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7:01 PM
Well, I'm consistent and complete.
 
@Gregor Percic, But are you one of the very, very few humans who was able to tell my argument was invalid without the extra premise "H is not me".
 
(Whether or not I can do arithmetic remains to be seen)
 
@user107952 Pretty sure you are more rigorous than yourself
 
$>$, not $\ge$
 
>>>
 
7:02 PM
You can't be strictly more rigorous than yourself, but you sure are more rigorous than yourself
 
Also, I must admit, I was surprised when I looked at the questions you asked on this site and saw that you looked pretty sane
6
 
I always mean more as in strictly more.
 
Not everyone does, so your statement isn't too true
 
@user107952 Okay then. H is not me. Gödel's theorems still apply. Also, reflexivity axiom has been broken. Hence This statement cannot be proven from axioms horror is upon us.
 
@GregorPerčič Gödel doesn't say that everything can't be proven from the axioms
It just says that there exist things that can't be proven from the axioms
 
7:04 PM
@user107952 Strictly you are rigorous. But are you less so than the more?
If yes, what is more rigorous than lesser?
I am sure you will agree.
 
Actually guys, how is rigor defined as a mathematical object?
 
@AkivaWeinberger True, but that makes the whole complex in which provable statements are entangled as well in mathematical sense incomplete.
 
(For reference) First Incompleteness Theorem: "Any consistent formal system F within which a certain amount of elementary arithmetic can be carried out is incomplete; i.e., there are statements of the language of F which can neither be proved nor disproved in F."
 
Don't you need an extra hypothesis
Like, you need that it should be decidable whether or not a given statement is an axiom
Otherwise, you could make every true statement an axiom, which would make it complete trivially
 
That's from Wikipedia, to clarify, citing from somewhere else
 
7:07 PM
@AkivaWeinberger Well, no, technically, it does not say that not everything that cannot be proven without axioms do not exist.
 
@user107952 That's what I said!
 
or more specifically, can we define a measure such that it spits out a value on how rigorous a given input mathematical object is?
 
@BalarkaSen not bothering to parse the negatives Right, it could be that it's all inconsistent, so you could prove everything
 
So, I'm playing with fire here, but basically I believe that if you think about it via models, it says that an axiomatic system powerful enough to contain arithmetic cannot define a single model
 
@user107952 Could you cite me a paper that describes a mathematical system of "true arithmetic"? Because Gödel clearly proved in the contex of Peano axioms that some statements in it are clearly true but cannot be proven from axioms.
 
7:09 PM
Either it's empty, meaning inconsistent, or it defines a class of mutually incompatible models
 
@Daminark Model categories can be embedded in $\infty$-topoi. Your argoment is invalid.
 
Ohi @Dami
 
Foolish fools!
 
@Daminark Define being the key word. Seems fine, though I think you need some amount of choice in the metalogic
 
@user107952 Stop trolling. Have a timeout.
 
7:10 PM
Also, decidability
 
@user107952 Also don't insult people.
 
Thanks, @SevenSidedDie.
 
[insert absurdly pretentious claim here]
 
@BalarkaSen *hat tip*
 
@Daminark What did you learn today? :)
 
7:10 PM
@Semiclassical Oh, that's the word I was looking for
Thanks
 
Pretentious?
 
Claim
 
Here?
 
Haven't started yet, this came from my just staring at the word model, not at all official banter
 
That's just absurd
 
7:12 PM
(Or insert)
@Semiclassical Yes.
 
I'll ask my rigor definition question later...
 
@Daminark Ah OK
 
It's a good word for this moment, yeah
 
@SevenSidedDie Your username is messing with my brain a little
 
Kek @Balarka and @Astyx
 
7:13 PM
@AkivaWeinberger Is it? How? :)
 
I mean, I could have a hexagonal pyramid, but that's not much good as a die
 
neh it's biased
heavily
 
Today I'll have number theory and... Something else
There was probability this morning (read 1:15PM) that I didn't go to
 
@Daminark Why do you not have an REU talk on infinity-categories?
 
Gödel basically killed mathematical utopism and is now dancing on its corpse.
 
