gonna go ahead and doubt that

Howdy @Mike bad boy

Hi

hi @Ted

Hi @user127001

@Ted I heard someone saying about Prof. Farris's class the other day that they wished the lecture pertained more to the homework. I got all grr inside.

Yeah, all upper-level math classes should hold hands like Calc I does ...

I wish lol

Not really

I'm not too popular after my exam today in differential geometry. I won't grade until the weekend.

Was it hard?

I never give joke exams, but they're fair and no twists and turns if one understands the material reasonably well. The problem is that most students don't.

Why don't they? Do you know the main reason(s) why

25 points were to state definitions/theorems. 10 points for a one-line argument. 35 points for computations largely based on an example done carefully in class and required homework problems. 15 points to prove a theorem done in class and in the text. And 15 points to put concepts together a bit, but most had done homework based on this.

Their homework was entirely doable with methods from the calculus sequence. An example question was that to show that a logarithmic spiral has constant angle between the position vector and tangent vector.

I have been over basic concepts (like parallel translation) numerous times in class, and done the same basic examples numerous times, but students haven't internalized it.

You grading for it, @Mike? You might learn something later in the course :P

What you/they discover, @Mike, is that most senior math majors (indeed, most grad students) don't really know multivariable calculus/linear algebra well enough.

I'm not, he found another person to grade. I'm grading linear algebra in exchange for being allowed to sit in and watch that professor teach :P

(And also, money.)

Is that a great professor?

Yay another Ted test.

mr @Pedro !

In my opinion, yes.

Boo mike boo

So you'll learn something about how to be a great teacher, yourself, since you're watching him for pedagogy rather than to learn stuff you already know. :)

I wish I had paid more attention to how he taught when I had taken topology from him. But I'm learning now, and it's not linear algebra :)

@Pedro: I won't send you this one, as you don't know diff geo yet :)

Oh snif

@Ted So I've gotta prove the spectral theorem tonight.

In functional analysis, @Mike?

Yah.

On other news my comm alg. course is ging quite well

You'll only get upset if I send it to you, @Pedro ... You won't know the terms.

Using the tools we developed in that class so far.

I'm glad you're taking it, @Pedro. Finally a course you have to work hard at.

Well, @Mike, this is stuff I never did, so you're ahead of me.

I amswered an Apóstol question.

To start, I need to prove that the spectrum of a compact operator accumulates at 0 and is isolated elsewhere.

Take look fuys

Guys

normal* compact operator

The one with $(n+1)^{p+1}-n^{p+1}$?

I think I'll be satisfied if I can do that. :P

Accumulates at most at $0$? @Mike

Compact operators aren't invertible...

Huh? What about finite dimensions?

Boo to you.

Horrible when I'm correct ? :P

Compact operator on an infinite dimensional Hilbert space.

Well, ok, then.

@Ted Howdy! Hope you're doing well.

heya, @Kevin ... haven't seen you in months ... good for you! :)

@Ted Anyway, it's going to approach differently than I know it. I have to use the C* algebra techniques we've done.

sounds cool, @Mike. You'll have to teach me.

@Ted Indeed. That's because I'm about 80% of the way to my next paper. Been doing actual work.

Good for you, @Kevin.

@Ted Now I'm trying to finish up by doing some integrals by the residue theorem, hence my presence here. I managed one so far. The others are proving to not have enough symmetry.

Well, work harder, then :P

I guess I need to!

Im thinking that in this case it isn't possible because the function is like $\frac{1}{z-1} \text{trig functions}(z)$

That should be easily done, if you reduce to the standard problems. You might need a CPV, though.

I should also mention that the contour is a vertical line int he plane

oh, hmm ... maybe not.

with $1>Re(z) > 0$

Of course, you can reduce to the real axis by rotating, etc.

I forgot that some people need to integrate explicit functions sometimes.

Doing the integrals numerically is good enough, but I was hoping to get an analytical answer.

Yup, even me years ago with singular Chern classes.

I'm not convinced you can't, @Kevin.

Im basically out of ideas so I'll write up the question, post it and toss the link in here

@Ted I'm not 100% convinced either. Its one of those things where its simple enough to think you might be able to do it, but nothing I try manages to work. There's one where I'm leaning towards can't be done right now and theres another where I'm pretty sure it can be done and I just cna't figure it out.

