@Chris'ssis There's a better solution.

:14855198 math.stackexchange.com/users/97378/cleo

Just differentiate $$\sum\limits_{n \geqslant 0} {{{\cos }^n}x\cos nx} = f\left( x \right)$$

@PedroTamaroff Sure, I didn't say my solution is the best.

Then you get $f'(x)=0$.

@Chris'ssis I am thinking whether you are Cleo, lol

It's rather informal.

But works,.

The series has lots of poles.

@PedroTamaroff lots

Oops.

@WillHunting I'm not Cleo. :-)

@WillHunting I know, but my point is i'm not a CS nerd :)

Ok @Pedro I think I got it

@WillHunting wish i could be. i like logic, but that's about it

ORLY

@meer2kat Never thought about doing maths?

Suppose we have a cover $X_g \subseteq \cup X_{f_i}$

@PedroTamaroff tis my new major actually.

wlog we can say $X_g = \cup X_{f_i}$

@FernandoMartin Because...? (I said the same)

@meer2kat Ah! That's pretty cool.

Guess you'll be around for a while.

@PedroTamaroff indeed. math and education

(via intersecting each $X_{f_i}$ with $X_g$ and using that is equal to $X_{f_ig}$)

Yes, said the same thing.

Cool, so then we basically follow the same argument as the previous exercise

Taking complements we get

$V(g) = \cap V(f_i) = V(\cup f_i)$

yas

so we have $\sqrt{(g)}=\sqrt{(\cup f_i)}$

Ok, here comes the ugly part

ORLY

ya rly

well then!

we have that $g\in\sqrt{(\cup f_i)}$

yas

so $g^n = \sum a_i f_i$

yiss

that sum is finite, let's say it ranges over some finite index set $F$

I claim $\sqrt{g} = \sqrt{(\cup_F f_i)}$

Yes, indeed.

If that's correct, then we're done since that would imply $X_g$ is covered by $\cup_F X_{f_i}$

$\supseteq$ is obvious since taking radicals preserves order

On the other hand, if $x\in\sqrt{(g)}$, then $x^k= ag$, but then $x^{kn}=ab$ with $b$ the finite sum I mentioned above

so $x\in$ ugly radical and then we're done

woo

how much wood would a woodchuck chuck if a woodchuck could chuck wood?

I will be haunted until I know why it's 25 :(

@FernandoMartin YAS.

@meer2kat Err... that's just a tongue-twister.

How many pets would @Pedro pat if Pedro pats pets?

@WillHunting grammar!

@WillHunting

@user127001 Yes?

I just wanted to say your name

Haha, you and I, we are really lonely people

@PedroTamaroff no there's a real answer.

@PedroTamaroff ...well, kinda.

@user127001 Have you watched the movie Good Will Hunting?

@meer2kat Lies. =)

@WillHunting Yes

@user127001 Good. Matt Damon is hot

@PedroTamaroff i'll find it again one day

@WillHunting No thanks

with the same context, has anybody seen The Proof ?

Yes

I try to watch every movie, documentary, and TV show that has math in it

dang vba

Including Numb3rs

fd.ReplaceReference (NewFileName) ARGH

Which I am ashamed of

@GabrielR. I only watched Proof, not The Proof

@meer2kat Mind if I ask your age?

@WillHunting the rapper?

@PedroTamaroff 19

@user127001 What do you mean? Proof has Gwyneth Paltrow in it

@PedroTamaroff soon to be 20

@meer2kat Are you taken? =)

@WillHunting Nope. ditcched that guy

@WillHunting Gwyneth is wonderful. =)

@WillHunting anyways, be back soon guys. heading home.

Also Julie Bowen.

@meer2kat Ah I see. @Pedro stands a chance then, lol

@WillHunting hahahahaha

@WillHunting sorry for that ... What about the movie about Jon Nash ?