7:14 PM
@AkivaWeinberger The most common d7 is a hex pentagonal prism of just the right proportions. It's not the most fair die when made with typical manufacturing tolerances, but it's actually not bad. There are regular polyhedrons manufactured as dice that are less fair.
 
Little puzzle : does there exist an uncountable amount of infinite parts of $\Bbb N$ $(A_i)_{i\in I}$, where if $i,j \in I^2, i\ne j$ then $A_i\cap A_j$ is finite ?
 
Before asking why, ask if
 
Though I'm fairly mellow about logic mumbo-jumbo. (Quantum mechanics nonsense, by contrast, gets under my skin something fierce.)
 
@Daminark Yeah screw probability
 
I think based on the unexpected response on that set theory integral [random], it tells me that some time later I need to tell you guys how to figure out which [random] you should ignore and which don't
 
7:15 PM
(Not really; I'm just bad at it so I am entitled to hate on it)
 
@SevenSidedDie Why not just get a d8 of some sort and roll again if you land on the extra side
 
Actually Lawler seemed to have made it rather dank. He introduced some kinda distribution via the probability of a random element of $S_n$ having no fixed points and all
 
@AkivaWeinberger That's the most practical way of doing it, but it doesn't satisfy the urges for collecting unusual dice. :) I have two, but I've honestly never used them as randomisers.
 
@Astyx By infinite parts, you mean $\cup A_i = \Bbb N$ where $A_i$ is an infinite family?
 
But perhaps, I think next time if similar trolls are encountered, the better solution is to flag them...
 
7:17 PM
Also, if you require dice to have symmetry to be labeled "fair," I think the maximum possible amount of sides is 120
which is sad
(Not counting dipyramids, which are harder to use as dice)
 
I think my favorite bit of combinatorics is generating functions
 
No, I mean $A_i \subset \Bbb N$ is infinite for all $i\in I$ and $I$ is uncountable
 
If only because it bridges counting, algebra, and complex analysis
 
@Semiclassical Does not saying so violate the rule of stackexchange? I meant typing typing "Foolish fools"
 
Also I was kidding earlier there are no talks on category stuff. Gotta go because number theory
 
7:19 PM
@Astyx Ah, I see.
 
@AkivaWeinberger Barrel dice are symmetrical and theoretically have no upper limit, but eventually you have a practical problem of a) it rolls for an unreasonable amount of time, b) it's hard to tell which side is up in order to read the result.
 
I'm acting under the presumption that no one would take that statement seriously
 
What's a barrel die? Is it a prism not counting the bases?
 
@Daminark See ya. Tell me about it after you get back. Ideally on the NT room.
 
7:20 PM
Also, fun fact, there exists just one more 7-sided convex polyhedron with regular polygons as sides
 
or use a deck of cards.
 
The elongated triangular pyramid (a tetrahedron = triangular pyramid glued to a triangular prism)
Not good for dice either, though
 
Count out seven cards, shuffle them and pick one.
 
So it's that, the hexagonal pyramid, and the pentagonal prism
 
@AkivaWeinberger It's just an n-gonal prism.
Long dice (sometimes oblong or stick dice) are dice, often roughly right prisms, designed to land on any of several marked lateral faces, but not either end. Landing on end may be rendered very rare simply by their small size relative to the faces, by the instability implicit in the height of the dice, and by rolling the long dice along their axes rather than tossing. Many long dice provide further insurance against landing on end by giving the ends a rounded or peaked shape, rendering such an outcome physically impossible (at least on a flat solid surface). Design advantages of long dice include...
 
7:21 PM
4 hours ago, by Akiva Weinberger
user image
do dice roll in hyperbolic space?
 
Why wouldn't they ?
 
Presumably so
 
@Astyx I think that's a suggestion of action rather than question of possibility. :)
 
I still love that gif, though I wish the animator had shown more sides
Maybe through an irrational-degree rotation
 
Irrational would make the video infinitely long wouldn't it ?
 
7:24 PM
Actually, I am not sure how rolling look like in hyperbolic geometries, so...
 
Fair. But something with a large denominator, at least
Like $\frac{13}{21}\cdot360^\circ$ or something
 
Anyway, 5:25 here, I am out to sleep
 
Good night
 
That's not really a dice though
Doesn't that have a puncture or something
 
7:31 PM
Heh, I just saw this pointed out
Gabriel's horn has infinite surface area but finite volume
 
Yeah
 
The vuvuzela, on the other hand, has finite surface area but infinite volume
(ba-dum tish)
 
If it's a joke, I don't get it.
 
