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00:00
heya tern ...
There was a basic Galois theory question that I thought I could answer but couldn't coherently manage. Maybe you can help? @arctictern
hi @Semiclassic ... Exam all printed?
as far as I know, yeah.
Oops, tern, the coefficient of $x^3$ is $-1/6$ :P Maybe $\tan$ has $+1/3$.
I took another look at the first problem I was worried about and I feel more okay with it.
The second one was a problem, but he did replace it. So that's fine.
00:03
@TedShifrin err, yes. arctan
I remain a bit annoyed with the timing of it.
@TedShifrin I am very good at maths
Grad students are supposed to be annoyed, @Semiclassic :P
Especially since his second message with the new draft was just: "Hi Michael,
Thanks, I have made some changes. Now it is off to [the front office]. Cheers, X."
Well, he did say "Thanks" and "Cheers."
00:05
Yeah. It's not the gratitude aspect of it that bothers me, mind, more the fact that that was the only message he sent.
But eh. I'm past being grumpy.
Well, it's been clear for a while he isn't the most dedicated of teachers. Don't expect a change at the end. Is this the end of his out-of-retirement teaching?
Not sure. I don't think so?
But I really don't know.
anyways, best to move on.
Indeed.
@AkivaWeinberger Were you speaking... to me?
Hey everyone!
00:10
@SimplyBeautifulArt That may be, but the hint is totally wrong. What's $\arccos(0)$?
Rehi Demonark
@TedShifrin It was quite sarcastic :P
I don't think there's need for purposeful garbage in here.
@user379685 Notice that $\arccos(1/n^2)\to\pi/2$, so the limit is $+\infty$.
@TedShifrin Sorry.
I don't advocate just doing people's homework for them, either, as I'm sure you know. But writing down garbage to be "sarcastic" without an indication that you're lying is very misleading.
@TedShifrin Well, it started off as a genuine mistake on my Taylor series. And I didn't think much of it when responding towards you, not the original asker. But of course, I retract my earlier statement as being un-well.
00:16
As young people say these days, "Whatever."
@TedShifrin Actually, I tend to say ._.
I tend to favor 'mmkay', though that serves a variety of different purposes.
@ForeverMozart You have changed profile images
no
i dont think so
00:41
Hm, then it is my imagination :-/
God it's really annoying to spend so long trying to figure something and then it turns out to be something stupid
Welcome to the world of mathematics, Demonark.
I think Forever's the same as he's been forever.
There was an assumption we were supposed to show we could assume without loss of generality, and a friend of mine and I were trying to figure it out with Besicovitch
Turned out there was absolutely nothing to it
An assumption used in the proof of Besicovitch? :D
Oh no
Like, the problem is to show that any union of closed disks is measurable
And the hint was to assume, without loss of generality, that the radii were all at least 1
00:47
Isn't a closed disk measurable?
Oh, why that hint?
Well, potentially uncountable union
Oh.
In what space?
The plane
The plane has a countable basis.
Hmm ... my argument for open doesn't quite work for closed.
Yeah, turns out an arbitrary union of closed disks is not necessarily Borel
00:50
So you were trying to figure out how to justify the hint?
Yeah
Now, the answer is, just prove the problem for disks of radius at least $\frac{1}{n}$
If the diameters are bounded below by a positive number, it's immediate.
So that you could just scale all the balls to have radius at least 1, and then take a union over $n$
Oh, I see. That's clever.
Now, instead of that, I was trying to use Besicovitch covering theorem and was getting nowhere
00:52
That's how you learn. I can't even begin to imagine how many hundreds of hours I wasted on dead ends as an undergrad (or grad).
Oh hell, or in research as a faculty member ... Make that thousands of hours :P
The old canard applies: "If we knew what we were doing, it wouldn't be research!"
Yeah I've started feeling some of that this quarter. Up until now I tended to cop out on problems very quickly
If you're copping out on problems quickly, you don't really want to do mathematics.
Now I've started being more willing to actually spend time on things until it works without jumping to Google
This is what annoys me so much about the people who insist on posting answers on MSE so quickly ... especially when I'm engaged in hinting.
G'night @MikeM.
00:56
But before, frustration would mount very quickly, if it didn't respond in 30 minutes tops I'd just be done with it
Hey @Mike!
well, I'm out for the evening ... maybe back late. Bye all.
See you @Ted
Keep working, Demonark.
01:05
Will do
Mike's blue avatar is the most beautiful autogenerated one on SE.
The 3rd chapter is out.

