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00:07
Yeah... I believe the director of Star Wars IX said he wanted to expand on her role or so, which would've been great... and then this :(
00:26
They killed off the wrong character in the new movie they made last year.
hi chat
I saw Rogue One this weekend. It was good. Not much more to say.
I liked it.
@JessyCat How did your PDE final go?
@Semiclassical still haven't taken it.
00:35
I'm expecting to before the term starts, but I have been unable to get ahold of my prof
It's a weird situation.
I got an incomplete for the semester, and technically I have until the 31st of January to complete all my work.
Ah. That is a bit odd.
There's been a lot going on with my family - my dad has been very ill, and this has been the hardest I've ever had to work just to keep myself from coming apart at the seams.
Not to mention my depression and ADHD.
Have you heard from any of the other students about the final?
No. But the day after they took it, the professor sent out an email to the whole class telling them they all did well.
00:39
What?
Nah, didn't mean anything by it.
Just that it's hard to gauge things from that.
Not that there's much else than to do than dive in and do the best you can.
Remember 4 people in the class, one ofthem complained to me one day he didn't think he was cut out for grad school, and I of all people had to talk himdown off the ledge, so to speak.
I've been reading the chapters, taking notes, doing exercises with answers in the back.
Trying to focus on the things he mentioned on the last day would be on the exam.
We'll see how it goes from there.
00:54
@MikeMiller I am inclined to agree with Andrew Hwang on the description of the $S^1$-bundle on $S^3 \times S^3$.
*with total space, not "on"
(Because if I project to the first and second factors I get the 2nd and 3rd iterations of the Hopf bundle)
01:10
Hello chat!
Hi @Fargle
How goes it @Balarka?
Just woke up; it seems like I fixed my sleep schedule for one.
Awesome! Just as I broke mine...lol
Plan to do something constructive today
@Fargle Oops
01:14
Yeah, no kidding, haha. Oh well. Just got a Raspberry Pi from my brother for Christmas, I think I'll spend the day (er, night) messing with that.
huh, neat
I suggest using the Lambert W function to draw the fringing fields of a parallel plate capacitor :)
I don't mean that entirely seriously, but there are some neat physics of that here: ir.lib.uwo.ca/cgi/…
that's cool
I'm just graphing ordinary functions for proofwiki though
I figured.
Have you drawn any doubly periodic ones yet?
I don't know of any doubly periodic functions
let me google some
basically, elliptic functions and Jacobi theta functions
02:20
haven't used either of those
$\Bbb C(\wp,\dot{\wp})$
Yup.
I don't know elliptic functions all that well, if I'm honest.
@arctictern all of doubly periodic functions are contained in that field, yeah?
do you want me to graph them for you?
@BalarkaSen with a given pair of periods
02:22
Eh, don't feel obliged on my part. I've got Mathematica.
I don't have elliptic functions installed but I can install them if you'd like
I can probably graph them more easily than you, given what I've got access to.
well I'm doing this partly because I need to know R better
imgur.com/a/EH07y I'm not sure how to interpret this actually but my ritalin's run out and if I take more it will keep me up so
it's pretty
Yeah, makes sense with the periodicity.
02:24
Does Mathematica allow you to customize every part of the graph?
Pretty much.
scale, color, font, number of contours, resolution,
it's still good for me to know a language really really well, that's what I feel
Yeah. I don't know how to use all of them, but the tools are definitely there.
a lot of this stuff isn't straightforward and the guides online aren't for exactly what I'm doing
after I get this right, I have to decide what I want to do next: contours in the domain and their images in a side-by-side plot? In which case I'd do two versions, one with constant horizontal and vertical, then polar constant radius and angles
or: rotating wireframe surfaces
animated gifs
though if the former I could graph any contour and see its image, if I wanted to
that's just the standard example I see used in literature, circles and lines
One thing I occasionally play around with when doing Arg contour plots is to do, say, $\text{Arg }(if(z))-\pi/2$ rather than $\text{Arg }f(z)$ directly.
02:30
why?
that would rotate it cc pi/2 and shift it up pi/2?
They give the same output when $f(z)$ has small argument, but the jump in the argument function now effectively occurs when $f(z)$ is positive imaginary rather than negative real.
interesting
Hence the jump cut is drawn a bit differently, occasionally more nicely.
e.g. it restricts the argument of $f(z)$ to $(-3\pi/2,\pi/2]$ rather than $(-\pi,\pi]$.
