg Top

yeah definitely

definitely hit the ground running

well, you can't be a pro at everything

you still need to learn bundles and Chern classes

v true

right now im starting a reading course with Danny Calegari and we're trying to figure out topics

but it looks like that will be fun

you need to learn bundles and complex geometry :)

this is true

sigh there's so much stuff

@ted FYI, I'm comparing Sacramento State to UCD or UC Berkely, where I hope to end up eventually. From what I've heard, Sac State is definitely "less push" from what I've heard, but I guess the question is pretty situation-specific

and there's so much stuff I know nothing about ... but I'm forgetting everything at my age, Eric :P

there's always grad school to learn more things i guess

@Cookie: Generally, my advice is that you're better off taking real math courses at real places. You need to be competing with more serious students than you would otherwise ... if you want to be serious.

Even for stuff like multivariable and linear algebra, Cookie, you should work on proofs (like with my books/videos) and deeper understanding.

@Ted And even then, I need to admit to myself that when I go through a series of your videos for example, though the lectures are quite rigorous and challenging, I'm not competing against anyone, and being physically in the lecture hall absolutely bring more competition and push me more

Well, I'm suggesting you practice doing some harder exercises and thinking about proofs in lectures — which you've never been exposed to. That's a reasonable first step that you can attain without external variables interfering.

There are some great people here to learn math from, too. They're young, but they've learned a lot.

looks for great people

Well, I guess Balarka and DogAteMy are currently absent :P

Honestly, made it a habit to just skim through math.SE or this chat for a half hour or so every day, and its already helped a lot

You're only good for advanced analysis, Eric :

This chat is a gold mine though, seriously

I'm still waiting for you to polish off that integral, @Cookie :P

looks around sheepishly who, me?

I'm such a procrastinator haha :P

I thought you made a new years resolution!

im only a fake analyst

a geometer in analyst's clothing

Yeah, but I decided it would be a better resolution for next new years!

Well, you're definitely more of a geometric analyst than I've ever been, Eric.

smacks @Cookie [notes that is the first smack]

I have to run! Gotta avoid more smacks haha!

I'll catch y'all later. Thanks for the advice as usual ted!

Bubye

i might yet pivot to some other part of geometry than the very analysis heavy variational stuff ive been exposed to

but iunno

mumbles Bryant

ya i have a very real interest in his schtick

he is soooo broad

yeah i got that impression

tbh i havent seen much geometry i didnt like a lot

im pretty excited for this quarter on the whole

im doing diff top, a reading course on something on the spectrum of topology/geometry, and attending Neves minimal surfaces deal

sounds like very geometry-centric, but I guess you just did a rep thy course ... definitely should learn some bundles

yeah ill pick some stuff up

Hey u

hmm, @EricSilva, this one doesn't leap out at me.

Morning Ted

heya Mr Faust ! HNY.

HNY?

Happy New Year

Lol happy New Year to you too

See, @Faust, I finally taught you something useful. :)

Given A vector space V and a Subspace U why must the additive identity be the zero vector

Of U

Because additive identities are unique.

(Prove that.)

I be live it must be the same as the one from V but I can't seem to think of why in a satisfactory manner

I just told you :P

@TedShifrin know any good graph theory books?

I know nothing about graph theory, @orbit.

Ah

Ah, okay

Enjoy it when I say things like that, @orbit :P

hahaha

Ted I still don't get it

I get that additive identies are unique

@Faust: Well, $0_V$ works.

If you've proved that the additive identity must be unique, then it can be the only one.

Omg so obvious 4th

Ty*

Most of these sorts of linear algebra questions/proofs are pretty much — do I know the definition?

I get to take topology this semester :) also thanks for this notes and problems on analysis I got 100% on my analysis final

Wow, great going!

I don't remember sending you anything analysis.

Was super stoked cause there was some hard problems there

Maybe that wasn't directed to me.

@TedShifrin you did

@Faust Were any of them about Stokes' theorem?

A problem set

That's not in a standard first analysis course, Demonark.

Often not in second.