@meer2kat I won't attempt, because I am too old

@GabrielR. Yes, I watched A Beautiful Mind

@WillHunting hahaha so many people say that

@WillHunting Don't know if laughing at you or me, or both? =O

@meer2kat I will be 33 this year

@PedroTamaroff Just laughing

@WillHunting wasn't it too theatrical ?

@Pedro: Was your solution the same?

I did fall deeply in love with a girl on SE. I shall keep her identity a secret...

@FernandoMartin Yup.

@WillHunting oldest i've dated was...27 i think

@GabrielR. It's more romantic than the real life version

At some moment I thought you were peeking through my webcam or something @FernandoMartin =D

@WillHunting so true. but yeah. bye guys!

@meer2kat I will give myself a limit of plus minus 10 my age

@GabrielR. It's John not Jon, lol

tpyo everywhere !

@pedro userXXX will be 50 this year, lol

@WillHunting which user

@user127001 Oops, sorry, that's a little secret between me and Pedro

Sorry

I won't butt in again

I apologize for being nosy

@user127001 Why don't you go find a gf? You are still young

Because I'm too busy studying and girls don't like me anyway

My own mom doesn't even like me

So I guess nobody will

It's OK to be single

I will probably stay single this life

I just wanna recover from my OCD and live a normal life

@user127001 "Girls don't like me, because I'm too busy studying."

See, it's the other way around.

Like that SMBC comic.

What's SMBC

@user127001 have you seen "how i came to hate maths" ?

@GabrielR. No I admit I haven't

@meer2kat A woodchuck would chuck as much wood as a woodchuck could chuck, if a woodchuck could chuck wood. About as many good cookies a good cook could cook, if a good cook could cook cookies good. About as many boards as the bored Mongol hordes would hoard if the bored Mongol hordes did hoard wooden boards in gourds .... now get 25 gourds from the nearest super market and be done with it ;)

@pedro The girl I like stays in another country and has a bf already, so she never replied to my email

you can stlil try this @user127001 wired.com/2014/01/how-to-hack-okcupid

Thank you @GabrielR.

Will this documentary help me get a girlfriend?

@user127001 this is not a documentary, but it might inspire you lol

@Chris'ssis That is how I would approach it except to use complex analysis to show $(1)$.

Wow... 22 people in the room!

@robjohn Glad to hear that. It's good we can finish $(1)$ by real methods too. :-)

@WillHunting an interesting observation ;-)

@robjohn the telescoping sums do a great job there :D

@Chris'ssis yeah, you usually can do anything with real methods. I remember integrating $\int_{-\infty}^\infty\sin(x^2)\,\mathrm{d}x$ using real methods

@robjohn Yeah, I usually do this by real methods.

@Chris'ssis Polar coordinates helped, so I did use 2 dimensions...

@robjohn I see.

prime avoidance lemma <3

@FernandoMartin Used it to prove something?

18?

yup

exactly

Which one?

i) and ii)

I don't understand how $X$ is always a $T_0$-space.

That follows easily from iii)

Isn't ${\rm Spec}\,\Bbb Z$ not a $T_0$ space...?

Since $\overline{(0)}=X$.

why not?

That's not contradictory

But that means $(0)$ is in every open set.

Or am I thinking $T_1$? Wait.

You're thinking T1 I guess

@robjohn Yes, I am ten per cent gay.

@FernandoMartin Right, I was thinking $T_1$.

If you pick $(0)$ and $(p)$, $(p)$ is closed

since it's maximal

so its complement is a neighborhood of $(0)$

so indeed $\text{Spec}\,\Bbb Z$ is $T_0$

@FernandoMartin Not $T_1$-

definitely not $T_1$

Boo ${\rm Spec}\,\Bbb Z$.

Boo.

$T_1$ implies points are closed

so in particular every prime ideal should be maximal

so for instance the only integral domains which have $T_1$ spectra are fields

since $(0)$ is prime but not maximal otherwise

well @r9m that settles it

@meer2kat I call bull on that.

Way too farfetched.

@PedroTamaroff bull on what?

r9m's paragraph above.