Well, I'll wait and see what you post.

Step 1. Order some dinner!

Noe I'm hungry.

it's almost dinnertime for ya, @mike.

@Mike: I proved the change-of-basis formula today in my multivariable math class ... back to linear algebra for the end of the semester/course.

Dinner ordered successfully. Time to LaTeX.

When I was a grad student, I could only afford to cook for myself :P

Ya honestly, I waste money on having other people make my food too much

Plus it's usually not good for ya.

Modern grad students are paid so much that we have to eat out daily.

Yeah right @Mike.

True that at Berkeley, my rent took a huge fraction of my stipend.

You can't live in the library and shower in the gym?

Personal experience: no

You there @Pedro?

Yas

Just finished tennis

@fernando

I have two questions about AC

Do you have some time?

They're easy though

@Ted Yeah my rent takes about 40% of my stipend. If I moved to a less-nice apartment I could same a good bit there as well but I like this place quite a bit and its super ocnvenient

@FernandoMartin OK.

I usually have time bro.

Thanks

Did you find the example of radical ideals which sum is not radical?

I absolutely love this typography. Does anyone know if it is TeX-available?

@FernandoMartin Yeah.

I chose (x+4) and (x)

did you choose the same?

No, I did $(2)$ and $(X^2+2)$.

@FernandoMartin Your sum is $(4,x)$, right?

Whats the hyperlink code in here again? I forgot....

[text](url)

@KevinDriscoll [altext](link)

Thanks!

@TedShifrin my integrals question

Yes, and 2 is there in the radical

but not on the ideal

Yas.

@PedroTamaroff alt in what way?

Tries something random [stuff](google.com)

@anon As in "alternative".

Doesn't really apply here though-.

then what was the original text?

@anon I know it is not right! =P

@Pedro: I saw you talking about this one the other day, so I wanted to ask if you know any geometric interpretation to that

@FernandoMartin Of what?

A sum of radical ideals not being radical

Or easier: what it means for an ideal to be radical, geometrically

@KarlKronenfeld @anon might be of use.

[[hmmmm](mathoverflow.net/)](http://math.stackexchange.com/)

@FernandoMartin Ah.

That the "curve" is irreducible?

Aw

It is an exercise in AM.

Oh that's cool then

Haven't reached it yet

Wait, no.

I am trippin balls.

@FernandoMartin It's a maximal set of polynomials which have their simultaneous solution set as a solution set.

The exercise is that $\sqrt{\mathfrak a}$ is prime iff $V(\mathfrak a)$ is irreducible.

Where irreducible means it cannot be written as the union of two proper closed subsets.

@FernandoMartin A simpler way to say the same thing: all coordinate rings are reduced.

Thanks @Karl

Well, I have another question @Pedro

@FernandoMartin Cool.

I was trying to write down the "there are minimal prime ideals" problem

Suppose I have a chain of prime ideals $J_i$

Yas.

I want to prove that their intersection is a lower bound

i.e. that it is prime

Let's call the intersection $J$

Suppose $xy\in J$ and $x\not\in J$; then $x\not\in J_i$ for some $i$

now, I want to pick the smallest $i$ such that that happens

but I can't do that in general

No, don't do that. =P

If $x\notin J$; there is some prime $p_x$ with $x\notin p_x$.

Then consider what lies over $p_x$ and what lies under $p_x$.

I claim all of those must contain $y$.

Since $J$ is a chain, that comprises all of $J$.

Wait a sec

I'm stupid

Thanks @Pedro

@FernandoMartin STAHP

For some reason I thought I needed to pick a first element

and then I'd need the chain to be well ordered

but that clearly isn't the case

@FernandoMartin How dangerous is 4chan?

It's a fucked up place.

Yes

Tell @Mike about it

?

WHOA! @Pedro The logarithmic divergences CANCEL!!!

@Pedro Ya its pretty f-ed up. Its sad but Im impervious to it now.

@KevinDriscoll What do you mean?

@KevinDriscoll WAT. =D

@Pedro I just computed an integral in a weird way using a convergence factor because it wasnt decreasing fast enough previously to be integrable

@Pedro so I made it convergent, then summed the residues and took the limit that the convergence factor approaches 1 so that I get back my original function

@KevinDriscoll You manipulating bastard!

and the logarithmic divergences magically cancelled

Im not sure why. I expected the regularization to fail and it to be divergent.

