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12:52 AM
@leslietownes more like cateboring
 
1:29 AM
@Obliv Figures that when I ask for help I'm kicked out where I got the question.
Anyways, I've got another one. Its involving finding difference, but as I've learned its a trick. I got this far with the problem, the invisible 1 changed some of the signs. $-2d^3 + 6d - 9 + 7d^2 - 5d - 13$
Originally it was $(-2d^3 + 6d - 9) - (-7d^2 + 5d + 13)$. I got this far alone and want to confirm my thought process, so I'd like a second opinion.
 
the core principle with polynomial arithmetic is recognizing what the monomials are. in one variable d, that's powers of d, in a multivariable setting it is subtler.
the addition is defined so that you can "collect" like monomials and "add their coefficients." but none of this is well-posed without a lot of background understanding. "6a^4 b^4 +(8a^4 b^4 - 9b)" is not in anyway disqualified as an answer to an ill-formed question that just asks you to "add" those polynomials
 
Agreed. That is an ill-posed question.
 
same reason that "31 - 26/2" is a perfectly good response to "what's 31 - 26/2." specifying the form of an answer to the question is 99% of the question.
 
Were I to give a similar question to my students, the instruction would likely be something like "Simplify the following polynomial expression so that there is only one of any term of the form $Ca^m b^n$, where $C$ is a real number, and $m$ and $n$ are natural numbers."
I would need to word smith that a bit more, as it is a bit hard to grok, but I think that the general intention of the question is to get you to "combine like terms".
In other news, it is currently 77°F inside. Is it time to turn on the AC?
 
@XanderHenderson THANK YOU! Thats what I thought but wasn't sure. As soon as I finish these, I'm goin to sleep.
 
1:42 AM
@YourLordJoyBoy Again, I don't know that this is what the question writer is asking, but that is my best guess.
 
@YourLordJoyBoy Yeah so you basically distribute the minus sign. Also the first parentheses is just another implicit $1(...)$ so we just have $-2d^3+6d-9-1(-7d^2+5d+13)=-2d^3+6d-9+(-1)(-7)d^2+(-1)(5)d+(-1)(13) = -2d^3+6d-9+7d^2-5d-13)$
 
@XanderHenderson @Obliv No worries guys, I got it. :D I combined em and it ended up being $āˆ’2d^3+7d^2+dāˆ’22$
 
Actually, I would probably give the instruction "Rewrite the polynomial in standard form," where "standard form" is something that would have been discussed previously in class---roughly, write the terms in order of decreasing degree, and if two terms have the same degree, write them in lexicographical order; no two terms should contain precisely the same combination of variables, i.e. collect "like terms").
 
@XanderHenderson I believe that was in standard form, the final answer I mean. And now I'm doing a similar one, and trying to combine $9/7z with 15z$. Other than that the problem is cake.
$9/7z with 15z$
 
yeah, generally speaking, you make a problem like that unambiguous by specifying the desired form of the answer
we can all enjoy the ambiguity of "9/7z"
 
1:49 AM
@leslietownes a fraction combining with a number, each with the variable z
$9/7$
 
That clearly means $$ \frac{9}{z}\cdot 7.$$ Doi.
Hrm... or maybe $$\left\{ \frac{9}{7}, \frac{9}{z}\right\}.$$ Yeah, okay. I agree. It's ambiguous.
 
@XanderHenderson How do you do the fraction bit? And I'm trying to combine the fraction that's multiplied by z with 15z.
 
@YourLordJoyBoy I think that, perhaps, the joke has flown past you a bit too quickly. Something that you should keep in mind is that Xander lies.
 
@XanderHenderson To be fair, any joke'll pass me quick when I'm this tired. XD I'm tempted to dub you Usopp now with the lying habit though.
 
@YourLordJoyBoy Sadly, your joke flew over my head.
Is Usopp some gen alpha thing?
 
1:56 AM
@XanderHenderson Consider it vengeance XD
I'll probably explain it in time
Unless someone else gets it and decides to.
 
Google says its some shonen BS.
 
@XanderHenderson Yes, its a character who is always telling tall tales. XD
 
(The question is, how many otaku / weebs do I piss off by calling One Piece 'some shonen BS'?)
 
@XanderHenderson Not me since I could care less who hates it XD
$$\frac{9}{7}{z} + {15}{z}$$ Wonder if I did this right
$$\frac{9}{7}{z} + {15}{z}$$
ABSOLUTELY SCREWED THAT UP! :o
FTR, only adding a star to those two messages.
 