That that settles it

Clearly he is BSing

@FernandoMartin How did you do $(\rm ii)$?

18)ii)?

$\subseteq$ is clear

Do you want a hint or do you want to compare solutions?

@Pedro

@FernandoMartin You mean $\overline{\{x\}}\subseteq V(x)$?

Yup

Since $V(x)$ is closed and contains x

oh ok

anyway

We want to see that $\overline{\{x\}}=\bigcap_{E: P_x\subseteq V(E)} V(E) \supseteq V(P_x) = \bigcup_{P_i\supseteq P_x} P_i$

that's almost unreadable

how can I do really big $\cap$s?

Well, anyway, it suffices to check that $P_i\subseteq V(E)$ for every prime $P_i$ that contains $P_x$

and for every $E$ in the intersection

if we write $P_x\subseteq V(E)=\cup P_j$ with $P_j$ primes containing $E$, we have that $P_x\cap P_i = P_x \subseteq \bigcup (P_j\cap P_i)$

Ahhh, I just found a mistake here.

Ugh please StackExchange, make the chat refresh automatically on mobile!!

@FernandoMartin Fernando.

@JasperLoy

@Pedro: I was going to use the prime avoidance lemma to see that $P_x\subseteq P_j\cap P_i$ for some $j$

and I concluded that $E\subseteq P_i$, but that's not true

I'm thinking how to fix it

@FernandoMartin I am trying the following.

Suppose $\mathfrak p\in V(\mathfrak p_x)$.

Let $O$ be an open set containing $\mathfrak p$, so $O^c=V(\mathfrak a)$ for some ideal $\mathfrak a$.

This means $\mathfrak a\not\subseteq\mathfrak p$.

We would like to show $\mathfrak a\not\subseteq \mathfrak p_x$.

Wow how do you guys memorize all that LaTeX syntax with the mathfraks and subseteqs

So $\{\mathfrak p_x\}\cap V(\mathfrak a)\neq\varnothing$.

and all that hodge podge

ok

When I used to play pretend Pokemon at the park, I would pretend I was Mewtwo and kept yelling "PHYSIC! PHYSIC!" Instead of the move, 'psychic'

@Chris'ssis: I have spent too much time searching for this theorem. Ceva's theorem is everywhere, but this theorem, whose statement and result seem so similar, are not.

@user127001 When I used to pretend Pokemon park at the play, I would pretend yelling Mewtwo was kept PHYSIC PHYSIC, instead of psychic 'move'.

@robjohn Agree. I didn't find anything on this subject either.

@meer2kat You look prettier in the current pic, the one with blue

@WillHunting Agree. I didn't find this subject on either anything.

@PedroTamaroff I don't understand your sentence

@WillHunting he sounds like a badly written bot

@robjohn Ah, I see. So now we are playing the bot game!

Pedro is a well known troll

@WillHunting I see now. Ah, so we are the bot playing game!

@Mike Maybe he is just drunk

@pedro Some day I will tell you which girl I like, hehe

@Mike Roll well, or pat own n' kis Ed.

@FernandoMartin It is friggin trivial, isn't it?

Yup, the way you wrote it is better. I still don't know how to fix my old proof sketch

@FernandoMartin I mean, if $\mathfrak a\not \subseteq\mathfrak p$ and $\mathfrak p_x\subseteq\mathfrak p$; then certainly $\mathfrak a\not\subseteq\mathfrak p_x$ =D

@FernandoMartin Also, $T_0$ iff "$x=y\iff \overline{\{x\}}=\overline{\{y\}}$".

That's kinda Yoneda-ish in this context

@FernandoMartin WHA...?

The closure of a point

is just the union of all the primes containing it

I don't know what Yoneda is.

I mean, I can google it.

No idea about it.

the space being $T_0$ means that you can recover what your prime ideal was but looking at its "order ideal" i.e. the primes that contain it

It's a lemma in category theory

when applied to a poset it just says that two elements are the same if their sections are the same

this is kinda similar

@FernandoMartin "sections"?