I think it's handy in the present case. Here's the two graphs, one with that prescription and one without.
how do you prove that $\sum_{h=0}^{k-1}\frac{(-1)^h(H_{k-1}-H_h)}{h!(k-h)!}=\frac{1}{k\cdot k!}$?
oh whoops, too many colors
02:35
Eh, you also plotted Re not Arg.
no its not that, it's that I did too few contour lines
um, according to your scale on the RHS, you're doing Re not Arg
oh thats just me forgetting to change the label
it's Arg
Actually, the issue is: You did $\text{Arg }(\cos(iz))-\pi/2$, not $\text{Arg }(i\cos z)-\pi/2$
02:38
I meant the latter, not the former.
oops
Though I'm a bit annoyed at Mathematica's plot as well: It's plotting the same contour lines but with slightly different colors and I don't see why.
so explain to me again what I did here
please
02:42
Well, remember how arg works as a complex function. You get it by first taking log of $f(z)$, and then taking the imaginary part mod $2\pi$ restricted to $(-\pi,\pi]$.
right
However, in the above we subtract by $\pi/2$ at the end. So now it maps to $(-3\pi/2,\pi]$.
can you mathjax that please its hard to read
thank you
ah, so that's why there's less white
(I mapped white to zero)
To counteract that, we multiply by $i$ inside. That ensures that it still acts like the usual Arg when $f(z)$ is real, since then $\text{Arg}(i f(z))-\pi/2=\pi/2-\pi/2=0.$
wait no
no there isnt
02:45
What you should mostly notice is that there's less black.
right
And none of the black lines cross one another.
If you want to get something really weird looking, try doing $e^{i\pi/4}$ inside and subtracting $\pi/4$ outside.
I don't see less black but that could be a problem with my code
Well, less of the thick black lines. Though I guess 'less' isn't the right word.
It's not actually so weird looking in this case, but in others it can be pretty goofy.
@BalarkaSen I agree. And that has total space $S^2 \times S^3$.
02:48
You can play around with a bunch of different angles, in any case.
On an entirely different note: I like how my gravatar complements my dog hat.
Namely, that it looks like he's in a furnace not just paws on fire :P
oh no poor dog
I may need to sample more data for these, the discontinuities are very intense
for the original meme
wow, I need to sample a lot more points to get it less jagged
and my computer is unhappy with my whipping it forward
let's try ... 500 x 500 samplings
computer blows up
i got the this is fine plush
lol its only marginally better except it took 3 times as long
02:56
it's good
@MikeMiller Huh, interesting
Saw your comment below; I see now
03:21
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be continuous and $A,B\subseteq \mathbb{R}$. I want to prove that "$A$ bounded $\Rightarrow$ $f(A)$ bounded"

I have done the following:

We also have that $f: D\rightarrow \mathbb{R}$ is bounded $\iff$ $f(D)$ is bounded. This is equivalent to $\exists c\geq 0 \ \ \forall x\in D : |f(x)|\leq c$.

Since $A$ is bounded, we have that it is upper and under bounded. There are $x,y$ such that $$x\leq a \leq y , \ \ \text{ for all } a\in A$$ Since $f$ is continuous we have that $$f(x)\leq f(a) \leq f(y) , \ \ \forall a\in A$$
Hi, I am wondering if my definition for products for families of objects correct.
Suppose the family is indexed by $\mathbb{J}$ consider the category $\mathcal{C}_{\mathbb{J}}$ defined as follows:
03:36
@MaryStar can't do that. as f may not be monotone
if A is bounded then it's contained in an interval I, which is compact. continuous images of compact sets are compact, so f(I) is compact, so f(I) is bounded, but f(I) contains f(A) so f(A) is bounded
Why is I compact? @arctictern
@MaryStar You might not have Heine-Borel yet, but in $\Bbb R$, a set is compact iff it is closed and bounded.
I should have said closed interval
So, we suppose that I is closed and bounded, right? How do we know that there is such a set? @arctictern @Fargle
$Obj(\mathcal{C}_{\mathbb{J}})$ are indexed morphisms indexed by $\mathbb{J}$ that is $f_j : Z \rightarrow A_j$. Morphisms are commutative diagrams. Those exist in Set as infinite products.