@YourLordJoyBoy I don't hate it. I simply have no feelings toward it at all.
I bounce off of most shonen BS pretty hard.
 
2:05 AM
In other words, @XanderHenderson just no care.
For me however, it heavily impacted meh life during hard times.
And you're still Usopp to me XD
But seriously its one of those things, you're going through crap times irl and something crazy pops up that proves to be a fantastic escape from the reality bs in your life.
 
2:20 AM
@XanderHenderson I don't care about One Piece, but I'll take issue on general grounds of reductiveness
 
3:08 AM
@Thorgott Good. That was the reaction I was hoping for. :D
 
me: modifies someone else's Tikz code to make a mass spectrometer
result:
 
@Semiclassical I might try to learn Tikz someday. But I learned pstricks such a long time ago, and Tikz looks so very annoyingly not at all the same.
 
That looks like a butt.
 
i'd agree, except for the shaft going upward. but that makes it worse
 
3:11 AM
Oh, sure. It could be a dick'n'balls.
I see that, now that you mention it.
Maybe it's both things!
Right. I need to go to bed. It is so late.
It is after EIGHT PM!
 
 
2 hours later…
5:15 AM
when people say the killing form is "ad" invariant, do they mean little or big adjoint?
concretely, with matrices little ad is the commutator over lie algebra $[X, -]$ and big adjoint is lie group conjugation over lie algebra $g(-)g^{-1}$
 
 
5 hours later…
9:50 AM
modular form on disk
keyword: j-invariant
 
 
1 hour later…
11:04 AM
@SineoftheTime Do you like Caparezza?
 
 
2 hours later…
1:24 PM
What is this figure called?
 
@SoumikMukherjee A "book cover".
 
@XanderHenderson There goes Usopp again XD
 
@XanderHenderson Sorry about my behaviour the other day
Due to the current political scenario of my country, I get tilted very easily
 
@SoumikMukherjee don't discuss politics as we know it is total mess in our country. We have best laws but worst implementation.
@SoumikMukherjee @XanderHenderson doesn't care about it.
Have some Maaza!!
 
1:33 PM
@LuckyChouhan General people not discussing about politics enough is what that led to this situation
 
Three Potato
 
Also we don't have the best laws, eg uapa
 
@SoumikMukherjee its called a link
its a collection of knots
 
@LuckyChouhan Anyway I am sorry to you as well. I shouldn't have talked to you in that rude way.
 
1:36 PM
@SoumikMukherjee the name here has an error. Its Karol Sieklucki
 
I'm really frustrated with the political bs we're dealing with on my end. But yeah...
 
@Jakobian oh thanks
 
In mathematical knot theory, a link is a collection of knots which do not intersect, but which may be linked (or knotted) together. A knot can be described as a link with one component. Links and knots are studied in a branch of mathematics called knot theory. Implicit in this definition is that there is a trivial reference link, usually called the unlink, but the word is also sometimes used in context where there is no notion of a trivial link. For example, a co-dimension 2 link in 3-dimensional space is a subspace of 3-dimensional Euclidean space (or often the 3-sphere) whose connected components...
 
Government has nothing to do with climate change, biodiversity, education, etc. They just focus on Trading and GDP lol, I don't know where we're going. Do you know what recently happened in Dubai? and few months ago in Joshimath. Many more such incidence yet to come. Anyway, who cares?
 
@Jakobian Ooh, never noticed that
 
1:37 PM
@SoumikMukherjee Never mind,
 
@Jakobian It's more than just a link. A knot is homeomorphic to a circle, and that figure contains more than just circles. I looks more like the complement of a link in a handlebody.
 
@XanderHenderson šŸ¤—šŸ¤—
 
@XanderHenderson nah its a link in $\{x\in\mathbb{R}^3 : 1 < \|x\| < 2\}$
 
I don't buy that. Based on the overlaps, the bits between the two circles are meant to be embedded into a three dimensional space (there are four circles, linked together), and the surrounding annulus looks like a handlebody, i.e. a solid torus.
I think it is mean to be a link complement.
 
