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8:00 PM
@Daminark Okay, by now I understand Peter May has a giant boner for this Eilenberg guy but you don't need to bring him in everywhere :P
 
LOL, DogAteMy, you let me know.
 
I made some braised pork belly over rice the other day and it was g r e a t
 
Eilenberg was quite a character. He was around for the beginning of my career.
Neither of those is too good for me, Eric, but yum.
 
@Steamy for what it's worth, in his main atop class he doesn't emphasize that stuff as much
 
it's a traditional Taiwanese dish my girlfriend taught me and it's so easy but delicious
 
8:01 PM
He wanted to present things in a way that is both accessible to beginners and not redundant for those who already took his class
Solution: present cohomology via classifying spaces and axioms
Or I mean
 
I was mostly joking :P
 
"Solution"
(It is yet to be seen if this works nicely)
But yeah right now I'm just trying to figure out how simplices and CW complexes and all that good stuff works
 
Re: the link, "keto" is a type of diet I think
Low carb
So no sugar and all that
 
idt i could do a no sugar diet
i love making sweets
 
I eat relatively few sweets. I only bake a cake or tart if I'm doing a dinner party.
 
8:04 PM
There's a dessert on its front page right now @EricSilva
 
@Ted I bake something like once a week basically
 
If I did that, I'd be even more of a blimp.
 
it's bad for me but im still young so right now i can handle it
 
I'm just finishing the leftovers of an apricot tart I made when I had friends over the other night. Mostly just apricots, but a French short pastry and a bit of sugar ... :(
 
i love apricots
I've been on a kick with indigenous Brazilian traditional stuff lately so I've been mostly cooking weird things
 
8:09 PM
These were called black velvet apricots, so dark skin ... They must be a cross with plums. Really amazing.
Someone will flag us for culinary transgressions.
 
> Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.
Hmm
 
wait for Demonark to learn transgression from P. May so he can pun
 
What, there's an actual math thing called "transgressions"?
 
hi @MikeM
yes, it's a map on certain cohomologies, DogAteMy. The Chern-Simons invariant is one of the famous examples.
 
The hint for the above is "generalize"
I don't know how to solve it
 
8:11 PM
hi
@TedShifrin how is chern-simons that?
not combative, just not knowledgeable in that language
 
Wait hold on
 
I wasn't taking it as combative. I have to rethink everything to answer.
 
By that definition isn't every real number repetitive
 
DogAteMy: for every $k$
 
@TedShifrin Still
There's only finitely many possible strings of $k$ digits
 
8:13 PM
So what's your way of thinking of transgression, Mike?
 
And every real has infinitely many digits (I assume numbers that terminate are considered to end in infinitely many $0$s)
 
I think the map from $H^n(N,A)^{G/N} \to H^{n+1}(G/N,A^N)$ is called the transgression?
 
(otherwise $\sqrt2$ is an exception)
 
I guess Chern-Gauss-Bonnet is basically a transgressive form, too.
 
because you "transgress" from dimension $n$ to $n+1$
 
8:14 PM
Oh, that's not the way I think of it. I want to go down.
 
You're going down
I feel cheated by that problem…
 
Oh, it's stooopid, DogAteMy.
 
@TedShifrin I don't think much about transgressions.
 
I have at least once said "You're going down!" to a friend who entered a descending elevator,
and then realized I needed to get to a lower floor as well,
 
I guess I think of Chern-Simons as a functional that is naturally built so that its critical points are flat connections.
 
8:17 PM
I want to say that when a certain cohomology class vanishes on the total space of a bundle, I get a specific representative of a cohomology class of degree something less on the base space.
 
and entered the elevator and announced to him, "If I'm going down, I'm taking you down with me"
 
Oh, I'm thinking of the three-form, not the functional.
rolls 2 1/2 eyes at DogAteMy
 
lol
 
Can you give me 2 sentences more detail so I can try to reconstruct the idea?
 
Whoa whoa whoa this has gone to a whole new level
 
8:18 PM
It's like a derivative of curvature or something?
 
Non-integral eye counts
 
Not without thinking, Mike. But I'll prepare something for you.
 
