Oh dear g-d... we're doing category theory today. :/

what? no

i didn't vote for that

@leslietownes I know... the name of the seminar is "Mathematical Physics: Experimentation, Structure, and Frameworks".

WHY ARE WE DOING CATEGORY THEORY!?

that's one huge difference between now and when i was in school. nobody was doing that when i started grad school.

for a while i thought it was just that a lot of CT people were Really Online, but no, it's just a lot of places now.

Category theory is for organizing

@XanderHenderson But @robjohn and I have already established that cat(s) rule all.

@TedShifrin Honestly, if I am going to be picking up another creatures poop, I think I would prefer a cat to a dog, as cats hate you, but take care of themselves.

Dogs are so needy.

Though I really do miss my dane. :(

My cats never hated me. This one does need her claws trimmed, but she tries to be good when we're playing. The cats don't exactly "take care of themselves" — unless you know cats who clean out their own litter boxes and dispense their own food and water daily.

But cat(s) are a category unto themselves.

@TedShifrin So, I already indicated that I would be picking up anther creature's poop. :P

Cats require less maintenance, however, as they don't want to be walked, and generally don't need a lot of attention / snuggles / interaction.

They largely treat people with contempt.

livvy somehow jumped into my t-shirt drawer and had it close behind her. i heard confused meowing and couldn't place it for a while.

The hordes of parrots are right out of Hitchcock.

@leslie How do drawers close themselves?

MAGIC.

Holbrook High has a new principal.

Fun.'

And everyone in the district just got a 2.5% pay raise. But CPI is 7.5%. Yay. :/

(Local news. I know that y'all care.) :P

Does the principal have principles?

I hope so.

ted: she may have shifted slightly. it's tilted just slightly to be secured against the wall on account of munchkin. and there's a lot of play in this drawer for some reason. not a lot of friction keeping it open.

she was in there for about 30 minutes before meowing.

Interesting. Screech settled for crawling under the fitted sheet when I was making the bed. She never did meow, but was interested in exiting by the time I finished the entire process.

she does sometimes crawl into boxes of clothing and just curl up and sleep there.

thankfully the drawer was almost empty. i've opened it to find a layer of black cat hair all over my shirts.

Pizza for dinner: I am planning on pineapple (COME AT ME!), Canadian bacon, and jalepeños. My brother opted for mushrooms, black olives, and anchovies. The dough has been fermenting in the fridge for three days.

I think those of us with shedding cats should have magnetic fields cycling in our abodes.

I'm excited. As soon as this school board meeting is over, I can start working on that...

I HATE pineapple on pizza.

@TedShifrin Many people do.

I happen to like it.

I made a few pizzas a week ago. Homemade sauce, various vegetables. Including canned artichoke hearts the second pizza. I used some goat cheese, too, along with mozzarella.

I do love Canadian bacon and jalapeños, but will opt for your brother's pizza :)

Are you on the school board?

@TedShifrin No, but when the current rep from my area resigns, I plan to run.

Good for you!

introduce a bill or motion or whatever banning category theory in schools.

@leslietownes That is the first and most important plank of my platform. :P

Gosh, the school board here is f'n' efficient. One hour in, and they have dealt with four pages of agandæ and gone to executive session (the only thing left on the agenda is adjournment). I wish that the community college's District Governing Board were so efficient.

Too bad that the president of the DGB takes every possible opportunity to rant about how people with PhDs are all a bunch of ivory tower, leftist, commie pinko, Trump-hating parasites. :/

that sounds useful

Well, some of sure are. Especially Tromp-hating. I think I’m less of a parasite than he is, though.

Okay, we are adjourned. Imma go make dinner. 'night, all.

Bon appétit!

gordos

Maybe you can refer Gauss lemma.

Can I say that if this happens then $f(x)=h(x)g(x)$ would mean that the both h(x) and g(x) are not unit elements in D[x] and that atleast one of h(x) or g(x) is a unit element in F[x]?

One example to support what I'm saying is to consider Z and Q. Clearly, f(x)=2x+4 is irreducible over Q but reducible over Z.

@love_sodam.

I have not yet reached till Gauss' Lemma. :(

Notation shaming is a thing :(

@Koro If neither $h(x)$ nor $g(x)$ are non units in $F[x]$ then $f$ is reducible over $F$.

@ShaVuklia What do you mean, "$X$ is affine"? Isn't that a given, as $X = \operatorname{Spec}R$?

@TedShifrin Can you help me with this chat.stackexchange.com/transcript/message/60599992#60599992 ?

Thanks!

@love_sodam yeah, that's why I said non units in D[x].

hello

@S.M.T Usually not. It's healthier to not think about this stuff.

@Jakobian Hmm , I can feel it too. But , it’s reality as well right. We have have a logic or answer to it than ignoring it

That answer , I’m unable to find.

You won't find it

Enjoy your trip

@Semiclassical That's the generalized Coupon Collector's Problem. The plain version is fun, the generalized version gets a little tricky.

@Semiclassical I am not sure that has a finite expectation. The simple computation, which might not be exact, seems to indicate that it has an infinite expectation.

Hi all.

My question hasn't been answered for three days.

Is there anything I can clarify or add to the question to make it more likely to be answered?

Add a few months of time + a bounty and maybe you will get something. Open-ended questions like this don't really fetch answers in a week.

Is there an example of function $f$:[0,1]\to\mathbb R$ that $\int_0^1 f(x)dx=9$ but has no integral function?

how prove that open parallelepiped is open set in $R^m ?$ Saying open parallelepiped I mean $(x^1,..,x^m): a^i \leq x^i \leq b^i$

@unit1991 that's not open

@Jakobian that's my mistake I said open parallelepiped and wrote with $\leq$.