Sure its more probable that this is inside a cyllinder without the center of the cyllinder
but its a link, just embedded somewhere
that's how it looks like
$\{x\in \mathbb{R}^3 : 1 < x_1^2+x_2^2 < 2\}$ lets say
 
1:44 PM
One of the most interesting problems I've seen in AT is that $\Bbb CP^n$ is not a covering space of any manifold except itself. It's not very intuitive at first but if you know any continuous map on $\Bbb CP^n$ to $\Bbb CP^n$ has a fixed point (Lefschetz I guess) then it's believable
 
Given that knot complements are related to the classification of 3-manifolds, I would expect that the image is a link complement. But, at this point, we are each saying "No it isn't!" "Yes it is!" with no references, so this seems unproductive.
 
now it's time for Thorgott to say that it's a consequence of some more general fact
2
 
Also, I dare @XanderHenderson to look up what a Joy Boy is
 
there're two types of general methods to create closed 3-manifold, one is dehn filling link complement and the other is heegaard splitting. Any closed 3-manifold can be realized in both two ways
 
@XanderHenderson I mean it might be a solid torus but I'm not sure why you brought up link complements
 
1:49 PM
the latter is quite straightforward if you know that any 3-manifold has a triangulation
 
4 mins ago, by Xander Henderson
Given that knot complements are related to the classification of 3-manifolds, I would expect that the image is a link complement. But, at this point, we are each saying "No it isn't!" "Yes it is!" with no references, so this seems unproductive.
 
I mean there's no real way from a picture to tell between a set and its complement. Both are drawn, necessarily, if one is
 
I want to read the conformal dimension book you suggested @XanderHenderson long time ago but first the book is not very easy and second I'm studying different stuff so hard to make a spare time
 
Was working with Gwen, she's really helpful with the factoring :D
 
@SoumikMukherjee @XanderHenderson I found exactly what the picture represents
Its the construction of the so called Antoine necklace
In mathematics Antoine's necklace is a topological embedding of the Cantor set in 3-dimensional Euclidean space, whose complement is not simply connected. It also serves as a counterexample to the claim that all Cantor spaces are ambiently homeomorphic to each other. It was discovered by Louis Antoine (1921). == Construction == Antoine's necklace is constructed iteratively like so: Begin with a solid torus A0 (iteration 0). Next, construct a "necklace" of smaller, linked tori that lie inside A0. This necklace is A1 (iteration 1). Each torus composing A1 can be replaced with another smaller necklace...
so we were wrong that those were links
(the same illustration is under an exercise in the book which asks to verify details of this construction)
 
2:06 PM
seems the construction is not very hard
 
@YourLordJoyBoy its a relatively new user and no one will know who you are talking about when you say "Gwen". Its not a regular user in this chatroom for sure
 
Yes I'm aware
When I say Gwen, thats who I speak of @Jakobian
 
@LuckyChouhan discussing politics is allowed, its just about peace of the chatroom itself. Since this can lead to controversial opinions and arguing with each other, which is what the actual problem is.
 
@Jakobian Which is why I hardly speak of it. I'm surprised I didn't drive everyone into a frenzy when I talked about how humans and making stuff XD
I'm quite thankful for that actually.
 
how humans and making stuff
 
2:12 PM
How humans didn't make anything without taking something that already exists to make it.
We simply turned something existing into something else, but made nothing new except to us.
 
"but made nothing new except to us." this is where you're wrong
when an artist makes a stone statue, they only carve the stone that already existed
and its making something new. And "except to us" makes no sense
its new for us just as it is new for a pigeon or any animal
what is "except to us" supposed to mean here? Eh
 
It's new to us. But the materials were there to begin with.
We didn't just make something out of nothing, that would be mind blowing.
 
so what
 
Simply stating.
Nothing more or less.
 
In a sense we did make something out of nothing if we give some kind of structure to an already existing thing
 
2:20 PM
I suppose that's true/
I do wish I could post my study guide on here.
 
Not truly out of nothing since it requires resources. But the point is that changing something still counts as creating something
 
@Jakobian Yes. Agreed. But what if we could make something into existance without any previous resources?
 
why am I supposed to care about that
 
@Jakobian I just figured if it could be done, it would be AMAZING!
 
Basic question. Are singleton sets closed in $\overline{\mathbb{R}}^n$?
The reason I'm asking is because it's claimed $A=\{(x,x)\in\overline{\mathbb{R}}^2:x\in \overline{\mathbb{R}}\}$ is measurable since it's closed. If we replace $\overline{\mathbb{R}}$ with $\mathbb R$ in $A$, call it $A'$, I know how to show it's closed. However, I'm not sure about $A$. If the singleton sets are closed in $\overline{\mathbb{R}}^2$, then $A$ would simply be a finite union $(-\infty,-\infty)\cup A'\cup(\infty,\infty)$, which is indeed closed.
 