@Daminark :D
 
Antiderivative in some sense, @MikeM.
 
@Daminark At least we could justify an integral amount of eyes with Banach-Tarski
 
8:19 PM
Also I think I've heard the phrase "What goes up must come down," as well as "What goes down must come up," used way too often at my school when talking about our elevators
 
@TedShifrin No need to work too hard about it.
 
If some polynomial of curvature vanishes on the frame bundle, I get an actual form that it's d of that lives downstairs. Like $\Omega = d\omega_{12}$ for surfaces ... but if curvature vanishes, you can make a global $\omega_{12}$ on the manifold.
It's a more sophisticated version of that. I'll think a bit.
Anyhow, in that situation, I would call $\omega_{12}$ the transgression. Chern's proof of Gauss-Bonnet was a far more complicated version of this.
 
When you roll various amounts of eyes at us are you choosing a random number or are you rating the pun @TedShifrin
 
DogAteMy: How many times have you gotten stuck between floors?
It's perhaps correlated, DogAteMy, but not in a scientific way.
 
Nah, it's when someone goes into an elevator going the wrong direction @TedShifrin
 
8:23 PM
My question was independent of that. I have a phobia of dying elevators. Even in Evans Hall at Berkeley I often walked up and down either 7, 8, 9, or 10 floors because someone got stuck a few times.
 
Oh
I don't think I've ever encountered a broken elevator
Actually, maybe like once
(wasn't me in it)
 
Actually there was this one time I was gonna get into an elevator and someone pushed the button to keep it open, then it just got stuck closed
We were there for over an hour before fire department people fixed it
 
concerned look
 
It used to drive me nuts how many people (particularly college students) would insist on taking the elevator so as not to walk one flight of stairs ... at UGA.
heya @AndrewThompson!
 
Laziness is common
 
8:25 PM
By the way, I hate the symbols for "open" and "close" on elevators. They're way to similar.
 
Though we were all gonna go to the 8th floor
 
I propose X and O.
X for close, O for open.
 
Hello, @TedShifrin!
 
Easy to tell at a glance, is pictorial, and O is the first letter of "open"
 
>< and <> are too similar? I suppose bad for the dyslexic.
Your shortcut applies only to English speakers?
 
8:26 PM
Hi all
 
@TedShifrin Easy to tell at a glance and is pictorial, then
 
I'd actually back Akiva on that point. I have a really bad time telling < and > apart
No clue why it messes me up so much
 
Some sine problem
 
@TedShifrin Yeah
 
No wonder you like algebra and formulas.
 
8:26 PM
4
Q: Conjecture about arcsin and $\sqrt{\quad}$

mickLet $r(a,b)$ be a rational number depending on $a,b$ and nonnegative. For every $b$ there is an $a$ such that $r(a,b)$ is not $0$. Let $C(a,b)$ be a squarefree positive integer depending on $b$ and different for every $b$. Consider for positive integers $I,J$ : $$ S_J = \sum_{j=1}^J \sum_{i=1}...

 
Your left hand is the one that spells a L kiddies.
 
@PVAL-inactive does it with palm facing self
 
I never heard that before, PVAL, but it seems correct.
My right wrist has tendonitis, DogAteMy, so that hurts too much.
 
A bit late to the party, it seems; congrats a bunch, @Danu!
 
8:28 PM
But like, there was this one problem in high school which was multiple choice, each of them was a 4-way inequality, and I spent a full minute trying to figure out why all 4 answer choices were the same
 
Polling the audience: Which answer will the OP accept here?
I think @Danu is a much relieved, happier camper.
 
o/
thanks @Andrew
 
@Ted The OP will never respond, and never post anything else on this site.
 
Anyone have any idea why $Sp(n)\cdot Sp(1)/(U(2n)\cap Sp(n)\cdot Sp(1))\cong Sp(1)/U(1)$? Or why $U(2n)\cap Sp(n)\cdot Sp(1)\cong Sp(n)\cdot U(1)$? Perhaps... the magician @arctictern? :P
 
That may in fact be right, PVAL.
 