@Ted "super stoked"

@Daminark no we never covered that in this class I did in a calculus course with a reasonable amount of rigour though

as usual, ignores Demonark

Faust, he's punning. Ignore him or make him madder by ...

@Eric one of the problems in the functional pset is to show that a normed space has a completion using the equivalence classes of Cauchy sequences... Welp

Actually equals, as opposed to isomorphic?

yeah it says actually equals thats why i think its no

Do some examples.

Make sure you think $\oplus$ rather than just $+$.

Is that external or internal direct sum?

Internal, Demonark, as they're all subspaces of $V$.

Right, brain fart

No, that's a good question to think about.

I think I found a "new" (in the sense that I didn't know it) "topological" proof that a Noetherian commutative ring has only finitely many minimal primes that avoids talking about irreducible components

topological as in I'm using the Zariski topology without the structure sheaf

like if W = 0,x,y $ U_1 = z,l,0 U_2= z,0,0 $ i think its the same thing

I have no idea what you're doing, @Faust.

lol ok

Oh, you put the x,y in funny places.

Yes, you have a good example.

lol ok i still don't know how to actually compute it properly so ima go figure that out

I have (almost) finished developing a new field of math: portonmath.wordpress.com/2018/01/05/…

Congratulate me!

is the empty set finite?

Well, it's a set with 0 elements, so I would call that a finite set.

yes. Zero is a finite number

so it is a topology?

Is an arbitrary union of finite sets finite (or $\mathbb{N}$)?

infinite union seems to make sense and so do finitine intersection

i think so

Hmm. Well, every set is the union of its singleton subsets, right? So you could express (e.g.) the set of primes in $\mathbb{N}$ as the union of all singletons consisting of prime numbers. That would be an infinite subset of $\mathbb{N}$ which is a union of finite subsets of $\mathbb{N}$ but is not $\mathbb{N}$, no?

oh shit

thats tricky

thank you

Even numbers would have worked, too.

all evens would work as well?

No probs

Yep

Btw, hi @Ted :)

LOL ...

Sniped by Ted

hi @Alex :)

Ted's leaving to go cook dinner, so you're all safe.

Enjoy dinner!

See you Ted!

@Mathein: what's your proof?

If you're interested in sharing it, anyway :)

Bye @Ted

Let $R$ be a commutative Noetherian ring and $P$ be the set of minimal primes. Let $S=\displaystyle \cap_{\mathfrak{p}\in P} R\setminus \mathfrak{p}$ which is a multiplicatively closed subset. By some generalities on localizations and because $P$ consists of minimal prime ideals, $\operatorname{Spec}S^{-1}R$ is in an order-preserving bijection with $P$.

But the elements of $P$ don't have any non-trivial inclusions, thus every prime ideal of $S^{-1}R$ is maximal, thus $\operatorname{Spec}S^{-1}R$ is Hausdorff, but $S^{-1}R$ is also Noetherian as localization of a Noetherian ring, so $\operatorname{Spec}(S^{-1}R)$ is Noetherian

Now it remains to prove the purely topological fact that a Noetherian Hausdorff space is finite: in a Noetherian space, every subset is compact, but in a Hausdorff space, any compact subspace is closed, thus in a Noetherian Hausdorff space, every subspace is closed, thus it is discrete, thus finite by compactness

I was not sure if anyone was interested @AlexWertheim

@Matein: very neat proof! Thanks for sharing.

@Secret @LeakyNun consider all functions of the forc C1(x)x + C2(x)x^2 where Cn is piecewise constant

there are continuous solutions that wouldnt match with solutions of the form c1x + c2x^2

if you see them then you see the one great trouble

piecewise trivial solutions

have to exclude them

if we wish to somehow relate differential and implied differential equations

neat eh?

wat

im trying to learn topology but out book is really just a collection of problems

there no solutions or proofs for any of theorems

i meant what did that have to do with what im saying

contrarty to your belief this chat doesnt revole around what you say :P

who said I believed that?

your comment implied it

your post appeared like a reply thats all

oh no it was just a random question

fine ill leave though if it bothers you

:-)

lol idc what you do, though i still miss the turtle

i never had a turtle

@Daminark isnt this just a problem in rudin

If you know that $\Bbb R$ is complete, you can also embed a normed space into its double dual and take the closure

Faust: I think you're getting confused. The intersection of any subset of $\mathbb{N}$ with $\mathbb{N}$ is just that subset.