03:49
@MaryStar let $b = \sup A$, $a = \inf A$, then $A \subset [a,b]$.
@MaryStar we hypothesized A is bounded, so of course it's contained in some closed interval
I see!!
We have that $A\subseteq I$ and since $f$ is continuous we get $f(A)\subseteq f(I)$, right? @arctictern @Fargle
well, $f(A)\subseteq f(I)$ is automatic, doesn't depend on $f$ being continuous
$f(I)$ being compact just like $I$ is depends on $f$ being continuous
@arctictern How can we prove that if $f$ is continuous and $I$ is compact then $f(I)$ is also compact?
04:01
@Olivier Oloa Do you want to solve my problem here? math.stackexchange.com/questions/2081363/… Thanks in advance
I wanted to ask a question to all you mathematics people: in every-day life, when using Modus tollens, do you do it consciously or subconsciously?
@JesterTran yes
@Ross Millikan Do you want to solve my problem here? math.stackexchange.com/questions/2081363/… If you do had time, please reply. Thanks in advance
@arctictern consciously or subconsciously?
04:06
@arctictern which one? both?
HNY, tern :)
new year for sure
trying to find an old comment of mine, and that's impossible on this site
@Victor: Stop spamming everyone on this. Enough.
3
Hi @TedShifrin
04:09
@arctictern you are not the yes-man
@MikeM: So that $S^3\times S^3$ didn't work?
My argument didn't, but it's correct. Andrew Hwang essentially gave a proof.
Hi @Balarka ... HNY to you too.
@TedShifrin Alright
04:11
Andy's an old friend. Very smart.
@TedShifrin people outside this room with 10k is also being spammed by the flags being raised
@Victor Seriously, you posted it 59 minutes ago...
@DainIIIronfoot actually it's been reposted multiple times throughout the day
@Memor-X That's the point of the flag system.
4
04:15
@Ted Indeed. I had the last detail he was missing, but I feel weird posting it when it was essentially his idea.
Whenever I'm in that position, I tend to default to doing a CW answer.
Mike, I'm confused. How can a circle action on. 6-manifold result in a circle bundle over a 4-manifold?!
That still feels weird (I posted the answer and nobody got any rep whereas the person who actually had the idea should have)
@Ted: The point is that you have a circle action on both factors.
So you can bootstrap this up to a $T^2$ action with obvious quotient
Maybe I'll post an answer giving some more details and further ideas.
04:18
No, because you need the diagonal action.
I am confused.
Yes, but the actions only differ by an automorphism of T^2.
I am skeptical.
There's incentive to write a detailed answer, then.
Yeah, cuz Ted doesn't see how 6-1 legitimately becomes 6-2.
What are you talking about? The quotient of the circle action is a circle bundle over a 4-manifold.
6-1=4+1.
04:22
Oh, damn. It's past my bedtime.
Hiya, @Ted. Happy New Year.
@TedShifrin :(
...and I guess bye, too.
night, @ted
Hi @Fargle. I'm being a dummy.
04:23
Hey @arctictern I just posted a new answer.
No more than I am at least once a day, @Ted.
I am an expert in arithmetical coincidences.
OK @MikeM.
Translation: I can count good.
2
Night.
04:24
No, I slipped a cog and didn't see it.
@MikeMiller 196884 = 196883 + 1 is a pretty fun one.
I am getting stuck on a stupid problem
smacks Fargle
I will probably leave it and be back to it tomorrow.
since people don't like discussing category theory here :D.
Ow! What'd I do, @Ted? I just happen to think monstrous moonshine is neat even if I don't really get it.
04:26
@TedShifrin you have to teach me cooking.
You forgot to leap.
LOL, Karim. Not remotely.
I got +100 point in MSE why I got this point?
Cooking is better than category theory, for sure
You Don't like food, Balarka?
no @BalarkaSen :S. I saw ted's meat stew looks very nice.
04:28
Guys please tell me why I got +100 point in MSE
It says because they trust me...
@TedShifrin I don't eat much
@kayak why look a gift horse in the mouth?
It's because you reached a certain rep level on another site.
It's the bonus you get whenever you join a stack as a reputable user.
@kayak: You don't trust them when they said they trust you?
2
04:29
@BalarkaSen Why indian food is amazing :D
@JessyCat because I want more
I love orange chicken.