The Story of Large Numbers, a dive into the unknown: 3rd chapter:

23 mins ago, 22 minutes total – 3 messages, 1 user, 0 stars

Bookmarked 12 secs ago by Simply Beautiful Art

01:21
@Ted I just worked out what you were saying about integrating along the fiber for the Poincare lemma and I'm like struck by this argument. It's so much prettier and conceptual than what I got when I first saw the proof of the lemma
02:08
Is it proper... language to say "$H$ is a nullhomotopy of $f$"?
 
3 hours later…
05:10
Demonark: Sure. ... @Eric: Glad you're happier with this — glad to discuss it with you.
05:29
Thanks @Ted!
Are you doing some algebraic topology? @dami
Differential topology
We do a pretty reasonable amount of homotopy stuff there as well
hi @Alessandro
05:57
Schwarz's Lemma
$f(z) $ is analytic inside $|z| < R$
with $f(0) = 0$
and $|f(z)| \leq M \forall |z| < R $
Then
$|f(z)| \leq M|z| / R$
$|f'(0)| \leq M/R$
and
So we can say nothing about $|f(z)|$ when $|z| = R$
?
Since $f(0) = 0$ we can define our $f$ as , $f(z) = az + bz^{2} + cz^{3} + ...$
Next we define $g(z) : \triangle \rightarrow C$ by
$g(z) : = f(z)/z $ for $0<|z| < R$
and $g(z) := f'(0) , |z| = 0$ or $z = 0$
Now $g(z)$ is analytic.
in $|z| < R$
$\leq M/r$
and max($|g(z)|$,$|z| = r$)
$\leq M/r$
so $|g(z)| \leq M/r$ for all $0<r \leq R$
for all $|z| < R$
Thus proving that $|f(z)| \leq M|z| /R$
and also as $|g(0)| \leq M/R$
and as $|g(0)| = |f'(0)|$
so $|f'(0)| \leq M/R$
and $|f(z)| \leq M|z|/R$
$|f'(0)| \leq M/R$
so we have our two inequalities
06:34
But still we had assumed $|f(z)| \leq M$ and now the reslt of the theorem tells us again regarding $|f(z)|$ that $|f(z)| \leq M|z|/R$ for $|z| < R$ ?
Why so that there are two inequalities of $|f(z)|$ one at the starting and other at the end
06:52
Hello anyone here?
Any hints on how to solve this problem?
Next we also observe that $g(z)$ is a constant function
07:08
If we show that it is a subfield then I think we are done @SantoshLinkha
oh but we do not know about order of $F$ ,so I think we can't tell that one.
all we know about is it's characterisitcs
*has p ...
I think I should post a question on this
yes sure
1
Q: Proving the roots of a polynomial are a subfield

FireGardenDefine $f(X)=X^{p^r}-X \in \mathbb{F}_p[X]$ where $r\geq 1,\ p$ prime. I'm to show its roots are a subfield of its splitting field $E$ (I have already shown it has $p^r$ distinct roots in $E$). What I currently have: $0,1$ are clearly roots of $f$. Suppose $a,b$ are roots of $f$. Then $(a+b)^{...