8:30 PM
@PVAL-inactive OP never delivers
 
@TedShifrin Yours seems better, but OP said he had specifically had trouble with the differentiation so perhaps the other one
 
I'm just thinking about how the world prefers formulas to understanding ideas. :P
 
Triple plot twist: The OP is actually the second answer-er
 
Quadruple plot twist: Ted is OP
 
OP is actually Ted trying to drum up book sales.
 
8:31 PM
Um, yeah, right.
LOL
 
Too late, PVAL!
 
That's what y'all think of me ...
 
Sniped
 
I gave the crime a motive.
 
Better delivery, I must admit
 
8:32 PM
@Danu Good thing this conversation isn't out loud face-to-face
 
@Danu Were you the scotch drinker in this chat?
 
Lol, IRL sniping gets replaced with being jinxed
 
Someone was suggesting me scotches.
 
If I were in PVAL's position I wouldn't get a chance to speak until the joke was no longer relevant
 
@AkivaWeinberger therefore Pell's theorem is just a lemma given the properties of ring $\Bbb Z[\sqrt{n}]$?
 
8:34 PM
@PVAL-inactive Indeed.
Got any questions? :D
 
Oh what's your favorite brand of scotch tape?
chuckles
 
Well I'd like to blame you for some financial losses I've had to incur.
 
@LucasHenrique Pell's theorem says that you have infinitely many solutions to $x^2-ny^2=1$, right? The $\Bbb Z[\sqrt n]$ stuff tells you that if you have one, you have infinitely many. But you still need to show that at least one exists in the first place.
 
@PVAL-inactive Yikes!
So what'd you get :D
 
I've gotten into the peaty stuff.
 
8:36 PM
Ask a Scottish person if he knows the square root of negative one
 
I drank a bottle of Laphroaig 10
 
Awww.... yis.
 
@AkivaWeinberger im thinking mod 354 might have enough to solve collatz. Heh, I might end up requiring infinite cases.
 
and recently bought a bottle of Ardbeg 10
 
8:36 PM
maybe the colatz conjecture is unsolvable
 
which is like 2/3 gone
 
Noice
I recently started on a new bottle of good ol' Ardbeg 10 myself!
Classic!
 
<--- gin connoisseur but scotch ignoramus
 
@AkivaWeinberger I can actually prove the inductive step for all cases in mod 12 except 7 and 3. It's frustrating.
 
All the scotches I tried that weren't high smoke I had no interest in.
 
8:37 PM
Hey @TedShifrin, what's up?
 
Don't even try Lagavulin if you are already spending over your budget ;)
 
@Typhon Hey, can I share with you a problem?
 
howdy, Typhon
 
Have you had anything cask strength @PVAL? That's also something to try...
 
> Call a real number repetitive if for every k you can find a string of k digits that appears more than once in its decimal expansion. The problem is to prove that if a real number is repetitive then so is its square.
It's uh…
I shouldn't make any comment about it…
 
8:38 PM
@Typhon +1
 
@ArtEze hi
 
@Typhon Hello.
 
I just realized
 
?
 
Except for as a fixed phrase, how is "game over" grammatical
 
8:40 PM
@Typhon ¿?
 
@Danu not really.
 
or "level up"
 
...
 
Can you level down?
 
@AkivaWeinberger it's not a sentence.
 
Anonymous
8:41 PM
What does singular solution of a differential equation mean? I couldn't understand what the lecturer meant by "envelope of function..."
 
over is an adjective, not just a preposition?
 
@Blue singular makes me think there is a unique solution
 
it's usually an envelope of a family of curves (or surfaces), @Blue. I have no idea what the envelope of a function is.
 
but that isn't hardly ever true
 
two more two is four (sorry, i speak spanish only)
 
8:42 PM
No, a singular solution has a singularity (blows up somewhere).
 
@ArtEze uuuh...ok
 
Anonymous
@TedShifrin Could you give an example for "envelope of a family of curves" ?
 
("More" and "plus" are the same in Spanish)
(Más)
 
@AkivaWeinberger Exactly.
 
@PVAL-inactive If you can get your hands on anything it's pretty great. Ardbeg Uigeadail is a nice one.
 