@user21820 I don't. Even as a "trusted user" (20k+ rep) I got "are you a robot?" sometimes.
HAHAHA! Good answer.
@user21820 I don't trust them.
04:29
@kayak: Then join more SE sites.
LOL, tern
There should be a SE site on trust
@arctictern Hahaha I get that too, and I somehow take many tries to pass it.
@user21820 It says it is because site association bonus
Especially when it asks about houses and store-fronts.
04:30
And when it wants a copy of your tax returns and you're Donald Trump
Karim, that doesn't sound Indian.
Who's "Karim"?
me
04:31
OK, hi :)
I get in trouble with names ...
@TedShifrin I don't know. Whenever, I go to indian resturants they have something called orange chicken which is very nice.
Just got drawn here by the flags, don't know anywhere near as much math as you guys.
hi @DainIIIronfoot
Some of us are stooopid even when we know some.
04:32
@Dain Hi, son of Nain grandson of Gror greatgrandson of Dain.
@BalarkaSen Hi.
First of his name and rightful king of the andals and first men, and ruler of the Seven Kingdoms.
Breaker of chains and mother of dragons.
@Ted: One time in late high school when I was first coming upon higher math, I managed to convince myself that the Klein four-group is cyclic.
I am starting to wake up at 5 am. So happy with my current schedule. I am getting a lot of stuff done in the morning.
@Fargle how?
@Adeek: it was a gross misunderstanding of the definitions. It hurts my head to try to be that wrong again, haha.
04:34
@Adeek I've never heard of orange chicken.
What was your definition? Where's your element of order 4?
@BalarkaSen they put food coloring in the chickens' feed
It Sounds Chinese, Karim, @Balarka?
@TedShifrin I don't remember. The definition was right, I was just dead wrong for a good hour.
04:35
@arctictern was that you who upvoted?
It originated in UK...
smacks tern
haha @Fargle
@JessyCat yes. only glanced at it, but resembles the format an answer should take. can't read symbol salad inebriated.
04:36
@TedShifrin you're on a smack-roll today
You too, huh?
I'm not the one doing hw...
@Adeek Ahh, I've heard of that
@BalarkaSen It is pretty good.
I'm atoning for lost days and days to be lost, Balarka.
04:38
Yes, don't drink and derive...
this semester I am planning to cook. I have been spending a fortune on crappy food from University.
@arctic re:chicken. they give them norflox-tz as antibiotics nowadays so i won't be surprised.
I'll write an answer.
Really wish people would stop spamming the starboard
Want me to star that?
04:48
...
It's tempting.
The desire to be meta is practically an instinct for a mathematician.
That said, yeah, there have been a lot more stars in the last ten minutes or so than I may have liked to see.
good I figured out my problem that I was stuck on.
04:50
Blame me.
Well, not the top two.
If anyone is interested it is the following problem.
Xam
Xam
Hello people, what's up?
Ok, maybe it's time for me to work.
Let A, res B be sets, endowed with equivalence relation $\sim_A$, resp $\sim_B$. Define a relation $\sim$ on $A\times B$ by setting $(a_1,b_1) \sim (a_2,b_2) iff a_1 \sim_A a_2$ and $b_1 \sim_B b_2$. Use univeristal property for quotients and products to prove that $(A \times B) / \sim \cong (A / \sim_A) \times (B / \sim_B)$.
The part that I was missing is that I misread the question and thought that A and B are regular sets and I was getting confused. I need not to read only part of the question only to save time. It doesn't save time. (Note to self).
It is always good to step back from a problem to see if their is any error in your understanding as well.
Xam
Xam
Does anyone have some notes about elliptic curves?
04:57
hahaha @Xam
I just a project on elliptic curves.
I just did a project on elliptic curves I could send you a lot of links.
Xam
Xam
@Adeek omg, really? that's nice :D
do you want the project as well I could email it to you.
It has all the links as references
I have to keep an eye on this question
user228700
Hi, everyone :-) I'm having some trouble with piece-wise defined functions.
user228700
I've been asked to define the "pieces", so to speak, of the function $f(x) = | |x-3| -2|, 0 \le x \le 4$
user228700
04:59
First, I removed the inner mod to get:
user228700
$f(x)=$
Either $|x-5| ; 3 \le x \le 4$
Or $|x-1| ; 0 \le x < 3$

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