@JaakkoSeppälä what's integral function

@unit1991 do you know that $\mathbb{R}^m$ is topologically a product of $m$ copies of $\mathbb{R}$

taking $x$ and showing there is ball containing that $x$ in $I_{a,b}$

Given a dedekind cut $x$, If $x \gt 0$, I want to define the inverse of a cut $x^{-1}= \{ r \in \Bbb Q^{+}\mid \exists p \in \mathbb{Q}-X \ ( pr\lt1)\}$, it’s easy to check that it’s a cut, but how to prove that $xx^{-1}=1^{\ast}$ where $1^{\ast}$ is the usual dedekind cut cut for $1$, it’s easy to see that $xx^{-1}\subseteq 1^{\ast}$ but how to show the other direction?

I suppose it can be shown that there exists $a_{n}$ in $x$ such that $1/(a_{n}+1/n)$ is in $x^{-1}$ and then using the fact that $a_{n}/(a_{n}+1/n)$ eventually approaches $1$, but is there an easier proof especially one that avoided the Archimedean property and does my proof hold water?

@Jakobian Could you help me with this?

Hello i have to found the limited development of $\sqrt(ln(1+x))-\sqrt(ln(x))$ at infinity

I take u=1/x

To find the limited development at 0

Of ln(1+u)

$u-u^2/2+o(u^2)$

Then at infinity it is $1/x-1/x^2+o(x^2)$

But it is not the result that gives wolfram alpha

Someone here?

@Astyx Sorry I wasn't phrasing things correctly yesterday. So the question was: can the cirkel $S^1$ be the underlying space of a scheme. Assume this is true. Then there exists a non-empty open $U\subset S^1$ such that $U$ is affine. Since $U$ is affine, it is compact, and hence closed as $\mathbb S^1$ is Hausdorff, and since $\mathbb S^1$ is connected, it follows that $U$ must be $\mathbb S^1$, as it is a non-empty clopen, so $\mathbb S^1$ is affine.

Since $\mathbb S^1$ is affine and each principal open is affine (and must be either empty or the whole space), it follows that $\mathbb S^1$ is indiscrete, which is a contradiction.

Looks good to me

If I'm trying to calculate the fourier coefficients for $cos(x)$, I evaluate the integral $$\int_{0}^{2\pi} \cos(x) e^{-ikx} dx$$

But when I do so I just get $0$.

Well I get $\frac{i}{4\pi}[\frac{e^{-i(k-1)x}}{k-1} + \frac{e^{-i(k+1)x}}{k+1}]^{2\pi}_0$

for $k\notin \{1,-1\}$

Ah

Thank you

I'm a bit surprised by the factor you have

$i/4\pi$

I changed variables $z = e^{-ix}$, I got an $i$ from that and I got a $1/2$ from the exponential def of $\cos$

I think

OH

Sorry theres meant to be a $1/2\pi$ in the integral

Apologies

It's your work ;)

Hello someone can help me on taylor series

Of $\sqrt{ln(1+x)}$ at 0

@Vrouvrou have you tried computing a few example derivatives?

I start bu ln(1+1/u)

Actually...

@Vrouvrou where did you get this question? I don't think you can find the Taylor series at x=0.

At infinity sorry

Not at 0

It's been awhile since I considered taylor series, but is that possible still? You can't even evaluate sqrt(ln(1 + 1/x)) at x=0.

Is this an exercise from a certain class, @Vrouvrou?

Does this always happen?

E always has only two prime factors?

139××113×5=78535

78535 - 140=78395

5×15679=78395

Two factors only

How delete my message?

@unit1991 I don't see a point in this. The metric $d(x, y) = \max\{|x_i-y_i|: i = 1, ..., m\}$ is equivalent to Euclidean metric - this is a standard fact. Taking $x$ in the open parallelepiped $P$, we can obviously take $r>0$ such that $d(x, y)<r \implies y\in P$. Since those metrics are equivalent, there is $\delta > 0$ such that $B_{d_e}(x, \delta)\subseteq B_d(x, r)\subseteq P$, where $d_e$ is the Euclidean metric

actually, we can take $\delta = r$ since $d(x, y)\leq d_e(x, y)$.

@Jakobian I suspect you're being way too fancy, using the notion that different metrics induce the same topology, etc.

Yeah, kind of squashing a bug with a tank.

If $(e_n)_n$ is an orthonormal sequence in a hilbert space $H$, can we extend it to a basis to all of $H$?

What do you get when you cross an elephant and a rhinoceros?

the extension may be uncountable, so i'm not assuming separability or that it has a schauder basis

Elefino. But I know how big it's gonna be: $\|\text{elephant}\| \|\text{rhinoceros}\| \sin(\theta)$.

(where $\theta$ is the angle between them).

Emojify it again, it will be a hit.

@mathsresearcher I think this is true, but you might need AC?

$\|🐘\| \|🦏\| \sin(\theta)$ ?

Yeah, I assume you would need zorn's lemma or something of that sort

Yes, Xander.

@mathsresearcher Yeah, exactly what I was thinking.

@mathsresearcher The Axiom of Choice is obviously true, the Well-Ordering Principle is clearly nonsense, and no one actually understands what the hell is going on with Zorn's Lemma.

Speaking of which, what is sour, yellow, and equivalent to the Axiom of Choice?

Zorn's Lemon. (or should I say Zorn's 🍋?)

Alright, you've been a lovely crowd, but I gotta jet! Make sure to tip your waiter!

tips Xander $e^{\pi i/2}$ dollars.