Well, that's not 100% right. The integral was convergent, it just wasnt quickly decreasing enough to be able to close a semi-circle contour and sum the residues to find the oriinal integral. So I modified it so I could do that.

@Perdo HOLY SHIT!!! My hacked together method agrees with a numerical result (up to a minus sign I mustve dropped somewhere)

@KevinDriscoll Show me the integral.

@Pedro its the one in my question I posed earlier

the first one

@Pedro

@KevinDriscoll UGH. =)

my eyes

@Pedro What I did was multiply inside the integral by $e^{- b z}$, close the contour and sum the residues, then take the limit that $b \rightarrow 0$

@KevinDriscoll I don't know contour integrationes.

@Pedro Wat. Shit blew my mind when I learned it. Integrals are hard. Here's a way to do them just by adding! YAY!

@KevinDriscoll Yeah, I will prolly take CA next semester.

What does that course cover?

@Pedro I dunno if youll find it as cool as I did cuz you know all sorts of other cool stuff. But I think its amazing.

@Mike My CA course?

Going to the complex plane seems to imbue functions with mystical properties

@Mike cms.dm.uba.ar/academico/programas/Analisis_Complejo

Can someone explain the Transitive property of sets?

What's confusing you @rubito?

@KarlKronenfeld How would you draw ${\rm Spec}\; \Bbb Z$? a big blob and little dots around it, one for each prime?

@PedroTamaroff Drawing topological spaces is not my specialty. :P I kinda view $(0)$ as a point looming above the others.

gtg

@KarlKronenfeld $(0)$ will be the Flying Spaghetti Monster.

Why are the relations R_2R_4 transitive?

@Pedro That's a good topics list.

@rubito because they satisfy the definition of transitive

@Mike Ja.

That's the problem @Mike, I don't understand why it does.

I'm glad they cover Riemann Mapping.

@rubito Because if aRb and bRc, then aRc.

@Mike. I just found a huge problem in trying to show that my $f(x)$ converges.

uniformly.

If I try to use the W-M test on this definition, I get like a double sum.

How would you suggest to prove the definition I have shown.

@eXtremiity The way that is written is very confusing.

What they mean is that the sequence is "uniformly summable" if the partial sums $s_n(x)$ are "uniformly convergent."

Uniformly summable.

@PedroTamaroff. Am I on the right track, or ?

That I am sure that does not converge.

So I can't use WM test.

Ayo

i am stupid. who can help me with commutative algebra

guys, guys. don't get so excited. form a line, take a number.

@Pedro: I can provide some motivation for the boolean ring problem

(I think)

Alas I cannot, @AlexanderGruber.

@Alexander: Shoot, though I don't think I can be of much help

it has been claimed in Matsumura that it is "immediately obvious from the definitions" that an extension $K/k$ is separable if and only if $K\otimes_k \overline{k}$ is reduced (has no nilpotent elements)

where $k$ is a field and $\overline{k}$ is an algebraic closure

I don't know about field extensions, sorry

and i can't see it

I have that a function is differentiable on an open interval, but not that it's continuous, can I still use MVT?

How did he get this [here]()

@Anthony why don't you have that it's continuous?

i'm pretty sure differentiable $\implies$ continuous. am i misreading your question?

Well differentiable implies continuous, but it's only differentiable on the open interval.

Does that imply continuity on the closed interval?

@Anthony oh i see, no it doesn't

take $1/x$ and $[0,1]$ for example

for a counterexample to your MVT thing, how about the function $$f(x)=\left\{\begin{array}{rcl}0&:&0\leq x < 1 \\ 1 & : & x=1\end{array}\right.$$ on the interval $[0,1]$?

After you're done, @AlexanderGruber. I have a question that I hope you can help me on. It's on uniform convergence

and I'm just stuck on a little thing.

@eXtremiity ok, you can try, i warn i'm not great at analysis though.

I am trying to show that my function, $f(x) = \sum_{k=1}^{\infty} \frac{x^{k}+sin(k)}{k^{2}}$ converges uniformly for its domain $[-1.1]. I am trying to use the W-M Test but I'm stuck here:

i.sstatic.net/J9eY9.jpg My working out.