Wow
 
Volume also means loudness
 
7:32 PM
I'm embarrassed and amazed at the same time
 
Oh, I see.
lol
 
So do you guys want the solution to that graph problem ?
 
7:48 PM
Not in this room plz
 
...?
 
lol
 
My face was actually pretty unimpressive
 
I laughed.
 
And unimpressed
 
7:56 PM
Okay, I've caught up to where everyone else is.
and now I give Akiva a star
 
@user107952 I think you're having a hard time determining what is and isn't appropriate content for chat rooms at SE. Gratuitous and contextless sexual content is one of them that isn't.
3
 
@Dodsy Thanks
 
Looks like someone flagged that pretty quick.
 
Could someone help me finding the radius of convergence of the sum from 1 to infinity $n^2 * ln^2(x)$ ?
 
@Dodsy I just happened to spot it while flipping through rooms. (Is user107952 just having an off day or are they always like this? They've been around long enough to know better…)
 
8:02 PM
Oh I have no idea, I have never spoken to them.
 
I haven't seen them around here.
I have disagreed with your decisions on appropriateness before but agree with this one. It's not particularly offensive, just sorta trollish and distracting.
 
@user379685 $*$ means convolution product ? or regular product ?
 
Balarka did you hear my news :C
 
@astyx regular
 
@Dodsy Yeah. I am sorry to hear it. Have you decided what you're going to do?
Apply for next year or something?
 
8:06 PM
@BalarkaSen (I agree it's borderline. These are the hard judgement calls. In the end I look at it and see something not contributing nor enhancing the appeal of the room. It's the sort of thing that would tend to drive away some lurkers too, when what SE wants is for good lurkers to become productive contributors, and for the ones who would drive them off to be instead driven off. It's a fine balance to strike, I agree.)
 
So you're looking at $\sum{n^2\ln^2(x)x^n}$ ?
 
I'm between going to Trent for a year, applying to UWO for part time, or trying to get into the social science program and just pretend I am a math student.
 
@Astyx oops my mistake it's $n^2 ln^n(x)$
 
I still don't follow
 
8:09 PM
@Dodsy Hmm. What do you mean by pretending you are a math student? Are you going to apply for the math course in UWO part-time?
 
hello guys
 
Hi
 
@Astyx sorry it's not this
 
@SevenSidedDie I agree with that philosophy, and understand it's hard for room moderators to process all the things that's going on. Conversations and room cultures tend to be rather varied after all.
 
@BalarkaSen so the science program is basically full. Unless i do part time.
But the social science program isn't.
But you can take courses in science.
 
8:10 PM
@astyx here prnt.sc/fmkjla
 
So i could take all of the maths I want to take
and then ask to switch to math in second year.
And never take a social science course.
 
Can anyone tell me an interesting topic for real-world applications of integrals? I made an algorithm that solves integrals and would like to apply to some real theme
 
What do you mean by radius of convergence then ? @user379685
 
Yeah but that's no power series
 
8:11 PM
part time wouldn't be bad though. I'd literally only have to take 2 math courses and a computer science or physics course.
and it turns out that the funding would still think I was full time.
 
@Dodsy Ok, I'm trying to understand. Why do you want to have social science in the option at all then?
So you can basically take part-time for a year and all the math courses you want, and switch full-time?
 
Because I could still take all of the same programs that I would've taken as a full time science student.
 
@Astyx well i still need to find the set of x where this sum converges
 
Yeah, so if I do part-time I can only take 3 courses a semester.
 
Hey, that's not bad.
 
8:13 PM
but if I did full-time social science, I know that I won't graduate as a social science student so then i can just take all the science courses
so it'd be like i was accepted to the full-time science program.
 
@astyx do you have a clue?
 
Anyone here with full access to researchgate?
 
@BalarkaSen one of the reasons I don't want to go to trent is that they have a calc 1 course, calc 2 course and then a linear algebra course.
 