nope .. there already seems to be answer :D the world is big
but mathematically world is small :) six degrees
07:24
six degrees?
Six Degrees: The Science of a Connected Age (2004 in paperback, ISBN 0-393-32542-3 and 2003 in hardcover, ISBN 0-393-04142-5) is a popular science book by Duncan J. Watts covering the application of network theory to sociology. The book covers Watts' own work on small-world networks, and continues on to cover scale-free networks, network searching, epidemics and network failures, social decision making, thresholds in networks, and innovation in large organizations and its lack. In addition to covering the theoretical models and empirical case studies, the book also includes several stories about...
haha .. thanks, I did not know it
^^ anyway that question shows $X^q-X$ where $q=p^n$ as $p^n$ roots. Now I am asked to show that it is subfield with elements of form $p^s$ where $s \le n$, is this even possible?
First I thought of this
but that was possible when $|F| = p^{n}$
07:40
anyway I posted a question
0
Q: showing roots of polynomial $X^q-X$ is subfield

Santosh LinkhaI have an assignment question, Let $F$ be a field of characteristics $p$ prime (b) Show that roots of $X^{p^r}-X$ is a subfield of $F$ with, with $p^s$ elements where $s \le r$. (Hint example: assume F is finite field with $4$ elements, take $p=2, r=3$ and compute $s$. The part to show th...

08:05
@TedShifrin I am well, well aware. Didn't you realize I was joking? :P
08:49
Hello, is the sum from i=2 to n of $1/(i(i-1))=1-1/n$?
thanks
If you want to prove it: use partial fractions and cancel terms
i proved it, just wanted to check if it was right
for the series 1/(1+a^i), a>0 how can i show if it converges or diverges?
i tried 1/(1+a^i) < 1/a^i converges for |1/a|<1 but this isn't exactly right is it?
hello?
 
1 hour later…
10:22
It seems fine , what makes you think that you are missing some points@user379685
@BAYMAX shouldn't the 1+a^i in the denominator influence |1/a|<1
It is also intuitive
like if $|1/a|<1$
then $a$ is larger
and as you increase the power of $a$ in $1/(1 + a^{i})$
the terms becomes much smaller and smaller?
@user379685
ok i get it
ok.
do you have an idea how to show whether or not this converges $1/(|a|+(|a|+1)i)^ε$
10:33
I think it is of the order like
$1/i^{\epsilon}$
and if you are talking of the series - then it follows from $1/n^{p}$ test?
Like as $i$ increases then $|a| + (|a| +1)i$ will be roughly $(|a| + 1)i$
as $|a|$ can be neglected as compared to $(|a| + 1)i$ for large $i$
isn't that a bit handwavy, $1/(|a|+(|a|+1)i)^ε > 1/((|a|+1)i)^ε)$
derp nevermind
and last one
$(|a|^i+1)/(|b|^i-1)$
${1\over a^i}\sim {1\over 1+ a^i}$ if $a\le 1$, 1 otherwise
For the last one what do you see ?@user379685
The numerator is equivalent to either a^i or 1 and same for the bottom
in that series,like how can you approach it!
well now you would have got it !
10:46
thanks
glad to help
number of distinct sub-rings with uniy of cardinality a prime number $p$ ,does the field $F_{p^2}$ have ? I don't remember any theorem ?
which hints this?
I'd say at most two by lagrange since a sub ring is a subgroup
Wait I might be saying nonsense
11:15
@SimplyBeautifulArt No
Just a pun on "beta"
Hi @Akiva.
 
1 hour later…
12:33
oh hi
Is, why and how is √-1 called 'i'?
.
@RaisingAgent Yes. The full name is imaginary unit, so "i" for short.
 