Probably, I've done it before it's just like, ugh again

but that uses Hahn-Banach

@Secret @LeakyNun pretty sure for the example I gave we can state that if C1(x) and C2(x) are never zero then there are no piecewise trivial solutions but that might be a bit premature

@Daminark: story of my life

@Mathein yeah normally that's what I'd do but shrug

@Faust wanna see a paper im writing?

@AlexWertheim facepalm

But yeah the difference in this case is that Smart's gonna want us to verify the universal property of the completion but that should be easy

@AlexWertheim for things in general or have you specifically had to do this problem a bunch?

Mathematics in general, lol

Lmao

@Daminark you can verfiy the universal property using the fact that a space is densely and isometrically embedded in its completion alone

Oh this is true, you just take the closure of the image

how is smart's lecturing

I've only seen a day of it but I rather like it

he's cool

drive.google.com/file/d/1uH23x_ivfqAYYvobULqPWWtX52_Bt9Gz/view

Yeah, his class will be good. Also I sat in on Rohit and I like him as well

@Faust

rohit is great

@Daminark how's your ring theory course so far?

i havent seen his lectures but it was a pleasure to be his grader, he assigned good problems and made my job not annoying

@AlexWertheim could i then talk about the finite union of finite sets is always finite and its compliment so its compliment is always infinite implying that a finite intersection of infinite sets whos compliment is finite is always infinite?

rohit?

Only had a day of that as well, taught by the guy who's gonna do Galois theory, but it was pretty good. Though my actual algebra prof may assign psets which are... Less than fun

@Eric he's doing something different in commutative, the book has all the answers at the back so he'll assign some, but he said that most of the test will be from the book problems so he recommends solving them all anyway

what book does commutative use

@Daminark Galois theory is the theory of appending solutions of integer coefficient polynomials to the integers and rationals right?

But in addition he's having everyone write (I think a page) about something, maybe an interesting follow-up or filling in details from class

such as Z[sqrt(3)]?

cant do that

Z isnt a field

its an intedral domain

but yeah

@Faust: a lot of words in there, but you have the right idea. Another way of phrasing things, which you might like: for any subset $X$ of $\mathbb{N}$ with finite complement, there is an integer $n_{X} \in \mathbb{N}$ which is maximal among integers which are not in $X$.

Hence, if $X_{1}, \ldots, X_{k}$ are subsets of $\mathbb{N}$ with finite complement, then if $n$ denotes the maximum of $n_{X_{1}}, \ldots, n_{X_{k}}$, $n$ is the largest integer not in $X_{1} \cap \cdots \cap X_{k}$. Hence, the intersection has finite complement.

@Faust Z[sqrt(3)] is a thing

its a ring

@Typhon Galois theory includes, but is not limited to studying fields which you obtain by adjoining roots of integer polynomials to the rationals.

ah

but you cant take the polynomial $x^2 -3 $ and try and add solution to that polynomial over Z it has to be Q

so he meant that isnt in galois theory

If you want to study things like Z[sqrt(3)], that's already in the realm of algebraic number theory

ok

the thing is any polynomial with rational coeficeints can be made into a polynomail with integer coeficeints but Z(\sqrt 3 ) is still not a field

@AlexWertheim i really like that solution

Thanks

@Faust the introduction to abstract math class I had last spring literally used Z[sqrt{3}] as the subject of 75% of the proofs of the class. It exists.

@Faust wrong and yes I said it was a ring.