Then look at the radius of convergence of $\sum n^2 x^n$ and see when $\ln(x)$ is in the disk of convergence maybe
 
UWO has enriched calculus
 
8:15 PM
Doesn't make much sense to me
 
and then an extra course on writing proofs.
 
@Dodsy Ah, ok. So let's tick off the Trent option outright for a moment.
 
Right.
 
So you got (1) part-time math, (2) full-time social science?
And in (1) and (2) both you can take the same science courses?
 
Right.
except I would have to take 2 less courses part-time.
Either way, for second year I would be a full time science student.
 
8:17 PM
Do you like social science? :)
 
also I can take courses at the campus during the summer.
No I hate it.
I only like math.
 
No physics ?
 
Physics is alright.
 
Or anyone here who can read JapanesE?
 
Kanji?
 
8:18 PM
I read that as "japaneasy"
 
So why the appeal for (2)? Because you can say you got accepted for full-time? :)
 
@BalarkaSen right, so I wouldn't have to do courses at all next summer.
but I guess I'm leaning more towards the part-time.
seems risky to start in social science
 
@Dodsy No idea
 
and just hope to get into math.
 
Yup, me too.
 
 
Take part-time for a year. You'll be fine.
 
True.
it costs an extra 58 bucks.
@Krijn just translate it
Spaces of rational functions on curves over finite fields
 
Yeah I know that
That's the paper I've been trying to get for the past hour, but it's much too hard to get
 
@Dodsy
 
lmfao
what does it mean to pickle the owls
 
8:22 PM
Left to your imagination. I would love to get some pickle made of owls.
 
Where is that from? Where did you find that?
 
Ryan Armand
from Mike
 
I remember someone here linked a professors page
and that was on it
 
it's on his website and I read a bunch of Ryan Armand's comics on his suggestion
@Dodsy a grad student's, yes
 
@BalarkaSen Are you becoming Mike 2?
 
The last one Johnson solids are pretty cool
 
hebesphenorotunda
really
 
The rotunda has pentagons in it, though
 
3 pentagons, huh
 
What?
The rotunda has six pentagons
The hebesphenomegacorona has zero
 
8:33 PM
in the net description of hebesphenorotunda in wikipedia i see 3 pentagons
am i deranged?
 
Oh, the triangular hebesphenorotunda
Sorry, I got confused
I thought you were saying that the hebesphenomegacorona should really be called the hebesphenorotunda or something
 
I didn't know these names existed before I googled them
I was just doing a dumb "really?" at that heck of a name
hebesphenomegakegaunicorn
 
Got it
Disphenocingulum
(That one has the most faces out of the final nine)
 
Balarka
I will take your advice
thanks for sharing it with me.
 
@Dodsy No problem. I think it's a good choice. If you feel uncertain, talk to a few more people maybe.
Either way you'll learn math.
 
8:45 PM
 
looks a lot like icosahedron
if you pinch the topmost and the bottommost edge you should get an icosahedron
the two squares become triangles
 
@BalarkaSen If you punch the edge into a vertex, that vertex has six neighbors
This polyhedron actually has 24 faces
The icosahedron only has 20
 
ohh
breaks my heart
So you'd have to pinch one of those squares onto an edge I suppose, on both top and bottom
does that work?
 
Doubt it
 
Akiva when you learn piano
 
8:59 PM
Each pair of green triangles is so close to coplanar, by the way
 
do you memorize the pieces?
I can't site read for the life of meself.
 
The green and blue make a belt twelve long, and the red and hello cap them off
@Dodsy A few years ago, but I've fallen out of practice
 
I just started this year.
my girlfriends mom has a piano so I play it every once and a while.
 
I'm in choir, which gives me sight-reading practice I guess
 
oh nice.
boychoir
with dustin hoffman
 
9:01 PM
??
 
the movie
nevermind.
 
@BalarkaSen The owls are not what they seem.
 
what does it mean to pickle an owl
 
Oh, I am back. I am sorry for saying those offensive things. I was having a bad day. I also asked similar things on math stack exchange and philosophy stack exchange. Both those posts were deleted. I am seriously worried that I might get suspended from all of stack exchange for simply having a bad day.
 
You made my day, though, with your comments about being rigorous
I had a good laugh.
 
9:09 PM
@user107952 Hey, no worries. I am also having a terrible day!
 
hope you two get to feel better soon
 
I hope I won't get banned from asking questions on stack exchange.
 