1 hour later…
13:49
A popular meme recently
Now to check how general this phenomenon is:
@ShaVuklia Hey, awesome @ShaVuklia, how are you doing? :P
@Hippalectryon hey! How is it going? :-)
Consider the derivative
$$\frac{d}{dx}f$$
The equation that give the required symmetry is:

$$\frac{d}{dx}f=\frac{d}{d}\frac{f}{x}=\frac{f}{x}$$

Now to solve for all $f$ that satisfy this equation
Sorry typo, forgot a minus sign:

$\frac{d}{dx}f=\frac{d}{d}\frac{f}{x}=-\frac{f}{x}$
Solving this ODE gives $f=\frac{1}{x}+C$ therefore, this meme is indeed only possible for $f=\frac{1}{x}$
14:09
@Waiting heyyy:)
I'm doing pretty well! I am going to self-learn surface integrals the physics way... so I'm a little bit nervous :P
but apart from that, life's good:P how are you?
@Secret lol that might be a meme, but what physicists actually do isn't that much different :P
indeed, loosely using differentials for example in change of variables formulae of integrals
yes.. but I finaaaalllyyyy understand it. like, I have the mathematical tools to know what they're actually doing
so now I can do it too, though I still find it scary :P
@ShaVuklia Cool, you seem to have some fun there :P Well, I've been navigating on the see of some possibly strange mathematical problems. Apart from that, I'm thiking to go out this evening for some jogging.
It's relaxing :P
haha, sure it is!
I'll be studying till 21.00 at the university, and then I'll go home
I have not done any maths recently, the research group that I am in found some interesting results and we have been spending the past few days calculting intensely to figure out what the chemistry is telling us
14:17
which might be meaningless to you, because for all i know we live in different timezones :P
@ShaVuklia Oh, so late? You stay a lot there! :P
You're a chemist? @Secret
Lol yea, every day :P @Waiting
I am a chemistry PhD, thus yeah, I am a chemist
I just happened to like abstract algebra
which is why I hang around in maths chat a lot (well, used to be)
haha cool :P
Nowadays, the chemistry research started to get busy thus I am struggle to squeeze time to satisfy my maths curiosity
But the research progress is quite good
14:20
@ShaVuklia I love to study things alone, in these conditions I can access my little powers and get the problems and solutions I dream of. :P
ah that's good! I still have no idea what direction I'll go, math, physics, or a combination of both
@Waiting me too! I skip school 80% of the time, and I just study here at the library of the university. I prefer this forum over our school :P
@ShaVuklia like me :P However I'm not that much in here these days.
Wow I am opposite to you guys then, back in my undergrad, I am well known among my classmates to never have skipped a lesson nor lecture
@ShaVuklia In general my mind works like that: I have to understand things in a particular way that defines me. I cannot follow the teaching methods usually used today.
@ShaVuklia Lemme know if you want some problems to cut your teeth on, I've got plenty
14:25
For me, I have a certain way to understand things, and via interacting with the class, somehow I can shape the class dynamics towards the thinking process that is similar to mine
@Secret hm, yea I have a very "stubborn" nature of mind, so when someone tells me to do something, I oppose mentally :P I have actually "failed" a lot this year because I didn't follow the university program, but the things that I do do, I do well :P
@Secret hehe
@Semiclassical ohh, yesterday I got such a headache (I was tired, it was 24.00 after all:P) so I couldn't entire follow what you said about the envelope and the maxima
I am going to read it today though
I am equally fine following the curriculum or not following it, but I tend to work a lot better in groups than alone (where I tend to easily get distracted)
@Waiting well as you can read, we are alike!!:)
14:27
Mmkay.
@ShaVuklia yeah, we have a lot of things in common, that's obvious! :P
If you've got access to graphing software, you might try plotting $e^{-\gamma t}C\cos(\omega t+\phi)$ for some specific $\gamma, \omega, C$ but varying $\phi$.
I am also famous for asking the most questions in class. That's really how I can learn from my undergrad so far, and the professors are often nice to tell us more about off curriculum stuffs
(Actually, you can drop the C in there since it's just an overall constant. can't affect the overall shape of the graph.)
If you don't, I can mock something up quick.
@Secret I highly believe some people are meant to educate themselves, to learn things in a very particular way which cannot be obtained by the teaching ways (in uni) known today.
14:30
That's true. When one self learn, the thrill of "aha" moments is very satisfying
Perfectly true!
(e.g. I self learn abstract algebra so far (modulo discussion with maths chat people))
My self learning abilities, however are not as good as my peers, which is why I often rely a little bit on the curriculum to give basic tools for me to get started, which I can then use them and the mindset the developed when I interact with them to generate my own solutions
If you stick around in grad school, you definitely have to be good at self-learning. That's what research is, really.
Which can be a bit of an awkward transition, because if you go to grad school you were probably also good at taking classes.
For example, if someone tell me to learn classical mechanics from scratch, I will struggle (although given *sufficient* time, I will get there), but a little bit of crash course on that, I can pick up with the self learn afterwards