2:26 PM
that will never happen so its pointless and meaningless to ponder it
@psie yes
 
ok, thanks
 
$\overline{\mathbb{R}}^n$ is a metrizable space
 
@Jakobian Or say, if we could make stars. Or a planet.
 
@psie you mean one-point compactification here by the way, right?
 
yeah, probably
 
2:28 PM
oh you mean product of extended real lines
@psie no you don't mean that
 
yeah, I'm unsure
I can post a screenshot
 
$\overline{\mathbb{R}}$ is a metrizable space
it fact its homeomorphic to $[0, 1]$
and the product you wrote is homeomorphic to $[0, 1]^n$
still metrizable
all good properties you expect from metric spaces hold in this case
 
ok
 
@psie $A$ is closed because $\overline{\mathbb{R}}$ is Hausdorff
having closed singletons is not enough
your argument is wrong here
For a space $X$, the diagonal $\Delta_X = \{(x, x) : x\in X\}$ is closed in $X\times X$ iff $X$ is Hausdorff
 
ok, I should look up a proof of that
 
2:34 PM
@Jakobian Thanks
 
@psie Its pretty simple, if $x, y$ are distinct points find open disjoint $U, V$ with $x\in U$ and $y\in V$, then $U\times V\subseteq (\Delta_X)^c$, the complement of the diagonal
and this shows that the diagonal is closed
and converse is not needed for you anyway
(but the converse is also really simple, just take $(x, y)\in (\Delta_X)^c$ and then there is a basic open set $U\times V$ with $(x, y)\in U\times V\subseteq (\Delta_X)^c$)
basic means element of the basis
 
alright, neat, thanks
 
you know for product topology on $X\times Y$ those are products of open sets
 
indeed
 
3:03 PM
Another day, another attempt at editing posts.
 
 
2 hours later…
4:39 PM
@onepotatotwopotato I don't know any!
all arguments I've ever seen involved in showing that a space does not cover other spaces non-trivially were usually rather ad hoc
 
5:23 PM
how to solve 2?
 
 
2 hours later…
7:20 PM
0
Q: analytically continue functions on independent planes in $\Bbb R^3$

John ZimmermanI know techniques to analytically continue some functions off the positive real line. I am less sure given I have 2 independent euclidean planes sitting in $\Bbb R^3$ with some function $f$ defined on the positive real lines of the respective planes. For ex. I can continue $f$ on each plane indep...

I still don't get this
 
7:38 PM
I know the concept exists already but I don't know where to find it online
 
7:50 PM
Let $n \ge 4$. Take $X=(0,1)^n.$ Fix all points $p,q$ s.t. sup $\text{dist}_n(p,q)=\sqrt{n}.$ Does there exist a smooth regular codim. one foliation of $X$ whose leaves are topologically $J=(0,\sqrt{n})\times S^{n-2} $ accumulating to $p,q$? For all $n\ge 4$ is $J$ convex?
 
8:12 PM
@robjohn @XanderHenderson why Michael Hardy was suspended? He has answered more than 7000 questions on MSE.
Can moderators suspend themselves?
 
you can request to be suspended
 
@LuckyChouhan We don't publicly discuss the reasons for suspensions.
@LuckyChouhan In theory, sure, I guess? Maybe? I have no idea.
 
8:43 PM
Anyone ever try to recall information from earlier in the year when studying only to struggle? That's what's happening to me.
 
9:38 PM
@YourLordJoyBoy I can't remember
 
9:58 PM
@LuckyChouhan people with a lot of reputation are usually suspended for answering low quality questions
 
10:19 PM
@YourLordJoyBoy not surprising
knowledge is not eternal
 
@Jakobian The more recent stuff is especially easy to remember because we just did a quiz on it yesterday.
And I know, but not having said knowledge can and will cost me if I don't do something.
 
only way to remember something good is to reinforce that knowledge
 
That can be a challenge when you're in college.
 
10:36 PM
I'm pretty much having to go back and rewatch previous videos of my class from this year, all of which feature me.
@Jakobian On another note, picture this. The right answer to a quiz or exam question is not available on said exam/quiz.
 
@YourLordJoyBoy what are you talking about
 
@Jakobian Something that happened on one of my quizzes.
The right answer wasn't a choice!
Oh it made me so mad.
 
11:10 PM
Um guys, does this meet site standards and ruling? math.stackexchange.com/questions/4905709/…
 

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