Glad to hear it!

you sure it wasn't $\Bbb Z_p [ \sqrt 3] $ where p is a prime?

no

@Daminark: are you taking the grad algebra sequence?

it was a ring not a firld

@Alex undergrad, but I'm auditing commutative algebra to see

@Typhon wierd must be a diffrent subject

Nice

(It's actually undergrad commutative anyway)

@Faust it was algebraic number theory not galois theory

texhnically he never named what subject it was

the class was "introduction to abstract math"

and the class was on proof theory

we just used that material as our proofs

(There's also grad AG/CA but... That might be much)

i just galois theory was the whole thing in general

not just field appendings

isnt nori teaching that @Daminark

I believe it - I've read materials from iterations of those classes, and they look pretty intense

Commutative algebra is beautiful stuff (though I know precious little). It gets an undeserved bad rap, IMO

(I say this, of course, having gotten fed up myself with how hard it can be!)

Yeah, he is @Eric

@Alex lmao, what stuff within algebra do you focus on?

to be fair, you might study some properties of $\Bbb Z[\sqrt{3}]$ in other classes, but if you want to study similar rings systematically, then that's done in algebraic number theory

I didn't know commutative algebra has a bad reputation

im almost definitely taking it next year i guess

unless something better happens

@Daminark: broadly, algebraic geometry. More specifically: cohomological invariants, algebraic groups, Galois cohomology.

@Eric same

@AlexWertheim that sounds neat!

jesus my internet spazzed out

@Mathein: maybe it doesn't. Over the years, I've heard various people (jokingly) poke fun at how ugly and difficult it can be, but I don't think any of it is too serious.

Press F to pay respects

@AlexWertheim really cool. I gave a talk on Galois cohomology not too long ago

@Daminark: thanks! I think so too, lools.

commutative algebra seems pretty cool and im a (fake) analyst

@Mathein: very neat. I've heard that you are an algebra machine - it seems the rumors were true!

Reasonable question closed by high-school homework helpers : math.stackexchange.com/questions/2585422/…

"Off-topic"

Commutative algebra is at least from my experience really beautiful and useful (depending on what you do, of course). I'd rate the hours that I spent doing exercises in Atiyah-Macdonald and Matsumura among the most fun and instructive

Last night dream:

The paradox is then resolved by incoporating signed distance to indicated travelling forward and travelling back. This results in $d_1,d_2=6,15$ and thus there are no more logical paradox

Main point: There are zero divisors of transcendentals in last night dream

I wonder what important rings has that property...

I dare someone to download this

uh

@Typhon check if the trivial solution is also the solution of the ordinary diff eqt . The existence of $C_1 (x) = C_2(x) = 0$ to the implied DE could mean the trivial solution is indeed a solution

Daminark: Lol, I think later versions of Linux already had some protections written to it whenever you try to execute rm -rf /?

or maybe that's just mac

I don't think I've ever typed a correct command in Linux

I just type the fewest letters so it autocompletes what I want.

you have to do "rm -rf / --no-preserve-root"

(try at your own risk)

[Evil overlord mode] In theory, if anyone wrote a virus that causes all target linux systems to execute the above, you can crash the financial sector

Does the financial sector run on linux?

actually not, most are windows if I recal

I'm having a really tough time with an exercise from Spivak's Calculus on Manifolds. It's asking to show that there are no injective $C^1$ maps from $R^n$ into $R^m$ when $m<n$. It's a pretty straightforward application of the inverse function theorem, but only so long as I know that the partials $D_jf_i$ for $n-m+1 \leq j \leq n$ and $1 \leq i \leq m$ form a non-singular matrix at least at one point. This is tricky and I'm having difficulties.

Not looking for a solution. Maybe a nudge in the right direction. Also, I'm trying not to appeal to any fancy theorems. I'm trying to stick to the sequence of the book.

All linear maps R^n to R^m are singular by the rank-nullity theorem.

I'm not sure if I understand. Perhaps I misused the term singular, but I simply meant that the determinant is nonzero. I'm not sure how that applies in general cases since $n$ need not be the same as $m$, so I'm not sure if there's some generalization of the determinant that works on non-square matrices. The other thing though, is that technically the matrix that needs to be non-singular is a square matrix ($m\times m$).