Semiclassical: That's true, take my current PhD for example, I need to learn bash just to get my chemistry calcculation jobs to run and to navigate the computer clusters on terminal
@Semiclassical Heya, you here?
14:34
well, yes.
@Secret To let your mind run wildly through all the apparently craziest ideas. One of the things I hear very often from those I share my problems with is: How on earth could you come up with that? The problem is that I'm only natural in my own way, no special kind of effort. All these are a result of self-education.
Great! Mind briefly looking at an integral? =)
Maple keeps giving me the wrong answer :p
@Waiting Every people has their own natural way to solve problems, some of these can be described and shared easily to others, others are more artistic and thus is entirely personal
14:36
Well, I'll probably go and check it via Mathematica, but go ahead.
For me, I tend to have explosion of ideas like a ordinal hydra growing out of control. It is literally chaos, because to outsiders it seems very incoherent and nonsensical, there's an order that only I can understand
That, however does not stop me from trying to explain what they are, with mixed success
@ShaVuklia did you watch The Man Who Knew Infinity eventually? :-)
$$ \frac{1}{(2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} | e^{-ix} + e^{-iy} | \,\mathrm{d}x \, \mathrm{d}y = \frac{1}{2\pi}\int_{0}^{2\pi} | 1 + e^{-ix} |\,\mathrm{d}x = \frac{4}{\pi}$$
For example, it has been 3 years since I have met you guys here, an I do noticed some of you are "partly assimilated" into my weirdness as you guys become more readily able to understand my thoughts
The second equality is straight forward just rewrite the integrand as $|2\cos(x/2)|$, however I can not for the life of me see the first equality.
14:39
Hmm, yeah.
I confirmed the result with Wolfram Alpha
It's definitely not 'obvious' to me.
While for me I have yet to came up with a truly artistic problem solving method (because I am a very analytic personality more than emotional, thus all problem solving methods I came up so far must have the property to be describable or i will be confused), I believe it will soon come t me given how in past years I have been self studying fine art and talking with artists
I guess the first thing I'd do is rewrite the first integrand as $\sqrt{(e^{-i x}+e^{-i y})(e^{i x}+e^{i y})}=\sqrt{2+2\cos(x-y)}$.
Actually, no, there's a simpler observation.
First, note that $|e^{-i x}+e^{-i y}|=|e^{-i x+i y}+1|$. So the integral is equivalent to $$\frac{1}{(2\pi)^2}\int_0^{2\pi}\int_0^{2\pi}|e^{-i (x-y)}+1|\,dxdy$$
@Semiclassical Because $|e^{-iy}|=1$?
14:44
Right.
We then redefine the $x$ variable as $x\mapsto x+y$. (If you're not comfortable with that, let $u=x-y$ and use $u$ instead.)
@Secret it's hard to describe accurately and entirely how the artistic part come to the surface. First it is about very hard work and a burning passion, and then there is something special within hard to define (think of Ramanujan, he had something special within, like a gift).
(You can check via the Jacobian that you don't have to include a substitution rule factor, though.)
@Semiclassical It is just a linear shift all after all
Right. But this does change the integrand and the limits.
$$\frac{1}{(2\pi)^2}\int_y^{2\pi+y}\int_0^{2\pi}|e^{-i x}+1|\,dxdy$$
Should not not just equate to a rotation around the unit circle, eg in the end the limits stay the same? <- some handwaving in that argument
14:48
Should, yeah.
Argue by periodicity that integrating from y to 2pi+y is the same as from 0 to 2pi
And then it simplifies down into the written form, thanks. Since the inner integrl does not deppend upon y
Once you've done that step, yeah.
Just becomes the fact that the average value of 1 on any interval is just 1.
Indeed thanks, I need this result in my thesis actually =)
Neat.