8:43 PM
Sure, @Blue. It's a curve that hits each curve in the family tangentially. For example, suppose I take the family of lines $(\cos t)x + (\sin t) y = 1$ as $t$ varies from $0$ to $2\pi$. Draw those. Can you guess the envelope?
 
@AkivaWeinberger well it is quite simple. I am nowhere near equipped to solve that, but here is a sketch of a proof. First off, the numbers will be rationals that are not integers. Then you just have to prove that all rationals that are not integers are repetitive. With those two tools in hand, the proof is trivial.
the non-integer rationals are closed with regards to squaring.
 
Me piro, bye bye.
 
However, I cannot prove the rationals are repetitive and vice versa. Sorry.
That's an infinite series problem and I hate infinite series with a fervor.
 
@Typhon Integers are repetitive, they end in infinitely many 0s
 
@AkivaWeinberger ok fine then.
all rationals
 
8:46 PM
Why no irrationals?
 
they don't have repeating digits.
 
only rationals have periodic expansions
 
Read the problem again
 
i know
listen
im getting there
 
Anonymous
8:47 PM
@TedShifrin $x^2+y^2=1$ ?
 
Yup. Good job. Can you see it geometrically?
 
in order to have a series that repeats for all n, you must have a series you can arbitrarily grow. This implies that the series is periodic at some point
therefore => rational
 
Anonymous
I guessed it from the graph (Yup, I "saw" it :) ). But how would I derive the equation mathematically? @TedShifrin
 
@Typhon I don't follow. Also, the conclusion is false
 
There are some interesting things besides smoky whiskies too, such as Bunnahabhain, @PVAL
 
8:48 PM
@AkivaWeinberger abababababab.... can allow me to find arbitrary k.
 
The condition is very tricky
 
but abcabdabeabf.... doesn't.
i get capped at 2
 
@Blue: This is actually an exercise in my multivariable math book. If the family of curves is given by $f(x,y,t)=0$, you solve $f(x,y,t)=0$, $\partial f/\partial t (x,y,t)= 0$.
 
that is the heart of my hypothesis
 
@Typhon There are only 10 digits in decimal
 
8:49 PM
"...."
^^^
 
What happens after abi
 
I wrote ....
 
Here's another one for you to try, @Blue: $y+t^2x=t$.
 
uuuh.... carry over?
I didn't think that through
 
Give a real example with real numbers
 
8:50 PM
sqrt{2}?
 
No reason to think it's not repetitive
 
Anonymous
@TedShifrin I'm a bit new to multivariable calculus. Could you please explain what $\partial f/\partial t (x,y,t)= 0$ means? Are you taking partial derivative w.r.t each of the variables x,y, t separately ?
 
No, no, partial with respect to $t$.
 
@AkivaWeinberger do you know the answer to this?
 
Evaluated at $(x,y,t)$
 
8:51 PM
Look, I said I shouldn't say more about the problem. But it's actually ridiculous
 
and does it relate to what I was saying before?
 
ly easy
 
is it all real numbers?
 
Losing internet connection for a bit, bye
 
Anonymous
@TedShifrin Okay. Got it. But why does solving those two set of equations work i.e. give the singular solution ?
 
8:53 PM
@AkivaWeinberger I have a counterexample.
 
Anonymous
Could you explain the logic?
 
No, this has nothing to do with singular solution. This is what an envelope is.
 
p = p_1p_2p_3p_4p_5...
@AkivaWeinberger let p be the appending of all prime numbers in order. That string never has repetition.
 
Anonymous
@TedShifrin Okay. But why does it give the equation of the envelope?
 
That's not trivial, @Blue. You need the multivariable chain rule to prove it, and you need to understand the geometry of the gradient vector. So maybe wait ...
 
Anonymous
8:57 PM
@TedShifrin Okay. I should try to cover up multi-variable calculus fully first before delving in differential equations maybe.
 
I still have no idea what the reference to envelope was in what you asked.
 
Anonymous
My uni has not started, so was reading random math topics :P
 
Anonymous
@TedShifrin It's on a youtube video. Trying to link it. One min
 
...
 

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