@AlexanderGruber So the full question is actually to show that if a function is differentiable on an open interval and the derivative is greater than zero it's strictly increasing, but I don't think I can use the MVT because the end points aren't stated to be continuous.

@AlexanderGruber I don't know about "immediately obvious from definitions" but here's an argument: $a\in K$ is inseparable $/k$ iff its minpoly $m(x)$ ($/k$) has repeated roots iff $k(a)\otimes{\bar k}\cong k[x]/(m(x))\otimes\bar{k}\cong\bar{k}[x]/(m(x))$ has a nilpotent (by CRT). now since $K\otimes_k\bar{k}$ is a union of its subalgebras $k(a)\otimes_k\bar{k}$ it has a nilpotent iff some $k(a)\otimes_k\bar{k}$ has a nilpotent iff $K/k$ is inseparable

Oh boy nevermind definition of a derivative.

@Anthony why do you think you can't invoke MVT?

I don't have that the interval is continuous on the end points.

Or rather on a closed interval.

It's continuous on an open interval.

what does "interval is continuous" mean?

@anon okay so I've gotten to $k(a)\otimes_k \overline{k}=k[x]/\left(\text{m}\right) \otimes_k \overline{k}$ but I don't understand why that must have a nilpotent element.

@AlexanderGruber

Hello there.

@Pedro

@AlexanderGruber and $k[x]/m(x)\otimes_k\bar{k}\cong \bar{k}[x]/m(x)$ (this is just extension of scalars)

I am leaving now.

But I extend my most pleasant helloes to your persona.

then decompose $\bar{k}[x]/m(x)$ by using $m(x)$'s factorization into linears in $\bar{k}$ and chinese remainder theorem

@anon right, the CRT thing though

right, but, i'm not seeing the last part. where is the nilpotent element?

@anon Goodbyes hardworking stranger.

if $m(x)=(x-a_1)^{e_1}\cdots(x-a_r)^{e_r}$ then $\bar{k}[x]/m(x)\cong\bigoplus_{i=1}^r \bar{k}[x]/(x-a_1)^{e_1}$

@PedroTamaroff later

@anon I'm sorry, the function is continuous on the interval.

so the element(s) is/are $x-a_k$?

if all of the $e_i=1$ then each summand is $\cong \bar{k}$ so each summand is reduced hence the sum is reduced. else if $e_i>1$ for some $i$ then the $i$th summand has a nilpotent (namely $x-a_i$) hence the sum has a nilpotent

okay... i believe that. thanks.

i have one more thing

regarding $k[x]/(m)\otimes_k \overline{k}\cong \overline{k}[x]/(m)$

obvious

(:

i have seen this in the form of $A/a\otimes_A M \cong M/aM$

but that's not exactly what we have here

well, we're doing algebras not modules

if $A$ is an $R$-algebra and $S$ a ring that contains $R$ then $A\otimes_R S$ is known as "extension of scalars"

this turns $A$ into an $S$-algebra

(everything commutative bc I cba to think about noncommutative)

commutative is fine

Pedddrooo por favor amigo, ayuda me !

@Anthony I still have no idea why MVT doesn't apply. within the interval the function is differentiable, so assume there are (a,f(a)) and (b,f(b)) with a<b in the interval and f(a) not <f(b) and apply MVT

@AlexanderGruber I think @AlexanderGruber gave a counterexample a ways up.

:14834366

Uh how does this work.

i was just saying that MVT doesn't work when the function isn't continuous at the endpoints, you may still be able to apply it to prove your theorem

just not to the whole interval

How can I apply it if it doesn't hold?!

still have no idea why you think it doesn't hold

@anon I guess it does, but why does the theorem require a continuous interval?

Oh because you're normally interested in that.

hi anthony

@Mike Hi Mike!

didn't i decide to give you a nickname? i forget.

«continuous interval»

@anon :( I'm sorry.

@Mike No, you didn't :(

With the definition of a derivative, why don't we need to specify a direction we approach from?

no, because that's not what a limit is

Well I mean, isn't it normally a limit of a sequence?

fweg2

Limit of a function.

Sorry.

remember that the limit of a function isn't defined by sequences, but rather by deltas and epsilon

Yep yep yep