It may be worthwhile to mention one other thing.
Suppose you were to make a contour plot of $|e^{-i x}+e^{-i y}|$ over a few unit cells.
Guys, say we have an 8-regular graph (say with $k$ vertices), could someone give me hint how I can show that there is a 4-regular subgraph with all the $k$ vertices?
I was thinking about using the fact that we know that there is an Euler cycle for this graph
So I should be able to "half" this cycle twice, I guess
14:52
What the above change of variables does is basically redefine your unit cell to be oblique rather than just a square. Once you've done that, the integrand doesn't change along the oblique axis and you can get rid of that integration trivially.
Makes sense!
Looking forward to finishing my thesis so I can back into translating my notes on integration
Due to the symmetry
$$\frac{1}{(2\pi)^2} \int_0^{2\pi} \int_0^{2\pi} | e^{-ix} + e^{-iy} | \,\mathrm{d}x \, \mathrm{d}y =\frac{2}{(2\pi)^2} \int_0^{2\pi} \int_0^{y} | e^{-ix} + e^{-iy} | \,\mathrm{d}x \, \mathrm{d}y = \frac{1}{2\pi}\int_{0}^{2\pi} | 1 + e^{-ix} |\,\mathrm{d}x$$
I decided to do the plot in Mathematica :P
Then integrate by parts for $$ \int_0^{2\pi} y' \int_0^{y} | e^{-ix} + e^{-iy} | \,\mathrm{d}x \, \mathrm{d}y$$
Q.E.D.
I think the observation of @Waiting amounts to the fact the function is symmetric about $x=y$, so one can restrict the integration to the triangle overlap between the two functions.
15:02
Yeap
@N3buchadnezzar Out of curiosity, what's the context of this computation? Makes me think of Brillouin zone stuff in solid state physics.
@Semiclassical It is the $H_1$ norm of the function $\varphi = z_1 + z_2$, where $H$ stands for the Hardy space.
Huh.
I only vaguely know about that stuff.
$\|f\|_{H_p(\T^d)} = \frac{1}{(2\pi)^d} \int_0^{2\pi} \cdots \int_0^{2\pi} | f(z_1,z_2, \ldots , z_d) |^p \, \mathrm{d}z_1 \, \cdots \mathrm{d}z_d$
Gotcha.
In this case $p=1,d=2$ and $f(z_1,z_2)=z_1+z_2$.
15:10
More succincly written as $$\|f\|_{H^p(\mathrm{T}^d} = \int_{\mathrm{d}^T} |f|^p \mathrm{d}m(z) $$
@Semiclassical Precicely.
I think this could be translated into solid state physics terms, and I wouldn't be surprised if there's a source out there that does just that.
I study some properties of some functions that live in this space. Eg the analytic functions whose average on the unit disk is finite.
But hardy spaces aren't something that a physicist would usually recognize.
I do, a little, but that's extracurricular.
Well they are working pretty much over Banach spaces.
Right. Not that, again, we'd ever bother to call them Banach spaces.
The farthest a physicist would usually go is Hilbert space and that's not the same.
15:12
In the little physics we had in quantum theory we worked with Hilbert spaces, and to some extent Banach. Mostly just noting that things where in $L^p$ and converged to a satisfying degr3ee.
I think you do run into this stuff a bit when you do index stuff, though.
Something something Wiener algebra.
Most physicists at my uni know Banach spaces, because they need Hilbert spaces and Banach is the stepping stone from from vector spaces to Hilbert.
I think it depends on whether you learned about Hilbert spaces in a math course or in a physics course first.
If you learn about it in a math course, then probably you do learn about Banach spaces.
If you just pick it up from quantum, though, then I could see Banach being skipped over.
I have to go. Take care @ShaVuklia :P
Yeah, that's true.
It also reminds me of the nuclear physicist teaching my course on quantum, who called a Hilbert space a "vector space that's closed under addition".
15:15
Byee! @Waiting
Hey @Steamy are you good at graph theory?:P (and do you have time XD)
One could clearly tell from the faces of the student if they were mathematicians or not
Not really, I'm afraid. Only really know stuff about Cayley graphs
hmmm:(
@SteamyRoot what do mathematician faces look like?