If you have a poset, the section of $x$ is just the elements $y$ such that $y\leq x$

it's more or less standard terminology

@FernandoMartin (Weak) initial segment.

different authors use different terminology

@FernandoMartin Just telling you mine.

@FernandoMartin stacks.math.columbia.edu/download/topology.pdf

Dis seems Alg. Topo related topology.

I know :)

What's that?

@FernandoMartin What I think is Alg. Topo related topology.

I know, but how did you find it?

@FernandoMartin I think I googled "Kolmogorov iff"

It doesn't look very algebraic though

@pedro The only book you need for topology is Bredon's Topology and Geometry, lol

@FernandoMartin For E14, I think one should first prove 19.

Something of the sorts.

This question again, lol

SIGH

Jasper forgeting to do [text](link) again.

@PedroTamaroff I'm sorry, but I don't follow that rule. No other room has that rule on SE

@WillHunting It's a good rule.

@PedroTamaroff I think it's stupid

@WillHunting That big ass stupid question is stupid.

In the Eng room, people love making links expand

It's just so beautiful when it expands

for E14 I first proved A-M's 19

@WillHunting Disgusting.

and then used that the contraction of a prime ideal in $A/I$ is contained in $V(I)$

@FernandoMartin OK. Let me try.

and that if two ideals in the quotient have non-trivial intersection, then its contractions have non-trivial intersection as well

YAS.

A-M 19 should definitely come before 14

Heyas @Pedro @Fernando

Hi @Ted

Hi @ted have you finished grading?

Well, I guess that killed off the conversation.

LOL @Will ... Have a whole pile of exams to grade now.

@TedShifrin Do you know who I am?

@TedShifrin Hello.

Presumably you're Jasper with your blue.

@TedShifrin Yes you are right, the old regulars would know it's me

Why the incognito?

SIGH dis questions

@TedShifrin Just for fun. Sometimes I call myself Jason Bourne, or Jacob Black

He's doing differential geometry/differential forms and he doesn't know what inner products are? Oh oh @Pedro

@TedShifrin Yep.

LE BIG SIGH.

Although some people don't know "inner product" but know "dot product."

The inner product is the one inside, the outer product is the one outside, lol

well, some people call exterior product "outer" :P

I hate tetration so much.

It is so horrible.

I hate titration

Jasper beat me to the obvious pun.

@TedShifrin I had to google titration.

You're not much of a chemist, @Pedro :D

Not at all.

My interesting in other sciences is vewwy small.

School was a waste of time, I only needed to study Eng and Math

I've always loved science ...

My best teacher in high school was a guy who taught a senior honors course half chemistry/half physics.

The math teachers were horrid.

My high school math teachers sucked

I took two physical chemistry courses in college and three physics courses. Loved them all.

@FernandoMartin My uni lecturers suck, most of them

And one biology.

vimeo.com/91425662

Well, Jasper, you'd say I suck, too :D

That sucks, so far I only had one lecturer that sucked. Almost all of them are great

@TedShifrin I don't know, I haven't taken your class

@FernandoMartin Names on FB? =D

I'm sure you'd complain, Jasper. I know you well enough :P

I really wonder how they teach at Cambridge

Well, there are great and horrid, just like most top universities.

I think no matter how bad it is it must be good at Cambridge, because in their syllabus they say they will spend X weeks on subtopic X

No other place I have seen breaks it down into that level of detail

What the official syllabus says is usually irrelevant.

They also say that the syllabus is minimal for lecturing and maximal for examining, lol

But Sir Michael Atiyah, who taught both at Oxford and at Cambridge, was, when I saw him lecture, captivating.

Hi @TedShifrin

heya @Mike

Have you come to whine at me again? @Mike

When did I do that?

You posted recently your disappointment that I wasn't here to whine at.

@TedShifrin I think Matt Damon is hot, lol

@robjohn When did you become skullpatrol? =)

Jasper, you renouncing girls for Matt?

@WillHunting I figured it was already on the screen twice, why not make it three? :-p