Nah, you should be fine.
Just a temporary chat ban.
 
@user107952 you will get perma banned eventually. We all will be.
 
deep
 
9:13 PM
@user107952 Chatbans are a pretty light consequence; it's harder to earn a suspension on the main sites. If one's having a rough day and seeing mainsite mods intervening though, it can be prudent to do something else until having a better day.
 
@BalarkaSen ah, capacitors.
 
@PVAL-inactive Oh indeed?
 
What's funny about capacitor problems is that they're boundary value problems in disguise
 
@Semiclassical Enlighten me!
 
9:28 PM
Well, the prototypical setup is two parallel conductors, one carrying a positive charge and the other a negative charge
Each plate will then carry a local surface charge density, one being $+\sigma$ and the other being $-\sigma$
An application of Gauss's law to each conductor individually then shows that the electric field at the surface of a conductor is $E=\sigma/\epsilon_0$. (That's the magnitude; the direction is dictated by the signs of the charges)
 
Right, agreed.
I can prove that.
 
However, the electric field isn't really the easiest thing to work with due to it vector nature
So let's switch to the electric potential $V$.
 
Ok, $V = \sigma d/\epsilon_0$ if $d$ is the spacing between the parallel plates.
 
well, we'll get there
 
Alright.
 
9:36 PM
The more important point is that $\vec{E}=-\nabla V$
 
I remember doing that a couple of years ago.
 
Ah yes I knew that's what you were getting at.
@Avantgarde You're a physics student, right?
 
yes
 
grad/undergrad?
 
Mmkay. Do you know the link to Laplace 's equation? (Typing on phone so if I can skip steps that's easier)
 
9:37 PM
grad (soon)
 
@Semiclassical Laplace's equation is $\nabla^2 f = 0$, right?
 
right
To get that here, take the divergence of the electric field
In terms of Gauss's law, that's zero in the absence of free charge between the sheets
 
Is $\text{div} E = 0$?
 
In the absence of charge, yes. Field lines begin and end at charges
 
One of my friends told me that if I can rigorously verify at least 20 percent of the abc proof by Mochizuki, he will call me the most rigorous human alive. So, I consider this my challenge. It will probably take many years, but I will do a service for the mathematical community.
 
9:40 PM
You mean, when the plates are not charged?
 
No, I mean in the region between the plates
 
Good old differential version of Gauss's law
 
Hm, oh, I see, Gauss's lemma really implies $\text{div} E = \rho/\epsilon_0$, right? Where $\rho$ is the volumetric charge density.
 
Right
 
Ok, I agree.
So $\nabla^2 V = 0$.
 
9:43 PM
right. But that's subject to a boundary condition on E, i.e. on the derivative of V
 
Yeah. The boundary condition being $\nabla V = -\sigma/\epsilon_0$ at the surface of the conductors.
 
Hence, boundary value problem
 
This is cool!
I never thought about this perspective before.
@Avantgarde Neat. What do you work on?
 
Well, that's not quite right as stated. $\nabla V$ is vectorial, but the density is scalar
 
@BalarkaSen Gravity and QFT
 
9:46 PM
The fix is that it's the component perpendicular to the surface
 
the essentials
 
@Semiclassical Right, thanks.
@Avantgarde aka the good stuff
 
So it's really a condition on the normal derivativr
 
yup
 
So it's really a condition on the normal derivative
This specifically corresponds to Neumann boundary conditions
 
9:48 PM
@BalarkaSen really good
 
To close the link, we expect the conductors to be equipotential surfaces
Hence the charge distribution will correspond to some well-defined potential difference between the two conductors
 
Of course.
 
And $Q/\Delta V$ defines the capacitance
 
agree
 
so you could say that the capacitance is telling us about how Neumann boundary conditions dictate Dirichlet boundary conditions and vice versa
(That might be a bit overboard since Dirichlet and Neumann aren't required to be uniform in the way the physics problem usually is
i.e. one takes the potential on the surface to be constant rather than an allow an arbitrary function)
This mindset makes some problems really easy, e.g. the capacitance of a cylindrical capacitor
Just solve Laplace's equation in cylindrical coordinates and impose boundary conditions
 

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