@SteamyRoot Oof.
Setting aside the patent silliness of that statement, the fact that a Hilbert space is equipped with an inner product is a Big Deal in physics.
At that moment, it was something between pure horror, anger, and disappointment
15:18
Without it, you wouldn't be able to compute matrix elements / transition amplitudes.
It was a mandatory course, and that situation happened the very first lecture... No chance to drop the course.
Reminds me of going to an art museum in college, and having our guide show us this painting:
yea that's what I saw too, but we haven't had these theorems yet :(
15:20
And somehow describing Euclid as being from the Renaissance. :/
and some of the concepts he uses we're not familiar with
so I was hoping there is a more "elementary" approach
(I may not be remembering what was said properly, but I remember it was just terribly off and I was the only one of us who knew enough math history to be properly insulted.)
Looking at a description of the painting, maybe the point they were intending was that the notion of perspective drawing was a Renaissance invention. So I may be being unfair, but I remember being annoyed at the time.
@Semiclassical Hehe Wiener
You can always be annoyed at the lack of rigour and unambiguity
15:25
Do you mean ambiguity?
I think that should be read as lack of (rigour + unambiguity)
Yeah...
Stupid languages being ambiguous too.
lol.
probably simplest to say: annoyed at ambiguity and lack of rigor.
Depends on what you're doing, though. For stories, I don't mind ambiguity.
Double negatives for the win
For things that purport to be authoritative, though, ambiguity is frustrating.
15:28
Exactly
Hullo thar
Hello Sweden
About the graph proof; I don’t understand what they mean by: “For each directed arc $(u,v)$ in our digraph $D(G)$, draw the edge $(u_r,v_b)\in E(G’)$.
How do we know we might not be having $(u_r,v_r)$? Are they saying we’re only going to consider the edges between red and blue vertices?
Every vertex is split, so there must exist both a red and blue one
@Semiclassical I was first suprised when I thought that clip seemed to float in midair without any suport, then the transparent canvas impression materlised and finally reading the description told me it is a painting
15:35
@Daminark yea I understand, but they say they do it for each directed arc $(u,v)$. Shouldn't they then say, only for the $(u_r,v_b)$ and $(u_b,v_r)$ we draw the associated vertices in $E$?
Well, the idea is that $u$ and $v$ are vertices in your original graph (which is now a digraph), so if there's a path from $u$ to $v$, draw an edge between $u_r$ and $v_b$
Like you manufactured a new graph and are building it up on basis of information from the first one
At least this is how I parse it out
Last night dream seemed to play with a pretty grim type of eternity that existence itself will eventually reach: In a white void, no matter which direction you look, all you see is a countable lattice of $\infty$ symbols
@Daminark ohhh, I thought a directed arc was an edge:P thanks!
15:50
The dream also showed that this $\infty$ acts like some kind of ultimate maximal element of the ordinals since it will absorb any ordinal number under any type of operations, i.e. it is a fixed point to all conceivable maps
@Secret Yeah, it's a neat painting. Especially if you see it in person and up close.
@Daminark wait, I think you're wrong:P
it's not about a directed path, but about a directed edge
I mean I guess it's how use parse the term "arc"
I've never seen that word before, but arc feels more like a path than an edge
Either way, though, I think there shouldn't be a problem
yea I thought it was a path too, but then I googled it:P
and it makes more sense that it's an edge
(Why can't they just say edge tho?)
(Don't make our lives confusing o writer of the textbook)

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